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INDEX TOPIC NAME 1. Basic Calculations

Page No. 2-9

2. Number System

10-21

3. L.C.M. & H.C.F

22-29

4. Percentages

30-38

5. Average

39-48

6. Ratio and Proportion

49-59

7. Partnership

50-65

8. Mixtures (or) Alligations

66-71

9. Profit and Loss

72-81

10. Problems on Ages

82-86

11. Time and Work

87-94

12. Pipes and Cisterns

95-101

13. Time and Distance

102-111

14. Problems on Trains

112-118

15. Boats and Streams

119-124

16. Simple Interest

125-133

17. Compound Interest

134-142

18. Clocks

143-148

19. Calendars

149-153

20. Mensuration - 2D

154-164

21. Mensuration - 3D

165-170

Key to Assignments

171-173

2

1. Basic Calculations VBODMAS The order of various operations in exercises involving brackets and functions must be performed strictly according to the order of the letters of the word VBODMAS. Each letter of the word VBODMAS stands as follows: V for Vinculum B for Bracket O for Of D for Division M for Multiplication A for Addition S for Subtraction

: : : : : :

: - (bar) [{( )}] of ÷ x + -

Note: There are three brackets. 1. ( ) 2. { } They are removed strictly in the order ( ), { } and [ ].

3. [ ]

Solved Example: 1.

  1  1 1 1  1 5     3  4 of5  1 1  3  1    2  5 2 3  4 8       Sol: Given expression   9 1 6 9 16  5 5    =   of  1 1  3     2 5 2 3 4 8        Simplify: 4

=

9 1 6 9 16  5      of  1 1  3    2  5 2 3 8    

=

9 1 6 9 16  1 9   of  1 1   2 5 2 3 8  

=

9 1 6 9 1 6 6 9   of  2  5 2 3 8 

=

9 16 9 16 69       2  5 2 3 8

=

9 1 6 1 6 9    2  5 24 8 

9 1 6  1 0 3 5  2  1 2 0  9 1051 5 4 0 1 0 5 1 = =  2 120 120 511 =  120 =

Square Root And Cube Root Square: A number multiplied by itself is known as the square of the given n umber. E.g. square of 3 is 3 x 3 = 9 Square Root: Square root of a given number is that number which when multiplied by itself is equal to the given number. It is denoted by the symbol 2

E.g. square root of 16 is 4 because 4 = 4 x 4 = 16

.

3

Thus,

16

= 4.

Methods of finding the Square Root: I. 1. 2. 3.

Prime Factorization Method: This method is used when the given number is a perfect square or when every prime factor of that number is repeated twice. Follow the ste ps as mentioned below. First find the prime factors of the given number. Group the factors in pairs. Take one number from each pair of factors and then multiply them together. This product is the square root of the given number.

E.g. Find the square root of 225. Sol: 225 = 5 x 5 x 3 x 3 So, √225 = 5 x 3 = 15. II. E.g.

Method of Division: This method is used when the number is large and the factors cannot be easily determined. Find the square root of 180625.

So, the square root of 180625 i.e. √180625 is 425. Explanation: 1. 2. 3. 4. 5. 6. 7.

First separate the digits of the number into periods of two beginning from the right. The last period may be either single digit or a pair. Find a number (here it is 4) whose square may be equal or less then the first period (here it is 18). Find the remainder (here it is 2) and bring down the next period (here it is 06). Double the quotient (here 4) and write to the left (here 8). The divisor of this stage will be equal to the above sum (here 8) with the quotient of th is stage (here 2) suffixed to it (here 82). Repeat this process till all the periods get exhausted. The final quotient is equal to the square root of the given number (here it is 425).

Square root of a Decimal: If the given number is having decimal, separate the digits of it into periods of two to the right and left starting from the decimal point and then proceed as followed in the example.

E.g. 1. Find the square root of 1.498176.

4

So, √1.498176 = 1.224 Note: The square root of a decimal cannot found exactly, if it has an odd number of decimal places. Try with finding the square root of 0.1790136 Square Root of a Fraction: Case 1: If the denominator is a perfect square, the square root is found by taking the square root of the numerator and denominator separately. 2601 E.g. Find the square root of . 49

2601 = 49

Sol:

2601 49

5 1 5 1

=

77

=

2 51 = 7 7 7

Case 2: If the denominator is not a perfect square, the fraction is converted into decimal and then square root is obtained or the denominator is made perfect square by multiplying and dividing a suitable number and then its square root can be determined. E.g. Find the square root of Sol:

461 = 8

461 . 8

4 6 1 2 = 82

922

=

3 0.3 6 4 4 = 7.5911 (nearly) 4

16 Cube: Cube of a number is obtained by multiplying the number itself thrice. E.g. 64 is the cube of 4 as 64 = 4 x 4 x 4. Cube Root: The cube root of a number is that number which when raised to the third power produces the given number, that is the cube root of a number a is the number whose cube is a. The cube root of a is written as

3

a.

Methods to find Cube Root: 1. Method of Factorization: a. b.

First write the given number as product of prime factors. Take the product of prime numbers, choosing one out of three o f each type. This product gives the cube root of the given number.

E.g. Find the cube root of 9261. Sol:9261 = 3 x 3 x 3 x 7 x 7 x 7 so,

3

9261  3  7 = 21

2. Method to find Cube Roots of Exact Cubes consisting the numbers up to 6 Digits: Before we discuss the actual method it is better to have an overview of the following table. Sl. No 1 2 3 4 5 6 7

If the cube ends in … 1 2 3 4 5 6 7

then Cube root ends in 1 8 7 4 5 6 3

Example 1 8 27 64 125 216 343

5

8 9 10

8 9 10

2 9 0

512 729 1000

The method of finding the cube root of a number up to 6 digits which is actually a cube of some number consisting of 2 digits can be well explained with the help of the following examples. E.g. 1. Find the cube root of 19683. Sol: First make groups of 3 digits from the right side. 19,683 : 19 lies between 2 3 and 3 3 , so left digit is 2. 687 ends in 3, so right digit is 7. [See the table.] Thus, the cube root of 19683 is 27. E.g. 2. Find the cube root of 614125. 614 125 : 614 lies between 8 3 and 9 3 , so left digit is 8. 125 ends in 5, so right digit is 5. [See the table.] Thus, the cube root of 614125 is 85.

Brainstorming 1.

Let „a‟ and „b‟ be two integers such that a + b = 10. Then the greatest value of a x b is ___ 1. 20 2. 100 3. 21 4. 24

2.

If a factory A makes x cars in an hour and another factory B, makes y cars every half an hour, how many cars will both factories make in 4 hours? 1. 4x+4y 2. 4x+8y 3. 8x+4y 4. 4x_2y

3.

Which of the following is the same as 50+12? 1. 10(5+3) 2. (60  5)+(100  2) 3. 25  12x2

4.

If x*y = xy1. 16

5.

6.

7.

If 4 x  1. 2

9.

3 2 , then the value of x is _______ 2. 3 3. 4

If the 23x5x = 5x103, then the value of x is ________ 1. 4 2. 3 3. 2 If 5x =

1 , then the value of x is ________ 25

1 2

2. -2

1.

8.

x 1 then the value of 6* is _______ y 3 2. 17 3. -16

3. 2

5 1 xy = 17, then the value of (x, y) is ________ 4 x 1. (6, 2) 2. (7, 2) 3. (9, 2)

4. 50  4  3

4. -17

4. 5 4. 1

4. -4

If 7

3 4 23 =? 4  32 3 9 1. 4

2.

3 2

3.

11 18

4. (8, 2)

4.

7 36

6

10.

11.

1 1 of 111  of ? = 0 3 37 1. 1 2. 3 Which of the following fraction is greater than 1.

12.

19 39

14.

17.

18. 19.

4.

33 49

2. 2.6

3. 100

4. 10

3. 11

4. 10

3. 2.3

4. None of these

If 102y = 25 then what is the value of 10y ?

1 25 b 2a  b 2 If = 0.25 then what should the value of  ? a 2a  b 9 4 5 1. 1 2. 3. 9 9 0.0 1  0.1 Which number is equal to     0.0 1 0.1  1. 1.01 2. 1.1 3. 10.1 1. -5

16.

2. 0.1

The number 1027-1 is not divisible by ____ 1. 9 2. 90 3 2.3  0.0 2 7 = (2.3)2  0.6 9  0.0 9 0 1. 0

15.

23 47

1 3 but less than ? 2 2 29 3. 57

4. 12

(0.0 1)2  (0.1 1)2  (0.0 1 4)2 =? (0.0 0 4)2  (0.0 1 1)2  (0.0 0 1 4)2 1. 0.01

13.

2.

3. 9

2. 5

3.

4.

1 25

4. 2

4. 10.01

What is the value of [0.3+0.3-0.3-0.3 x (0.3 x 0.30)] 1. 0.09 2. 0.27 3. 0.60

4. None of these

Find the number which is equal to (50)3 + (-30)3 + (-20)3 1. 3x50x30x(-20) 2. 30x50x3x20 3. 3x50x(-30)

4. 3x(-30)x(-20

Find the value of the following: 20. 21. 22. 23.

24.

25.

111111x11 = _______ 1. 122221 2. 1222221

3. 222221

4. 12222221

5776800x11 = ___________ 1. 65344800 2. 63544800

3. 62544800

4. 63545800

12369x11 = ________ 1. 135069 2. 136059

3. 136069

4. 135059

15.60x0.30 = ? 1. 4.68

3. 0.468

4. 0.0468

3420 ? x7 =?  19 0.0 1 35 1. 9

2. 0.458

2.

63 5

3.

18 7

If 2276  155 = 79.2, the value of 122.76  15.5 is equal to 1. 7.092 2. 7.92 3. 79.02

4. None of these

4. 79.2

7

26.

27.

28.

29.

1 7.2 8  ? = 200 3.6  0.2 1. 120

3. 12

4. 0.12

12  0.09 of 0.3x2 = ? 1. 0.80 2. 8.0

3. 80

4. None of these

2 0  8  0.5 20 ? 1. 8

3. 2

4. None of these

If

12

31.

32.

0.2 8 9 =? 0.0 0 1 2 1 170 1. 11

34.

35.

3 5

2 5  0.4 8 2. 1.68

2.

will be __________

17 110

5555+6666-9999-1111 = ? 1. 1001 2. 1011 Which is greater among 1.

33.

2. 18

5 = 2.24, the value of

1. 0.168

30.

2. 1.20

5 9

5 15 7 8 , , and , 9 19 8 9 15 2. 19

3. 16.8

3.

0.1 7 11

3. 1111

3.

7 8

4. 168

4.

17 11

4. 1221

4.

8 9

(17)2+(23)2 = ? 1. 718

2. 818

3. 988

4. 8283

132-123 = ? 1. 369

2. 396

3. 496

4. 469

1 1 2 1 =?  3  13  8 2 7 7 4 11 13 1. 5 2. 5 28 28 4

3. 6

11 28

4. 6

15 28

36.

What approximate value should come in place of the question mark (?) in the following equation? 2 66 % of ? = 32.78x18.44 6 1. 900 2. 880 3. 920 4. 940

37.

What should come in place of the question mark (?) in the following equation? 85.147+34.912x6.2+? = 802.293 1. 8230 2. 8500 3. 8410 4. 8600

38.

What should come in place of question mark (?) in the following equation? 5679+1438-2015 = ? 1. 5192 2. 5012 3. 5102 4. 5002

39.

Four of the five parts numbered (1), (2), (3), (4) and (5) are exactly equal. Which of the parts is not equal to the other four? The number of that part is the answer.

8

1. 40% of 160+

1 of 240 3

2. 120% of 1200

1 of 140-2.5x306.4 2 In the following equation what value would come in place of question mark(?)? 5798-? = 7385-4632 1. 3225 2. 2595 3. 2775 4. 3045 3. 38x12-39x8

40.

4. 6

41.

What should come in place of question mark (?) in the following equation? 197x?+162 = 2620 1. 22 2. 12 3. 14 4. 16

42.

Which of the following numbers are completely divisible by seven? A. 195195 B. 181181 C. 120120 1. only A & B 2. only B & C 3. only D & B

D. 891891 4. All are divisible

43.

what should come in the place of the question mark (?) in the following equation 21 9 5 10 =?    25 20 12 17 77 119 9 29 1. 2. 11 3. 4. 1 10 90 125 450

44.

What should come in the place of the question mark (?) in the following equation? 28 ?  ? 112 1. 70 2. 56 3. 48 4. 64

45.

What should come in the place of the question mark (?) in the following equation? 48 ? +32 ? = 320 1. 16

2. 2

3. 4

4. 32

46.

What should come in the place of the question mark (?) in the following equation? (7  ? )2  81 49 1. 9 2. 2 3. 3 4. 4

47.

What should come in the place of the question mark (?) in the following equation? 4 52  2 72 1 3 52 1. 81

48.

49.

50.

. 2. 1

3. 243

Simplify: 18  1 0  4 +32  (4+10  2-1) = ___________ 1. 5 2. 9 3. 8

  

4. 9

4. 7



Solve: 4- 6  1 2  5  4  3 . 1. 5 2. 4

3. 6



4. 8



If x=4, y=3, then the value of x  y  x  2 is _________ 7 1 4 1. 2. 1 3. 4 2 5

4.

5 4

9

1) 4 2) 2 3) 2 4) 3 5) 2 6) 1 7) 2 8) 3 9) 2 10) 2 11) 4 12) 1 13) 3 14) 4 15) 2

16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

1 3 2 2 2 2 2 1 4 2 4 4 2 2 1

31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) 42) 43) 44) 45)

3 4 2 4 3 1 2 3 2 4 2 4 4 2 1

46) 47) 48) 49) 50)

3 1 4 3 3

2. NUMBER SYSTEM Natural Numbers (N): Counting numbers 1, 2, 3, . . . . . are called Natural numbers. They are also called Positive Integers. N = {1, 2, 3, . . . . . . . .} Whole Numbers (W): All the natural numbers including 0 together constitute the set of Whole numbers. W = {0, 1, 2, 3, . . . . . . . .} Integers (I or Z): All the whole numbers including negative counting numbers together constitute the set of Integers. I or Z = {. . . . . . ., -3,-2,-1, 0, 1, 2, 3 . . . . . . . .} p Rational Numbers (Q): Numbers which are in the form of , where p, q are integers and q ≠ 0, are q called Rational numbers. p Q = { , (q  0)/ p, q  Z} q E.g. -3, 1, 3.2,

1 22 , , etc. 3 7

Note: 1. Rational numbers are divided into two groups, namely integ ers and non-integers. 2. Non-integer belonging to the set of rational numbers is called fraction. p Fraction: A number expressed in the form of is also called fraction, where „p‟ is the numerator and q „q‟ is the denominator. Fraction is a part of an integer. 6 2 1 E.g. , ,  , etc. 6 5 7 Proper Fraction: Fractions in which Numerator < Denominator are called Proper Fractions. 1 3 7 E.g. , , , etc. 5 7 9 Improper Fraction: Fractions in which Numerator > Denominator are called Improper Fractions. E.g. 8/3, 7/5, 9/4, etc. Mixed Fraction: It has two parts. One is integer and the other is a fraction. E.g. 1 1/3, 2 3/5, 5 4/3, etc. Note: 1. All the mixed fractions can be converted into improper fractions. 2. A rational number can be expressed in the decimal form. 3. The decimal form of a rational number is either recurring or a terminating decimal. E.g. 10/3 = 3.3333… (recurring)

10

3/4 = 0.75

(terminating)

Irrational Numbers (Q’): A number which cannot be expressed in the form of rational number is called an Irrational number. For an irrational number, the decimal part is non -recurring and non-terminating. E.g. √2 = 1.414…. It is non-recurring and non-terminating. Even number: An integer divisible by 2 is called an Even number. E.g. 2, 4, 6, 8,…….. Odd Number: An integer not divisible by 2 is called an Odd number. E.g. 1, 3, 5, 7,……. Prime Numbers: Numbers which are not divisible by any number other then 1 and itself are called Prime numbers. E.g. 2, 3, 5, 7,……. Composite Numbers: Except 1, the numbers which are not prime are called Composite numbers E.g. 4, 6, 9, 12,…… Co-prime Numbers: Numbers which do not have any common factor other than 1 are called Co-prime numbers. E.g. (4, 15), (9, 22), (12, 29),…… Note: 1. 1 is neither prime nor composite. 2. 2 is an even prime number. 3. Co-prime numbers can be prime or composite numbers. 4. Any two prime numbers are always Co-prime numbers. 5. Any two consecutive positive integers are always co-primes. Place Value of a Digit in a Numeral: The value of where the digit is in the number, such as units, tens, hundreds, etc. Face Value: Face Value of a number is the number itself. Consider the number 12654: Place Place Place Place Place

Value Value Value Value Value

of of of of of

4 5 6 2 1

= = = = =

4 5 6 2 1

ones = 4, Face Value of 4 = 4 tens = 50, Face Value of 5 = 5 hundreds = 600, Face Value of 6 = 6 thousands = 2000, Face Value of 2 = 2 ten thousands = 10,000, Face Value of 1 = 1

Perfect Numbers: If the sum of the factors of a given number is twice the number, the number is said to be a Perfect number. E.g. Factors of 6 = 1, 2, 3, 6 and 1 + 2 + 3 + 6 = 12 28, 496, etc….are the other examples of perfect numbers. MULTIPLICATION TIPS: 1. For multiplication of a given number by 9, 99, 999, etc., that is by 10 n – 1, the easy way is: Put as many zeros to the right of the multiplicant as there are nines in the multiplier and from the result subtract the multiplicant and get the answer. E.g. Multiply 2893 by 99. Sol:2893 x 99 = 2893 (100 – 1) = 289300 – 2893 = 286407. 2. For multiplication of a given number by 11, 101, 1001, etc., that is by 10 n + 1, the easy way is: Place n zeros to the right of the multiplicant and then add the multiplicant to the number so obtained. E.g. Multiply 3782 by 11. Sol:3782 x 11 = 3782 (10 + 1) = 37820 + 3782 = 41602. 3. For multiplication of a given number by 15, 25, 35, etc. Double the multiplier and then multiply the multiplicant by this new number and finally divide the product by 2. E.g. Multiply 5054 x 15 = ½ (5054 x 30) = ½ (151620) = 75810

11

4. For multiplication of a given number by 5, 25, 125, 625, etc., that is, by a number which is some power of 5. Place as many zeros to the right of the multiplicant as is the power of 5 in the multiplier, then divide the number so obtained by 2 raised to the same power as is the power of 5. E.g. 2982 x 5 = 29820/2 = 14910 5739 x 25 = 573900/2 2 = 143475

p q r a) No. of factors of a given number: If N  a  b  c ..... then the number of factors of N = (p + 1) (q + 1) (r + 1)…………., where a, b, c are prime factors of N and p, q, r,………. are positive integers.

E.g. Find the number of factors of 24. Sol:24 =



23  31 The number of factors of 24 = (3 + 1) (1 + 1) = 8.

b) Sum of the factors of a given number: If p 1

q 1

a

N  a p  b q  c r .....

then the sum of the factors of N =

r 1

1 b 1 c 1   .......... ......... where a, b, c are prime factors of N and p, q, r,………. are a1 b 1 c 1 positive integers. E.g. Find the sum of the factors of 24. Sol:24 = 23  31

23  1  1 31  1  1   6 0. 2 1 3 1 c) No. of ways of expressing a given number as a product of two factors:

 Sum

If

of the factors of 24 =

N  a p  b q  c r .....

where a, b, c are prime factors of N and p, q, r,………. are positive integers then 1 the number of ways in which N can be expressed as product of two factors = (p  1)(q  1)(r  1)..... . 2 E.g. Find the number of ways of expressing 48 as a product of two factors. Sol:48 =

2 4  31

1 (p  1)(q  1)  1 (4  1)(1  1)  5 . 2 2 d) No. of ways of expressing a given number which is a perfect square as a product of two factors: No. of ways =

If N  ap  b q  cr ..... where a, b, c are prime factors of N and p, q, r,………. are positive integers then the number of ways in which N can be expressed as product of two factors = 1 (p  1)(q  1)(r  1)........  1. 2 E.g. Find the no. of ways of expressing 36 as a product of two factors. Sol:36 =

2 2  32

No. of ways =

1 (p  1)(q  1)  1  1 (2  1)(2  1)  1  5 . 2 2

TIPS ON SQUARES: Condition

Method

To square any number ending with 5.

(a5) 2 = {a(a+1)}25

To square a number in which every digit is one. To square a

Count the number of digits in the given number and start writing numbers in ascending order from one to this number and then in descending order up to one. Use the formula:

Example (35) 2 = {3(3+1)}25 = 1225 (11) 2 = 121, (111) 2 =12321 (1004) 2 =

12

number which is nearer to 10 x.

x2  (x2  y2 )  y2  (x  y)(x  y)  y2

(1004 – 4) (1004 + 4) + (4) 2 = 1000(1008) + 16 = 1008016

DIVISION: Dividend = (Divisor x Quotient) + Remainder E.g.

TESTS OF DIVISIBILITY Divisibility by…

Rule

2

Unit‟s digit of the number should be zero or divisible by 2.

3

Sum of the digits in the number should be divisible by 3.

4

Number formed by the last two digits should be divisible by 4 or are both zero.

5

Unit‟s digit of the number should be 0 or 5.

6

Should satisfy divisibility rules of 2 and 3.

7

The unit digit of the given number is doubled and then it is subtracted from the number obtained after omitting the unit digit. If the result is divisible by 7, then the given number is also divisible by 7.

Example 4, 2, 102, etc. 1782

4784, 300, etc.

4784  Since 84 is divisible by 4, 4784 is also divisible by 4.

4518

448

8 9

Sum of the digits in the number should be divisible by 9.

1395

10

Number should end in zero.

1000

11

1+7+8+2 = 18 which is divisible by 3 hence 1782 also divisible by 3.

120, 625, etc.

Number formed by the last three digits should be divisible by 8. or zero‟s

Sum of digits at odd places – Sum of digits at even places should be 0 or divisible by 11.

Explanation

1576

38797

12

Should satisfy divisibility rules of 3 and 4.

156

14

Should satisfy divisibility rules of 2 and 7.

322

448  44 – 8(2) = 44 – 16 = 28 which is divisible by 7 and hence 448 is also divisible by 7. 1576  576 is divisible by 8 and hence 1576 is also divisible by 8. 1395  1+3+9+5 = 18 is divisible by 9 and hence 1395 is also divisible by 9. 38797  Sum of digits at odd places = 3+7+7 = 17 Sum of digits at odd places = 8+9 =17 and 17 – 17 =0, hence 38797 is divisible by 11. 156 is divisible by 2 and 3 hence 156 is also divisible by 12. 322 is divisible by 2 and 7 hence 322 is also divisible by 14.

13

Last two digits in the number should be 0 or divisible by 25.

25

Last three digits in the number should be 0 or divisible by 125.

125

175 2250

175  75 is divisible by 25 and hence 175 is also divisible by 25. 2250  250 is divisible by 125 and hence 2250 is also divisible by 125.

Steps to find whether a given number is prime number or not: 1. Find the least positive integer, a such that a 2 > given number. 2. Test the divisibility of given number by every prime number that is les s than a. 3. The given number is a prime number only if it is not divisible by any of these primes. E.g. Check whether 923 is a prime number or not? 1. 923 lies between 900 and 961 which are perfect squares having square roots 30 and 31 respectively. 2. Prime numbers less than 31 are 2,3,5,7,11,13,17,19,23,29. 3. 923 are not divisible by any of these numbers and hence it is a prime number. a) To find the number in the unit place for odd numbers: When there is an odd digit in the unit place except 5, multiply the number by itself until you gets 1 in the unit place.

(...1) n

= (… 1)

4n

(...3) = (...7) 4 n = (...9) 2 n =

(… 1) (… 1) (… 1) where n = 1, 2, 3, . . .

b) To find the number in the units place for even numbers: When there is an even digit in the unit place, multiply the number by itself until you gets 6 in the unit place.

(...2) 4 n = (… 6) (...4) 2 n = (… 6) (...6) n = (… 6) (...8) 4 n = (… 6)

where n = 1, 2, 3, . . .

c) If there is 1, 5 or 6 in the units place of the given number: If there is 1, 5 or 6 in the unit place of the given number, then after any times of its multiplication, it will have the same digit in the unit place.

(...1) n

= (… 1)

(...5) n (...6) n

= (… 5) = (… 6) where n = 1, 2, 3, . . .

Solved Examples 1.

On dividing 64652 by a certain number, the quotient is 101 and the remainder is 12. Find the divisor. Sol: Here, the required number is divisor. Divisor =

=

Dividend- Remainder Q uotient

6 4 6 5 2 1 2 6 4 6 4 0   640 101 101

14

2.

A number when divided by 160 leaves a remainder 52 and the quotient is 15. Find the number. Sol: Here, the required number is dividend. Dividend = (Divisor x Quotient) + Remainder = (160 x 15) + 52 = 2452

3.

Find the least number of 5 digits which is exactly divisible by 642. Sol: The least number of 5 digits is 10,000. Dividing this number by 642, the remainder is 370. So, the required number is 10,000 + (642 - 370) = 10272.

4.

Find the greatest number of 5 digits which is exactly divisible by 642. Sol: The greatest number of 5 digits is 99,999. Dividing this number by 642, the remainder is 489. So, the required number is 99,999 - 489 = 99510.

5.

Find the number nearest to 14800 which is exactly divisible by 245. Sol: The remainder on dividing 14800 by 245 is 100. So, the number required number = 14800 – 100 = 14700 which is exactly divisible by 245.

6.

Find whether 577 is a prime number. Sol: (24) 2 = 576 < 577 and (25) 2 = 625 > 577  n = 25 Prime numbers less than 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23. Since, 577 is not divisible by any of these numbers, it is a prime number.

7.

How many numbers up to 531 are divisible by 15?

Sol: Divide 531 by 15. 531 = 35 x 15 + 6 The quotient is the required number and here it is 35. So, there are 35 numbers up to 531 are divisible by 15. 8.

How many numbers up to 200 are divisible by 5 and 7 together? Sol: L.C.M. of 5 and 7 = 35. Divide 200 by 35. 200 = 5 x 35 + 20 The quotient is the required number and here it is 5. So, there are 5 numbers up to 200 are divisible by 35. Find the number in the unit place in (7 2 9)59 .

9.

Sol: (7 2 9)59  (7 2 9)58  7 2 9 (...1)  7 2 9 9 in the unit place. 10.

Find the number in the unit place in (9 8)42 .

Sol: (9 8)42  (9 8)4 10  (9 8)2  (...6)  (...4)  4 in the unit place. 11.

Find the number in the unit place in (6 3 6)36 .

Sol: (636)

36

 (...6) 36  6

in the unit place.

12. Convert 0.4444…….. into a rational number. Sol: Let x = 0.4444……..(1) Since 1 digit (4) is repeating multiply equation 1 on both sides by

101 .

15

10 x = 4.4444…….(2) Subtract Equation 1 from 2 on both sides 10 x = 4.4444……. - x = 0.4444……. ---------------------9 x = 4.0000……. ---------------------4  9x  4  x  . 9 13. Convert 5.626262…….. into a rational number. Sol: Let x = 5.626262…….. (1) Since 2 digits (62) is repeating multiply equation 1 on both sides by 100 x = 562.6262…….(2) Subtract Equation 1 from 2 on both sides

10 2 .

100 x = 562.6262……. - x = 5.6262…….. --------------------------99 x = 557.0000……. ---------------------------

 99x = 557  x =

557 99

Brainstorming 1.

2.

O1 . + 8 ÷ 9 x11  (3x3)+6x7  14+7  3x3 = ________ 89 45 1. 2. 1 3. 9 9 The number divisible by 99 is_________ 1. 3572404 2. 135792 3. 913464

4.

101 9

4. 114345

3.

The smallest number which when subtracted from 43079 makes it exactly divisible by 9 is 1. 4 2. 5 3. 6 4. 7

4.

The number 111, 111, 111, 111 is divisible by 1. 9, 11 2. 5, 11 3. 5, 9

4. 3, 11

5.

The largest natural number by which the product of three consecutive even natural numbers is always divisible is 1. 16 2. 24 3. 48 4. 96

6.

The number of divisors of 120 individuality is 1. 8 2. 12

7.

3. 14

The greatest number which exactly divides 826 is 1. 2 2. 7 3. 59

4. 16 4. 413

8.

The least number which is a perfect square as well as a factor of 936 is 1. 36 2. 9 3. 4 4. 16

9.

Which of the following is a perfect number? 1. 28 2. 30

10. 11.

3. 14

Which of the following numbers is divisible by 11? 1. 21434799 2, 74325566 3. 85437657

4. 16 4. 93825677

The value of 320.16x320.04 is __________ 1. 2. 2

2

3. 2

4. 4. 2

16

12. 13.

14.

Find the sum of 1+2+6+12+………………..+90. 1. 550 2. 440

3. 330

4. 388

How many numbers are there between 200 and 300 in which 0 occurs only once? 1. 18 2. 885 3. 115 4. 1000

8 8 5 8 8 5 8 8 5 1 1 5 1 1 5 1 1 5 = 8 8 5 8 8 5 1 1 5 1 1 5 8 8 5 1 1 5 1. 770 2. 885

3. 115

4. 1000

15.

The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are 1. 628, 372 2. 600, 400 3. 640, 630 4. None

16.

A number is as much greater than 21 as is less than 71. The number is 1. 39 2. 46 3. 41 4. 49

17.

The sum of three numbers is 102. If the ratio between first and second be 2:3 and that between second and third be 5:3, then the second number is _____ 1. 30 2. 45 3. 27 4. 48

18.

The value of 1+

1 1

1

1. 19.

29 19

= _________

1 1 9

2.

10 19

The value of 9+9x9-9  9 is ___________ 1. 17 2. 89

3.

29 10

3. 9

4.

10 9

4. None

20.

On dividing a number by 999, the quotient is 377 and the remainder is 105. The number is 1. 476727 2. 376538 3. 376728 4. 359738

21.

0.77777…………………… is equal to 77 99 1. 2. 83 77

22.

23.

3.

77 99

199+299+399+………………………..9999 is divisible by 1. 99 2. 98 3. 100 If the fraction number? 143 1. 156

4.

77 109

4. 101

143 136 63 17 and are arranged in ascending order, then which is the third , , 156 153 60 16 2.

136 153

3.

17 16

4.

63 60

24.

How many three-digit numbers, when divided by 15, leave the remainder 15? 1. 49 2. 50 3. 60 4. 62

25.

The value of (112+122+132+…………+202) is 1. 3854 2. 2485

26.

27.

3. 3485

4. 2870

A boy was asked to multiply a certain number by 25. He multiplied by 52 and got his answer more than the correct one by 324. The number to be multiplied was 1. 12 2. 15 3. 25 4. 52

1  3  5 9 9 7    2  3  2  5  2  7 .......... ...... 2  9 9 9 is equal to      

17

1.

5 999

2.

1001 999

3.

1001 3

4.

3 1001

28.

A number consists of two digits whose sum is 15. If 9 is added to the number, then the digits change their places. The number is 1. 69 2. 78 3. 87 4. 96

29.

One of the two consecutive positive integers, the sum of whose squares is 761, is 1. 15 2. 24 3. 20 4. 25

30.

The denominator of a rational number is 3 more than its numerator. If the numerator is increased by 7 and the denominator is decreased by 7, we obtain 2. The rational number is 15 1 7 8 1. 2. 3. 4. 18 4 10 11

31.

Find the value of 1.

3n  3n  1 3n  1  3n

1 2

2.

is

4 3

3.

2 3

4.

3 4

32.

A number N when divided by 120 gives a remainder 76. What is the remainder obtained when the same number is divided by 8? 1. 2 2. 3 3. 4 4. 5

33.

What is the value of 1+

1 1

?

1 1

1 7

19 13 18 7 2. 3. 4. 13 19 13 6 The sum of two consecutive prime numbers is 30. The two numbers are 1. 18, 12 2. 7, 23 3. 11, 19 4. 13, 17 1.

34. 35.

By how much is 11% of 22.2 less than 10% of 24.4? 1. 0.2 2. 0.02 3. 0.002

4. 0.0002

36.

The sum of two numbers is 29 and the difference of their squares is 145. The difference between the number is 1. 13 2. 5 3. 8 4. 11

37.

The ratio between a two digit number and the sum of the digits of that number is 4:1. If the digit in the units place is 3 more than the digit in the tenth place, what is the number? 1. 24 2. 63 3. 36 4. Can‟t be determined

38.

The difference between a two digit number and the number obtained by interchanging the digits is 27. What is the sum of the two digits of the number? 1. 3 2. 6 3. 9 4. Can‟t be determined

39.

0.02% of a number is 20. What will be 20% of the number? 1. 200 2. 4000 3. 20,000

4. 2000

40.

A number whose fifth part increased by 5 is equal to its fourth part diminished by 5 is 1. 160 2. 200 3. 180 4. 220

41.

What will be the remainder when 2975 is divided by 30? 1. 28 2. 27 3. 1

42.

4. 29

If a number is divided by 527, the remainder 42. What will be the remainder if it is divided by 17? 1. 8 2. 7 3. 16 4. 15

18

43.

Two different numbers when divided by 47, leave remainders 13 and 23 respectively. If their sum is divided by the same number 47, what will be the remainder? 1. 35 2. 36 3. 20 4. 25

44.

The units digit in the product (2767)135x(576)298 is 1. 2 2. 6 3. 8

4. 3

Find the number of divisors of 240? 1. 20 2. 10

3. 24

4. 18

Find the sum of the divisors of 360? 1. 1107 2. 1170

3. 1175

4. 1180

45. 46. 47.

What is the total number of possible values of „p‟ so that the number 3p8p4709 is divisible by 9? 1. 0 2. 1 3. 2 4. 3

48.

A man travels

49.

What fraction of a week is a second? 1 1 1 1. 2. 3. 60480 86400 604800 4 of a number exceeds its 2/3 by 8. The number is 5 1. 30 2. 60 3. 75

50.

1 1 1 of a distance by car, of the remaining by bus and of the still remaining 3 3 3 by train and the remaining 72 km by scooter. Find the distance he traveled? 1. 216 2. 243 3. 336 4. 576

4. 604800

4. 90

51.

The expression 1/1.2 + 1/2.3 + 1/3.4 + ………. + 1/n(n+1) for any natural number n, is _____ 1. Always greater than 1 2. Always less than 1 3. Always equal to 1 4. Not definite

52.

A two digit number becomes five-sixth of itself when its digits are reversed. The difference of the two digits is 1. The number is ______ 1. 45 2. 54 3. 56 4. 65

53.

The sum of the squares of two numbers is 3341 and the difference of their squares is 891. The numbers are _______ 1. 25, 46 2. 35, 46 3. 25, 36 4. None of these

54.

Of the three numbers, the sum of the first two is 45, the sum of the second and third is 55 and the sum of the third and thrice the first is 90. The third number is _______ 1. 30 2. 27 3. 39 4. 52

55.

Three numbers are in the ratio 3 : 4 : 5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is _______ 1. 20 2. 27 3. 39 4. 52

56.

Of the three numbers, the first is twice the second and is half of the third. If the average of these numbers be 56, the numbers in order are _________ 1. 48, 96, 24 2. 48, 24, 96 3. 96, 24, 48 4. 96, 48, 24

57.

The number of prime factors in the expression (6)10 x (7)17 x (11)27. 1. 54 2. 64 3. 71

4. 81

The number of prime factors in 2222 x 3333 x 5555 is _______ 1. 3 2. 1107 3. 1110

4. 1272

58. 59.

The total number of prime factors of the product (8)20 (15)24 (7)15 is _______ 1. 59 2. 98 3. 123 4. 138

19

60.

24 is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. The first part is _________ 1. 11 2. 13 3. 16 4. 17

61.

The sum of squares of the numbers is 80 and the square of their difference is 36. The product of the two numbers is ______ 1. 22 2. 44 3. 58 4. 116

62.

The product of two numbers is 120. The sum of their squares is 289. The sum of the two numbers is ________ 1. 20 2. 23 3. 169 4. None of these

63.

A fraction becomes 4 when 1 is added to both the numerator and denominator and it becomes 7 when 1 is subtracted from both the numerator and denominator. The numerator of the given fraction is ________ 1. 2 2. 3 3. 7 4. 15

64.

Three numbers are in the ratio 5 : 4 : 3. The sum of the largest and the smallest equals the sum of the third and 72. The largest number is ________ 1. 60 2. 27 3. 90 4. 65

65.

The sum of three numbers is 132. If the first number be twice the second and third number be one-third of the first, then the second number is _______ 1. 32 2. 36 3. 48 4. 60

66.

If the unit digit in the product 75 ? x 49 x 867 x 943 be 1, then the value of ? is ______ 1. 1 2. 3 3. 7 4. 9

67.

What is the value of the digit in the place of $ in 3875 $ 622, if the number is divisible by 9? 1. 4 2. 3 3. 2 4. 1

68.

What is the least value of A + B, if 407A234B is exactly divisible by 9? 1. 3 2. 4 3. 5 4. 7

69.

If a number is subtracted from the square of its one half, the result is 48. The square root of the number is __________ 1. 4 2. 5 3. 6 4. 8

70.

Two numbers are such that the ratio between them is 3 : 5, but if each is increased by 10, the ratio between them becomes 5 : 7. The numbers are _________ 1. 3, 5 2. 7, 9 3. 13, 22 4. 15, 25

1) 2) 3) 4)

4 4 2 4

16) 17) 18) 19)

2 2 1 2

31) 32) 33) 34)

3 3 1 4

46) 47) 48) 49)

2 2 2 3

61) 62) 63) 64)

1 2 4 3

5) 3 6) 4 7) 3

20) 3 21) 3 22) 3

35) 2 36) 2 37) 3

50) 2 51) 2 52) 1

65) 2 66) 4 67) 2

8) 3 9) 1 10) 3

23) 1 24) 3 25) 2

38) 4 39) 3 40) 2

53) 2 54) 1 55) 3

68) 4 69) 1 70) 4

11) 3 12) 3 13) 1

26) 1 27) 3 28) 2

41) 4 42) 1 43) 2

56) 2 57) 2 58) 3

14) 4

29) 3

44) 3

59) 3

15) 1

30) 2

45) 1

60) 2

20

3. L.C.M. AND H.C.F. Common Multiple: A common multiple of two or more numbers is a number which is exactly divisible by each one of them. E.g. 32 is a common factor of 8 and 16 Least Common Multiple (L.C.M): The least multiple among all the common multiples of given numbers is called Least Common Multiple. Methods of finding L.C.M. 1.

Method of Prime Factors a. Resolve each given number into prime factors. b. Take out all the factors with highest powers that occur in given numbers. c. Find the product of these factors. This product will be L.C.M. E.g. Find the L.C.M. of 12, 14 and 20. 12 = 2 2 x 3 14 = 2 x 7 20 = 2 2 x 5 So, the L.C.M. = 2 2 x 3 x 5 x 7 = 420

2.

Method of Division E.g. Find the L.C.M. of 12, 15, 18 and 20.

So, the L.C.M. = 2 x 2 x 3 x 5 x 3 = 180 Common Factor: A common factor of two or more numbers is a number which divides each of them exactly. E.g. 4 is a common factor of 8 and 12 Highest Common Factor (H.C.F): Highest common factor of two or more numbers is the greatest number that divides each one of them exactly. It is also called Greatest Common Divisor or Greatest Common Measure. Methods of finding H.C.F. 1.

Method of Prime Factors E.g. Find the H.C.F. of 50 and 70 Sol: 50 = 2 x 5 x 5 70 = 2 x 5 x 7 Common factors are 2 and 5. So, H.C.F. = 2 x 5 = 10

2.

Method of Division E.g. 1. Find the H.C.F. of 3332, 3724. Sol:

21

So, the H.C.F. of 3332, 3724 is 196. E.g. 2. Find the H.C.F. of 10, 15 and 23. Step 1: First find the H.C.F. of 10 and 15. It is 5 Step 2: Then find the H.C.F. of this 5 and 23. It is 1. So, the H.C.F. of 10, 15 and 23 is 1. Note: 1.

L.C.M. and H.C.F. of fractions

2.

Product of two numbers = L.C.M. of two numbers x H.C.F. of two numbers.

3.

To find the greatest number that will exactly divide a, b and c, simply find the H.C.F. of a, b and c.

4.

To find the greatest number that will divide a, b and c leaving remainders x, y and z respectively, find the H.C.F. of (a – x), (b – y) and (c – z).

5.

To find the least number which is exactly divisible by a, b and c, simply find the L.C.M. of a, b and c.

6.

To find the least number when divided by a, b and c leaving remainders x, y and z respectively, find the (L.C.M. of a, b and c) – k, where k = (a – x) = (b – y) = (c – z).

7.

To find the least number which when divided by a, b and c leaves the same remainder r in each case, find (L.C.M. of a, b and c) + r.

8.

Two numbers when divided by a certain divisor give remainders sum is divided by the same divisor, the remainder is

r3 .

r1

and

r2 .

When their

Then the divisor is given by

r1  r2  r3 . 9.

A number on being divided by d1 and d2 successively leaves the remainders r1 and r2 , respectively. If the number is divided by d1 x d2 , then the remainder = ( d1 x r2 + r1 ).

10.

To find the greatest number that will divide x, y and z leaving the same remainder r in each case:

Case 1:

When the value of remainder r is given Required number = H.C.F. of (x – r), (y – r) and (z – r).

Case 2:

When the value of remainder is not given Required number = H.C.F. of |(x – y)|, |(y – z)| and |(z – x)|.

11. To find the n-digit greatest number which when divided by x, y and z: a. Leaves no remainder i.e. exactly divisible

22

Step 1: Find the L.C.M. of x, y and z. Let it be L. Step 2: Divide the n -digit greatest number by this L. Let the remainder be R. Step 3: Required number = (n-digit greatest number – R). b. Leaves remainder k in each case: Required number = (n-digit greatest number – R) + k. 12. To find the n-digit smallest number which when divided by x, y and z: a) Leaves no remainder i.e. exactly divisible Step 1: Find the L.C.M. of x, y and z. Let it be L. Step 2: Divide the n -digit smallest number by this L. Le the remainder be R. Step 3: Required number = n-digit smallest number + (L – R). b) Leaves remainder k in each case: Required number = n-digit smallest number + (L – R) + k.

Solved Examples 1.

Find the greatest number that will exactly divide 200 and 310. Sol: The required number = H.C.F. of 200 and 310 = 10.

2.

Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively. Sol: The required number = H.C.F. of (148 – 4), (246 – 6), and (623 – 11) = H.C.F. of 144, 240, 612 = 12.

3.

Find the smallest number that is exactly divisible by 45, 63 and 50. Sol: Required number = L.C.M. of 45, 63 and 50 = 3150.

4.

Find the least number which when divided by 36, 48 and 64 leaves the remainders 25, 37 and 53 respectively. Sol: (36 – 25) = (48 – 37) = (64 – 53. = 11 Required number = (L.C.M. of 36, 48 and 64) – 11 = 576 – 11 = 565

5.

Find the least number which when divided by 12, 16 and 18, will leave the remainders 7 in each case. Sol: Required number = (L.C.M. of 12, 16 and 18) + 7 = 144 + 7 = 151

6.

Find the greatest number which will divide 772 and 2778 so as to leave the remainder 5 in each case. Sol: Required number = H.C.F. of (772 – 5) and (2778 – 5) = H.C.F. of 767 and 2773 = 59.

7.

Find the greatest number which on dividing 152, 277 and 427 leaves equal remainder. Sol: Required number = H.C.F. of |(152 – 277)|, |(277 - 427)|, |(427 – 152)| = H.C.F. of 125, 275 and 150 = 25.

8.

Find the greatest number of 4 digits which, when divided by 12, 18, 21 and 28 leaves 4 as a remainder in each case. Sol: L.C.M. of 12, 18, 21 and 28 = 252. Greatest 4-digit number = 9999.

23

The remainder when 9999 is divided by 252 = 171 So, the required number = (9999 – 171) + 4 = 9832. 9.

Find the greatest number of 4 digits which, when divided by 12, 15, 20 and 35 leaves no remainder. Sol: L.C.M. of 12, 15, 20 and 35 = 420. The remainder when 9999 is divided by 420 = 339 So, the required number = (9999 – 339) = 9660.

10. Find the least number of 4 digits which is divisible by 2, 4, 6 and 8. Sol: L.C.M. of 2, 4, 6 and 8 is 24. The least number of 4 digits = 1000 The remainder when 1000 divided by 24 = 16. So, the required number = 1000 + (24 – 16) = 1008. 11. Find the smallest number of 4 digits when divided by 12, 18, 21 and 28 leaves remainder 5 in each case. Sol: L.C.M. of 12, 18, 21 and 28 = 252 The least number of 4 digits = 1000 The remainder when 1000 divided by 252 = 244. So, the required number = 1000 + (252 – 244) + 5 = 1013. 12. Two numbers when divided by a certain divisor give remainders 16 and 12 respectively. When their sum is divided by the same divisor, the remainder is 4. Find the divisor. Sol: Required divisor = 16 +12 – 4 = 24. 13. A number on being divided by 10 and 11 successively leaves the remainders 5 and 7, respectively. Find the remainder when the same number is divided by 110. Sol: Required remainder = 10 x 7 + 5 = 75. 14. Find the least number which when divided by 8, 10 and 15 leaves the remainders 3, 5 and 10, respectively. Sol: Here, 8 – 3 = 10 – 5 = 15 – 10 = 5 L.C.M. of (8, 10, 15) = 120  The required least number = 120 – 5 = 115.

Brainstorming 1.

Find the L.C.M. of 12, 25, 30 and 20. 1. 240 2. 300

3. 320

2.

Find the L.C.M. of 2 x 3 x 52 x 7, 23 x 32 x 5 and 22 x 33 x 52. 1. 23 x 53 x 32 2. 23 x 32 x 5 x 7 3. 23 x 32 x 5 x 7

3.

Find the L.C.M of

4. 5.

4 3 2 8 . , , and 5 10 15 15 4 4 3 1. 4 2. 3 3. 4 5 5 5 Find the L.C.M. of 0.01, 0.1, 0.001 and 0.0001. 1. 0.1 2. 0.01 3. 0.001 Find the H.C.F. of 300, 450 and 525. 1. 125 2. 50

3. 75

4. 150 4. 23 x 33 x 52 x 7

4. 3

3 5

4. 1 4. 25

24

6.

Find the H.C.F. of 32 x 52 x 7 , 3 x 52 x 7, 33 x 72 x 52. 1. 3 x 5 x 72 2. 3 x 5 x 7 3. 3 x 52 x 7

7.

Find the G.C.D. of

8. 9.

3 9 6 12 . , , and 5 20 25 25 3 3 1. 2. 10 100 Find the G.C.D. of 0.01, 0.1, 0.001, 0.0001. 1. 0.1 2. 0.01 Find the G.C.D. of 0.2, 2.0, 0.02, 20.0. 1. 0.2 2. 0.02

3.

3 1000

4. 32 x 52 x 72

4. 3

3. 0.001

4. 0.0001

3. 0.002

4. 2

10.

L.C.M. of two numbers is 192 and their H.C.F. is 16. If one of them is 48 then find the other. 1. 32 2. 64 3. 48 4. 68

11.

The product of two numbers is 1575 and their L.C.M. is 315. Find the numbers. 1. 30 and 45 2. 35 and 45 3. 45 and 65 4. 45 and 75

12.

The L.C.M. of two numbers is 12 times their H.C.F. The sum of L.C.M. and H.C.F. is 65. If one of them is 20. Find the other. 1. 15 2. 18 3. 25 4. 16

13.

The L.C.M. of two numbers is 4 times their H.C.F. The product of numbers is 900. Find their L.C.M. 1. 64 2. 48 3. 45 4. 60

14.

Two numbers are in the ratio 5 : 8 and their H.C.F. is 4. Find the numbers. 1. 25 and 40 2. 20 and 32 3. 30 and 48 4. 15 and 24

15.

Three numbers are in the ratio 3 : 6 : 8 and their L.C.M. is 120. Find the smaller number. 1. 16 2. 18 3. 15 4. 12

16.

Find the greatest number which will divide 1026, 1215 and 2349 exactly. 1. 24 2. 27 3. 29 4. 37

17.

Find the greatest possible length which can be used to measure the length 6 m 76 cm, 4 m 81 cm and 7 m 67 cm exactly. 1. 13 cm 2. 15 cm 3. 23 cm 4. 17 cm

18.

Find the greatest number that will divide 1657 and 2037 to leave remainders 6 and 5 respectively. 1. 121 2. 123 3. 129 4. 127

19.

Find the greatest number that will divide 520, 1140 and 1220 leaves the remainders 7, 6 and 5 respectively. 1. 29 2. 23 3. 27 4. 21

20.

Find the greatest number which divides 208, 1194 and 1449 leaving 4 as the remainder in each case. 1. 27 2. 19 3. 17 4. 13

21.

Find the greatest number which divides 284, 678 and 1618 leaves the same remainder in each case. 1. 2 2. 3 3. 4 4. 6

22.

Find the least number which is exactly divisible by 10, 12, 14, 16 and 18. 1. 5010 2. 5030 3. 5040 4. 5020

23.

Find the least number which when divided by 12, 16, 18 leaves 5 as remainder in each case. 1. 143 2. 149 3. 139 4. 144

25

24.

Find the smallest number which when decreased by 3 is exactly divisible by 9, 12, 15, 18 and 21. 1. 1253 2. 1263 3. 1267 4. 1243

25.

Which of the following when increased by 4 is exactly divisible by 12, 14, 16, 28 and 20?. 1. 5034 2. 5006 3. 5026 4. 5036

26.

Find the least number which when divided by 35, 45 and 55 leaves remainders 20, 30 and 40 respectively. 1. 3450 2. 3250 3. 3420 4. 3410

27.

Find the least number of five digits which is exactly divisible by 4, 6, 8, 12 and 16. 1. 10025 2. 10024 3. 10032 4. 10034

28.

Find the greatest number of five digits which is exactly divisible by 16, 24, 28 and 32. 1. 99456 2. 99556 3. 99446 4. 99566

29.

Find the greatest number less than 800 which is exactly divisible by 12, 15, 20 and 30. 1. 740 2. 720 3. 780 4. 760

30.

Find the greatest number between 600 and 700 which is exactly divisible by 20 and 30 1. 660 2. 640 3. 620 4. 480

31.

Find the least number of four digits which when divided by 4, 5 and 6 leaving the remainder 2 in each case. 1. 1022 2. 1012 3. 1032 4. 1002

32.

Find the greatest number of four digits which when divided by 10, 15 and 20 leaves remainder 3 in each case. 1. 9983 2. 9663 3. 9963 4. 9960

33.

Find the least number which when divided by 8, 12, 15 and 18 leaves a remainder 3 in each case and exactly divisible by 11. 1. 360 2. 361 3. 363 4. 383

34.

Find the least number which when divided by 15, 20, 24 and 30 leaves 1 as remainder but when divided by 19 leaves no remainder. 1. 228 2. 361 3. 380 4. 399

35.

Five bells toll at an interval of 6, 9, 12, 15 and 18 seconds respectively beginning together. After what interval of time will they toll again together? 1. 1 min 2. 4 min 3. 2 min 4. 3 min

36.

4 traffic signals at four different places change at an interval of 5 sec, 10 sec, 15 sec and 25 sec respectively. If they change simultaneously at 10 : 20 A.M., at what time will they again in change simultaneously? 1. 10 : 22: 30 A.M. 2. 10 : 24: 30 A.M. 3. 10 : 21: 30 A.M. 4. 10 : 24: 40 A.M.

37.

4 wheels moving 24, 32, 48 and 52 revolutions in a minute starting at a certain point on the circumference downwards. After what interval of time will they come together again in the same position? 1. 15 sec 2. 30 min 3. 30 sec 4. 15 min

38.

Four men start together to travel the same way around on a circular path of 36 km with speeds 2 km/h, 3 km/h, 4 km/h and 6 km/h respectively. When they meet at the same point again after? 1. 24 hours 2. 36 hours 3. 12 hours 4. 18 hours

39.

Find the least number of square tiles required to pave the floor of a room of 5 m 44 cm long and 3 m 74 cm broad. 1. 178 2. 164 3. 176 4. 172

26

40.

Find the least number of square tiles required to pave the floor of a room of 8 m 99 cm long and 6 m 67 cm broad. 1. 713 2. 723 3. 717 4. 719

41.

A trader has three kinds of liquids of first kind 117 lit, second kind 130 lit and the third kind 143 lit. Find the least number of full casks of equal sizes which this can be stored without mixing them. 1. 28 2. 24 3. 36 4. 30

42.

The H.C.F. of two numbers is 11 and their product is 1452. How many pairs of such numbers possible? 1. 1 2. 2 3. 3 4. 4

43.

The H.C.F. of two numbers is 15 and their sum is 195. How many pairs of such numbers possible? 1. 1 2. 3 3. 4 4. 6

44.

The H.C.F. of two numbers is 23 and their sum is 184. How many such pairs of numbers are possible? 1. 0 2. 2 3. 3 4. 4

45.

Find the least perfect square which is exactly divisible by 20, 30, 40 and 60. 1. 2500 2. 3600 3. 900 4. 1600

46.

Find the least number by which 240 must be multiplied in order to produce a multiple of 300. 1. 2 2. 4 3. 5 4. 6

47.

64 mango trees, 48 apple trees and 80 orange trees have to plant in rows such that each row contains the same number of trees of one variety only. Find the least number of rows in which the trees may be planted. 1. 18 2. 12 3. 16 4. 24

48.

Find the greatest number of five digits which can be divisible by to 15, 20, 24, 27, 32 and 36. 1. 90039 2. 90139 3. 86400 4. 90019

49.

Find the smallest number of five digits which when divided by 52, 56, 78 and 91 leaves remainders 28, 32, 54 and 67 respectively. 1. 10896 2. 10196 3. 10496 4. 10696

50.

When finding the H.C.F. of two numbers by division method, the quotients obtained are 1, 2 and 5 respectively and the last divisor is 34. Find the numbers. 1. 374 and 544 2. 324 and 646 3. 314 and 624 4. 616 and 348

51.

When finding the H.C.F. of two numbers by division method, the quotients obtained are 2, 2, 4 and 3 respectively and their H.C.F. is 23. Find the smallest number which is less than 100. 1. 686 2. 667 3. 647 4. 657

52.

Four bells first begin to toll together with an interval of 5, 10, 15 and 20 seconds. How many times do they toll together in an hour? 1. 60 2. 59 3. 61 4. 58

53.

3 men start together to walk along a circular track at the same rate. The length of their tracks is 56 cm, 64 cm and 80 cm. How far will they be in step again? 1. 21 m 20 cm 2. 22 m 40 cm 3. 20 m 20 cm 4. 21 m 40 cm

54.

Find the least number of soldiers in a regiment such that they stand in rows of 10, 15 and 20 and form a perfect square. 1. 400 2. 1600 3. 900 4. 2500

55.

A trader has three kinds of sugar of 77 kg, 147 kg and 252 kg. Find the least number of bags of equal size required to pack them without mixing. 1. 64 2. 68 3. 66 4. 60

27

56.

The sum of two numbers is 721 and their H.C.F. is 103. Find the numbers. 1. 319, 103 2. 309, 412 3. 103, 512 4. 520, 201

57.

The L.C.M. and H.C.F. of two numbers are 693 and 11 respectively. One of them is 77. Find the other. 1. 91 2. 98 3. 97 4. 99

58.

Find the least number of boys so that they can be arranged in the groups of 20 or 25 or 40. 1. 200 2. 250 3. 240 4. 216

59.

Find the least perfect square which is divisible by 12, 15 and 20. 1. 225 2. 900 3. 1600

60.

4. 2500

The product of two numbers is 768 and their H.C.F. is 8. What are the numbers? 1. 31 and 28 2. 32 and 24 3. 33 and 26 4. 35 and 24

1) 2 2) 4 3) 3 4) 1 5) 3 6) 3 7) 2 8) 4 9) 2 10) 2

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

2 1 4 2 3 2 1 4 3 3

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

1 3 2 2 4 1 3 1 3 1

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

1 3 3 2 4 1 1 2 3 1

41) 42) 43) 44) 45) 46) 47) 48) 49) 50)

4 2 4 2 2 3 3 3 1 1

51) 52) 53) 54) 55) 56) 57) 58) 59) 60)

2 3 2 3 2 2 4 1 2 2

4. PERCENTAGE Percent: The term per cent means per hundred or for every hundred. The word is derived from the Latin word per centum. Percentage: A fraction whose denominator is 100 is called a percentage. Rate per cent: The numerator of the fraction is called rate per cent. 5 E.g. and 5 percent means the same thing i.e. 5 parts out of every hundred parts. 100 Basic Formulae:

1 1 × 1 0 0% into a rate per cent, multiply it by100 and put % sign i.e. n n 3 E.g. What percentage is equivalent to ? 4 3 Sol: x 100 = 25% 4 2. To convert a per cent into a fraction, drop the per cent sign and divide the number by 100. 1.

To convert any fraction

1 %? 3 1 1 25 x = 100 12 3

E.g. What fraction is 8 Sol: 8

3.

25 1 % = = 3 3

x% of a given number (N) =

x x N 100

28

E.g. 75 % of 800 = ? 75 Sol: 75 % of 800 = x 800 = 600 100 4.

x   If A is x % more than that of B, then B is less than that of A by   1 0 0 %. 1 0 0  x  

5.

x    1 0 0 %. If A is x % less than that of B, then B is more than that of A by  1 0 0  x  

6.

If A is x % of C and B is y % of C, then A =

7.

If two numbers are respectively x % and y % more than a third number, then the first 1 0 0 x  number is   1 0 0 % of the second and 1 0 0  y  

x y

x 100% of B.

8.

1 0 0 y   1 0 0 % of the first. the second number is  1 0 0  x   If two numbers are respectively x % and y % less than a third number, then the first 1 0 0 x  number is   1 0 0 % of the second and 1 0 0  y  

9.

1 0 0 y   1 0 0 % of the first. the second number is  1 0 0  x   If the price of a commodity increases by N%, then the reduction in consumption so as not to N   increase the expenditure is   1 0 0 %. 1 0 0  N  

10. If the price of a commodity decreases by N%, then the increase in con sumption so as not to N    1 0 0 %. decrease the expenditure is  1 0 0 N  11. If a number is changed (increased/decreased) successively by x % and y % then net % xy   change is given by x  y  % which represents increase or decrease in value a ccording 1 0 0  as the sign is +ve or –ve. Note: If x and y indicates decrease in percentage, then put –ve sign before x and y else +ve sign. 12. If the population of a town (or the length of a tree) is P and its annual increase is r%, then: n

r   (i) Population (or length of tree) after n years = P1   . 1 0 0  P (ii) Population (or length of tree) n years ago = . n r   1  1 0 0   13. If the population (or value of a machine in rupees) is P and annual decrease (or depreciation) is r%, then n

(i) (ii)

r   Population (or value of machine) after n years = P1   . 1 0 0  P Population (or value of machine) n years ago = . n r   1  1 0 0  

14. If a number K is increased successively by x % followed by y % and value of K will be x   y   z   1 1 K 1  1 0 0  1 0 0  1 0 0 

z %, then the final

29

15. In an examination, the minimum pass percentage is x%. If a student scores y marks and fails 1 0 0(y  z) by z marks, then the maximum marks in the examination is . x 16. In an examination a % and b % students respectively fail in two different subjects while c % students fail in both the subjects, then the percentage of students who pass in both the subjects will be (100 – ( a + b - c ))%.

SOLVED EXAMPLES 1.

If Srujana salary is 20% more than that of Deepa, then how much percent is Deepa‟s salary less than that of Srujana? Sol: Here, x = 20. Required answer

x    1 0 0 % =  1 0 0 x  20   =   1 0 0 % 1 0 0  2 0 

2.

4 2 1 0 0 =  % = 1 6 % = 1 6 %.  3 6  6  If Anita‟s income is 30% less than that of Saritha, then how much percent is Saritha‟s income more than that of Anitha? Sol: Here, x = 30. Required answer

x    1 0 0 % =  1 0 0 x  =

3.

30 ×1 0 0 % 100 30

6  3 0 0 =  % = 4 2 %.  7  7  If A is 25% of C and B is 30% of C, then what percent age of A is B? Sol: A = 25% of C and B = 30% of C B 30 = A 25 30 B = x 100% of A 25 = 120% of A

4.

Two numbers are respectively 25% and 50% more than a third number. What percent is the first of the second? Sol: Here, x = 25 and y = 50 So, First number

1 0 0 x   1 0 0 % of the second =  1 0 0 y  1 0 0  2 5   1 0 0 % of the second =  1 0 0 5 0 

500 % of the second 6 2 1 = 8 3 % = 8 3 % of the second. 3 6 Two numbers are respectively 20% and 32% less than a third numb er. What percent is the second of the first? =

5.

Sol: Here, x = 20 and y = 32

30

1 0 0 y  So, Second number =   1 0 0 % of the first 1 0 0  x  

1 0 0  3 2  =   1 0 0 % of the first 1 0 0  2 0   = 85% of the first. 6.

If the price of a commodity increases by 50%, find how much percent its consumption be reduced so as not increase the expenditure. N    1 0 0 % Sol: Reduction in consumption =  1 0 0  N  

50    1 0 0 % =  1 0 0  5 0  

7.

1 1 0 0 =  % = 33 %.   3  3 If the price of a commodity decreases by 50%, find how much percent its consumption be increased so as not decrease the expenditure. N    1 0 0 Sol: Increase in consumption =  1 0 0  N    50  =   1 0 0 % 1 0 0  5 0   = 100%.

8.

If the salary of Mr. Shashi is first increased by 18% and thereafter decreased by 15%, what is the net change in his salary? Sol:

Here, x = 18 and y = -15

xy   So, the net % change in the salary = x  y  % 1 0 0 

(1 8)(1 5)   = 1 8  1 5  % 1 0 0   (1 8)(1 5)   = 1 8  1 5  % 1 0 0   = 0.3%. Since the sign is +ve, there is an increase in the salary of person by 0.3%. 9.

The population of a town is decreased by 20% and 40% in two successive years. What percent population is decreased after two years? Sol: Here, x = - 20 and y = - 40

xy   So, the net % change in population = x  y  % 1 0 0  (2 0)(4 0)   =  2 0  4 0  % 100  

8 0 0  =  6 0  % 1 0 0  = - 52%. Since the sign is -ve, there is decrease in population after two years by 52%. 10. If the side of a square is increased by 10%, its area increased by k%. Find the value of k. Sol: Area of square = side x side

31

So, net % change in area

xy   = x  y  % 1 0 0  (1 0)(1 0)   = 1 0  1 0  % 1 0 0  

[Take x, y = 10]

= 21% Hence, the area is increased by 21%. Here, k = 21. 11. The radius of a circle is increased by 4%. Find the percentage increase in its area. Sol: Area of circle = ∏ x radius x radius xy   So, net % change in area = x  y  % 1 0 0  (4)(4)   = 4  4  % 1 0 0  

[Take x, y = 4]

16  4  = 8   % = 8 2 5 %. 1 0 0   12. The tax on a commodity is diminished by 12% and its consumption increases by 10 %. Find the effect on revenue. Sol: Revenue = tax x consumption

xy   So, net % change in revenue = x  y  % 1 0 0  (1 2)(1 0)   =  1 2  1 0  % 1 0 0  

1 2 0  =  2  % 1 0 0  So, the revenue decreases by 3.2%.

[Take x = - 12, y = 10]

= - 3.2%.

13. The population of a town increases by 4% annually. If its present population is 12500, what will it be in 2 years time? Sol: Here, P = 12500, r = 4 and n = 2. n

Population after n years

r   = P1   . 1 0 0 

Population after 2 years

4   = 12500 1  1 0 0  

2

1   = 12500 1   2 5 

 2 6 = 12500    2 5 = 1 2 5 0 0

2

2

26 26  25 25

= 13520. 14. The population of a town increases by 10% annually. If its pre sent population is 12100, what it was 2 years ago? Sol: Here, P = 12100, r = 10 and n = 2. Population n years ago =

P r   1   1 0 0 

n

32

Population 2 years ago =

12100 2

10   1   1 0 0  12100 12100 = = 2 1 1 11 1 1      10 10  1 0 =

12100  1 0  1 0 = 10000. 1 1 1 1

15. The population of a town is 100000. It increases by 5% during the first year. During the second year, it decreases by 10% and increases by 15% during the third year. What is the population after 3 years? Sol: Here, P = 100000, x = 5, y = - 10 and z = x   So, Population after 3 years = P 1  1 0 0 

15 y   z   1  1 0 0 1  1 0 0    

5   10   15   = 1 0 0 0 0 01   1  1 0 0 1  1 0 0 1 0 0       100000  1 0 5 9 0 1 1 5 =  108675 1 0 0 1 0 0 1 0 0

16. In an examination, a student must get 50% marks to pass. If a student, who gets 110 marks, fails by 50 marks, find the maximum marks. Sol: It is clear, Minimum Marks = 110 + 50 = 160 Let the maximum marks = x So, 50% of x = 160 x = 320 17. In an examination, 40% students failed in Mathematics and 51% failed in Science. If 16% failed in both the subjects, find the percentage of those who passed in both the subjects. Sol: Here, a = 40, b = 51 and c = 16 Percentage of students passing both the subjects = (100 – ( a + b - c ))% = (100 – (40+51–16))% = 25%.

Brainstorming 1.

What percentage is 150 ml of 3 litres? 1. 2% 2. 5%

3. 8%

4. 50%

2.

If 40% of a number added to 1800, it gives the number itself. Find the number? 1. 3000 2. 257 3. 1500 4. 4500

3.

The difference between 12 ½% and 37½% of a number is 750. Find the number? 1. 1500 2. 4500 3. 3000 4. 257

4.

A number exceeds its 75% by 50. Then, find the number? 1. 175 2. 500 3. 750

5.

4. 200

If 30% of a number added to 441, it gives the number itself. Find the number? 1. 350 2. 630 3. 580 4. 250

33

6.

7.

The price of an article X is 10% less than that of article Y. By how much percent of the price article Y is more than that of the price of X? 1 1 1. 9 % 2. 11 % 3. 90% 4. 10% 9 11 In a library 50% of the books in Hindi, 40% of the remaining is English, and 30% of the remaining is Telugu. The remaining 18,900 books are Marathi. Find the total number of books? 1. 3,15,000 2. 45,000 3. 63,000 4. 90,000

8.

The cost of 1 kg of sweets increases every year by 10%, After two years the cost of 1 kg of sweets will increase by 1. 10% 2. 20% 3. 21% 4. 25%

9.

The marks obtained by two candidates A and B in arithmetic are 6.25 and 5.75 respectively. Marks percentage of A exceeds that of B by 1. 7.8% 2. 8% 3. 8.7% 4. 8.8

10.

Gopal spends 30% of his monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. If his monthly salary is Rs.18,400, how much money does he save every month? 1. Rs.3864 2. Rs.4264 3. Rs.3624 4. Rs.5888

11.

A man spends 25% of his income on house rent, 45% of this income on food and 40% of the balance on conveyance. If he is left with Rs.540 his income is 1. Rs.4500 2. Rs.3000 3. Rs.1350 4. Rs.900

12.

In an examination 450 candidates were boys and 550 were girls. If 32% of the boys and 38% of the girls passed the examination, the percentage of failed candidates is 1. 35.3% 2. 62% 3. 64.7% 4. 68%

13.

After having spent 25% of his money on machinery, 30% on raw material and 10% on staff, a person is left with Rs.70,000. The amount spent on raw material is 1. Rs.64000 2. Rs.60000 3. Rs.48000 4. Rs.40000

14.

The price of kerosene increases by 25%. By how much percentage a family must reduce consumption of kerosene so as not to increase the monthly expenditure on kerosene? 1. 15% 2. 20% 3. 25% 4. 30%

15.

Fifty percent students in a class are N.C.C cadets and 40 percent of the rest are scouts. Find the percentage of the total who are neither N.C.C. nor Scouts. 1. 18% 2. 15% 3. 25% 4. 30%

16.

A monthly return railway ticket costs 25 percent more than a single ticket. A week‟s extension can be had for the former by paying 5 percent of the monthly ticket‟s cost. If the money paid for the monthly ticket (with extension) is Rs.84, the price of the single ticket is 1. Rs.48 2. Rs.80 3. Rs.64 4. Rs.72

17.

Mahesh gets a certain sum as pocket money every month. He spends 20 percent on purchasing books and 25 percent of the remainder on stationary, 10 percent of what is left is given as charity. The rest he uses for purchasing sweets etc, and this works out to be Rs.108. His monthly pocket money is 1. Rs.200 2. Rs.250 3. Rs.350 4. Rs.500

18.

The population of a city increases at the rate of 10% annually. Its present population is 90.51 lakhs. The population 3 years ago was nearly ________ 1. 60 lakhs 2. 68 lakhs 3. 71 lakhs 4. 72.8 lakhs

19.

8% of the people eligible to vote are between 18 and 21. In an election, 85% of those eligible to vote, who were between 18 and 21, actually voted. In that election, the number of persons between 18 and 21, who actually voted, was what percent of those eligible to vote? 1. 4.2 2. 6.4 3. 6.8 4. 8

34

20.

When the price of wheat is increased by 20%, a family has to reduce the consumption by 4 kg keeping the expenditure fixed at Rs.80. The increased price per kg is 1. Rs.20 3. Rs.5 3. Rs.4 4. Rs.10

21.

In an examination every candidate took History or Geography of both, 64.8% took History and 60.2% of candidates took Geography. If the total number of candidates is 2000, the candidates who took both are 1. 500 2. 400 3. 375 4. None of these

22.

In a school of 400 students, 40% of girls, 20% of the boys and 30% of the students failed in the exam. The % of students passed in the exam is: 1. 75% 2. 76% 3. 45% 4. None

23.

In an election, the winning candidate received 10400 votes, which represented 52% of the electorate. The only other candidate secured votes from 23% of the electorate, calculate the number of people who did not cast their votes? 1. 3500 2. 4000 3. 4800 4. 5000

24.

In a competitive examination, a candidate obtained 45% marks and failed to get selected by 120 marks. Another candidate obtained 58% marks and failed to get selected by 16 marks. Find the maximum marks and the percentage of marks for selection? 1. 600, 60% 2. 750, 55% 3. 800, 65% 4. 800, 60%

25.

Three persons A, B and C whose salaries together amount to Rs.2560, spend 90, 85 and 80 percent in their salaries respectively. If their savings are 5:6:14, find their respective salary? 1. 800, 640, 1120

2. 840, 600, 1120

3. 780, 760, 1020

4. 800, 760, 1000

26.

In an examination a candidate attempts 90% of the total questions at of these 80% his answers are correct. Each question carries 2 marks for correct answer and negative (-1) mark for wrong answer. If the marks secured by the candidate is 189 what is the total number of questions?] 1. 250 2. 150 3. 100 4. 350

27.

The sales of a company, reduced to 20%. After how much percentage increase the sales of the company be in original? 1. 20% 2. 25% 3. 10% 4. 30%

28.

In a town 60% of people read Hindu, 50% of read the Times of India, 30% read both. Find the percentage of the people who read neither? 1. 80% 2. 20% 3. 10% 4. 40%

29.

The population of the town 20,000. If the number of males are increased by 40% then at the end of the year the population will be 23,200. Find the number of females, males in the town at the beginning? 1. 12000, 8000 2. 18000, 12000 3. 10000, 10000 4. None

30.

In an election where there are two candidates only contested. One candidate who gets 70% of the votes is elected by a majority of 2000 votes. What is total number of votes? 1. 5000 2. 6000 3. 3000 4. None

31.

A candidate has to obtain 33% of the total marks to pass the examination. Ram got 25% of the total marks and failed by 40 marks. What are the maximum marks? 1. 400 2. 750 3. 500 4. 5000

32.

In an examination A gets 28% of the total marks and fails by 24 marks. While B gets 31% of the total marks and got 48 marks more than the pass mark. Find the maximum marks and pass %? 1. 3000, 29% 2. 2400, 29% 3. 2500, 29% 4. 3000, 30%

33.

In an examination there were 5000 candidates out of which 3000 were boys. If 60% of the boys and 20% of the girls passed. Find the percentage of failed candidates? 1. 60% 2. 76% 3. 56% 4. 62%

35

34.

35.

36.

If the price of sugar decrease by 25%. Find by how much percent must a householder increase his consumption of that article so as not to reduce his expenditure? 1 1 3 1. 33 % 2. 11 % 3. % 4. None 9 3 100 If the price of wheat decreased by 20%, find by how much percent must a householder increase his consumption of that article so as not to reduce his expenditure? 2 2 2 2 1. 11 % 2. 13 % 3. 15 % 4. 16 % 3 3 3 3 The population of a town is 1,00,000 now. If it increases at a rate of 10% p.a. find the population after 2 years? 1. 112000 2. 121000 3. 120100 4. None

37.

If the population of a town is 5000 now and it increases at 10% first year 20% in the second year then its population after 2 years? 1. 6600 2. 5500 3. 6000 4. 5000

38.

Two numbers are 20% and 40% of the third number. How much % of the first number in the second number? 1. 20% 2. 200% 3. 10% 4. 100%

39.

A reduction of 20% in the price of oranges would enable a person to get 2 dozen more for Rs.50. Find the reduced price per dozen. 1. Rs.3 2. Rs.5 3. Rs.4 4. Rs.4.5

40.

The tax on a commodity is diminished by 15% and its consumption increased by 10%. What is the effect on the revenue derived from it? 1. 5.5% increase 2. 5.5% decrease 3. 6.5% decrease 4. 6.5% increase

1) 2 2) 1 3) 3 4) 4 5) 2 6) 2 7) 4 8) 3 9) 3 10) 1

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

2 3 2 2 4 3 1 2 3 3

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

1 4 4 4 1 2 2 2 1 1

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

3 2 3 1 4 2 1 2 2 3

36

5. AVERAGES Average: The average of a number of quantities of the same kind is equal to their sum divided by the number of those quantities. It is also called mean or arithmetic mean. 16 1357 For example: The average of 1, 3, 5, 7 is = = 4 4 4 Basic Formulae:

1. 2. 3. 4.

Sum of quantities N umberof quantities Sum of quantities = Average  Number of quantities s umof quantities Number of quantities = A verage If the number of quantities in two groups be a 1 and a 2 and their average is x and y, respectively, then the combined average (average of all of them put together) is given by

Average =

a1 x  a2 y . a1  a2 5.

6.

If the average of a 1 quantities is x and the average of a 2 quantities out of them is y, the a x  a2 y average of remaining group (rest of the quantities) is given by 1 . a1  a2 If

x

is the average of x 1 , x 2 ,…….., x n , then a.

The average of x 1 + a, x 2 + a,……., x n + a is

x+

b.

The average of x 1 - a, x 2 - a,……., x n - a is

c.

The average of ax 1 , ax 2 ,…….., ax n is ax , where a ≠ 0.

d.

The average of

x-

a.

a.

x1 x2 x x . , ,......... ., n is a a a a

7.

The average of n quantities is equal to x. If one of the given quantities whose value is p, is replaced by a new quantity having value q, the average becomes y, then q = p + n (y – x).

8.

The average of n quantities is equal to x. If a quantity is removed, the average becomes y. The value of the removed quantity is n (x – y) + y.

9.

The average of n quantities is equal to x. If a quantity is added, the average becomes y. The value of the new quantity is n (y – x) + y. n1 10. The average of first n natural numbers is . 2 (n  1)(2n  1) 11. The average of squares of natural numbers till n is . 6 12. The average of cubes of natural numbers till n is

13. The average of odd numbers from 1 to n is 14. The average of even numbers from 1 to n is

n(n  1)2 . 4

(las toddnumber+ 1) 2

(las tevennumber+ 2)

2 15. If n is odd: The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers is always the middle number.

37

16. If n is even: The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers is always the average of the middle two numbers. 17. The average of first n consecutive even numbers is (n + 1). 18. The average of first n consecutive odd numbers is n.

2(n  1)(2n  1) . 3 (n  1)(n  2) 20. The average of squares of consecutive even numbers till n is . 3 n(n  2) 21. The average of squares of consecutive odd numbers till n is . 3 22. If the average of n consecutive numbers is m, then the difference between the smallest and the largest number is 2 (n – 1). 19. The average of squares of first n consecutive even numbers is

23. Geometric Mean or Geometric Average: It is useful in calculating averages of ratios such as average population growth rate, average percentage inc rease and so on. Geometric mean of x 1 , x 2 ,…….., x n is denoted by G.M. =

n

x1.x2.......... .xn

24. Harmonic Mean or Harmonic Average: It is useful for finding average speed of a vehicle, average production per day and so on. 1 H.M. = 11 1 1  ..........    n  x1 x2 xn  25. If a certain distance is covered at a speed of x km/h and the same distance is covered at  2xy  a speed of y km/h, the average speed during the whole journey is   km/h. x  y 26. If a person or a motor car covers three equal distances at t he speed of x km/h, y km/h, z km/h, respectively, then for the entire journey average speed of the person or motor car   3xyz is   km/h. xy  yz  zx   27. If a person covers A km at a speed of x km/h, B km at a speed of y km/h and C km at a    A BC   km/h. speed of z km/h, then the average speed during the whole journey is  A B  C x y z   th th 28. If a person covers A part of the distance at x km/h, B part of the distance at y km/h and the C th part at z km/h, then the average speed during the whole journey is     1   km/h. A B  C x y z  

SOLVED EXAMPLES 1.

Sunil purchased 4 toys at the rate of Rs.100 each, 6 toys at the rate of Rs.150 each and 8 toys at the rate of 200 each. What is the average cost of one toy? Sol: Cost of 4 toys Cost of 6 toys Cost of 8 toys Total number of toys = 4 +

= 100 x 4 = Rs.400 = 150 x 6 = Rs.900 = 200 x 8 = Rs.1600 6 + 8 = 18

38

Average price of 1 toy

=

4 0 0+ 9 0 0+ 1 6 0 0 18

= Rs.161.11 2.

The average marks obtained by 100 students in a competitive examination is 50. Find the total marks. Sol: Total marks

3.

= Average marks x Number of students = 100 x 50 = 5000.

The average weight of 40 students of section A of I-CET class is 55 kg and that of 50 students of section B is 60 kg. Find the average weight of all the 90 students of the class. Sol: Here, a 1 = 40, a 2 = 50 and x = 55, y = 60 a x  a2 y Average weight = 1 a1  a2 =

4.

(4 0)(5 5)  (5 0)(6 0) 40 50

= 57.78 kg. Average salary of all the 30 employees including 5 officers of a company is Rs.750 0. If the average salary of the officers is Rs.15000, find the average salary of the remaining staff of the company. Sol: Here, a 1 = 30, a 2 = 5 and x = 7500, y = 15000 a x  a2 y Average salary of the remaining staff = 1 a1  a2 =

3 0(7 5 0)  5(1 5 0 0) 3 0 5

= 6000. 5.

The average value of five numbers 7, 10, 16, 24 and 28 is 17. If 6 is added to each number, what will be the new average? Sol: The new average =

6.

+ a = 17 + 6 = 23.

The average of x numbers is 4x. If x – 3 is subtracted from each given number, what will be the new average? Sol: The new average =

7.

x

x

- a = 4x – (x – 3) = 3x + 3.

The average of 8 numbers is 20. If each of the numbers is multiplied by 8, find the average of a new set of numbers. Sol: The average of a new set of numbers = a

x

= 8 x 20 = 160.

8.

The average weight of 20 persons is increased by 3 kg when one of them whose weight is 50 kg, is replaced by a new person. What is the weight of th e new person? Sol: The weight of the new person, q = p + n (y – x). = 50 + 20(3) = 50 + 60 = 110 kg.

9.

The average age of 24 students and the Maths teacher is 16 years. If the Maths teacher‟s age is excluded, the average age reduces by 1 year. Wha t is the age of the Maths teacher? Sol: The age of Maths teacher

= n ( x - y) + y = 25 (16 – 15) + 15) = 40 years.

39

10. Find the average of first 79 natural numbers.

n1 2 7 9 1 = = 40. 2 11. Find the average of squares of the natural numbers from 1 to 47. Sol: The required average =

(n  1)(2n  1) 6 (4 7  1)2(4 7)  1 = 6 4 8 9 5 = = 760. 6 12. Find the average of cubes of the natural numbers from 1 to 15. Sol: The required average =

Sol: The required average =

n(n  1)2 4

1 5(1 5  1)2 4 1 5 1 6  1 6 = = 960. 4 13. Find the average of odd numbers from 1 to 50. =

l as toddnumbe r 1 2 4 9 1 = = 25. 2 14. Find the average of even numbers from 1 to 61. Sol: The required average =

las te v e nnumbe r 2 2 60 2 = = 31. 2

Sol: The required average =

15. Find the average of 5 consecutive numbers 4, 5, 6, 7, 8. Sol: The required average = middle number = 6. 16. Find the average of consecutive odd numbers 21, 23, 25, 27, 29, 31. Sol: The required average

= average of middle two numbers =

25  27 = 26. 2

17. Find the average of first 25 consecutive even numbers. Sol: The required average = (n + 1) = 25 + 1 = 26. 18. Find the average of first 30 consecutive odd numbers. Sol: The required average = n = 30. 19. Find the average of squares of first 16 consecutive even numbers. Sol: The required average =

2(n  1)(2n  1) 3

40

2(1 6  1)2(1 6)  1 3 2  1 7 3 3 = = 374. 3 20. Find the average of squares of consecutive even numbers from 1 to 35. =

(n  1)(n  2) 3 (3 4 + 1)(3 4 + 2) = 3 3 5× 3 6 = = 420 3 21. Find the average of squares of consecutive odd numbers from 1 to 44. n(n + 2) Sol: The required average = 3 4 3(4 3 + 2) = 3 4 3× 4 5 = = 645 3 22. If the average of 6 consecutive numbers is 48, then find the difference between the smallest and the largest number? Sol: The required average =

Sol: The required difference = 2 (n – 1) = 2 (6 – 1) = 10. 23. The production of a company for three successive years has increased by 10%, 20% and 40% respectively. Find the average annual increase of production. 1

Sol: Average annual increase = G.M. of x, y and z = (x × y × z) 3 % 1

= (1 0× 2 0× 4 0) 3 % = 20%.

24. The population of a city in two successive years increases at the rates 12% and 3% respectively. Find the average increase of two years. Sol: Average population increase = G.M. of x and y 1

= (x  y) 2 % 1 (1 2  3) 2

= = 6%.

%

25. A man runs 2 km at 20 km/ph and another 2 km at 10 km/ph. Find the average speed for the whole distance in covering 4 km. 2xy Sol: Required formula = and here, x = 20 and y = 10 x+y

 2(2 0) (1 0)  =   km/ph  20 10  1 40 = = 1 3 km/ph. 3 3 26. A train covers the first 150 km at a speed of 100 km /ph, another 150 km at a speed of 120 km/ph and the last 150 km at 80 km/ph. Find the average speed of the train for the entire journey.   3xyz Sol: Required Average speed =   km/ph xy  yz  zx   So, Average speed

41

  3(1 0 0)(1 2 0)(8 0) =   km/ph  (1 0 0 1 2 0)  (1 2 0 8 0)  (1 0 0 8 0)   2 8 8 0 0 0 0 =   km/ph  29600 

= 97.3 km/ph (app) 27. A person covers 8 km at a speed of 4 km/ph, 15 km at a speed of 5 km/ph and 20 km at a speed of 10 km/ph. Find the average speed of the whole journey.    A BC   km/ph. Sol: The required average speed =  A B  C x y z   =

8 + 1 5+ 2 0 8 15 20 + + 4 5 10

km/ph.

 8  1 5  2 0 =   km/ph.  232  =

43 1 km/ph. 6 7 7

1 th 3 th of the distance at 5 km/ph, the next at 8 km/ph and the 4 5 remaining distance at 10 km/ph. Find the average speed during the entire journey.

28. A person covers the first

Sol: The required average speed

=

=

=

=

=

1 A B C + + x y z

km/ph

1 1 4 3 5 3 20 + + 5 8 10

1 1 3 3 + + 20 40 200

km/ph

km/ph

1 km/ph 1 0+ 1 5+ 3 200 1 200 km/ph = 7 km/ph. 7 28

29. A train covers 50% of the journey at 25 km/ph, 25% of the journey at 15 km/ph and the remaining at 10 km/ph. Find the average speed during the entire journey.     1 0 0  km/ph Sol: The required average speed =  A B C x  y  z  

    100 =   km/ph 5 0 2 5 2 5      2 5 1 5 1 0 

42

    100 =   km/ph  1 2  1 0  1 5 6   =

600 8 km/ph. = 16 37 37

Brainstorming 1.

The average marks obtained by 144 candidates in a certain examination is 55. Find the total marks? 1. 7290 2. 7920 3. 7930 4. 7390

2.

Ram travels half of a journey at the speed of 24km/hr and the next half at a speed of 16km/hr. What is the average speed of the Ram during the whole journey? 1 1 2 1. 19 km/hr 2. 20km/hr 3. 19 km/hr 4. 16 km/hr 5 5 5 A person divides his total route of journey into three equal parts and decides to travel the three parts with speeds of 20, 25 and 40km/hr respectively. Find his average speed during the whole journey? 2 1 2 1 1. 26 km/hr 2. 26 km/hr 3. 25 km/hr 4. 25 km/hr 23 23 23 23 1 th A person runs the first of the distance at 2km/hr, the next one half at 3km/hr and the 5 remaining distance at 1km/hr. Find the average speed. 11 5 1. 2.02 km/hr 2. km/hr 3. km/hr 4. None 5 11 A man covers first 20% of the distance at 10km/hr, next 50% at 5km/hr and the remaining at 20km/hr. Find the average speed of the train during the whole journey? 25 5 23 1. 25 km/hr 2. 25 km/hr 3. 25 km/hr 4. None of these 47 47 47 The average salary of the entire staff in a office is Rs.220 per day. The average salary of officers is Rs.650 and that of non-officers is Rs.170. If the number of officers is 25, then find the number of non-officers in the office. 1. 215 2. 315 3. 250 4. 350

3.

4.

5.

6.

7.

The average age of all the students of a class is 22 years. The average age of boys of the class is 26 years and that of the girls is 19 years. If the number of girls in the class is 16, then find the number of boys in the class. 1. 12 2. 10 3. 6 4. 8

8.

There were 45 students in a hostel. If the number of students increases by 9, the expense of the mess increase by Rs.25 per day while the average expenditure per head diminishes by Rs.1. Find the original expenditure of the mess. 1. Rs.390 2. Rs.295 3. Rs.395 4. Rs.400

9.

There were 42 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs.32.5 per day while the average expenditure per head diminishes by Rs.1.5. Find the original expenditure of the mess. 1. Rs.636 2. Rs.536 3. Rs.630 4. Rs.656

10.

The average weight of 20 balls is 2gm. If the weight of the bag be included the average weight increases by 0.02gm. What is the weight of the bag? 1. 2.04gm 2. 2.42gm 3. 3.42gm 4. 3.04gm

11.

Find the average of first 14 consecutive even numbers. 1. 15 2. 14 3. 210

4. 28

43

12. 13.

Find the average of first 17 consecutive odd numbers. 1. 18 2. 289 3. 17

4. 34

Find the average of squares of first 23 consecutive even numbers. 1. 750 2. 754 3. 725

4. 752

14.

Find the average of squares of consecutive even numbers from 1 to 22. 1. 184 2. 174 3. 182 4. 186

15.

Find the average of squares of consecutive even numbers from 1 to 35. 1. 484 2. 445 3. 408 4. 420

16.

Find the average of squares of consecutive even numbers from 1 to 62. 1. 1280 2. 1281 3. 1821 4. 1218

17.

The average weight of a group of 20 boys was calculated to be 89.4kg and it was later discovered that one weight was misread as 78kg instead of the correct one of 87kg. The correct average weight is ____________ 1. 88.95kg 2. 89.25kg 3. 89.55kg 4. 89.85kg

18.

The average weight of 15 students was calculated to be 52kg and it was later discovered that one weight was misread as 21kg instead of the correct one of 12kg. The correct average weight is __________ 1. 51.4kg 2. 50.6kg 3. 52.4kg 4. 51.6kg

19.

The average of the first and the second of three numbers is 10 more than the average of the second and the third of these numbers. What is the difference between the first and the third of these three numbers? 1. 40 2. 10 3. 20 4. None

20.

The average of the first and the second of three numbers is 13 more than the average of the second and the third of these numbers. What is the difference between the first and the third of these three numbers? 1. 25 2. 24 3. 26 4. 19

21.

The average marks scored by in English, Science, Mathematics and History is less than 15 from that scored by him in English, History, Geography and Mathematics. What is the difference of marks is Science and Geography score by him? 1. 40 2. 50 3. 60 4. Data inadequate

22.

The average temperature for Monday, Tuesday and Wednesday was 40oC. The average for Tuesday, Wednesday and Thursday was 41oC and that of Thursday being 45oC. What was the temperature on Monday? 1. 48oC 2. 41oC 3. 46oC 4. 42oC

23.

If average of 6 consecutive even numbers is 48, what is the difference between the smallest and the largest numbers? 1. 10 2. 12 3. 9 4. Data inadequate

24.

The average attendance of a college for the first three days of a week is 325, and for first four days it is 320. How many were present on the fourth day? 1. 305 2. 350 3. 530 4. 503

25.

A car runs for t1 hours at v1 km/hr, t2 hours at v2 km/hr. What is the average speed of the car for the entire journey?

26.

1.

t1  t 2 km/hr v1 t 1  v 2 t 2

2.

v1 t 1  v 2 t 2 km/hr t1  t 2

3.

v1 t 2  v 2 t 1 km/hr v1  v 2

4.

v1  v 2 km/hr v1 t 2  v 2 t 1

The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 years, the average age of the family at the birth of the youngest member, was _____________

44

1. 15 years

2. 17 years

3. 17.5 years

4. 18 years

27.

5 years ago, the average of A, B, C and D was 45. With E joining them now, the average age of all the five is 49 years. How old is E? 1. 25 years 2. 40 years 3. 45 years 4. 60 years

28.

The average height of 40 students is 163cm. On a particular day, three students A, B, C were absent and the average of the remaining 37 students was found to be 162cm. If A, B have equal heights and the height of C be 2cm less than that of A, find the height of A? 1. 176cm 2. 166cm 3. 180cm 4. 186cm

29.

Of the three numbers, second is twice the first and is also thrice the third. If the average of the three numbers is 44, the largest number is ____________ 1. 24 2. 36 3. 72 4. 108

30.

The average weight of 3 men A, B and C is 84kg. An other man D joins the group and the average now becomes 80kg. If another man E, whose weight is 3kg more than that of D, replaces A, then average weight of B, C, D and E becomes 79kg. The weight of A is ____ 1. 70kg 2. 72kg 3. 75kg 4. 80kg

31.

A cricketer has completed 14 innings and his average is 30 runs. How many runs must he make in his next innings so as to raise his average to 32? 1. 60 2. 55 3. 65 4. 50

32.

The average of marks obtained by 102 candidates in a certain examination is 18. If the average marks of passed candidates is 21 and that of the failed candidates is 15, what is the number of candidates who failed the examination? 1. 51 2. 52 3. 61 4. 50

33.

The average of marks obtained by 65 candidates in a certain examination is 25. If the average marks of passed candidates is 27 and that of the failed candidates is 14, what is the number of candidates who passed the examination? 1. 55 2. 65 3. 60 4. 75

34.

The average of marks obtained by 115 candidates in a certain examination is 36. If the average marks of passed candidates is 40 and that of the failed candidates is 17, what is the number of candidates who failed the examination? 1. 30 2. 25 3. 20 4. 34

35.

On a journey across Mumbai, a taxi averages 30km/hr for 60% of the distance, 20km/hr for 20% of it and 10km/hr for the remainder. The average speed for the whole journey (in km/hr) is _______________ 1. 20km/hr 2. 22.5km/hr 3. 25km/hr 4. 24.635km/hr

36.

The average of marks obtained by 108 candidates in a certain examination is 20. If the average marks of passed candidates is 28 and that of the failed candidates is 16, what is the number of candidates who failed the examination? 1. 70 2. 78 3. 81 4. 72

37.

In a class there are 16 students whose average age is decreased by 3 months, when 2 students aged 24 and 26 years respectively are replaced by the same number of students. Find the average age of the new students. 1. 23 years 2. 21 years 3. 18 years 4. None of these

38.

The average age of 44 boys in a class is 26 years. If the teacher‟s age is included the average age of the boys and the teacher becomes 27 years. What is the teacher‟s age? 1. 69 years 2. 71 years 3. 59 years 4. Data inadequate

39.

The average age of 30 students in a class is 12 years. The average age of a group of 5 students is 10 years and that of another group of 5 students is 14 years. Find the average age of the remaining students. 1. 14 years 2. 10 years 3. 12 years 4. Data inadequate

45

40.

The average age of 15 students and the class teacher is 15 years. If the class teacher‟s age is excluded, the average reduces by 1 year. What is the age of the class teacher? 1. 30 years 2. 31 years 3. 29 years 4. 28 years

1) 2 2) 3 3) 1 4) 4 5) 4 6) 1 7) 1 8) 3 9) 1 10) 2

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

1 3 4 1 4 2 4 1 3 3

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

3 4 1 1 2 4 3 1 3 3

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

1 1 1 3 1 4 1 2 3 3

6. RATIO AND PROPORTION Ratio: A ratio is a comparison of two quantities by division. If a and b are two numbers, then the ratio of a to b is

a or a ÷ b. It is denoted by a : b. b

Note: 1. The two quantities in a : b that are being compared are called terms. 2. The first term is called antecedent and the second term is called consequent. 4 Ex: The ratio 4:5 represents with antecedent 4 and consequent 5. 5 3. A ratio is a number, so to find the ratio of two quantities; they must be expressed in the same units. 4. A ratio does not change if both of its terms are multiplied or divided by the same number. Thus, 8 12 4 = = etc. 5 10 15

TYPES OF RATIOS: 1. 2.

Duplicate Ratio: If a : b is the given ratio then its duplicate ratio is a2 : b2. Triplicate Ratio: If a : b is the given ratio then its triplicate ratio is a3 : b3.

3.

Sub-duplicate Ratio: If a : b is the given ratio then its sub-duplicate ratio is

4. 5. 6.

a : 3

3

b.

Sub-triplicate Ratio: If a : b is the given ratio then its sub-triplicate ratio is a : b . Inverse Ratio or Reciprocal Ratio: If a : b is the given ratio, then it‟s inverse or reciprocal ratio is b : a. Compound Ratio: If a : b and c : d are two given ratios, then ac : bd is the compound ratio of the given ratios.

Proportion: The equality of two ratios is called proportion. Note: a c 1. If = , then a, b, c and d are said to be in proportion. d b 2. We write it as a : b :: c : d and read as “a is to b as c is to d”. a c 3. Each term of and is called a proportional. d b

46

4.

In the a : b :: c : d proportion, a, d are known as extremes and b, c are known as means.

5.

If four quantities are in proportion, then

Product of Means = Product of Extremes

3 6 = , we write as 3:2 :: 6:4 and say 3, 2, 6 and 4 are in proportion. 2 4 Here, 3 and 4 are extremes and their product = 3 x 4 = 12 2 and 6 are means and their product = 2 x 6 = 12 Ex:

6.

7.

Fourth Proportional: If a : b :: c : x, then x is called the fourth proportional of a, b, c. a c bc We have, = , or x  . b x a bc Thus, fourth proportional of a, b, c is . a Third Proportional: If a : b :: b : x, then x is called the third proportional of a, b.

8.

Mean Proportional: If a : b is the given ratio, then the mean proportional of a and b is

ab .

Formulae: 1.

If

a c = , then d b ab cd = d b ab cd b) Dividendo Rule: = d b a) Componendo Rule:

ab cd = cd ab If two numbers are in the ratio of a : b and the sum of these numbers is x, then these ax bx numbers will be and respectively. ab ab If in a mixture of x litres two liquids A and B are in the ratio of a : b, then the quantities of ax bx liquids A and B in the mixture will be litres and litres respectively. ab ab If three numbers are in the ratio a : b : c and the sum of these numbers is x, then these ax bx cx numbers will be , and respectively. abc abc abc If two numbers are in the ratio a : b and the difference of these numbers is x, then these numbers will be ax bx a) and respectively, (where a > b). ab ab ax bx b) and respectively, (where b > a). ba ba The ratio between two numbers is a : b. If x is added to each of these numbers, the ratio bx(c  d) ax(c  d) becomes c : d, then the two numbers are and . ad  bc ad  bc The ratio between two numbers is a : b. If x is subtracted from each of these numbers, the bx(d  c) ax(d  c) ratio becomes c : d, then the two numbers are and . ad  bc ad  bc If the ratio of two numbers is a : b, then the number that should be added to each of the ad  bc numbers in order to make this ratio c : d is given by . cd c)

2.

3.

4.

5.

6.

7.

8.

Componendo and Dividendo Rule:

47

9.

10.

11.

12.

13.

14.

15.

If the ratio of two numbers is a : b, then the number that should be subtracted from each of bc  ad the numbers in order to make this ratio c : d is given by . cd There are four numbers a, b, c and d. The number that should be subtracted from each of ad  bc the numbers so that the remaining numbers may be proportional is given by . (a  d)  (b  c) There are four numbers a, b, c and d. The number that should be added to each of the bc  ad numbers so that the new numbers may be proportional is given by . (a  d)  (b  c) The incomes of two persons are in the ratio of a : b and their expenditures are in the ratio of c : d. If the savings of each person be Rs.X, then their aX(d  c) bX(d  c) Incomes are Rs. and Rs. and their ad  bc ad  bc c X(b  a) dX(b  a) Expenditures are Rs. and Rs. . ad  bc ad  bc If in a mixture of x litres of two liquids A and B, the ratio of liquids A and B is a : b, then the x(ad  bc) quantity of liquid B to be added in order to make this ratio c : d is . c(a  b) In a mixture of two liquids A and B, the ratio of liquids A and B is a : b. If on adding x litres of liquid B to the mixture, the ratio of A to B becomes a : c, then in the beginning the quantity ax bx of liquid A in the mixture was and that of liquid B was litres. c b c b When two ingredients A and B of quantities q1 and q2 and cost price per unit c1 and c2 are mixed to get a mixture c having cost price per unit cm , then a)

q1 c  cm = 2 and cm  c1 q2

b) c m 

c1  q1  c 2  q2 q1  q2

16. If a : b  n1 : d1 and b : c  n2 : d2 , then

a : b : c  (n1  n2 ) : (d1  n2 ) : (d1  d2 ) 17. If

a : b  n1 : d1

and

b : c  n2 : d 2

and

c : d  n3 : d 3 , then

a : b : c : d  (n1  n2  n3 ) : (d1  n2  n3 ) : (d1  d2  n3 ) :(d1  d2  d3 ) 18. If a mixture contains two ingredients A and B in the ratio a : b, then a a) Percentage of A in the mixture =  1 0 0% . ab b b) Percentage of B in the mixture =  1 0 0% . ab 19. If two mixtures M1 and M2 contain ingredients A and B in the ratios a : b and c : d respectively, then a third mixture M3 obtained by mixing M1 and M2 in the ratio x : y will contain: cy  ax  ab  c  d   1 0 0% ingredient A, and a)  xy    

dy  bx  ab  c  d   1 0 0% ingredient B or b)  xy    

48

  1 0 0   

 cy   ax    a  b c  d     1 0 0% ingredient B xy       

SOLVED EXAMPLES 1. The duplicate ratio of 4 : 5 is ______. Sol: Duplicate ratio of 4 : 5 is 42 : 52 = 16 : 25. 2. The triplicate ratio of 2 : 3 is _______. Sol: Triplicate ratio of 2 : 3 is 23 : 33 = 8 : 27. 3. The sub-duplicate ratio of 9 : 16 is ________. Sol: Sub-duplicate ratio of 9 : 16 =

9 : 1 6 = 3 : 4.

4. The sub-triplicate ratio of 8 : 27 is _________. Sol: Sub-triplicate ratio of 8 : 27 is

3

8 : 3 2 7 = 2 : 3.

5. The inverse ratio of 4 : 5 is _________. Sol: Inverse ratio of 4 : 5 is 5 : 4. 6. The compound ratio of 3 : 4, 4 : 5 and 5 : 6 is _________. Sol: Compound ratio of 3 : 4, 4 : 5 and 5 : 6 = (3 x 4 x 5) : (4 x 5 x 6) = 60 : 120 = 1 : 2. 7. Find a fourth proportional to 2, 5, 6. Sol: Let x be the fourth proportional to 2, 5, 6. 2 6 Then, 2 : 5 :: 6 : x or = 5 x So, x = 15. 8. Find a fourth proportional to 5, 4. Sol: Let x be the third proportional to 5, 4. 5 4 Then 5 : 4 :: 4 : x or  4 x 16 x  3.2 . 5 9. Find the mean proportional between 32 and 8. Sol: The mean proportion between 32 and 8 =

3 2 8  2 5 6  1 6.

10. Two numbers are in the ratio 3 : 4 and the sum of these numbers is 21. Find the two numbers. Sol: Here, a = 3, b = 4 and x = 21

49

3  21 ax = = 9 ab 34 4 21 bx The second number = = = 12 ab 34 (or) = 21 – 9 = 12. The first number =

11. Two numbers are in the ratio 3 : 5 and the difference of these numbers is 36. Find the two numbers. Sol: Here, a = 3, b = 5 and x = 36 336 ax The first number = = = 54 ba 53 536 bx The second number = = = 90 ba 53 (or) = 54 + 36 = 90. 12. If a : b = 3 : 4 and b : c = 5 : 6, find a : b : c. Sol: Here, n1  3, n2  5, d1  4, d2  6

a : b : c  (n1  n2 ) : (d1  n2 ) : (d1  d2 )

 (3  5) : (4  5) : (4  6) = 15: 20: 24 . 13. Given two numbers which are in the ratio of 2 : 3. If 6 is added to each of them, their ratio is changed to 4 : 5. Find the two numbers. Sol: Here, a : b = 2 : 3 and c : d = 4 : 5 and x = 6 2(6)(4  5) ax(c  d) The first number = = ad  bc 2(5)  3(4) 1 2 = 6 2 3(6)(4  5) bx(c  d) The second number = = 2(5)  3(4) ad  bc 1 8 =  9. 2 14. The ratio of two numbers is 5: 9 If each number is decreased by 5, the ratio becomes 5 : 11. Find the numbers. Sol: Here, a : b = 5 : 9 and c : d = 5 : 11 and x = 5 5(5)(1 1  5) ax(d  c) The first number = = 5(1 1)  9(5) ad  bc 2 5 6 =  15 10 9(5)(1 1  5) bx(d  c) The second number = = 5(1 1)  9(5) ad  bc 4 5 6 =  2 7. 10 15. Find the number that must be subtracted from the terms of the ratio 5 : 6 to make it equal to 2 : 3. Sol: Here, a : b = 5 : 6 and c : d = 2 : 3 bc  ad Required number = cd 6(2)  5(3)  3. = 23

50

16. Find the number that must be added from the terms of the ratio 11 : 29 to make it equal to 11 : 20. Sol: Here, a : b = 11 : 29 and c : d = 11 : 20 1 1(2 0)  2 9(1 1) ad  bc Required number = =  1 1. cd 1 1 2 0 17. Find the number subtracted from each of the numbers 54, 71, 75 and 99 leaves the remainders which are proportional. Sol: Here, a = 54, b = 71, c = 75 and d = 99 5 4(9 9)  7 1(7 5) ad  bc   3. Required number = (a  d)  (b  c) (5 4  9 9)  (7 1  7 5) 18. Annual income of Shashi and Ravi is in the ratio of 4 : 3 and the ratio of their annual expenses is 3 : 2. If each of them saves Rs. 500 at the end of year, then find their annual income. Sol: Here, a : b = 4 : 3 and c : d = 3 : 2 and X = 500 aX(d  c) So, Annual income of Shashi = ad  bc 4(5 0 0)(2  3) 2 0 0 0  = 4(2)  3(3) 1 = Rs.2000 Annual income of Ravi

bX(d  c) ad  bc 3(5 0 0)(2  3) 1 5 0 0  = 4(2)  3(3) 1 = Rs.1500. =

19. The incomes of Madhav and Kiran are in the ratio 9 : 4 and their expenditures are in the ratio 5 : 2. If each saves Rs.1000, find their expenditures. Sol: Here, a : b = 9 : 4 and c : d = 5 : 2 and X = 1000 c X(b  a) So, the expenditure of Madhav = ad  bc 5(1 0 0 0)(4  5) 5 0 0 0  = 9(2)  4(5) 2 = Rs.2500 dX(b  a) Expenditure of Kiran = ad  bc 2(1 0 0 0)(4  5) 2 0 0 0  = 9(2)  4(5) 2 = Rs.1000. 20. 630 ml of a mixture contains milk and water in the ratio 7 : 2. How much more water is to be added to get a new mixture containing milk and water in the ratio of 7 : 3. Sol: Here, a : b = 7 : 2 and c : d = 7 : 3 and x = 630 x(ad  bc) Required quantity of water to be added = c(a  b) = =

630 7(3)  2(7) 7(7  2) 630(21  14) 63

= 70 ml

51

21. A mixture contains milk and water in the ratio 8 : 3. On adding 6 litres of water, the ratio of milk to water becomes 8 : 5. Find the quantity of water in the mixture. Sol: Here, a : b = 8 : 3 and a : c = 8 : 5 and x = 6 bx So, the quantity of water in the mixture = c b 3(6) =  9 litres. 53 22. In what ratio the two kinds of tea powder (Ex: general tea powder with lams a) must be mixed together into one at Rs.10 per kg and another at Rs.15 per kg, so that the mixture may cost Rs.12.2 per kg? Sol: Here, c1  1 0 , c 2  1 5, cm  1 2.2 Required ratio,

q1 q2

=

c2  cm cm  c1

1 5  1 2.2 2.8   1.4 : 1.1 1 2.2  1 0 2.2 Thus, the two kinds of tea powder are mixed in the ratio 1.4 : 1.1. =

23. In a mixture of two types of liquids L 1 and L 2 , the ratio L 1 : L 2 is 4 : 3. If the cost of liquid L 1 is Rs.5 per litre and that of L 2 is Rs.10 per litre, then find the cost per litre of the resulting mixture. Sol: Here, q1  4, q2  3, c1  5, c2  1 0 So, the cost of the resulting mixture, c m 

c1  q1  c 2  q2 q1  q2 =

c1  q1  c 2  q2 q1  q2

5  4  1 0 3 5 0  43 7 = Rs.7.14 (app). =

24. If a mixture contains water and alcohol in the ratio 3 : 7, what is the percentage quantity of alcohol in the mixture? Sol: Water (a) : alcohol (b) = 3 : 7

b  1 0 0% ab 7 =  1 0 0% 37 = 7 0%. Note: Similarly you can find the percentage of water. Percentage of alcohol in the mixture =

25. Two alloys contain gold and copper in the ratio 3 : 2 and 5 : 2. In what ratio the two alloys should be added together to get a new alloy having gold and copper in the ratio of 2 : 1. Sol: Here, a : b = 3 : 2 and c : d = 5 : 2 Let the two alloys be mixed in the ratio x : y. Quantities of Gold and Copper should be added in the ratio ax cy + a + b c +d = bx dy + a+b c + d

52

=

3x 5y + 5 7 2x 2y + 5 7

=

2 1x + 2 5y 1 4x + 1 0y

Since, the ratio of gold and copper in the new alloys is 2 : 1

2 1x + 2 5y 1 4x + 1 0y

=

2 1

21x+25y = 28x+20y 7x = 5y x 5  y 7 Hence, the two alloys should be mixed in the ratio 5 : 7.

Brainstorming

1. 2.

Find the ratio of 6 hours to a day. 1. 1 : 4 2. 4 : 12

3. 2 : 3

4. 1 : 3

Find the ratio of 2.5 and 2.05. 1. 41 : 50 2. 50 : 41

3. 15 : 14

4. 21 : 23

3.

Two quantities are in the ratio 3 : 5. If first quantity is 150 kg then find the other quantity. 1. 240 kg 2. 150 kg 3. 250 kg 4. 200 kg

4.

The ratio between two quantities is the value of x. 1. 5 2. 4

x : 21. If the original quantities are 200 m and 840 m. Find 3. 3

4. 6

53

5.

The antecedent of a ratio which is equal to 7 : 9 is 70. Find its consequent. 1. 80 2. 90 3. 75 4. 60

6.

The compound ratio of 2 : 3 and the inverse ratio of 7 : 9 is 54 : x. Find x. 1. 63 2. 48 3. 54 4. 60

7.

Two horse riders travel 150 km and 200 km taking 5 hours each. Find the ratio of their speeds. 1. 4 : 3 2. 5 : 4 3. 3 : 4 4. 4 : 5

8.

If A : B = 2 : 3 and B : C = 4 : 5, then find A : B : C. 1. 12 : 8 : 15 2. 8 : 12 : 15 3. 15 : 8 : 12

4. 12 : 15 : 8

If 2A = 4B = 5C then find A : B : C. 1. 2 : 4 : 5 2. 10 : 4 : 5

3. 10 : 5 : 4

4. 5 : 4 : 3

a b c then find a : b : c. = = 2 3 4 1. 3 : 4 : 2 2. 2 : 3 : 4

3. 2 : 4 : 3

4. 2 : 5 : 4

9.

10.

If

1 2 3 : : . The difference between the greatest and the 2 3 4 smallest numbers is 36. Find the numbers. 1. 72, 84, 108 2. 60, 72, 96 3. 72, 84, 96 4. 72, 96, 108

11.

The three numbers are in the ratio

12.

If (2x + 5y) : (3x + 5y) = 9 : 10 then find x : y. 1. 5 : 7 2. 7 : 5 3. 4 : 5

4. 5 : 4

13.

What least number must be subtracted from each term of the ratio 6 : 11 to make it equal to 4:9. 1. 1 2. 2 3. 3 4. 5

14.

The monthly salaries of two persons A and B are in the ratio 4 : 5. If each salary is decreased by Rs.500 then the ratio becomes 3 : 4. Find the salary of A. 1. Rs.3000 2. Rs.2500 3. Rs.2000 4. Rs.2400

15.

Find the first part if Rs.1024 is divided in the ratio of 7 : 9. 1. Rs.540 2. Rs.576 3. Rs.448

4. Rs.60

16.

In a mixture of 28 litres, the ratio of milk and water is 5 : 2, if 2 litres of water is added to the mixture, find the ratio of milk and water in the new mixture. 1. 2 : 3 2. 1 : 2 3. 2 : 1 4. 3 : 2

17.

The incomes of A and B are in the ratio 4 : 5 and the incomes of B and C are in the ratio 2 : 3. If C‟s income is Rs.1500 then find A‟s income. 1. Rs.750 2. Rs.650 3. Rs.400 4. Rs.800

18.

Two numbers are in the ratio 3 : 5. If 10 is subtracted from each of them then the ratio becomes 1 : 5. Find the greater number. 1. 24 2. 20 3. 25 4. 16

19.

Find the third proportion to 24 and 36. 1. 68 2. 60

3. 54

4. 48

Find x, if 1.68 : 2.52 = 1.46 : x. 1. 1.19 2. 3.19

3. 2.11

4. 2.19

20.

21.

What least number must be subtracted from each of 13, 16, 17, 21 to make them in proportion. 1. 2 2. 1 3. 0 4. 3

22.

20 pumps can empty a reservoir in 12 hours. In how many hours can 45 such pumps do the same work?

54

1. 4 1/3 hours

2. 5 1/3 hours

3. 3 1/3 hours

4. 6 1/3 hours

23.

75 men can dig a canal in 90 days. How many more men are required to dig the canal in 18 days? 1. 250 2. 375 3. 200 4. 300

24.

In a 40 kg of mixture of sugar and sand, the ratio of sugar to sand is 5 to 3. How many kg of sugar are to be added into the mixture so that the ratio becomes 3 : 1? 1. 14 kg 2. 18 kg 3. 20 kg 4. 25 kg

25.

A contractor under took to finish a certain work in 124 days and employed 180 men. After 64 days he found that 2/3 of the work completed. How many men can he discharge now so that the work may finish in time? 1. 88 2. 84 3. 60 4. 52

26.

The sum of the cubes of three numbers is 584 and the ratio of first two numbers and last two numbers is 1 : 2. Find the middle number. 1. 2 2. 3 3. 4 4. 5

27.

If 3 chairs cost as much as 5 benches, 8 benches cost as much as 4 tables, 6 tables cost as much as 4 cots. If one cot cost is Rs.3600 then find the cost of one chair. 1. Rs.2000 2. Rs.1500 3. Rs.2400 4. Rs.2500

28.

A bag contains Re.1, 50 p, 25 p and 10 p coins in the ratio of 3 : 5 : 7 : 8 and together worth of Rs.80.50. Find the value of 25 p coins. 1. Rs.16.50 2. Rs.14.50 3. Rs.17.25 4. Rs.18.50

29.

Rs.3630 has been divided among A, B and C in such a way that if their share be diminished by Rs.5, Rs.10 and Rs.15 respectively, the remainders are in the ratio 3 : 4 : 5. Find the share of A. 1. Rs.900 2. Rs.805 3. Rs.905 4. Rs.1005

30.

A sum of Rs.7000 is divided among A, B and C in such a way that shares of A and B are in the ratio 2 : 3 and those of B and C are in the ratio 4 : 5. Find the amount received by C. 1. Rs.2600 2. Rs.2800 3. Rs.3000 4. Rs.3900

31.

Men, women and children are employed to do a work in the proportion of 1 : 2: 3 and their wages are in the ratio 6 : 3: 2. When 50 men are employed, total wages of all the hands amount to Rs.450, find the wages paid to a man, a women and a child? 1. Rs.3, Rs.1.50, Rs.1 2. Rs.1.50, Rs.2, Rs.3 3. Rs.4, Rs.3, Rs.1 4. None

32.

Rs.4250 is divided among 4 men, 5 women and 6 boys so that the share of a man, a woman and a boy are in the ratio 9 : 8 : 4. What is the share of a boy? 1. Rs.150 2. Rs.170 3. Rs.200 4. Rs.270

33.

A man divides his property so that his son‟s share to his wife‟s and wife‟s share to his daughter are both in the ratio 3 : 1. If the daughter gets Rs.10,000 less than the son, then find the total worth of the property. 1. Rs.16,000 2. Rs.15,000 3. Rs.16,250 4. Rs.17,000

34.

Pavan is younger than Murali by 10 years. If 5 years back their ages were in the ratio 1 : 2, how old is Murali? 1. 20 2. 15 3. 25 4. Data Inadequate

35.

If 40% of a number is added to another number then it becomes 125% of itself. What will be the ratio of first and second number? 1. 8 : 5 2. 5 : 7 3. 5 : 8 4. None

36.

The ratio of present ages of Sunil and Sunny is 7 : 8 respectively. Four years hence this ratio become 9 : 10 respectively. What is Sunil‟s present age in years? 1. 14 2. 17 3. 18 4. Data Inadequate

37.

Three numbers are in the ratio 1 : 2 : 3 and their product is 20250. The numbers are _______

55

1. 15, 30, 45

2. 25, 50, 75

3. 35, 70, 105

4. 45, 90, 135

38.

A varies inversely with B. When B is 64 units, A is 36 units. Find A when B is 48 units. 1. 54 units 2. 94 units 3. 48 units 4. 72 units

39.

If

40.

2pa 3qb 4 a p 7 = and = then find . q pa+ qb 3 b 5 1. 116:31 2. 19:31

3. 11:43

4. 18:35

If x and y have the same sign and 4 (8x2 + xy)= 15 y2 then find Y:X = ____ 1. 5:8 2. 8:5 3. 11:43 4. 8:3 1) 1 2) 2 3) 3 4) 1 5) 2 6) 1 7) 3 8) 2 9) 3 10) 2

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

4 1 2 3 3 3 4 2 3 4

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

2 2 2 3 2 3 1 3 3 3

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

4 2 3 3 4 1 1 3 3 2

56

7. PARTNERSHIP Partnership: Two or more persons carry on a business and share the profits of the business at an agreed proportion. This is called partnership. Partners: Persons who have entered into partnership with one another are individually called partners. They are of two types. a) Sleeping Partner: A person who invests the capital in the business but does not actively participate in the conduct of business is called sleeping partner. b) Working Partner: A person who takes part in running the business besides investing the capital is called working partner. He gets salary for his work or some per cent of profit, in addition. Firm name: The name under which the business is carried on is called firm name. Note: The partnership may be simple or compound. Simple Partnership: It is one in which the capital of each partner is in the business for same time. Compound Partnership: It is one in which the capitals of partners are invested for different periods. Results: 1. Three partners A, B and C invested their capitals in a business in the ratio of C1 : C2 : C3 and their profits are in the ratio P1 : P2 : P3 , then the ratio of timing of their investments i s

P1 P2 P3 : : . C1 C2 C3 2. Three partners A, B and C invested their capitals in a business. If the timing of their investments is in the ratio t1 : t2 : t3 and their profits are in the ratio P1 : P2 : P3 , then the ratio of their capitals invested is

P1 P2 P3 : : . t1 t 2 t 3

3. If capitals of two partners A, B be C be Rs. C1 and Rs. C 2 respectively for the periods t1 ,

t2 and t3 respectively, then Profit of A : Profit of B : Profit of C = C1t1 : C2t2 : C3t3 4. If there is a loss in the business, then Loss of A : Loss of B : Loss of C = C1t1 : C2t2 : C3t3 (Note: Investment Ratio = Profit Ratio)

SOLVED EXAMPLES 1.

Shashi, Ravi and Kiran invested Rs.2000, Rs.5000 and Rs.4000 respectively in a business. The net profit for the year was Rs.1210 which was divided in proportion to investments. Find the profit of each.

57

Sol: Here, C1 = 2 0 0 0, C2  5 0 0 0 and C3  4 0 0 0 and P = 1210 Ratio of capitals = 2:5:4 2 Shashi‟s profit = x 1210 = 220 254 5 Ravi‟s profit = x 1210 = 550 11 4 Kiran‟s profit = x 1210 = 440 11 Note: You can solve in this way also. Profit of Kiran = Total Profit - (Profit of Shashi + Profit of Ravi) 2.

A and B are two partners in a business. A contributes Rs. 1000 for 6 months and B Rs. 600 for 5 months. If total profit is Rs.600, find the profits of A and B. Sol: Here, C1  1 0 0 0, C2  6 0 0 and t1  6 , t 2  5 and P = 600 Profit of A : Profit of B

= C1t1 : C 2t2 = 6000 : 3000 = 2 : 1

2 x 600 = 400 3 1 Profit of B = x 600 = 200 3 Profit of A =

3.

A, B and C are three partners in a business. A contributes Rs.1500 for 6 months and B Rs.1200 for 4 months and C Rs.2000 for 2 months. Find the ratio of their shares in the profit. Sol: Here, C1  1 5 0 0, C2  1 2 0 0 and C3  2 0 0 0 and

t1  6 , t 2  4 and t 3  2 Profit of A : Profit of B : Profit of C = C1t1 : C2t2 : C3t3 = 1500(6) : 1200(4) : 2000(2) = 9000 : 4800 : 4000 = 45 : 24 : 20. 4.

Raju, Suman and Sunil invested capitals in the ratio of 3 : 4 : 7. At the end of the business term, their profits are in the ratio 1 : 3 : 5. Find the ratio of time for which they invested the capitals. Sol: Here, C1 : C2 : C3 = 3 : 4 : 7 and P1 : P2 : P3  1 : 3 : 5 Required ratio =

5.

P1 P2 P3 1 3 5 : :  : :  2 8: 6 3: 6 0. C1 C2 C3 3 4 7

Phani, Deepa and Bindu start a business. If the ratio of their periods of investments are 1 : 4 : 5 and their profits are in the ratio 4 : 8 : 10, find the ratio of their capitals. Sol: P1 : P2 : P3  4 : 8 : 1 0 and t1 : t 2 : t 3 = 1 : 4 : 5 Required ratio =

P1 P2 P3 4 8 10 : :  : :  4 : 2 : 2  2 : 1 : 1. t1 t2 t3 1 4 5

Brainstorming 1.

A and B started a business by investing Rs.8000 and Rs.6000 respectively. Find the share of each, out of an annual income of Rs.4200. 1. Rs.2400; 1800 2. Rs.1800; 1500 3. Rs.1900; 1200 4. Rs.1550; 1200

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2.

A started a business by investing Rs.4500 after 6 months B joined him by investing Rs.6000. Find the share of each, out of an annual profit of Rs.5000. 1. Rs.2000; 3000 2. Rs.3000; 1500 3. Rs.3000; 2000 4. Rs.1500; 3000

3.

A, B and C started a business by investing Rs.30,000, Rs.20,000 and Rs.10,000 respectively. After 6 months, A withdraws Rs.10,000 and B withdraw Rs.5000 and C invests Rs.15,000 more. Find the share of B out of an annual profit of Rs.4800. 1. Rs.1500 2. Rs.1400 3. Rs.2200 4. Rs.2500

4.

A and B started a business by investing Rs.4000 and Rs.6000 respectively. After 5 months A invested Rs.3000 more and B withdraws Rs.3000. Find the share of B out of a total annual profit of Rs.4000. 1. Rs.2300 2. Rs.1700 3. Rs.1500 4. Rs.2000

5.

Sharma and Shastri entered into a partnership investing Rs.12,000 and Rs.9,000 respectively. After 3 months Sharma withdraws Rs.4,000 and Rechal joined them with an investment of Rs.16,000. Find the share of Rechal out of total annual profit of Rs.10,000. 1. Rs.5000 2. Rs.3000 3. Rs.2000 4. Rs.4000

6.

A started a business with Rs.2400 and he is joined afterwards by B with Rs.3600. After how many months did B joined if the profits at the end of the year are divided equally? 1. 8 months 2. 4 months 3. 1 month 4. None

7.

A started a business with Rs.16,000 and is joined afterwards by B with Rs.12,000. After how many months did B joined, if A‟s annual profit is double that of B. 1. 4 months 2. 8 months 3. 1 month 4. None

8.

A, B, C subscribed Rs.46,000. A subscribes Rs.12,000 more than B subscribes and B Rs.2000 more than C. Find the share of A out of a total profit of Rs.23,000. 1. Rs.6000 2. Rs.5000 3. Rs.12000 4. Rs.24000

9.

A, B, C subscribed Rs.26,000. A subscribes Rs.4000 more than B and B Rs.2000 more than C. Find the share of B out of a total profit of Rs.5200. 1. Rs.800 2. Rs.1200 3. Rs.1600 4. Rs.1000

10.

A, B, C subscribed Rs.30,000. A subscribes Rs.4000 more than B and B Rs.2000 less than C. Find the share of A out of a total profit of Rs.6000. 1. Rs.2400 2. Rs.3000 3. Rs.2000 4. Rs.400

11.

A, B, C subscribed Rs.38,000. A subscribes Rs.2000 less than B and B Rs.4000 less than C. Find the share of C out of a total profit of Rs.950. 1. Rs.450 2. Rs.400 3. Rs.200 4. Rs.275

12.

A and B invests in a business in the ratio 4 : 5. If 10% of the total profit goes to a charity. If A share is Rs.4000 then find the total profit. 1. Rs.10000 2. Rs.9000 3. Rs.12000 4. Rs.14000

13.

A is a working and B is a sleeping partner in a business. A puts in Rs.8000 and B Rs.7000. A receives 10% of the profits for running the business the rest being divided in proportion to their investments. Find the total amount received by A out of a total profit of Rs.500. 1. Rs.240 2. Rs.450 3. Rs.290 4. Rs.400

14.

A is a working and B is a Sleeping partner in a business. A puts in Rs.5000 and B Rs.4000. A receives 2/11 of the profit for managing and the rest being divided in proportion to their capital. If A receives a total profit of Rs.2800. Find the total profit. 1. Rs.4400 2. Rs.4000 3. Rs.3300 4. Rs.5500

15.

A, B and C contract a work for Rs.810 together. A and B are to do 5/9 of the work. Find the share of C. 1. Rs.400 2. Rs.360 3. Rs.200 4. Rs.519

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16.

A and B started a business. A invested thrice as much as B and the period of his investment is two times the period of investment of B. If B got Rs.5000 as profit, then find the total profit in the business. 1. 35000 2. 38000 3. 22000 4. None

17.

Padmanabhan and Sharavana started a joint firm. Padmanabhan‟s investment was 2/3 of Sharavana and his investment period is ¾ that of Sharavana‟s. If the difference between their profits is Rs.4000 then find the total profit. 1. Rs.54000 2. Rs.30000 3. Rs.20000 4. Rs.12000

18.

5 3 6 after 4 months A increases : : 2 4 5 his share by 50%. If the total profit at the end of the year Rs.3170 then find the profit share of B. 1. Rs.800 2. Rs.450 3. Rs.1700 4. Rs.1540 A, B and C enter into a partnership with shares in the ratio

19.

A and B enter into a partnership with the capitals in the ratio of 6 : 7. After 3 months A 1 1 withdraws of his capital and B reinvested of his capital more. If the total profit at the end 6 7 of a year is Rs.2600. Find the share of each. 1. Rs.1050; 1550 2. Rs.1800;800 3. Rs.1300; 1300 4. None

20.

A and B enter into a partnership with the capitals in the ratio of 4 :5. After 5 months A invested 25% of his capital more and B withdraws 40% of his capital. The total gain at the end of 10 months was Rs.850. Find the share of A. 1. Rs.150 2. Rs.300 3. Rs.200 4. Rs.450

21.

A and B started a business with Rs.2000 and Rs.3000 respectively. If the total profit was Rs.1500 find the share of A. 1. Rs.500 2. Rs.600 3. Rs.900 4. Rs.1000

22.

A and B started a business with Rs.1500 and Rs.2500 respectively. 20% of the total profit was given to A as salary. At the end of the year, if the total profit was Rs.4000, find the total amount received by A. 1. Rs.2000 2. Rs.1200 3. Rs.800 4. Rs.3200

23.

A and B started a business with Rs.2000 and Rs.3000 respectively. 10% of the total profit was given to A as salary. At the end of the year, if B got Rs.1080 as his profit, then find the total profit. 1. Rs.5000 2. Rs.3000 3. Rs.2000 4. Rs.4000

24.

A and B started a business with Rs.3000 and Rs.5000 respectively. 20% of the total profit was given to A as salary. At the end of the year, if A got a profit of Rs.1760, then find the total profit. 1. Rs.2000 2. Rs.3520 3. Rs.1800 4. Rs.3600

25.

A started a business with Rs.6000. After 4 months B joined with Rs.8000. At the end of the year, if the total profit was Rs.5100, find the share of A. 1. Rs.2700 2. Rs.2500 3. Rs.2000 4. Rs.2400

26.

A started a business with Rs.4000. After 4 months B joined with Rs.5000. Again after 2 months, C joined with Rs.6000. At the end of the year if C got Rs.1557 as his share, find the share of B. 1. Rs.2076 2. Rs.1730 3. Rs.2000 4. Rs.5363

27.

A and B started a business with Rs.6000 and Rs.8000 respectively. After 4 months A invested Rs.1000 more and B withdraws Rs.1000. At the end of the year if B got Rs.3300 as his share, find the share of A. 1. Rs.3000 2. Rs.6600 3. Rs.2700 4. None

28.

P started a business with Rs.7000. After a few months Q joined with Rs.10,000. At the end of the year, if Q got Rs.2500 as his share out of a total profit of Rs.5500, after how many months Q joined? 1. 7 months 2. 5 months 3. 10 months 4. 2 months

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29.

R and S started a business with Rs.7000 and Rs.8000 respectively. If the total profit at the end of the year was Rs.4500 then find the R‟s share. 1. Rs.2500 2. Rs.2250 3. Rs.1500 4. Rs.2100

30.

M and N started a business with Rs.4500 and Rs.5500 respectively. At the end of the year, if N got Rs.4400 as profit then find M‟s share of profit. 1. Rs.2500 2. Rs.3600 3. Rs.4500 4. Rs.8000

31.

Ram and Dolly started a business with Rs.6300 and Rs.7200. 10% of the profit was given to Ram as salary. At the end of the year if Dolly got Rs.2592 as profit then find the total profit. 1. Rs.3240 2. Rs.2268 3. Rs.5000 4. Rs.5400

32.

Rambabu started a business with Rs.12,000, 4 months later Nagarjuna joined with a capital of Rs.15,000. At the end of the year total profit was Rs.5500. Find Rambabu‟s share. 1. Rs.3000 2. Rs.2500 3. Rs.4000 4. Rs.4800

33.

Srihari and Pratap started a business with Rs.3000 and Rs.4000 respectively. 6 months later, Sudheer joined with a capital of Rs.5000. At the end of the year, if Srihari got a profit of Rs.3600 then find the total profit. 1. Rs.3200 2. Rs.3000 3. Rs.11400 4. Rs.9000

34.

A, B and C enter into partnership with investments in ratio 3 : 5 : 7 respectively. The period of investments is 10 months, 4 months and 5 months respectively. Find A‟s share of profit out of the total of Rs.4250. 1. Rs.1500 2. Rs.850 3. Rs.1625 4. None

35.

A started the business with Rs.1.5 lakhs. After one year he allowed 10 people to join the business as silent partners with each investing Rs.18,000. A claimed 1/5th of the profit as his salary. After another one year, if the total profit earned is Rs.1 lakh, find A‟s total share in the profit. 1. Rs.30,000 2. Rs.50,000 3. Rs.40,000 4. Rs.70,000

36.

A invested Rs.1 less than B for every Rs.8 invested by B. If A gets Rs.150 less than B, how much is the total profit earned by them? 1. Rs.1200 2. Rs.1050 3. Rs.2250 4. None

37.

A‟s investment is 20% more than B‟s investment. If the total profit is Rs.1155, how much is B‟s share of profit less than that of A? 1. Rs.105 2. Rs.924 3. Rs.231 4. None

38.

A, B and C rented a pasture. A puts 90 sheeps for 20 days. B puts 75 sheeps for 18 days and C puts 60 sheeps for 30 days. If the total rent is Rs.2640, B‟s share of rent is? 1. Rs.264 2. Rs.720 3. Rs.810 4. None

39.

A and B rent a pasture. A puts 30 horses for 60 days while B puts some number of horses for 50 days. If the rent paid by them are in ratio 9 : 10, how many horses did B put in pasture? 1. 60 2. 40 3. 20 4. 10

40.

Two partners invest Rs.3500 and Rs.4500 respectively in business. They agree that 50% of the profit would be distributed equally between them and remaining profit in ratio of capital. If the difference between their shares of profit is Rs.100, then find the total profit. 1. Rs.800 2. Rs.1600 3. Rs.2400 4. None

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1) 1 2) 3 3) 2 4) 2 5) 4 6) 2 7) 2 8) 3 9) 3 10) 1

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

2 1 3 1 2 1 4 2 1 4

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

2 1 3 2 1 2 1 2 4 2

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

4 1 3 1 4 3 1 2 2 2

8. MIXTURES (OR) ALLIGATION Alligation: It means linking. a) It is a rule to find the ratio in which two or more ingredients at their respective prices should be mixed to give a mixture at a given price. b) It is a rule to find the average price of a mixture when the prices of two or more ingredients which may be mixed together and the proportion in which they are mixed are given. Mean Price: The cost price of a unit quantity of mixture is called the mean price. Alligation Rule:

1.

Let us suppose Rs.a per unit be the price of the first ingredient (superior quality) is mixed with another ingredient (cheaper quality) of price Rs.b per unit to form a mixture whose mean price is Rs.m per unit, then the two ingredients must be mixed in the ratio:

Q uantityof c heaper C .P .of s uperior- M eanpric e a  m   Q uantityof s uperior M eanpric e- C .P .ofc heaper m  b

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2.

This means, the two ingredients are to be mixed in the inverse ratio of the differences of their prices and the mean price. Note: This can be remembered easily through the diagram below:

Formulae: 1.

A vessel contains a litres of wine. From it b litres are withdrawn. The vessel is then filled with water. Next b litres of the mixture are withdrawn and again the vessel is filled with water. If this process is repeated n times then

Wine left in the ves s elafter nth operation  a  b     O riginalquantityof wine in the ves s el  a 

n

n   b  After n th the quantity of wine left in the vessel = a 1    a     

2.

There are n vessels of equal size filled with mixtures of liquids A and B in the ratio a 1 : b 1 , a 2 : b 2 ,………,a n : b n , respectively. If the contents of all the vessels are poured into a single large vessel, then  a1 a2 an      .......... ......   an  bn  Q uantityof liquidA  a1  b1 a2  b2  Q uantityof liquidB  b1 b2 bn     a  b  a  b  .......... ......  a  b  1 2 2 n n  1

3.

There are n vessels of sizes c 1 , c 2 , ..…… , c n filled with mixtures of liquids A and B in the ratio a 1 : b 1 , a 2 : b 2 ,………,a n : b n , respectively. If the contents of all the vessels are poured into a single large vessel, then  a1c1 a2c2 ancn      .......... ......  an  bn  Q uantity of liquid A  a1  b1 a2  b2  Q uantity of liquid B  b1c1 b2c2 bncn     a  b  a  b  .......... ......  a  b  1 2 2 n n  1

SOLVED EXAMPLES 1. In what ratio the two varieties of coffee one costing Rs.30 per kg and the other Rs.35 per kg should be blended to produce a blended variety of coffee worth Rs.32 per kg. How much should be the quantity of second variety of coffee, if the first variety is 72 kg.

63

Sol:

The required ratio of coffee is 3 : 2. Q uantityof c heaperquality 3  Q uantityof s uperiorquality 2 So, Quantity of superior coffee = 7 2 

of the two varieties i.e.

2  4 8 kg. 3

2. Salt at Rs.10 per kg is mixed with salt at Rs.15 per kg in the ratio 3 : 4. Find the price per kg of mixture. Sol: Let the mean price of the mixture be Rs.x.

Q uantityof c heapers alt 1 5  x  Q uantityof s uperiors alt x  1 0 3 1 5 x 90 6    3x  3 0  6 0  4x  7x  9 0  x   Rs.1 2 . 4 x 10 7 7 3. A vessel contains 100 litres of wine. 20 litres of wine was taken out and replaced by water. Then, 20 litres of mixture was withdrawn and again replaced by water. The operation was repeated for fourth time. How much wine is now left in the vessel? 4

4

20   4 Sol: Amount of wine left in the vessel = 1    1 0 0     1 0 0  4 0.9 6litres. 1 0 0  5 4. Two equal glasses are filled with mixture of milk and water. The proportion of milk and water in the first glass is 4 : 3 and in the second glass is 5 : 2. The contents of the two glasses are emptied into a single vessel. What is the proportion of milk and water in it? 4 5  Q uantityof milk 4  3 5  2  9  9 : 5.  Sol: 3 2 Q uantityof water 5  43 52 5. Three glasses of sizes 1 litres, 2 litres and 3 litres contain mixture of milk and water in the ratio 2: 3, 3 : 7 and 4 : 11 respectively. The contents of all the three glasses are poured into a single vessel. Find the ratio of milk to water in the resulting mixture.

64

43  2 6 1 2  2 1 3  2 12 18 24         Q uantityof milk 54 2  3 3  7 4  1 1 5 1 0 1 5    30 Sol:     1 8  4 2  6 6 3  1 7  2 1 1  3 3 1 4 3 3 126    Q uantityof water          30  2  3 3  7 4  1 1  5 1 0 1 5  = 54 : 126 or 27 : 63.

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Brainstorming 1.

How many kilograms of rice costing Rs.12/- per Kg should be mixed with another rice of 20 Kg at Rs.16/- per Kg to obtain the mixture costing Rs.13/- per Kg? 1. 50 Kg 2. 60 Kg 3. 45 Kg 4. 40 Kg

2.

In what proportion must water be mixed with spirit to gain 16 2/3% by selling at its cost price? 1. 6:1 2. 5:2 3. 1:6 4. 2:3

3.

How many kilograms of rice costing Rs.12/- per Kg should be mixed with another rice of 20 Kg at Rs.16/- per Kg to obtain the mixture costing Rs.13/- per Kg? 1. 50 Kg 2. 60 Kg 3. 45 Kg 4. 40 Kg

4.

Two equal glasses filled with mixtures of alcohol and water in the proportions of 2:1 and 1:1 respectively were emptied into a third glass. What is proportion of alcohol and water in the third glass? 1. 5:7 2. 3:5 3. 5:3 4. 7:5

5.

A vessel of 90 litres is filled with milk and water. 55% of milk and 40% of water is taken out of the vessel. It is found that the vessel is vacated by 45%. Find the initial quantity of water? 1. 30 litres 2. 35 litres 3. 40 litres 4. 60 litres

6.

9 litres are drawn from a cask full of wine, and it is then filled with water. Nine litres of the mixture are drawn and the cask is again filled with water. The ratio of the wine now left to that of the water is 16:9. How much does the cask hold? 1. 40 litres 2. 45 litres 3. 50 litres 4. 55 litres

7.

How much water must be added to 60 litres of milk at 1 mixture worth Rs.10 1. 15 litres

2 a litre? 3 2. 10 litres

1 litres for Rs.20 so as to have a 2

3. 5 litres

4. 12 litres

8.

In what proportion must a grocer mix tea at Rs.1.02 per Kg and Rs.1.44 per Kg so as to make a mixture worth Rs.1.26 per Kg? 1. 4:3 2. 2:3 3. 3:2 4. 3:4

9.

A vessel of 80 litre is filled with milk and water. 70% of milk and 30% of water is taken out of the vessel. It is found that the vessel is vacated by 55%. Find the initial quantity of milk? 1. 30 Kg 2. 25 Kg 3. 35 Kg 4. 50 Kg

10.

300 grams of sugar solution has 50% sugar in it. How much sugar should be added to make it 40% in the solution? 1. 25 gm 2. 75 gm 3. 50 gm 4. 35 gm

11.

A trader has 50 Kg of rice, a part of which he sells at 14% profit and the rest at 6% loss. On the whole his loss is 4%. What is the quantity sold at 14% profit? 1. 45 Kg 2. 25 Kg 3. 20 Kg 4. 5 Kg

12.

Vessel A contains milk and water in the ratio 7:3 and vessel B contains the same in the ratio 3:1. In what proportion should quantities be taken from A and B to form a mixture in which milk and water are in the proportion 11:4? 1. 2:1 2. 3:2 3. 1:2 4. 2:3

13.

In what proportion must a grocer mix tea at Rs.6 per Kg and Rs.8 per Kg so as to make a mixture worth Rs.7 per Kg? 1. 1:1 2. 4:3 3. 2:3 4. 3:2

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14.

In a Zoo, there are Monkeys and Parrots, If heads are counted, there are 450 and if legs are counted there are 1000. How many Monkeys are there? 1. 150 2. 50 3. 200 4. 250

15.

How many kilograms of rice costing Rs.8/- per Kg should be mixed with another rice of 60 Kg at Rs.15/- per Kg to obtain the mixture costing Rs.11/- per Kg? 1. 40 Kg 2. 20 Kg 3. 80 Kg 4. 60 Kg

16.

A man lend Rs.1500/- in 2 parts one at 5% and another at 8%. If at the end of the year he received Rs.93/- as interest. Then the sum lent at 8%? 1. 900 2. 700 3. 600 4. 500

17.

400 grams of sugar solution has 30% sugar in it. How much sugar must be added to make it 60% in the solution? 1. 80 2. 120 3. 300 4. 180

18.

A man buys two buildings for Rs.45000/- and sells one so as to lose 12% and the other so as to gain 15% and on the whole be neither gains no loses. What is the cost of building sold at 12% lose? 1. Rs.20000/2. Rs.22000/3. Rs.25000/4. Rs.23000/-

19.

Three vessels of equal capacity contain alcohol and water in the ratio of 3:2, 5:4 and 1:4 respectively. It all the solutions are mixed, what is the ratio of alcohol to water in the resultant mixture? 1. 61:74 2. 5145 3. 50:73 4. 74:45

20.

Two litres of pure milk is added to 6 litres of a milk solution containing 16 2/3% milk. What is the concentration of milk in the resultant solution? 1. 12:5% 2. 25% 3. 33 1/3% 4. 37.5%

21.

In what ratio must 82 % of wine solution be mixed with pure wine to get a resultant solution of 6% of water? 1. 3:1 2. 2:1 3. 1:3 4. 1:2

22.

In what proportion must water be mixed with the spirit to gain 26 2/3% by selling it at cost price? 1. 4:15 2. 15:4 3. 1:12 4. 15:7

23.

Three equal glasses are filled with a mixture of spirit water. The proportion of spirit to water in the first glass is 2:3, in the second glass is 3:4 and the third glass is 4:5. The contents of the three glasses are emptied into a single vessel. What is the proportion of spirit and water in it? 1. 544:401 2. 401:544 3. 410:544 4. 401:454

24.

A grocer mixes 2 varieties of tea, one costing Rs.75/- per Kg and another Rs.100/- per Kg. In what proportion he should mix these so by selling at Rs.88/- per Kg 10% is gained? 1. 1:4 2. 4:1 3. 3:4 4. 4:3

25.

What weight of rice worth Rs.4.20 per Kg should be mixed with 60 Kg of rice worth Rs.2.70 per Kg so that when the mixture is sold at Rs.3.30 per Kg. There may be neither gain nor loss? 1. 50Kgs 2. 45Kgs 3. 55Kgs 4. 40Kgs

26.

A mixture of milk and water contains 10% water. How much water must be added to make 20% water in the new mixture? 1. 5Ltrs 2. 6Ltrs 3. 8Ltrs 4. 10Ltrs

27.

In 70 litres of a mixture of milk and water, the quantity of water is 10%. should be added so that new mixture may contain 25%? 1. 7 2. 14 3. 21 4. 9

28.

A sum of Rs.39/- was divided among 45 boys and girls. Each girl gets 50paise, whereas a boy gets one rupee. Find the number of girls?

How much water

67

29.

1. 33 2. 32 3. 22 4. 12 A sum of Rs.34.000 was invested partly at 5% p.a. and the remaining at 3% p.a. simple interest for a period of 2 years. The total interest after the end of two years is Rs.2. 680. What was the amount interest at 5%? 1. 19,000 2. 14,000 3. 13,000 4. 16,000

30.

Goods worth Rs.400 are purchased. One fourth of them are sold at 5% loss. At what profit percent should the remaining be sold such that 10% profit is made on the whole? 1. 5% 2. 15% 3. 25% 4. 20%

31.

A container has 40 litres of whisky. From this 4 litres of whisky is replaced by water. This process was replaced two times. How much whisky is now contained in t he container? 1. 26.34 litres 2. 27.36 litres 3. 28 litres 4. 32.4 litres

32.

Two qualities of tea are mixed in the ratio of 4:1 and the mixture is sold at Rs.72 per Kg for a profit of 12.5%. If the tea of the second quality costs Rs.3.25 more per Kg than the tea of the first quality. What is the cost per Kg of the tea of first quality? 1. 65.53 2. 63.35 3. 65.35 4. 63.53

33.

In what ratio must a solution of milk and water containing 42% milk be mixed with pure milk to get a resultant solution containing 29% water? 1. 1:2 2. 2:1 3. 3:2 4. 1:1

34.

How many litres of pre water must be mixed with 144 litres of pure spirit in order to gain 11½% by selling the resultant mixture at cost price of pure spirit? 1. 32 2. 48 3. 18 4. 16

35.

A rice trader mixes two verities of rice. The cost price of the first variety rice is twice the cost price of the second variety. If the trader sells the mixture at the cost price of the other variety and makes a profit of 35% then what is the ratio in which the verities of the rice are mixed? 1. 7:13 2. 13:14 3. 1:3 4. cannot be determined

36.

In what ratio a grocer mix two varieties of tea worth Rs.60 per kg and Rs.65 Per Kg so that by selling the mixture at Rs.68.20 per Kg he may gain 10% 1. 3:2 2. 3:4 3. 3:5 4. 4:5

37.

A barrel has 50 litres of pure spirit. 5 litres of spirit is removed and replaced with the same amount of water. This process is repeated two more times, what is the amount of spirit in the resultant mixture? 1. 36.45 2. 34.45 3. 36.54 4. 35.45

38.

A grocer purchased 20 Kg of sugar at the rate of Rs.15/- per Kg and 30 Kg of sugar at the rate of Rs.13/-. At what price per Kg should be sell the mixture to earn 33 1/3% profit on the cost price? 1. Rs.28.00 2. Rs.20.00 3. Rs.18.40 4. Rs.17.40

39.

How many kilograms of wheat at Rs.9.30 per Kg must be mixed with 16 Kg of wheat at Rs.13.80 per Kg such that the mixture when sold at Rs.12.43 per Kg gives a profit of 10%? 1. 20 2. 33 3. 30 4. 25

40.

A trader mixes two types of rice, one costing Rs.18/- per Kg and the other costing Rs.15/- per Kg. so that the mixture is worth Rs.16.25 per Kg. In what ratio does he mix the two types of rice? 1. 5:3 2. 5:7 3. 3:2 4. 8:5

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1) 2 2) 3 3) 2 4) 4 5) 4 6) 2 7) 1 8) 4 9) 4 10) 2

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

4 3 1 2 3 3 3 3 3 4

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

4 1 2 2 4 1 2 4 4 2

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

4 2 4 4 2 1 1 3 1 2

9. PROFIT AND LOSS 1. 2. 3.

In any business transaction, it is common to have either profit or loss. But the aim of any business is to earn profit. The most commonly used term involving sale and purchase of any business are Cost Price and Selling Price.

Cost Price: The price at which an article has been purchased is called the cost price. It is denoted by C.P. Selling Price: The price at which an article has been sold is called the selling price. It is denoted by S.P. Profit or Gain: If S.P. > C.P., then there is a gain or profit. Thus, Profit or Gain = S.P. – C.P. Loss: If C.P. > S.P., then there is a gain or profit. Thus, Loss = C.P. – S.P. Note: Profit and Loss are always calculated w.r.t. C.P. of the item. Formulae: 1. 2. 3.

4.

Gain 1 0 0 C .P . Los s 1 0 0 Loss % = C .P . When S.P. and gain% are given, then 100   C .P .    S.P . 1 0 0  Gain%   Gain % =

When C.P. and gain% are given, then

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 1 0 0 Gain% S.P .    C .P . 100   When S.P. and loss% are given, then 100   C .P .    S.P .  1 0 0 Los s %

5.

6.

When C.P. and loss% are given, then  1 0 0 Los s % S.P .    C .P . 100  

7.

If a man buys a items for Rs.b and sells c items for Rs.d, then the ad  bc Gain or loss % =  100 bc

Note: 8.

a) In case of gain percent, the result will be positive. b) In case of loss percent, the result will be negative.

If the C.P. of x articles = S.P. of y articles, then x  y   1 0 0% Gain or loss % =   y  Note: a) If x > y, it is % gain. b) If x < y, it is % loss.

9.

The Cost Price of an article is C.P. If it is sold at S.P1., then gain% or loss% is x and if it is sold at a price S.P2., gain% or loss% is y then S.P1 S.P2 C .P S.P1  S.P2   (Or) 100 xy 1 0 0 x 1 0 0 y Note: a) If x or y is negative it indicates a loss. b) If x or y is positive it indicates a gain.

10.

A sells an article to B at a gain or loss of x%, and B sells it to C at a gain or loss of y%. If C pays Rs.z for it to B, then   1 0 02  z C.P. for A =    (1 0 0 x)(1 0 0 y)  Note: a) If x or y is negative it indicates a loss. b) If x or y is positive it indicates a gain.

11.

If A sells an article to B at a gain or loss of x%, and B sells it to C at a gain or loss of y% then the xy   Resultant Profit% or Loss % =  x  y   1 0 0  Note: 1. If x or y is negative it indicates a loss. 2. If x or y is positive it indicates a gain. 3. This expression represents resultant profit% or loss% according as it is positive or negative.

12.

If two different articles are sold at the same S.P., getting gain or loss of x% on the first transaction and gain or loss of y% on the second transaction, then the  1 0 0(x  y)  2xy  % Overall % gain or % loss =   (1 0 0 x)  (1 0 0 y)  Note: This expression represents overall gain or loss according as its sign is positive or negative.]

13.

If two different articles are sold at the same S.P., getting gain of x% on the first transaction and gain or loss of x% on the second transaction, then the

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2

 x  Overall % loss =   %  1 0 Note: In this type of questions there is always a loss. 14.

If a shopkeeper uses faulty measure and sells his goods at a gain or loss of x% then the 1 0 0 g T ruemeas ure  Overall % gain or % loss is 1 0 0 x Faultymeas ure Note: If the merchant sells his goods at C.P., then x = 0.

15.

If a merchant uses y% less weight or length and sells his goods at a gain or loss of x% then the  y  x     1 0 0% Overall % gain or loss =   1 0 0 y  

16.

If a person buys two items for Rs.A and sells one at a loss of l% and other at a gain of g% and if each item was sold at the same price, then A(1 0 0 gain% ) a. The C.P. of the item sold at loss = (1 0 0- los s % ) (1 0 0 gain% ) A(1 0 0 los s % ) b. The C.P. of the item sold at gain = (1 0 0- los s % ) (1 0 0 gain% )

17.

If two successive discounts on an article at x% and y% respectively, then the xy   Overall Discount =  x  y  % 1 0 0  If three successive discounts on an article at x%, y% and z% respectively, then the  (xy  yz  zx) xyz  % Overall Discount =  x  y  z   100 1 0 02  

18.

19.

A shopkeeper sells an item at Rs.x after giving a discount of d% on labeled price. Had he not given the discount, he would have earned a profit of p% on the C.P. then the C.P. of each item is given by   1 0 02 x  C.P. =   (1 0 0 d)(1 0 0 p)   

SOLVED EXAMPLES 1.

If C.P. = Rs.440, S.P. = Rs.480 then find the profit. Sol: C.P. = Rs.440, S.P. = Rs.480 Profit = S.P. – C.P. = 480 – 440 = Rs.40.

2.

If C.P. = Rs.135, S.P. = Rs.120 then find the loss. Sol: C.P. = Rs.135, S.P. = Rs.120 Loss = C.P. – S.P. = 135 – 120 = Rs.15.

3.

The cost price of a pen is Rs.400 and the selling price is Rs.460. Find the gain %. Sol: C.P. = Rs.400, S.P. = Rs.460 Gain = S.P. – C.P. = 460 – 400 = Rs.60 Gain 1 0 0 6 0 Gain % =   1 0 0  1 5 %. C .P . 400

4.

Rajani bought a jewel for Rs.825 and sold it for Rs.750. Find the loss%. Sol: C.P. = Rs.825, S.P. = Rs.750 Loss = C.P. – S.P. = 825 – 750 = Rs.75

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Loss % = 5.

Los s 1 0 0 7 5 1 0 0 1 0 0 1   9 %. C .P . 825 11 11

Shashi buys a T.V. set for Rs.9500. For how much should he sell in order to gain 4%? Sol: C.P. = Rs.9500, gain% = 4%  1 0 0 Gain% S.P .    C .P ., if C.P. and gain% are given 100  

 1 0 0 4  S.P .    95 0 0 1 0 4 9 5  Rs . 9 8 8 0.  100  6. Hari loses 8% by selling a Cooler for Rs.5520. Find the C.P. of the cooler. Sol: S.P. = Rs.5520, loss% = 8% 100   C .P .    S.P ., if S.P. and loss% are given  1 0 0 Los s %

100  100  C .P .   5 5 2 0 Rs .6 0 0 0.   55 2 0  92  1 0 0 8  1 7. By selling a toy for Rs.422, Kiran gains 5 % . Find the C.P. of the toy. 2 1 11 Sol: S.P. = Rs.422, gain% = 5 % = %. 2 2 100   C .P .    S.P ., if S.P. and gain% are given 1 0 0  Gain%  

8.

    1 0 0   4 2 2  2 0 0 4 2 2  Rs .4 0 0 C .P .  11   211  1 0 0  2   Venu buys oranges at the rate of Rs.10 per dozen and sells them at rate of 16 for Rs.11. Find his % gain or loss. Sol: a = 12, b = Rs.10, c = 16, d =  ad   1  1 0 0%  Gain or loss% =  bc  

Rs.11 132   1 6 0  1  1 0 0%  

28  1 0 0%  1 7.5% 160 Since, the result is negative; there is a loss of 17.5% 

9.

11 apples are bought at Rs.10 and sold at 10 for Rs.11. What is gain or loss%? Sol: a = 11, b = Rs.10, c = 10, d = Rs.11

 ad  121   1  1 0 0%    1  1 0 0% Gain or loss% =  bc   100  21 =  1 0 0%  2 1% . 100 10. The selling price of 18 articles is equal to the cost price of 10 articles. What is the profit%? Sol: x = 18, y = 10 x y 1 8 1 0   1 0 0%   1 0 0%  8 0% . Gain% =  10  y 

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11. By selling a mobile for Rs.2400, Phani lost 10%. What percent shall he gain or lose by selling it for Rs.2800. Sol:S.P1.= Rs.2400, x = - 10%, S.P2.= Rs.2800, y =? Here negative sign for x indicates loss. S.P1 S.P2 We have,  1 0 0 x 1 0 0 y 2400 2800 6 7     6 0 0 6y  6 3 0 1 0 0 1 0 1 0 0 y 9 0 1 0 0 y 6y  3 0  y  5 Since the result is positive, Phani can have a gain of 5% by selling it for Rs.2800. 12. Avinash sells a bike to Shireesh at a gain of 15% and Shireesh again sells to Teja at a profit of 10%. If Teja pays Rs.37950 to Shireesh then what is the cost price of the bike for Avinash? Sol: Here, x = 15, y = 10 and z = Rs.37,950   1 0 02  z C.P. for Avinash =    (1 0 0 x)(1 0 0 y) 

=

1 0 0× 1 0 0× 3 7 9 5 0 (1 0 0+ 1 5)(1 0 0+ 1 0)

=

1 0 0× 1 0 0× 3 7 9 5 0 1 1 5× 1 1 0

= Rs .3 0,0 0 0. 13. Mounika sells an i-pod to Bindu at a gain of 12% and Bindu again sells it to swapna at a loss of 20%. If swapna pays Rs.2240 to Bindu then what is the cost price of i-pod to Mounika? Sol: Here, x = 12, y = - 20 and z = Rs.2240   1 0 02  z C.P. for Mounika =    (1 0 0 x)(1 0 0 y) 

 1 0 0 1 0 0 2 2 4 0  1 0 0 1 0 0 2 2 4 0   1 1 2 8 0  (1 0 0 1 2)(1 0 0 2 0)  1 0 0 1 0 0   Rs .2 5 0 0. 4 14. Sachin sells a machine to Krishna at a profit of 3% and Krishna sells it to Dhruva at a profit of 5%. Find the resultant profit%. Sol: Here, x = 3, y = 5 xy   Resultant Profit% =  x  y   1 0 0 

3(5)  15 3  8 %. =  3  5    8  1 0 0 1 0 0 2 0   15. Pratap sells a trouser to Rajesh at a profit of 10% and Rajesh sells it to Chandu at a loss of 8%. Find the resultant profit or loss%. Sol: Here, x = 10, y = - 8

xy   Resultant Profit or loss% =  x  y   1 0 0 

1 0(8)  80   1 0  (8)    2  1 0 0 1 0 0   120 1   1 %. 100 5 This represents profit as the sign is positive. 16. Manoj sold two laptops, each for Rs.30000. If he makes 15% profit on the first and 10% loss on the second, what is his gain or loss% on the whole transaction? Sol: Here, x = 15, y = - 10

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 1 0 0(x  y)  2xy  % =   (1 0 0 x)  (1 0 0 y) 

Overall % gain or % loss

 1 0 0(1 5  1 0)  2(1 5)(1 0)  % =   (1 0 0 1 5)  (1 0 0 1 0)  200 40  5 0 0 3 0 0 % %. =  %  2 0 5  205 41 

This represents profit as the sign is positive. 17. Chary sold two bicycles for Rs.1000 each, gaining 20% on one and losing 20% on the other. Find his total gain or loss%. Sol: Here, x = 20 2

Overall % loss

 x  =   %  1 0 2

20 % = 4% . 10 18. A dishonest shopkeeper professes to sell his goods at C.P. but uses a 950 g for kg weight. Find his gain%. =

Sol: Here, True measure = 1000 g, False measure = 950 g, x = 0 Overall gain% is given by 1 0 0 g T ruemeas ure  1 0 0 x Faultymeas ure 1 0 0 g 1 0 0 0 1 0 0 g 2 0     100 950 100 19  1 9 0 0 1 9g  2 0 0 0 1 9g  1 0 0 100 5 g 5 %. 19 19 19. A shopkeeper sells the goods at 22% loss on the C.P. but uses 10% less weight. What is his gain or loss%? Sol: Here, x = - 22, y = 10  1 0  2 2   Overall gain or loss% =    1 0 0% 1 0 0  1 0   

120 40 12   1 0 0%    =  9 3  90  1 = 13 %. 3 This represents loss as the sign is negative. 20. Suresh buys two books for Rs.615 and sells one at a loss of 10% and other at a gain of 15%. If both the books are sold at the same price, then find the cost price of two books. Sol: The C.P. of the item sold at loss of 10% A(1 0 0 gain% ) = (1 0 0- los s % ) (1 0 0 gain% ) 6 1 5 (1 0 0 1 5 ) 6 1 5 1 1 5  = (1 0 0- 1 0 ) (1 0 0 1 5 ) 205 = 3 x 115 = Rs.345. The C.P. of the item sold at gain of 15% A(1 0 0 los s % ) = (1 0 0- los s % ) (1 0 0 gain% ) 6 1 5 (1 0 0 1 0 ) 6 1 5 9 0  = (1 0 0- 1 0 ) (1 0 0 1 5 ) 205

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= 3 x 90 = Rs.270. 21. Find a single discount equivalent to two successive discounts of 20% and 30%.

xy   Sol: Overall Discount =  x  y  % 1 0 0  2 0(3 0)   =  2 0  3 0  %  4 4% . 1 0 0   22. Find a single discount equivalent to two successive discounts of 20% and 30% and 40%. Sol: Overall Discount  (xy  yz  zx) xyz =  x  y  z   100 1 0 02 

 % 

 (2 0)(3 0)  (3 0)(4 0)  (4 0)(2 0) (2 0)(3 0)(4 0)  % =  2 0  3 0  4 0   100 1 0 02  

6 0 0 1 2 0 0 8 0 0 24000  2 6 0 0 2 4 0 0 0    =  9 0  %   9 0  1 0 0  1 0 0 0 0% 1 0 0 1 0 0  1 0 0     = (90  26  2.4)%  66.4%. 23. A shopkeeper sold the shirts at Rs.306 each after giving 15% discount on labelled price. If he had not given the discount, he would have earned a profit of 20% on the cost price. Find the C.P. of each shirt. Sol: Here, labeled price x = Rs.306, discount d = 15%, profit p = 20%   1 0 02 x  Required C.P. =   (1 0 0 d)(1 0 0 p)    1 0 0 1 0 0 3 0 6 1 0 0 1 0 0 3 0 6  = (1 0 0 1 5)(1 0 0 2 0) 8 5 1 2 0 = Rs.300.

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Brainstorming 1.

2.

Find the gain % if the C.P. is Rs.750 and S.P. is Rs.1000. 2 1 1 1. 1 6 2. 25 3) 3 3 4. 2 7 3 3 2 A watch is sold for Rs.1080 at a loss of 10%. The C.P. of the watch is ___ 1. 1125 2. 180 3. 1188 4)1200

3.

An article is sold at Rs.45 at a loss of 10%. If it is sold at Rs.65, the gain % is ___ 1. 10 2. 15 3. 20 4. 30

4.

A bicycle is sold at 20% gain. If it had been sold at 20% loss, the S.P. would have been Rs.120 less. The C.P. of the bicycle is ___ 1. 500 2. 300 3. 250 4)200

5.

A dishonest shopkeeper uses false weight to give 20% less quantity of food. His profit % is ___ 1. 80 2. 75 3. 25 4. 20

6.

A shopkeeper buys two types of tea one at Rs.700 for 10 kg and the other at Rs.770 for 10 kg. He mixed the two types and the mixture is sold at Rs.840 per 10 kg. His profit % is ___ 2 1 9 1 1. 1 4 2. 1 6 3. 1 2 4) 1 4 7 11 11 7 A man lost 10% by selling oranges at the rate of 11 for a rupee. How many should he sell them to gain 10% at a rupee? 1. 10 2. 9 3. 15 4. 12

7.

8.

9.

By selling 20 articles a man gains the S.P. of 5 articles. The gain % is ___ 1 2 1. 20 2. 25 3. 3 3 4. 1 6 3 3 A trader marks his goods at 20% above the C.P. and allows a discount of 10%. His gain % is __ 1. 8 2. 10 3. 15 4. 12

10.

A shopkeeper marks his goods in such a way that after allowing a discount of 5%, he gains 14%. How much % above the C.P. is marked price? 1. 18 2. 20 3. 15 4. 25

11.

By selling a radio for Rs.572, a shopkeeper earns a profit equivalent to 30% of the C.P. of the radio. The C.P. of the radio is ___ 1. 340 2. 400 3. 440 4. 404

12.

A man bought an article and sold it at a loss of 5%. If he had bought it for 10% less and sold it for Rs.15.30 more, he would have made a profit of 15%. The C.P. of the article is ___ 1. 320 2. 180 3. 260 4. 90

13.

A merchant makes a profit of 12% on his goods after giving 20% off on the printed price. He marked his goods by what % above his C.P.? 1. 40 2. 30 3. 25 4. 35

14.

Suresh bought 18 kg of rice @ Rs.4.50 per kg and 12 kg @ Rs.3.50 per kg. He mixes the two and sold the mixture @ Rs.5 per kg. What was his total gain in this transaction? 1. 25 2. 26 3. 27 4. 28

15.

The profit obtained by selling an article for Rs.560 is same as the loss when the same article is sold for Rs.506. The C.P. of the commodity is ___ 1. 510 2. 530 3. 533 4. 515

16.

A merchant professes to lose 5% on a certain tea powder, but he uses a weight of 900 g instead of 1 kg. His real gain% is ___ 5 6 1. 5 2. 5 3. 6 4. 6 9 7

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17.

If the profit % is numerically equal to the C.P. in rupees and the S.P. is Rs.39, what is the C.P.? 1. 21 2. 23 3. 30 4. 25

18.

A bicycle when sold for Rs.954 gave a loss of 10%. To earn 10% profit, it should be sold for Rs.___ 1. 1616 2. 1661 3. 1166 4. None

19.

The C.P. of four articles is same as the S.P. of three articles. What is the loss or gain %? 1 1. 20 % profit 2. 3 3 % profit 3. 25 % loss 4. 30 % profit 3 A man bought 11 mangoes at Rs.10 and sold 10 mangoes at Rs.11. Find the loss or gain %. 1. 21% gain 2. 20% gain 3. 25% gain 4. 31% loss

20. 21.

22.

A man sold 20 articles and got S.P. of 2 articles as the profit. What is his profit %? 1 1 1 1. 10 2. 1 2 3. 1 1 4. 1 0 3 9 9 Two successive discounts of 10% and 20% are equivalent to a single discount of ___ 1. 28% 2. 30% 3. 25% 4. 32%

23.

Three successive discounts 10%, 10% and 5% are equivalent to a single discount of ___ 1. 23.05% 2. 30% 3. 25% 4. 23.5%

24.

The marked price of an article is Rs.3600. What will be the S.P., if two successive discounts of 5% and 10% are offered? 1. 3780 2. 3078 3. 3708 4. 3807

25.

26.

27.

28.

2 1 The S.P. of an article is Rs.700 after offering two successive discounts of 1 2 % and 1 6 % . 2 3 What is the marked price? 1. 906 2. 1000 3. 960 4. 980 1 A man marks his goods 35% above the C.P. and allows 1 1 % discount. If he sells the goods for 9 Rs.2640 find the C.P. 1. 2200 2. 2400 3. 2020 4. 2600 At what price above the C.P. an article is marked so that it earns 12% after 8% discount? 2 1. 25 2. 12 3. 21.7 4. 1 6 3 A trader offers 30% discount on the Marked Price of an article yet makes 10% profit. If he gains Rs.140 by selling the article, then find the Marked Price. 1. 2200 2. 2300 3. 2400 4. 2000

29.

What profit % is made by selling an article at a certain price, if by selling at two-third of the price there would be a loss of 20%. 1. 20 2. 30 3. 15 4. 25

30.

A man sells two horses at Rs.990 each and gains 10% on one and loses 10% on the other. Find his gain or loss%. 1. 1 % loss 2. 1 % gain 3. No loss no gain 4. 2% loss

31.

A dealer sold two articles for Rs.396 each, gaining 20% on one and losing 20% on the other. What is his gain or loss%? 1. 4 % loss 2. 4 % gain 3. No loss no gain 4. 40% loss

32.

The C.P. of 15 pens was recovered by the sale of 12 pens only. Gain % is ___ 1 2 1. 25 2. 20 3. 1 6 4. 9 3 3 A tradesman by means of a false balance defrauds to the extent of 10% in buying goods and also defrauds to 10% in selling. Find his gain %. 1. 20 2. 21 3. 22 4. 18

33.

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34.

35.

By selling an article for Rs.450, a man loses 10%. Find his gain or loss % if he sells it for Rs.540. 1 1. 8 % loss 2. 8 % gain 3. 3% gain 4. 9 % gain 3 A trader marks his goods at 30% above the C.P. but constrained to give a discount of 30% to clear his stock. What is gain or loss%? 1. 9% loss 2. 9% gain 3. No loss no gain 4. 3% gain

36.

Oranges are bought at 2 for a Re.1 and sold at 5 for Rs.3. What will be the gain or loss% in the transaction? 1. 25% loss 2. 25% gain 3. 20% gain 4. 20% loss

37.

A dealer sold an article at a loss of 6.25%. If he had purchased it for 10% lesser price and sold it for Rs.42 more, he would have gained 10%. His original purchase price is ___ 1. 750 2. 800 3. 840 4. 880

38.

A dealer purchases 22 pencils for Rs.20 and sells them @ 10 pencils for Rs.11. His profit % is __________ 1. 10 2. 20 3. 21 4. 22

39.

A reduction of 20% enables a person to buy 5 more oranges for Rs.10. Find the price of each orange before reduction. 1. 0.5 p 2. 0.75 p 3. Rs.1 4)0.40 p

40.

A man bought a watch at 1. 15 1) 2 2) 4 3) 4 4) 2 5) 3 6) 1 7) 2 8) 3 9) 1 10) 2

19 20 2. 16 11) 3 12) 2 13) 1 14) 3 15) 3 16) 2 17) 3 18) 3 19) 2 20) 1

th

of its C.P. and sold at 14% above its C.P. What is his gain? 3. 17.5 21) 3 22) 1 23) 1 24) 2 25) 3 26) 1 27) 3 28) 1 29) 1 30) 1

4. 20 31) 1 32) 1 33) 3 34) 2 35) 1 36) 3 37) 2 38) 3 39) 1 40) 4

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10. PROBLEMS ON AGES In solving the problems related to ages, we come across three situations. 1. Age some years ago 2. Present Age 3. Age some years hence Models:

1.

If the age of A, t years ago, was n 1 times the age of B and at present the age of A is n 2 times that of B, then finding the present Age of A and the present Age of B.

2.

If the present age of A is n 1 times the present age of B. If t years hence, the age of A would be n 2 times that of B, then finding the present Age of A the present Age of B.

3.

If the age of A, t 1 years ago, was n 1 times the age of B and if t 2 years hence A‟s age would be n 2 times that of B, then finding the present Age of A and the present Age of B.

4.

The sum of present ages of A and B is S years. If t years ago, the age of A was n times the age of B, then finding the present Age of A and the present Age of B.

5.

The sum of present ages of A and B is S years. If t years hence, the age of A would be n times the age of B, then finding the present Age of A and the present Age of B.

6.

If the ratio of the present ages of A and B is a : b and t years hence, it will be c : d, then finding the present Age of A and the present Age of B.

SOLVED EXAMPLES 1.

The age of Mr. Bindra is 3 times the age of his son. If 10 years ago, his age was 8 times the age of his son, then at that time what would be the age of Mr. Bindra.

Sol: 10 years ago: present: Son = x son = x+10 Mr. Bindra = 8x Mr. Bindra = 8x+10 3(x+10) = 8x+10 3x+30 = 8x+10 x = 4 Mr. Bindra = 8x4+10 = 42 2.

The age of Mr. Sinha is 7 times the age of his daughter. After 15 years, the age of Mr. Sinha is only twice the age of his daughter. Find the present age of Mr. Sinha‟s daughter.

Sol: Present: After 15 years: Daughter = x Daughter = x+15 Mr. Sinha = 7x Mr. Sinha = 7x+15 7x+15 = 2(x+15) 7x+15 = 2x+30 x = 3 Mr. Sinha‟s daughter = 3 years 3.

12 years ago Lekha‟s father was 5 times older than her. After 12 years, the father will be 3 times older than her. Find the present age of Lekha.

Sol: 12 years ago: Lekha = x Father = 5x

Present: Lekha = x+12 Father = 5x+12 3(x+24) = 5x+24

After 12 years: Lekha = x+24 Father = 5x+24

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4.

3x+72 = 5x+24 x = 24 Lekha = 24+12 = 36 The sum of the ages of A and B is 36 years. 2 years back, the age of A is 4 times the age of B. Find the difference between the present ages of A and B.

Sol: 2 years back: B = x A = 4x

5.

The sum of the ages of father and a son is 44 years. After 8 years, the age of the father will be 5 times that of son. Find their respective ages.

Sol: After 8 years: Son = x Father = 5x

6.

Present: Son = x – 8 Father = 5x-8 x-8+5x-8 = 44 x = 10 Son = 10-8 = 2 Father = 50-8 = 42

The ratio of the present ages of father and son is 3 : 1. After 10 years, it will become 5 : 2. Find the present age of the son.

Sol: Present: Son = x Father = 3x

7.

Present: B = x+2 A = 4x+2 x+2+4x+2 = 36 x = 6.4 A-B = 25.6 – 6.4 = 19.2

After 10 years: Son = x+10 Father = 3x+10 3x  10 5  x  10 2 6x+20 = 5x+50 x = 30 son = 30 years

4 years ago Kalyani was thrice as old as Ramya. If the ratio of their present ages is 4 : 3 respectively. Find the difference between their ages.

Sol: Present: Kalyani = 4x Ramya = 3x

4 years ago: Kalyani = 4x-4 Ramya = 3x-4 3(3x-4) = 4x-4 9x-12 = 4x-4 x = 1.6 Kalyani – Ramya = 1.6

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Brainstorming

1.

The ratio of P‟s and Q‟s ages is 5:7. If the difference between the present age of Q and the age of P six years hence is 2, then what is the total of present ages of P and Q? 1. 52 yrs 2. 48 yrs 3. 56 yrs 4. None

2.

The ratio between the present ages of P and Q is 5:8. After 4 years, the ratio between their ages will be 2:3. What is Q‟s present age? 1. 36 yrs 2. 20 yrs 3. 24 yrs 4. None

3.

The ratio of present ages of Ram and Shyam is 7:8, respectively. Four years hence this ratio becomes 9:10, respectively. What is Ram‟s present age in years? 1. 18 2. 14 2. 17 4. None

4.

One year ago the ratio of Karan‟s and Ankit‟s ages was 6:7, respectively. Four years hence this ratio would become 7:8. How old is Ankit? 1. 35 yrs 2. 30 yrs 3. 31 yrs 4. None

5.

The sum of the ages of A & B is 50 years and the ratio between their ages is 7:3 then the age of A is how many years? 1. 15 2. 35 3. 20 4. 25

6.

The sum of the ages of P & Q is 72 years and the ratio between their ages is 2:7. Then the ratio between their ages 8 years ago is __________ 1. 2:5 2. 5:2 3. 10:15 4. 1:6

7.

The ratio between the present ages of A & B in 4:5 and the sum of their ages after 5 years is 100 years. What will be the ratio of their ages at that time? 1. 9:10 2. 10:9 3. 9:11 4. 11:9

8.

The ratio between the ages of a father and his son, 10 years ago was 9:2. The ratio between the present ages of the father and his son is 2:1. What is the ratio of the ages of Father & son 10 years hence? 1. 2:1 2. 19:12 3. 12:7 4. 21:11

9.

The sum of the ages of a mother and her daughter is 50 years. Also, 5 years ago, the mother‟s age was 7 times the age of the daughter. The present ages of the mother and daughter respectively are ________ 1. 35 & 15 yrs 2. 38 & 12 yrs 3. 40 & 10 yrs 4. 42 & 8 yrs

10.

A man is 4 times as old as his son. Four years hence the sum of their ages will be 43 years. The present age of son is _________ 1. 5 yrs 2. 7 yrs 3. 8 yrs 4. 10 yrs

11.

Kalyani got married 6 years ago. Today her age is 1 Her son‟s age is 1. 2 yrs

1 times her age at the time of marriage. 4

1 times her age. Her son‟s age is _________ 10 2. 3 yrs

3. 4 yrs

4. 5 yrs

12.

The difference between the ages of two persons is 10 years, 15 years ago, the elder one was twice as old as the younger one. The present age of the elder son is ________ 1. 25 yrs 2. 35 yrs 3. 45 yrs 4. 55 yrs

13.

The ratio between the ages of A and B at present is 2:3. Five years hence the ratio of their ages will be 3:4. What is the present age of A? 1. 10 yrs 2. 15 yrs 3. 25 yrs 4. None

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14.

Mahesh is as much younger to Anil as he is older to Prashant. If the sum of the ages of Anil and Prashant is 48 years, what is the age of Mahesh? 1. 20 yrs 2. 24 yrs 3. 30 yrs 4. None

15.

Five years ago, Vinay‟s age was one third of the age of Vikas and now Vinay‟s age is 17 years. What is the present age of Vikas? 1. 5 yrs 2. 36 yrs 3. 41 yrs 4. 51 yrs

16.

The age of Arivind‟s father is 4 times his age. If 5 years ago, father‟s age was 7 times of the age of his son at that time. What is Arivind‟s father‟s present age? 1. 35 yrs 2. 40 yrs 3. 45 yrs 4. 55 yrs

17.

10 years ago, Anitha mother was 4 times older than her daughter, after 10 years, the mother will be twice of the daughter. The present age of Anitha is ________ 1. 5 yrs 2. 20 yrs 3. 45 yrs 4. 30 yrs

18.

A is twice as old as B. 12 years ago, A was five times old as B. Find the present age of A? 1. 16 yrs 2. 32 yrs 3. 24 yrs 4. 28 yrs

19.

The age of the father 8 years ago was 5 times the age of his son. At present the father‟s age is 3 times that of his son. Find the age of father? 1. 48 yrs 2. 36 yrs 3. 46 yrs 4. 58 yrs

20.

At present the age of the father is five times that of the age of his son. Three years hence, the father‟s age would be four times that of his son. Find the present ages of the father and the son? 1. 8: 32 2. 45: 9 3. 12;57 4. None

21.

The age of Mrs. Anjali is 5 times the age of his son. After 12 years the age of Mrs. Anjali will be only twice the age of his son. Find the present age of Mrs. Anjalis‟s son? 1. 4 yrs 2. 16 yrs 3. 12 yrs 4. 18 yrs

22.

9 years ago Vimal was 5 times the age of Sudarshan. If the present age of Vimal is twice the age of Sudarshan, what will be the total of their present ages? 1. 66 yrs 2. 54 yrs 3. 36 yrs 4. 46 yrs

23.

At present the age of the father is 6 times the age of his son, 4 years hence the fathers‟ age would be 5 times that of his son. What is the sum of the present ages of father and his son? 1. 116 yrs 2. 112 yrs 3. 114 yrs 4. 111 yrs

24.

At present the age of the father is 7 times the age of the son, 4 years hence the fathers‟ age would be 4 times that of his son. What is the sum of the present ages of father and his son? 1. 21 yrs 2. 24 yrs 3. 28 yrs 4. 32 yrs

25.

At present the age of the father is 7 times the age of his son, 6 years hence the fathers‟ age would be 5 times that of his son. What is the sum of the present ages of father and his son? 1. 80 yrs 2. 64 yrs 3. 96 yrs 4. None of these

26.

The sum of ages of A and B is 60 years. After 10 years A will be thrice as old as B. Find the difference of their present ages? 1. 30 yrs 2. 40 yrs 3. 10 yrs 4. 50 yrs

27.

The sum of the ages of the father and his son is 41 years. After 17 years the father‟s age will be twice the age of his son. Then the respective ages of the father and son are ______ and ____ years. 1. 32, 9 2. 34, 7 3. 33, 8 4. 31, 10

28.

The sum of the ages of a son and father is 56 years. After four years, the age of the father will be three times that of the son. Their ages respectively are _____________ 1. 12 & 44 yrs 2. 16 & 48 yrs 3. 16 & 42 yrs 4. 18 & 36 yrs

29.

The present ages of the father and son are in the ratio 6:1. After 5 years, the ratio will be 13:3. Find the present age of the son?

82

1. 24 yrs

2. 28 yrs

3. 32 yrs

4. None

30.

The ratio of the present ages of P and Q is 8:5. After 6 years their ages are in the ratio of 3:2. Find the ratio of the sum and difference of the present ages of P and Q. 1. 13:3 2. 39:19 3. 13:2 4. 13:5

31.

8 years ago, the ratio of the ages of Nagendra and Sunil was 3:2. If the ratio of their present ages is 7:5 respectively, what is the sum of their present ages? 1. 96 yrs 2. 86 yrs 3. 76 yrs 4. 66 yrs

32.

The ratio of the ages of A and B at present is 4:2. 10 years earlier, the ratio was 3:2, then find the present ages of A and B. 1. 40 & 30 yrs 2. 48 & 36 yrs 3. 64 & 48 yrs 4. 20 & 15 yrs

33.

The ages of Kanchan is thrice the age of the Chandan. After 12 years the age of Kanchan will become twice the age of Chandan. Then the sum of their present ages is ____ years. 1. 42 yrs 2. 48 yrs 3. 46 yrs 4. 50 yrs

34.

The age of Jayshree is thrice the age of her younger sister. The product of their ages is 300 years. Then the Jayshree‟s present age is ___________ years. 1. 30 2. 20 3. 10 4. 40

35.

At the time of the marriage, a man was 6 years elder than to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was the wife‟s age at the time of marriage? 1. 18 years 2. 24 years 3. 30 years 4. 36 years

36.

The difference of present ages of A and B is 12 years. 6 years back their ages were in the ratio 3:2, how old is B? 1. 32 years 2. 35 years 3. 30 years 4. 45 years

37.

Siddhika‟s age is

38.

3 years ago, the father‟s age was twice the ages of his 4 daughters. In 3 years time, the father‟s age will be equal to sum of ages of the daughters. Find the present age of father. 1. 45 years 2. 36 years 3. 24 years 4. 39 years

39.

Ten years before, one is seven years more than the “half the age” of other. Now the bride‟s age is 30 years. And also one is nineteenths the age of other. What is groom‟s present age. 1. 35 2. 27 3. 40 4. 58

40.

When I add 4 times my age 4 years from now to 5 times my age 5 years from now, I get 10 times my current age. How old will I be 3 years from now? 1. 52 2. 40 3. 44 4. None

th

1 of her father‟s age. Siddhika‟s father‟s age will be twice of Kamal‟s age 6 after 10 years. If Kamal‟s 8th birthday was celebrated two years before, then what is Siddhika‟s present age? 1. 10 years 2. 15 years 3. 5 years 4. 3 years

1) 2 2) 4 3) 2 4) 4 5) 2 6) 4 7) 3 8) 2 9) 3 10) 2

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

2 2 1 2 3 2 2 2 1 2

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

1 3 2 4 3 2 3 1 4 1

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

1 1 2 1 1 3 3 4 2 3

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11. TIME AND WORK 1. 2. 3. 4.

Usually, we need to complete a particular job in a reasonable time. We need to complete any project earlier or later depending upon the requirements. Accordingly, the men on duty have to be increased or decreased. This means, the time allowed and the men engaged for a project are inversely proportional to each other. Again it means, more the number of men involved, the lesser is the time required to finish a job. 5. You also come across the situations where time and work or men and work are in direct proportion to each other. Formulae: 1. If A can do a piece of work in n days, then at a uniform rate of work ing, A will finish 1 th work in one day. n 1 2. If th of work is done by A in one day, then A will take n days to complete the full n work. 3. If A does

1 n hours. th of a work in m hours, then A will take m n

4. If A does three times faster work than B, then a) The ratio of work done by A and B is 3 : 1 b) The ratio of time taken by A and B is 1 : 3 5. A, B and C can do a piece of work in T 1 , T 2 and T 3 days respectively. If they have worked for D 1 , D 2 and D 3 days respectively, then D a) Amount of work done by A = 1 T1 b) Amount of work done by B =

D2 T2

c) Amount of work done by C =

D3 T3

D1 D2 D3 + + . T1 T2 T3 6. If A can do a piece of work in X days and B can do the same work in Y days, then both XY of them working together will do the same work in days. X+Y d) Amount of work done by A, B and C together =

7. If A, B and C, while working alone, can complete a work in X, Y and Z days XY Z respectively, then they will together complete the work in days. XY + Y Z+ ZX

84

8. A and B working together can complete a piece of work in X days. If A working alone can complete the work in Y days, then B working alone will complete the work in XY days. Y X 9. If A and B working together can complete a piece of work in X days, B and C in Y days, C and A in Z days, then 2XY Z a) A, B and C working together will finish the job in days. XY  Y Z  ZX 2XY Z b) A alone can finish the job in days. XY  Y Z  ZX 2XY Z c) B alone can finish the job in days. Y Z  ZX  XY 2XY Z d) C alone can finish the job in days. XY  Y Z  ZX 10. If A and B working together can complete a work in X days and B is k times efficient than A, then a) The time taken by A working alone, to complete the work is (k + 1) X. k  1 b) The time taken by B working alone, to complete the work is  x.  k 

 x  c) The time taken by both A and B working together to complete the work is  . 1  k  11. If A working alone takes a days more than A and B working together. If B worked alone, he takes b hours more to complete the job than A and B working together then both A and B working together can finish the job in ab days. a c 12. If A can complete part of a work in X days, then part of the work will be done in d b bcX days. a d 13. If a men and b women can do a piece of work in n days, then c men and d women can  nab  do the work in   days.  bc  ad  14. If A can complete a work in X days and B is k times efficient than A, then the time x taken by both A and B working together to complete the work is . 1k 15. If A is k times more efficient than B and hence able to finish the work in l days less than B, then l a) A working alone can finish the work in days. k 1 kl b) B working alone can finish the work in days. k 1 kl c) A and B working together, can finish the work in days. k2  1 16. There are two groups of people with same efficiency. In one M 1 persons can do W 1 works in D 1 time and in the other M 2 persons can do W 2 works in D 2 time. Then the relationship between the two groups is M1D1W2  M2D2W1 17. There are two groups of people with same efficiency. In one M 1 persons can do W 1 works in D 1 time working t 1 hours a day and in the other M 2 persons can do W 2 works in D 2 time working t 2 hours a day. Then the relationship between the two groups is M1D1t1W2  M2D2 t 2W1

85

SOLVED EXAMPLES 1. A can complete a piece of work by working alone in 5 days and B while working alone can finish the same work in 10 days. If both of them work together, then in how many days, the work will be finished. Sol: Here, X = 5, Y = 10 Working together, A and B will finish the work in

XY days XY

5(1 0) 10 days.  5  10 3 2. A, B and C can complete a piece of work in 5, 10 and 13 days respectively. In how many days would all of them complete the same work working together? =

Sol: Here, X = 5, Y = 10, Z = 13 So, the work will be completed in

3.

XY Z days XY  Y Z  ZX 5  1 0 1 3 32 2 = days. 5(1 0)  1 0(1 3)  1 3(5) 49

A and B working together take 10 days to complete a piece of work. If A alone can do this work in 15 days, how long would B take to complete the same work? Sol: Here, X = 10, Y = 15 So, B alone will complete the work in

1 0(1 5) XY days =  3 0 days. YX 1 5 1 0

4. Madhu and Anil can do a piece of work in 12 days, Anil and Sunil in 15 days, Sunil and Madhu in 20 days. How long would each take separately to do the same work? Sol: Here, X = 12, Y = 15 and Z = 20 2XY Z Madhu alone can do the work in days XY  Y Z  ZX 2  1 2  1 5 2 0 7200   3 0days. = 1 2(1 5)  1 5(2 0)  2 0(1 2) 240 2XY Z Anil alone can do the work in days Y Z  ZX  XY 2  1 2  1 5 2 0 7200   2 0days. = 1 5(2 0)  2 0(1 2)  1 2(1 5) 360 2XY Z Sunil alone can do the work in days XY  Y Z  ZX 2  1 2  1 5 2 0 7200   6 0 days. = 1 2(1 5)  1 5(2 0)  2 0(1 2) 120 5. Mukesh can do a piece of work in 16 days. If Nita works thrice as fast as Mukesh, how long would they take to finish the work by working together? Sol: Here, X = 16 and k = 3  x  Required time =   days 1  k 

 16  =    4 days. 1  3  6. A and B together can do a piece of work in 4 days. If A does twice as much work as B in a given time, how long A alone would take to do the work? Sol: Here, X = 4 and k = 2

86

2 1 k  1 Time taken by A, working alone =  x  2  4  6 days k   7. A alone would take 7 days more to complete the work than if both A and B worked 3 together. If B worked alone, he took 1 days more to complete the job both worked 4 together. What time would they take if both A and B worked together? 3 7 Sol: Here, a = 7, b = 1  4 4 7 7  days. 4 2 8. A is four times as good a workman as B and takes 18 days less to do a piece of work than B takes. Find the time in which B alone can complete the work.

Time taken by A and B working together =

ab days =

7

Sol: Here, k = 4 and l = 12 Time taken by B working alone = 9. A can do

4 12 kl days =  1 6 days. k 1 4 1

1 5 of a work in 15 days. In how many days he can finish of the work? 6 4

Sol: Here, a = 5, b = 4, c = 1, d = 6, X = 15 bcX 4  1  15 Required time =   2 days. a d 56 1 10. If 12 persons can complete th of a work in 4 days, then find the number of persons 5 required to complete the remaining work in 16 days. Sol: Here M 1 = 12, D 1 = 4, W 1 =

4 1 and M 2 =?, D 2 = 16, W 2 = 5 5

We have, M1D1W2  M2D2W1

 12 

x 4 x

4 1 = M 2 x 16 x 5 5

M 2 = 10.

11. If 5 persons can cut 15 trees in 6 days working 10 hours a day. Then, in how many days can 12 persons cut 36 trees working 6 hours a day. Sol: Here M 1 = 5, D 1 = 6, W 1 = 15, t 1 = 10 M 2 = 12, D 2 =?, W 2 = 36, t 2 = 6 We have, M1D1t1W2  M2D2 t 2W1

5 x  D2

6 x 10 x 36 = 12 x D 2 x 6 x 15 = 10 days.

12. 8 men or 12 women can do a work in 21 days. In how many days, 5 men and 3 women would complete the work? Sol: Here, a = 8, b = 12, c = 5, d = 3 and n = 21 2 1 8  1 2 2 1 8  1 2  nab   2 4days. Required no. of days =    1 2(5)  8(3)  bc  ad 84  

Brainstorming 1.

P can do a work in 12 days and Q can do it in 36 days. In how may days the work can be done if they do it together?

87

1. 24 2.

3.

2. 9

3. 10

4. 7

Madan and Raju can do a work in 15 days, if they work together. If Madan alone can do it in 20 days, in how many days Raju alone can do the work? 4 1. 8 2. 60 3. 10 4. 44 7 P and Q can do a work in 9 days. Q and R can do it in 12 days, R and P in 18 days. In how many days all of them can do the work? 1. 8 2. 9 3. 15 4. 5

4.

Shweta can do a work in 40 days. She worked for 5 days and Srikanth finished the remaining work in 21 days. In how many days can both of them do the work? 1. 18 2. 19 3. 16 4. 24

5.

Rajeev can do a work in 50 days, which Sanjay can do in 40 days. Rajeev started the work and was joined by Sanjay after 20 days. In how many days the total work is completed? 1 1. 10 2. 15 3. 33 4. 14 3 Pavan and Venu undertook a piece of work for Rs.600. Pavan alone can do the work in 24 days and Venu alone can do it in 12 days. What is the share of Venu? 1. Rs.360 2. Rs.400 3. Rs.200 4. Rs.240

6.

7.

P and Q can do a piece of work in 12 days. P, Q and R together can finish it in 8 days. In how many days will R finish that work? 24 1. 2. 24 3. 20 4. 4 5

8.

P can do a piece of work in 80 days. He works at it for 10 days and then Q alone finishes the work in 42 days. Both together could complete the work in how many days? 1. 30 2. 48 3. 24 4. 25

9.

P and Q together finish a work in 30 days. They worked for 20 days and then Q left. The remaining work was done by P alone in 15 days. In how many days P alone can finish the total work? 1. 35 2. 30 3. 45 4. None

10.

P can complete a job in 9 days, Q in 10 days and R in 15 days. Q and R started the work and were forced to leave after 3 days. Find the time taken by P to complete the remaining work. 1. 4½ 2. 5 3. 7½ 4. None

11.

P does half as much work as Q in three fourth of the time. If together they take 18 days to complete a work, how much time shall P take to do it? 1. 35 2. 30 3. 45 4. None

12.

P can complete a job in 15 days, Q in 10 days and R in 20 days. Q and R started the work together and were forced to leave after 4 days. Find the time taken by P to complete the remaining work? 1. 5 2. 12 3. 6 4. None

13.

A certain number of men complete a piece of work in 50 days. If there were 8 men more the work could be finished in 10 days less. How many men were there originally? 1. 30 2. 32 3. 36 4. 40

14.

If 20 men can build a wall of 120 meters long in 6 days, what length of a similar wall can be built by 25 men in 3 days? 1. 95 2. 60 3. 75 4. 70

15.

P and Q can do a piece of work in 12 days, Q and R in 15 days, R and P in 20 days. How long the work runs if they do it altogether. 1. 10 days 2. 15 days 3. 12 days 4. 5 days

88

16.

If 17 labourers can dig a ditch 26 meters long in 18 days, working 8 hours a day, how many more labourers should be engaged to dig a similar ditch of 52 meters length in 16 days, each labourer working 9 hours a day? 1. 34 2. 17 3. 30 4. None

17.

P and Q can do a piece of work in 30 and 70 days respectively. They worked together for 10 days and then Q goes away. In how many days will P finish the remaining work? 5 1. 11 2. 15 3. 16 4. 15 7

18.

If 5 men or 6 women can do a piece work in 38 days, in how many days will 3 men and 4 women will do the same work? 1. 20 2. 17 3. 30 4. None

19.

2 men and 3 boys can do a piece of work in 8 days, 3 men and 2 boys can do it in 7 days. In how many days can 1 man and 2 boys do the same work? 1. 24 2. 17 3. 12 4. 14

20.

P does two-fifth of work in 8 days. He then calls Q and they together complete the remaining work in 6 days. How long would Q alone take to complete the entire work? 1. 20 2. 17 3. 18 4. None

21.

A job can be completed by 10 men in 20 days and by 40 women in 15 days. How many days it takes for 5 men and 5 women to finish the work? 1. 40 2. 24 3. 30 4. 35

22.

After completing half of a work in 10 days P was joined by Q and they worked together for 3 days. P completed the remaining work in 4 days. In how many days Q alone can do the entire work? 1. 25 2. 20 3. 30 4. 18

23.

10 men can do a work in 40 days. They started together but 4 men left after a few days. As a result the work was completed in 60 days. After how many days the 4 men left? 1. 15 2. 20 3. 10 4. None

24.

80 men can finish work in 60 days? 1. 38

2 of work in 30 days. How many men are required to finish the remaining 5 2. 60

3. 30

4. 40

25.

The work done by 1 man and 3 boys are equal. A work can be done by 3 men and 5 boys in 8 days. How many boys should assist 9 men to finish the work in 4 days? 1. 4 2. 7 3. 1 4. None

26.

P and Q can do a job in 40 days and 36 days respectively. If they did a job for Rs.2660, what should be the share of Q? (in Rs.) 1. 1260 2. 1400 3. 1520 4. 1600

27.

8 men and 20 boys can construct a house in 5 days working for 6 hours a day. 5 men and 8 boys can construct two such houses in 16 days working for 7.5 hours a day. In how many days can 12 men and 6 boys construct 5 such houses working for 10 hours a day? 1. 12 2. 18 3. 24 4. 30

28.

P can do a piece of work in 30 days and Q alone can do it in 40 days. They begin together but 5 days before the completion of work, P leaves off. What is the total number of days to complete the work? 1. 20 2. 30 3. 25 4. 15

29.

P and Q can do a piece of work in 12 days and 20 days respectively. They started together and then P left after for some days. If Q finishes the remaining work in 12 days, how many days has P worked on it before leaving? 1. 3 days 2. 4 days 3. 6 days 4. 8 days

89

30.

P is twice as good a workman as Q. Working together they can finish a work in 16 days. How many days will P take to finish the work? 1. 4 days 2. 8 days 3. 24 days 4. 48 days

31.

P and Q can do a work in 10 and 15 days respectively. If they work at it on an alternate day, P starting first, in how many days the work will be completed? 1. 6 days 2. 10 days 3. 12 days 4. 15 days

32.

A contractor agreed to complete a work in 120 days. He engaged 150 men for the work and after 70 days, only half of the work was completed. How many additional men must employ to complete the work in agreed time? 1. 60 men 2. 90 men 3. 150 men 4. 210 men

33.

A camp has provision of food for 600 persons for 41 days. After 37 days, 360 persons left the camp. How long the food will last for the remaining persons? 1. 5 days 2. 8 days 3. 10 days 4. 12 days

34.

P can do a job in 10 days, Q can do it in 12 days and R can do it in 15 days. All begin together, but P leaves the work after 2 days and Q leaves 3 days before the work is completed. How long did the work last? 1. 9 days 2. 8 days 3. 7 days 4. 6 days

35.

P and Q can do a piece of work in 45 days and 40 days respectively. They began to work together but P leaves after „x‟ days and Q finishes the remaining work in (x + 14) days. After how many days did P leaves? 1. 19 days 2. 11 days 3. 9 days 4. 12 days

36.

Shekhar has done

37.

A garrison has provisions for certain number of days. After 15 days,

38.

If 4 labourer reap 40 acres in 30 days, how many acres will 18 labourer reap in 12 days? 1. 49 acres 2. 70 acres 3. 72 acres 4. 65 acres

39.

6 men can prepare 6 toys in 5 days working 4 hrs a day. In how many days can 10 men can prepare 15 toys working 5 hrs per day? 1. 4 2. 5 3. 6 4. 7

40.

2 men and 3 boys can do a piece of work in 10 days and 3 men together with 2 boys can do the same job in 8 days. In how many days can 2 men and a boy do the work? 1 1 1 1. 12 days 2. 18 days 3. 25 days 4. 25 days 2 2 2 1) 2 11) 4 21) 3 31) 3 2) 2 12) 3 22) 2 32) 1 3) 1 13) 2 23) 3 33) 3 4) 4 14) 3 24) 2 34) 3 5) 3 15) 1 25) 3 35) 3 6) 2 16) 2 26) 1 36) 4 7) 2 17) 4 27) 2 37) 2 8) 1 18) 3 28) 1 38) 3 9) 3 19) 4 29) 1 39) 3 10) 1 20) 1 30) 3 40) 1

1 rd of a job in 8 days, Kamalakar completes the rest of the job in 8 days. In 3 how many days would Shekhar and Kamalakar together had completed the work? 1. 24 days 2. 16 days 3. 20 days 4. 8 days

1 of the men left and if it 4 is found that the provisions will now last just as before, how long was that? 1. 50 days 2. 60 days 3. 70 days 4. 30 days

90

12. PIPES AND CISTERNS Pipes: Generally the pipes are connected to tank or cistern and are used to fi ll or empty the tank. Inlet: A pipe connected with a tank or a cistern that fills it is known as inlet. Outlet: A pipe connected with a tank or cistern emptying it is known as outlet. Note: 1. 2. 3. 4.

Pipes and Cistern problems are similar to those on Time and Wo rk. The only difference here is, the work done is in terms of filling or emptying a cistern and the time taken by a pipe or leak (Crack) to fill or empty a cistern respectively. Generally, the time taken to fill a cistern is taken as positive and the time taken to empty a cistern is taken as negative. The amount of work done i.e., filling or emptying a cistern is generally taken as unity, unless otherwise specified.

Formulae: 1. If an inlet can completely fill the empty tank in X hours, the part of the tank filled in 1 1 hour = . X 2. If an outlet can empty the full tank in Y hours, the part of the tank emptied in 1 hour 1 = . Y 1 1 3. If both inlet and outlet are open, net part of the tank filled in 1 hour =  . X Y 4. Two pipes A and B can fill or empty a cistern in X and Y hours respectively, while working alone. If both the pipes are opened together, then the time taken to fill or XY empty the cistern = hours. XY 5. Three pips A, B and C can fill a cistern in X, Y and Z hours respectively, while working alone. If all the three pipes are opened together, the time taken to fill the cistern = XY Z    hours. XY  Y Z  ZX   Note: This type of formulae can be generated by replacing negative sign wherever a pipe starts emptying the cistern instead of the standard positive sign. 6. Two pipes A and B can fill a cistern in X hours and Y hours, respectively. There is also an outlet C. if all the three pipes are opened together, the tank is full in Z hours. The XY Z   time taken by C to empty the full tank =  hours. Y Z  ZX  XY   7. A tank takes X hours to be filled by a pipe. But due to a leak, it is filled in Y hours.  XY  The amount of time in which the leak can empty the full tank =  hours. Y  X 8. A cistern has a leak which can empty it in X hours. A pipe which allows Y litres of water per hour into the cistern is turned on and now the cistern is emptied in Z hours.  XY Z  The capacity of the cistern is  litres. Z  X 9. One fill pipe A is k times faster than the other fill pipe B. a) If B can fill a cistern in x hours, then the time in which the cistern will be full,  x  if both the full pipes are opened together, is  hours. k  1 b) If A can fill a cistern in y hours, then the time in which the cistern will be full,  k  if both the full pipes are opened together, is    yhours. k  1

91

10. If one fill pipe A is k times faster and takes x minutes less time than the other fill pipe B, then

 x  a) A will fill the cistern in   minutes. k  1  kx  b) B will fill the cistern in   minutes. k  1 c) The time taken to fill a cistern, if both the pipes are opened together is  kx    minutes.  (k  1)2   

SOLVED EXAMPLES 1. A pipe can fill a tank in 4 hours. Find the part of tank filled in one hour. Sol: The part of tank filled in 1 hour =

1 . 4

2. A pipe can fill a tank in 20 minutes. Find the time in which

1 part of the tank will be 5

filled. Sol: The part of the tank filled in 1 minute = So,

1 20

1 1 part of the tank is filled in 2 0 = 4 minutes. 5 5

3. A pipe can empty a cistern in 20 minutes. Find the time in which

2 part of the cistern 5

will be emptied. Sol:

2 2 part of the cistern is emptied in 2 0   8 minutes. 5 5

4. A pipe can empty a cistern in 20 hours. Find the part of the cistern emptied in 4 hours. Sol: The part of the cistern emptied in 1 hour =

1 20

So, the part of the cistern emptied in 4 hours = 4 

1 1  . 20 5

5. A tap can fill a cistern in 8 hours and another can empty it in 12 hours. If both the taps are opened simultaneously, find the time in hours to fill the cistern. Sol: Here, X = 8 and Y = 12

1 1 1 1 1     . X Y 8 12 24 Total time taken to fill the cistern = 24 hours. Part of the cistern filled in 1 hour =

6. Two pipes A and B can fill a cistern in 20 and 30 minutes, respectively. If both the pipes are simultaneously then find how long will it take to fill the cistern? Sol: Here, X = 20 and Y = 30 Part of the cistern filled by (A + B) in 1 minute = 1 1 1 1 5 1      20 30 20 30 60 12 Hence, both the pipes together will fill the cistern in 12 minutes.

92

7. Two pipes A and B can separately fill a cistern in 4 hours and 6 hours re spectively, while a third pipe C can empty it in 3 hours. In what time will the cistern be full, if all the pipes are opened together? Sol: Here, X = 4, Y = 6 and Z = - 3   72 4  6  3   So, the cistern will be full in   1 2 hours. 6  (4  6)  (6  3)  (4  3)  8. Two taps A and B can fill a cistern in 15 and 30 minutes respectively. There is a third exhaust tap C at the bottom of tank. If all taps are opened at the same time, the cistern will be full in 25 minutes. In what time can exhaust tap C empty the cistern when full? Sol: Here, X = 15, Y = 30 and Z = 25 XY Z   C can empty the full tank in    Y Z  ZX  XY    1 5 3 0 2 5  minutes =  ( 1 5  3 0 )  ( 3 0  4 5 )  ( 1 5  4 5 )   1 5 3 0  2 5   1 0 minutes. 1125

9. A pipe can fill a tank in 6 hours. Due to leakage in the bottom, it is filled in 12 hours. If the tank is full, how much time will the leak take to empty it? Sol: Here, X = 6 and Y = 12

 XY  So, the time taken by the leak to empty the full tank =  hours Y  X

 6  1 2 =    1 2hours. 12  6  10. A leak in the bottom of a tank can empty the full tank in 3 hours. An inlet pipe fills water at the rate of 2 litres per minute. When the tank is full, the inlet is opened and due to leak, the tank is empty in 5 hours. Find the capacity of the tank. Sol: Here, X = 3, Y = 2 x 60 = 120 and Z = 5  XY Z   3  1 2 0 4  So, the capacity of the tank =  litres     7 2 0 litres. 53 Z  X   11. One fill pipe A is 2 times faster the second fill pipe B. If A can fill a cistern in 9 minutes, then find the time when the cistern will be full if b oth fill pipes are opened together. Sol: Here, k = 2 and y = 9 2  k   9  6 minutes. So, Cistern will be full in    y  2 1 k  1 12. One fill pipe A is 7 times faster the second fill pipe B takes 72 minutes less than the fill pipe B. When will the cistern be full if both fill pipes are opened together? Sol: Here, k = 7 and x = 72,

 kx Cistern will be full in   (k  1)2 

Brainstorming

   7  7 2  1 4 minutes.  (7  1)2 

93

1.

A tap can fill a cistern in 16 hours and another can empty it in 32 hours. If both the taps are opened simultaneously, find the time (in hours) to fill the tank. 1. 30 2. 32 3. 48 4. None

2.

A cistern normally takes 7 hours to be filled by a tap, but because of a leak it takes one hour more. In how many hours will the leak empty a filled cistern? 1. 56 2. 52 3. 50 4. 72

3.

Pipe P can fill an empty tank in 5 hours while pipe Q can empty the full tank in 8 hours. If both are opened in that empty tank, in how much time the tank will be filled? 1 1. 12 hrs 2. 13 hrs 3. 1 3 hrs 4. None 3 A tap can fill a cistern in 12 hours and another empties in 10 hours. If both are opened at the same time, how much time will it take to fill the cistern? 1. 3 hrs 2. 20 hrs 3. 5 hrs 4. Never

4.

5.

Two pipes P and Q can fill a cistern in 12 minutes and 18 minutes respectively, while a third pipe R can empty it in 9 minutes. In what time will the cistern will be full, if they all work simultaneously? 1. 30 min 2. 34 min 3. 36 min 4. 42 min

6.

A pipe can fill a cistern in 4 hours. What part of cistern is filled in 60 minutes? 1 1 1 1 1. 2. 3. 4. 6 4 2 3

7.

Two pipes P and Q separately fill a cistern in 7 ½ and 5 minutes respectively and drain pipe R can carry off 14 litres per minute. If all the pipes are opened, when the cistern is full, it is emptied in one hour. How many litres does it hold? 1. 40 litres 2. 55 litres 3. 45 litres 4. 60 litres

8.

Two pipes P and Q can fill a cistern in 60 minutes and 80 minutes respectively. Both the taps were opened simultaneously but Q is turned off after 20 minutes. Find the total time required to fill the cistern. 1. 20 min 2. 35 min 3. 45 min 4. 1 hr

9.

Three taps can fill a cistern in 10, 15 and 18 minutes respectively. In an empty cistern all the three taps are kept open and after three minutes the third tap is closed. In how many minutes more will the cistern be full? 1. 1 min 2. 3 min 3. 2 min 4. 4 min

10.

A tap can fill a cistern in 20 minutes. And another can fill it in 30 minutes. If both are opened simultaneously, find the time when the cistern will be full? 1. 10 min 2. 15 min 3. 20 min 4. 12 min

11.

Two pipes P and Q can fill a cistern in 10 minutes and 15 minutes respectively. But an empty pipe R can empty it in 5 minutes. The pipes P and Q are opened for 4 minutes and then the empty pipe R is also opened. In what time is the cistern emptied? 1. 10 min 2. 20 min 3. 15 min 4. 12 min

12.

A cistern is filled by two taps in A and B hours respectively. Then emptied by a tap in C hours. If all the three taps are opened simultaneously, the cistern is filled in F hours. The relation between F, A, B and C are given by _________. 1 A x b 1 1 1 1 1. F = (A + B) - C 2. F = 3. = 4. F = + + 1 1 1 c F A B C + A B C

13.

Two pipes P and Q can separately fill a cistern in 12 minutes and 15 minutes respectively. While a third pipe R can empty it in 6 minutes. The pipes P and Q are kept open for 5 minutes, then the pipe R is also opened. In what time the cistern be emptied? 1. 30 min 2. 35 min 3. 40 min 4. 45 min

94

14.

Two pipes P and Q can separately fill a cistern in 12 minutes and 24 minutes respectively. And a waste can drain off 20 gallons per minute. If all the three pipes are opened, the tank is filled in 24 minutes. What is the capacity of the tank? 1. 360 gallons 2. 380 gallons 3. 240 gallons 4. 320 gallons

15.

A tank can be filled separately by two pipes P and Q in 45 and 36 minutes respectively. A tap R at the bottom can empty the full cistern in 30 minutes. If the tap R is opened 7 minutes after the two pipes P and Q are opened, find when the tank becomes full. 1. 39 min 2. 45 min 3. 49 min 4. 59 min

16.

There is a leak in the bottom of the cistern. When there was no leak, the cistern was filled in 2

1 hours. It now takes half-an-hour longer. If the cistern is full, how long it would take the 2

leakage to empty the tank, if the water leaks out at double the rate after half the cistern becomes empty. 1. 10 hrs 15 min 2. 11 hrs 15 min 3. 12 hrs 15 min 4. 13 min 17.

An empty cistern has three taps P, Q and R. P and Q can fill it in 3 and 4 hours respectively. R can empty it in one hour. If P, Q and R are kept opened at 1 P.M., 2 P.M., 3 P.M., respectively, find when the cistern will be empty? 1. 3:20 PM 2. 5:20 PM 3. 5:12 PM 4. 6:20 PM

18.

An electric pump can fill a tank in 3 hours. Because of a leak, it took 3 ½ hour to fill the tank. In how much time the leak can drain the full tank? 1. 21 hrs 2. 20 hrs 3. 32 hrs 4. 25 hrs

19.

Two pipes P and Q can fill a cistern in 1 hour and 75 minutes respectively. There is also an outlet R. If all the three pipes are opened together, the tank is full in 50 minutes. How much time will be taken by R to empty the full tank? 1. 50 min 2. 70 min 3. 100 min 4. 120 min

20.

Two pipes P and Q can fill a cistern in 24 minutes and 32 minutes respectively. If both are opened simultaneously, after how much should Q be closed so that the tank is full in 18 minutes? 1. 8 min 2. 12 min 3. 15 min 4. 13 min

21.

Two pipes P and Q can fill a cistern in 15 and 20 hours respectively, while a third pipe R can empty the full tank in 25 hours. All the three pipes are opened. After 10 hours R is closed. The tank will be full in how many hours? 1. 10 hrs 2. 12 hrs 3. 15 hrs 4. 6 hrs

22.

A tank has a leak, which could empty it in 8 hours. A tap is turned on which admits 6 litres a minute into the cistern and it is now emptied in 12 hours. How many litres does the cistern hold? 1. 8840 litres 2. 8540 litres 3. 8000 litres 4. 8640 litres

23.

Two taps can fill a tank in 12 and 18 minutes respectively. Both are kept opened for 2 minutes and the first is turned off. In how many minutes more will the tank be filled? 1. 15 min 2. 20 min 3. 11 min 4. 13 min

24.

Two pipes P and Q can fill a cistern in 12 minutes and 18 minutes respectively. They are turned on at the same time. If the tap P is turned off after 4 minutes, how long will tap Q take to fill the rest cistern? 1. 8 min 2. 9 min 3. 7 min 4. 10 min

25.

If two pipes functions simultaneously, the reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe along to fill the reservoir? 1. 25 hrs 2. 20 hrs 3. 6 hrs 4. 30 hrs

95

26.

Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it was found that due to leakage in the bottom, it took 32 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it? 112 1. 112 hrs 2. 8 hrs 3. hrs 4. 7 hrs 28 min. 15

27.

Two pipes P and Q can fill a cistern in 24 minutes and 32 minutes respectively. If both are opened simultaneously, after how much time Q should be closed, so that the tank is full in 18 minutes? 1. 12 min 2. 10 min 3. 8 min 4. 6 min

28.

A cistern is 2/5th full. Pipe P can fill the total tank in 10 minutes and Pipe Q can empty it in 6 minutes. If both the pipes are opened, how long will it take to empty or fill the tank completely? 1. 6 min to empty 2. 6 min to fill 3. 9 min to fill 4. None of these

29.

Two pipes P and Q can fill a cistern in 12 minutes and 15 minutes respectively, while a third pipe R can empty the full cistern in 6 minutes. P and Q are kept open for 5 minutes in the beginning and then R is also opened. In what time the cistern is emptied? 1. 30 min 2. 45 min 3. 32 min 4. 33 min

30.

Three pipes P, Q and R can fill a tank in 6 hours. After working at it together for 2 hours, R is closed and P and Q can fill the remaining part in seven hours. The number of hours taken by R alone to fill the tank is ________. 1. 10 2. 12 3. 16 4. 14

31.

One pipe can fill a tank 3 times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in __________. 1. 81 min 2. 144 min 3. 108 min 4. 192 min

32.

Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for Bucket P to fill the empty drum. How many turns will it take for both the buckets P and Q having each turn together to fill the empty drum? 1. 30 2. 45 3. 40 4. 90

33.

Two pipes can fill a cistern in 20 minutes and 24 minutes respectively. And a drain pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is ________. 1. 60 gallons 2. 100 gallons 3. 120 gallons 4. 180 gallons

34.

Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectively. If P is opened all the time and Q and R are opened for 1 hour each alternatively, the tank will be full in __________. 2 1 1. 6 hours 2. 6 hours 3. 7 hours 4. 7 hours 3 2

35.

Two taps can fill an empty tank in 36 minutes and 60 minutes respectively. A third tap can empty the full tank in 90 minutes. At what time will an empty tank be filled if all the three are opened simultaneously at 8.50 A.M.? 1. 9:00 AM 2. 9:20 AM 3. 9:30 AM 4. 10:00 AM

36.

Two pipes can fill half of the tank in 20 minutes. If one of the pipe fills the whole tank in 60 minutes, find the time taken by the other pipe to fill the tank individually? 1. 90 min 2. 135 min 3. 100 min 4. 120 min

37.

Two pipes P and Q can fill a tank in 40 and 60 minutes respectively. When the tank will be filled, if the pipes are opened in alternative minutes starting with pipe P? 1. 24 min 2. 49 min 3. 47 min 4. 48 min

38.

Two pipes P and Q can fill a cistern in 20 minutes and 30 minutes respectively. Another leak pipe R can empty the whole tank in 60 minutes. A man opened the two filling pipes and came to stop them when the tank should be filled but he finds the leak pipe R is also opened and thus he stopped the leak pipe R. Find the time taken by the pipes to fill the tank after the leak pipe was closed.

96

1. 2

2 min 5

2. 3 min

3. 4 min

4. 3

1 min 3

39.

Three pipes P, Q and R are connected to a cistern. P and Q together can fill the cistern in 12 hrs, Q and R together in 20 hrs and R and P together in 15 hrs. In how much time will each pipe fill the cistern separately? 1. 10 hrs; 15 hrs; 30 hrs 2. 20 hrs; 15 hrs; 60 hrs 3. 20 hrs; 30 hrs; 60 hrs 4. 20 hrs; 30 hrs; 45 hrs

40.

A pipe can fill a tank in 20 minutes and another pipe in 30 minutes, but a third pipe can empty it in 10 minutes. The first two pipes are kept open for 8 minutes in the beginning and then the third pipe is also opened. In what time is the cistern emptied? 1. 35 min 2. 38 min 3. 40 min 4. 30 min

1) 2 2) 1 3) 3 4) 4 5) 3 6) 2 7) 1 8) 3 9) 3 10) 4

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

2 4 4 3 1 2 3 1 3 1

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

2 4 4 1 4 1 3 1 2 4

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

2 2 3 4 2 4 3 1 3 3

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13.TIME AND DISTANCE Speed: The speed of an object is the distance covered by it in a unit time interval. It is obtained by dividing the distance covered by the object, by the time it takes to cover the distance. Dis tanc etravelled Thus, Speed = time taken

Note: 1.

The terms, „time‟ and „distance‟ are related to the speed of a moving object.

2.

If the time taken is constant, the distance travelled is proportional to the speed i.e. more the speed; more the distance travelled in the same time.

3.

If the speed is constant, the distance travelled is proportional to the time taken i.e. more the distance travelled; more the time taken at the same speed.

4.

If the distance travelled is constant, the speed is inversely proportional to the time taken i.e. more the speed; less the time taken for the same distance travelled.

Formulae:

D is tanc e T ime

1.

Speed =

2.

Distance = Speed x Time

3.

Time =

4.

1 5  x km/hr =  x   m/sec 8  

5. 6.

D is tanc e Speed

8   x m/sec =  x   km/hr 1 5  If A covers a distance d 1 km at a speed s 1 km/hr and then d 2 km at s 2 km/hr, then the average speed during the whole journey is s s (d  d2 ) Average speed = 1 2 1 km/hr s1d2  s 2 d1

7.

If A goes from X to Y at s 1 km/hr and comes back from Y to X at s 2 km/hr, then the average speed during the whole journey is 2s1s2 Average speed = km/hr s1  s2

8.

A goes from X to Y at s 1 km/hr and returns back from Y to X at s 2 km/hr. If he takes T  s s  hours in all, the distance between A and B is T  1 2 km . s  s 2   1 A and B start at the same time from two points P and Q towards each other and after crossing they take T 1 and T 2 hours in reaching Q and P respectively, then speed

9.

A 's s peed  B's s peed

10.

T2 T1

a of the original speed, then the change in time taken to cover the b same distance is given by b  Change in time =   1  originaltime a  If the new speed is

98

11.

A body covers a distance d in time T 1 with speed s 1 , but when it travels with speed s 2 covers the same distance in time T 2 then

P roduc tof s peed s1 s Differenc eof s peed   2  d T2 T1 Differenc eof time Note: By equating any two of the above, we can find the unknowns as per the given question. 12.

A train travels a certain distance at a speed of s 1 km/hr without stopping and it covers the same distance at a speed of s 2 km/hr with stopping then  s  s2  Differenc eof s peed hr. The stopping time per hour =  1 s peedwithout s toppings  s1 

13.

If a train overtakes a pole or a man or a milestone, then the distance covered in overtaking = Length of the train.

14.

If a train overtakes a bridge or tunnel or a platform or another train, then the distance covered = Sum of the two lengths.

15.

Relative Speed: If two trains of lengths L 1 km and L 2 km, respectively are traveling in the same direction at s 1 km/hr and s 2 km/hr respectively such that s 1 > s 2 , then their a. Relative speed = s 1 – s 2 .  L  L2   hr. b. Time taken by the faster train to cross the slower train =  1   s1  s2 

16.

Relative Speed: If two trains of lengths L 1 km and L 2 km, respectively are traveling in the opposite direction at s 1 km/hr and s 2 km/hr respectively then their a. Relative speed = s 1 + s 2 .

 L  L2   hr. b. Time taken by the faster train to cross each other =  1   s1  s 2  17. Two trains of lengths L 1 m and L 2 m run on parallel tracks. When running in the same direction, the faster train passes the slower one in T 1 seconds, but when they are running in opposite directions with the same speeds as earlier, they pass each other in T 2 seconds then L  L2  1 1    a. Speed of the faster train = 1  T  T  m/s. 2 2  1  1 1     T  T  m/s. 2   1 A train starts from a place at s 1 km/hr and another fast train starts from the same place after T hours at s 2 km/hr in the same direction. Then,

b. Speed of the slower train = 18.

L1  L 2 2

a) The distance from the starting place at which both the trains will meet is given  s  s2  T   km. by  1   s 2  s1 

 s1 T   hr. b) The time after which the two trains will meet is given by    s 2  s1  19.

The distance between two stations A and B is d km. A train starts from A to B at s 1 km/hr. T hours later another train starts from B to A at s 2 km/hr. Then, a) The distance from the A at which both the trains will meet is given by  d  s2 T   km. s1    s1  s2  b) The time after which the two trains will meet is given by

 d  s 2T     s1  s 2 

hr.

99

20.

Two trains start simultaneously from the stations A and B towar ds each other each other with speed s 1 km/hr and s 2 km/hr, respectively. When they meet it is found that the second train had travelled d km more than the first. Then the  s  s2   km. Distance between two trains = d 1   s2  s1 

SOLVED EXAMPLES 1. Find the speed of a train which covers a distance of 160 km in 4 hours. Dis tanc e 1 6 0 Sol: Speed =   4 0km/hr. T ime 4 2. How long does a train 200 m long running at the rate of 80 km/hr take to cross a telegraphic pole? Sol: We know, in crossing the telegraphic pole, the train must travel its own length. So, Distance travelled = 200 m 8 0 1 0 0 0 200 Speed = 80 km/hr = m/s ec  m/s ec 6 0 6 0 9 200 So, time taken to cross the pole =  9 s ec onds . 200 9 3. A train running at a speed of 72 km/hr passes a pole on the platfor m in 15 seconds. Find the length of the train in metres. Sol: Speed of the train = 72 km/hr 5 = 7 2  2 0m/s e c 18 Length of the train = Speed of the train x time taken = 20 x 15 = 300 m. 4. A ship sails to Vizag at a speed of 10 knots/hr and sails back to the same point at the rate of 15 knots/hr. Find the average speed for the whole journey. Sol: Here s 1 = 10 and s 2 = 15 Average speed

5.

=

2s1s2 km/hr s1  s2

=

2  1 0 1 5  1 2k nots /hr. 10 15

Sheela started to a bakery with the speed of 5 km an hour and returns with a speed of 3 km/hr. if she takes 8 hours in all, find the distance in km between the bakery and her house. Sol: Here s 1 = 5, s 2 = 3 and T = 8

 s s The distance between the bakery and her house = T  1 2  s1  s 2

6.

   

53 = 8     1 5km . 5  3 Sujay starts his journey from Bombay to Kolkata and simultaneously Niteesh starts from Kolkata to Bombay. After crossing each other they finish their remaining journey 1 in 6 hours and 4 hours respectively. What is Niteesh‟s speed if Sujay‟s speed is 40 4 km/hr? Sol:

Sujay's s peed  N itees h's s peed

T2  T1

4 1 6 4



4 25 4



16 4  25 5

100

So, Niteesh‟s speed = 7.

5 5  Sujayss peed  4 0  5 0km/hr. 4 4

A bike during its journey travels 20 min at a speed of 15 km/hr, another 25 min at 30 km/hr and another 30 min at 15 km/hr. Find the average speed of the bike.

20 25 30 , T2 = , T3 = , s 1 = 15, s 2 = 30 and s 3 = 15 60 60 60 s T  s2T2  s3T3 So, Average speed of the bike = 1 1 T1  T2  T3

Sol:T 1 =

=

=

=

8.

1 5

1 5

20 25 30  3 0  1 5 60 60 60 20 25 30   60 60 60 20 25 30  3 0  1 5 60 60 60 20 25 30   60 60 60

3 0 0 7 5 0 4 5 0 1 5 0 0   2 0km/hr. 75 75

3 of her usual speed, Rani is 5 minutes late to the conference. Find her 4 usual time to cover the distance. By walking at

Sol: Here, change in time = 5 and

a 3  b 4

b  We have, change in time =   1  originaltime a 

 original

9.

time =

c hange in time  b   a  1  

5  1 5 minutes . 4   3  1  

Two scooterists do the same journey by traveling at the rates of 9 km/hr and 8 km/hr respectively. Find the length of the journey when one takes 20 minutes longer than the other. Sol: Here, change in speed = 9 – 8 = 1 Product of speed = 9 x 8 = 72 20 Difference of time = 20 min = 60 Length i.e. distance, d =? We have, P roduc tof s peed D ifferenc eof s peed 7 2 1    20 d D ifferenc eof time d 60 20  d  7 2  2 4km. 60

101

10.

Without stopping, a train travels certain distance with an average speed of 100 km/hr and with stoppages; it covers the same distance with an average speed of 80 km/hr. How many minutes per hour the strain stops? Sol: Here, s 1 = 100 and s 2 = 60  s  s2  1 0 0 8 0 1  Stoppage time/hr =  1  hr  1 2min.  100 5  s1 

11.

A train 500 m long crosses a pole in 6 seconds. Find the speed of the train in km/hr? Lengthof the train 500  m/s time takingin c ros s ingthe pole 6 500 18    3 0 0km/hr . 6 5

Sol: Speed of the train =

12.

A train 150 m long passes a bridge in 24 seconds moving with a speed of 54 km/hr. Find the length of the bridge.

Lengthof the train  Lengthof the bridge time takingin c ros s ingthe bridge 5 1 5 0 L engthof the bridge  54  18 24  1 5 0 L engthof the bridge  2 4  1 5  L engthof the bridge  3 6 0 1 5 0  2 1 0m.

Sol: Speed of the train =

13.

A train 120 m long is running with a speed of 68 km/hr. I n what time will it pass a man walking at 4 km/hr in the opposite direction to that of train? Sol: Here, L 1 = 120, L 2 = 0, s 1 = 68, s 2 = 4 L 1 + L 2 = 120 + 0 = 120 m 5 s 1 + s 2 = 68 + 4 = 72 km/hr = 7 2   2 0 m/s ec . 18  L  L2  1 2 0   6 s ec . Required time =  1  20  s1  s 2 

14.

Two trains of length 120 m and 80 m running on parallel tracks in the same direction with a speed of 48 km/hr and 50 km/hr respectively. In what time will they pass each other? Sol: Here, L 1 = 120, L 2 = 80, s 1 = 48, s 2 = 50 L 1 + L 2 = 120 + 80 = 200 m 5 s 1 - s 2 = 50 - 48 = 2 km/hr = 2  m/s ec . 18  L  L2  200 2 0 0 1 8    3 6 0 s ec . Required time =  1  5 25  s2  s1  2  18

15.

Two trains of lengths 213 m and 205 m run on parallel tracks. When running in the 1 same direction the faster train crosses the slower one in 9 seconds. When running in 2 opposite direction with the same speeds, they pass each other completely in seconds. Find the speed of each train. Sol: Here, L 1 = 213, L 2 = 205, T 1 = L 1 + L 2 = 213 + 205 = 418

19 2

and T 2 =

11 2

5

1 2

102

19 11   2 2 19 11 T1 - T2 =   2 2

T1 + T2 =

30  15 2 8 4 2

Speed of the faster train =

L1  L 2 2

 1 1    T  T  2   1

4 1 8 T1  T2  4 1 8 15     6 0 m/s .   1 9 11 2  T1T2  2  2 2 L  L2  1 1    Speed of the slower train = 1 T  T  2 2   1 =

=

16.

4 1 8 T1  T2  4 1 8 4     1 6 m/s .   1 9 11 2  T1T2  2  2 2

A train starts from Hyderabad at 9 A.M. with a speed of 50 km/hr and another train starts from there on the same day at 1 P.M. in the same direction with a speed of 70 km/hr. Find at what distance from Hyderabad both the trains will meet and also find the time of their meeting. Sol: s 1 = 50, s 2 = 70, T = time from 9 A.M. to 1 P.M. = 4 hours.  s  s2  T   km Distance of meeting point from Hyderabad =  1   s 2  s1 

 5 0 7 0 4  =    7 0 0km .  70 50   s T  5 0 4 hr  Time of their meeting =  1  1 0hr after 1 P.M.  20  s2  s1  i.e. at 11 P.M. on the same day. 17.

Vijay Wada is at a distance of 340 km from Hyderabad. A train starts from Hyderabad to Vijay Wada at 4 A.M. with a speed of 60 km/hr. Another train starts f rom Vijay Wada to Hyderabad at 5 A.M. with a speed of 80 km/hr. At what distance from Hyderabad will the two trains cross each other and also find the time when they cross each other? Sol:s 1 = 60, s 2 = 80, T = time from 4 A.M. to 5 A.M. = 1 hr  d  s2 T   km Distance of meeting point from Hyderabad = s1    s1  s2 

420  3 4 0 (8 0 1)   1 8 0km . = 6 0  km  6 0 6 0  8 0  140   d  s2 T  420  hr  Time of their meeting =   3hr after 4 A.M.  s  s 1 20 2   1 i.e. at 7 A.M. 18.

Two trains start at the same time from Chennai and Cochin and proceed toward s at a rate of 85 km/hr and 100 km/hr respectively. When they meet, it is found that one train has travelled 30 km more than the other. Find the distance between Chennai and Cochin. Sol: s 1 = 85, s 2 = 100, d = 30 Distance between Chennai and Cochin = d

185  8 5  1 0 0  2  1 8 5 3 7 0k m. = 3 0    3 0 15  1 0 0 8 5

s1 + s2 s2 s1

km.

103

Brainstorming

1. 2. 3.

4.

The speed of a bus is 54 km/hr. What is the speed in m/sec? 1. 18 m/sec 2. 15 m/sec 3. 5 m/sec

4. None

The speed of a train is 35 m/sec. What is the speed in km/hr? 1. 144 km/sec 2. 160 km/sec 3. 126 km/hr

4. None

A car covers a distance of 180 km in 8 hours. What is its speed in m/sec? 1 1 1. 6 m/sec 2. 8 m/sec 3. 4 m/sec 4. None 4 8 A bus covers a distance in 30 min, if it runs at a speed of 30 km/hr. The speed at which it must run to decrease the time of journey to 20 min will be ______ 1. 45 km/hr 2. 84 km/hr 3. 12.5 km/hr 4. None

5.

A car takes one hour to cover a distance at 80 km/hr. In how many minutes it cover the same distance when the speed is increased by 20 km/hr? 1. 80 2. 240 3. 4 4. None

6.

A bus covers half of the journey at 70 km/hr and the remaining journey at 30 km/hr. What is the average speed of the bus for the whole journey? 1. 50 km/hr 2. 82 km/hr 3. 42 km/hr 4. None

7.

Rita goes from P to Q at 60km/hr and returns from Q to P at 40 km/hr. If she takes 10 hours in all what is the distance between P and Q? 1. 400 km 2. 715 km 3. 580 km 4. None

8.

Shan takes 6 hours in walking to a certain place and riding back. However, he would have gained 2 ½ hours if he rides both the ways. How long would he take in walking both the ways? 1. 6 hrs 45 min 2. 8 hrs 3. 7 hrs 30 min 4. None

9.

An R.T.C. Bus between two towns reaches his destination 10 min early when it gives 25 km/hr and 15 min early when it gives 30 km/hr. What is the distance between the two towns? 1. 75 km 2. 98 km 3. 14 km 4. 12.5 km

10.

The average speed of a car is 40 km/hr without stoppages and with stoppage it is 26 km/hr. How many minutes per hour did the car stop? 1. 21 2. 16 3. 10 4. None

11.

Walking at 5/6 of his usual speed 12 min late. What is the usual time taken by him to cover the same distance? 1. 18 hr 2. 12 hr 3. 1 hr 4. None

12.

Two buses start at the same time from two stations towards each other. The speeds are 72 and 64 km/hr respectively. By the time the two buses meet, one bus had covered 72 km more than the other. What is the distance between the two stations? 1. 400 km 2. 216 km 3. 540 km 4. 450 km

13.

The speeds of two buses starting from P and Q at the same time are 28 kmph and 26 kmph. The distance between P and Q is 270 km. At what distance from P will the two buses meet? 1. 150 km 2. 178 km 3. 85 km 4. 140 km

14.

The distance between P and Q is 296 km. A train starts from P at 8 A.M. at 28 km/hr and another train starts from Q at 10 A.M. at 32 km/hr. When will they meet? 1. 12 Noon 2. 2:00 PM 3. 5:00 PM 4. 3:30 PM

15.

A car is 25 km ahead of a scooter. The car is traveling at 40 km/hr and the scooter at 50 km/hr. The scooter will overtake the car after _______ 1. 2 hrs 45 min 2. 2 hrs 30 min 3. 3 hrs 4. Never

104

16.

Rahul covers 1/4th of his journey at 25 km/hr, 1/5th of his journey at 12 km/hr and the remaining distance at 15 km/hr. If he takes 19 hours in all, what is the total distance traveled by Rahul? 1. 5 km 2. 6 km 3. 15 km 4. None

17.

The radius of a wheel is 1.75 m and it makes 3 revolutions per second. The speed of the wheel in km/hr is ________. 1. 118.8 2. 112 3. 33 4. None

18.

A man covers a certain distance on scooter. Had he moved 3 km/hr faster, he would have taken 40 min less. If he had moved 2 km/hr slower, he would have taken 40 min more. The distance is _________. 1. 20 km 2. 36 km 3. 37.5 km 4. 40 km

19.

A certain distance is covered at a certain speed. If half of the distance covered in double the time, the ratio of the speed is ____________. 1. 4:1 2. 1:4 3. 2:1 4. 1:2

20.

The ratio between the rates of walking of A and B is 2 : 3. Therefore, A takes 10 min more than the time taken by B, who reaches the destination. If A had walked at double the speed, he would have covered the distance in ___________. 1. 15 min 2. 20 min 3. 25 min 4. 30 min

21.

Raman has to cover a distance of 6 km in 45 min. If he covers half of the distance in

2 of the 3

time what should be his speed to cover the remaining distance in remaining time? 1. 12 km/hr 2. 16 km/hr 3. 1 km/hr 4. 8 km/hr 22.

The ratio between the rates of walking of P and Q is 3 : 4. If the time taken by Q to cover a certain distance is 48 min, the time taken (in min) by P to cover the distance is ____________. 1. 36 2. 64 3. 21 4. 27

23.

Kiran started cycling along the boundaries of a square filed from corner point A. After half-anhour he reached the corner point C diagonally opposite to A. If his speed was 8 km/hr what is the area of the field in sq. km? 1. 64 2. 8 3. 4 4. None

24.

Rahul goes to school with a speed of 3 km/hr and returns to the village with a speed of 2 km/hr. If he takes 5 hours in all, the distance between the village and the school is ____. 1. 8 km 2. 4 km 3. 5 km 4. None

25.

A bus completes a certain journey in 5 hours. It covers half the distance at 40 km/hr and the rest at 60 km/hr. The length of the journey is _________. 1. 240 km 2. 480 km 3. 225 km 4. None

26.

A girl could reach the school 5 min early if she walks at 6 km/hr but 5 min late if she walks at 5 km/hr. Find the distance to school? 1. 10 km 2. 12 km 3. 5 km 4. None

27.

P and Q covered a certain distance at 9 km/hr and 8 km/hr respectively. P reaches the destination 45 min early. Then the distance to destination is ________. 1. 54 km 2. 27 km 3. 20 km 4. None

28.

Harish started from his house at 8 A.M. and went to Bhasha‟s house at a speed of 30 km/hr. Bhasha started from his house at 8.30 A.M. and went to Harish‟s house at 40 km/hr. They would meet at the midway at __________. 1. 9:00 A.M. 2. 11:45 AM. 3. 1:30 P.M. 4. None

29.

Two men start together to walk a certain at 4 km/hr and 4.5 km/hr respectively. The former arrives 10 min before the later. Find the distance they walked? 1. 18 km 2. 6 km 3. 12 km 4. None

105

30.

A pick pocketer was 500 m ahead of police and two are running at 8 km/hr and 10 km/hr respectively. When can the police catch him? 1. 12 min 2. 30 min 3. 15 Min 4. 20 min.

31.

A man traveled from P to Q at a speed of 60 km/hr and return to P at a speed of 40 km/hr. Find the average speed? 1. 48 km/hr 2. 40 km/hr 3. 50 km/hr 4. None

32.

P and Q covered a certain distance at 9 km/hr and 10 km/hr respectively. If P took 32 min more to reach the destination find the distance covered? 1. 48 km 2. 72 km 3. 48 km 4. None

33.

A man misses a bus by 40 min if he travels at 30 km/hr. If he travels at 40 km/hr, then also he misses the bus by 10 min. What is the minimum speed to catch the bus on time? 1. 40 km/hr 2. 45 km/hr 3. 50 km/hr 4. None

34.

P and Q are running around a circular track of 600 m in opposite direction at speeds of 15 m/sec and 10 m/sec respectively starting at the same time from the same point. In how much time will they meet for the first time any where on the track? 1. 15 sec 2. 24 sec 3. 5 sec 4. None

35.

Trisha travels from P to Q by car and returns from Q to P by cycle in 7 hours. If she travels both the ways by car she saves 3 hours. What is the time taken by her to cover both the ways by cycle? 1. 12 hrs 2. 7 hrs 3. 10 hrs 4. 14 hrs

36.

Supriya reaches her office 1 hour late traveling 40 km/hr. If she travels at 50 km/hr, she is late by 40 min. What is the distance she has to travel to reach her office? 2 1 1. 70 km 2. 66 km 3. 58 km 4. None 3 2

37.

Saritha leaves from a certain point at 8.00 A.M. at a speed of 25 km/hr. Kavitha leaves from the same point at 9.30 A.M. at a speed of 37.5 km/hr in the same direction as that of Saritha. At what time do they meet? 1. 11:00 PM 2. 12:30 PM 3. 6:35 PM 4. 7:00 PM

38.

A bike run 300 m in 40 sec and another bike run it in 52 sec. In a 1950 m race, by how many metres does first bike beat the second one? 1. 450 m 2. 400 m 3. 45 m 4. 350 m

39.

In a 5000 m race, B beats C by 500 m or 50 seconds. What is the time taken by B to complete a 54000 m race? 1 1. 486 sec 2. 81 min 3. 1 hr 4. None 2

40.

Venu covered 22.5 km in 3 hours 20 minutes. He covered 10 km at 8 km/hr. At what speed did he cover the remaining distance? 1. 6 km/hr 2. 12 km/hr 3. 10 km/hr 4. None

14. PROBLEMS ON TRAINS A train is said to have crossed an object (stationary or moving) only when the last coach (end) of the train crosses the said object completely. Hence, the distance covered by the train = length of train + length of object Time to cross an object moving in the direction of train =

Length of train Length of objec t Speed of train Speed of objec t

106

DIFFERENT TYPES OF OBJECTS On the basis of various types of objects that a train has to cross, we find the following different cases: Case

Type of Object Object is stationary and is of negligible length, e.g., train crosses lamp post, pole, standing man, etc. Object is stationary and is of some length, e.g., train crosses bridge, a tunnel, platform, or another train at rest. Object is moving* and is of negligible length, e.g., train crosses a running man, a running car, etc. Object is moving* and has some length, e.g., train crosses another running train.

1. 2. 3. 4.

Time to cross + t =

t =

length of train s peedof train

length of (train objec t) s peedof train

t =

length of train s peedof (train objec t)

t =

length of (train objec t) s peedof (train objec t)

(* if the object is moving in opposite direction, then denominator becomes speed of (train + object).

TWO TRAINS CROSSING EACH OTHER IN BOTH DIRECTIONS Two trains are crossing each other Length of one train = L 1 Length of second train = L 2 They are crossing each other in opposite direction in t 1 seconds. They are crossing each other in same direction in t 2 seconds then,

1 1    t t 2  1

Speed of Faster train =

L1  L2 2

Speed of Slower Train =

L1  L2  1 1    2  t1 t2 

SOLVED EXAMPLES 1)

A train travelling with uniform speed crosses a platform of length 800 metres in 20 seconds and a tunnel of length 1.1 kilometer in 25 seconds. What is the length of the train? Sol: Let the length of the train be x metres. Speed while crossing the platform

=

Speed while crossing the tunnel

=

Since the train has uniform speed

=

5 (x + 800) = x =  The length of the train is 400 metres. 2)

(x  800) m/sec. 20 (x  1100) 25 (x  800) (x  1100)  20 25 4(x + 1100) 4400 – 400 = 400 metres.

If 460 poles are arranged such that the distance between any two successive poles is 20 metres, and a train 180 m long crosses them completely in three minutes, what is the speed of the train? Sol: The total distance covered = distance between 460 poles + length of the train

107

 Speed of the train = 3)

 459  20  180 3  60

= 52 m/s

The train crosses a man standing on a platform 150 metre long in 10 seconds and crosses the platform completely in 22 seconds. Find the length of train and speed of train. Sol: Let Lt and Vt be the length and the speed of the train L L Using the basic formula t = t Vt  V For standing man, put L = 0 and V = 0, then, we get L 10 = t _____________________ (1) Vt For the stationary platform, put V = 0, and L = 150 (given), then we get L  150 22 = t Vt



22 =

L t 150 ___________________ (2)  Vt Vt

From (1) & (2), 22 = 10 +

 

150 Vt

  Lt  10in1 Since V t  

Vt = 12.5 m/sec Speed of train = 12.5 metre/sec.

Lt 12.5 Lt = 12.5   length of train = 12.5 metres.

From (1)

4)

10 =

A toy train crosses 210 and 122 metre long tunnels in 25 and 17 seconds respectively. Find the length of train and speed of train. Sol: Let Lt and Vt be the length of train and speed of train respectively. L L Using the basic formula t = t Vt  V For tunnel, put V = 0



25

 Lt + 210

=

L t  250 Vt

= 25Vt L  122 = t ________________ (1) Vt

And

17



Lt + 122 = 17 Vt __________________ (2)

Subtracting (2) from (1) 210 – 122 = 8 Vt Vt = 11 m/sec  Putting Vt = 11 in equation (1), Lt + 210 = 25 x 11 Lt = 275 – 210 = 65 m

 Speed of the train Length of the train

= 11 m/sec. = 65 m

108

5)

A train 75 metres long overtook a man who was walking at the rate of 6 km/hr and crossed him in 18 seconds. Again, the train overtook a second person in 15 seconds. At what rate was the second person travelling? Sol: Let Vt be the speed of the train and V2 be the speed of the second person L L Using the basic formula t = t Vt  V For the first man, put L = 0, V = 6 km/hr = 6 







18 =

5 5  m/sec, t = 18 sec. 18 3

75 Vt 

5 3

5 75 Vt - 3  18 75 5 105 Vt = 18  3  18 m/sec.

105 For the second person, put L = 0, Vt = 18 m/sec. 

 

75 15 = 105  V2 18 105 18 - V2 = 5 15 18 V2 = 18  5 = 3 km/hr.

 The speed of the second person is 3 km/hr. 6)

Two trains of lengths 190 m and 210 m respectively, are running in opposite directions on parallel tracks. If their speeds are 40 km/hr and 32 km/hr respectively, in what time will they cross each other? Sol: Using basic formula t = Since

Lt  L Vt  V

Lt = 190 m, L = 210 m, Vt = 40, V = -32

 V – V = 40 – (-32) = 72 km/hr t

5 = 72  18 = 20 m/sec 190  210 Hence t= 20  t = 20 seconds  The two trains cross each other in 20 seconds. 7)

When two trains were running in the same direction at 90 km/hr and 60 km/hr respectively, the faster train crossed a man in the slower train in 27 seconds. Find the length of the faster train. Sol: It is given that the faster train crosses a man in the slower train in the same direction. It implies that a train crosses a man moving in the same direction. Speed of slower train = speed of the man (because the man is in the slower train)

109

Now, using the basic formula, we get t =

Lt  L , where L1 = length of faster train Vt  V

For the man, put L = 0 27 =

Lt (90  60) 

5 18

5   L = 30  t 18 27 = 225 Hence, the length of the faster train is 225 metres.

110

Brainstorming

1.

A train passes a 50m long platform in 14 seconds and a man standing on the platform in 10 seconds. The speed of the train is 1. 50 km/hr 2. 60 km/hr 3. 40 km/hr 4. 45 km/hr

2.

A train 160m long is running at 72 km/ph. In what time ti will cross a telegraph pole? 1. 15 sec 2. 7.5 sec 3. 8 sec 4. 10.5 sec

3.

A man is running at the speed 6 km/ph. A train runs at 54 km/ph in the same direction. If the length of the train is 240m. How much time will it take for the train to overtake the man? 1. 10 sec 2. 18 sec 3. 20 sec 4. 30 sec

4.

A train cross a bridge of 180 m in 27 seconds and it crosses a telegraph pole in 15 seconds. What is the length of the train? 1. 204 m 2. 225 m 3. 240 m 4. 215 m

5.

Two trains of 270 m, 150m are running in opposite direction. The speed of the two trains are 42 km/h and 30 km/hr. In what time they cross each other? 1. 10 sec 2. 21 sec 3. 17 sec 4. 9 sec

6.

Two trains of 550m and 250m are running in opposite direction can cross each other in 24 seconds. If the speed of the faster train is 78 km/hr. What is the speed of the slower train? 1. 112 km/hr 2. 52 km/hr 3. 42 km/hr 4. 68 km/hr

7.

The distance between A and B is 356 km. A train starts from A at 8 A.M at 28 km/hr and another train starts from B at 10 A.M at 32 km/hr. When will they meet? 1. 6 P.M 2. 3 P.M 3. 5 P.M 4. 4 P.M

8.

A train reaches its destination 10 minutes late when it drives at 40 km/hr and 16 minutes late when it drives at 30 km/hr. What is the distance between the two stations? 1. 10 km 2. 6 km 3. 12 km 4. 24 km

9.

A train 900m long is running at the speed of 90 km/hr. If it crosses a tunnel in 1 minute. Then the length of the tunnel is 1. 100 m 2. 800 m 3. 600 m 4. 1500 m

10.

A train 370 meters long is moving at a speed of 35 km/ph. It will cross a man coming from the opposite direction at a speed of 2 km per hour in 1. 36 sec 2. 18 sec 3. 12 sec 4. 6 sec

11.

A train passes a signal post in 6 seconds moving at the rate of 108 km per hour. The length of the train is 1. 90 m 2. 120 m 3. 150 m 4. 180 m

12.

A train 60 meters long takes 30 seconds in crossing a tunnel 240 meters long. The speed of the train is 1. 18 km/hr 2. 36 km/hr 3. 72 km/hr 4. 90 km/hr

13.

A train of length 250m crosses a bridge of length 400m in 52 seconds. How long does it take to cross a platform of length 600m? 1. 34 sec 2. 68 sec 3. 80 sec 4. 95 sec

14.

A train of length 300 m took 18 seconds to cross a pole. How long does it take to cross platform of length 240m? 1. 16.2 sec 2. 32.4 sec 3. 8.1 sec 4. 64.8 sec

15.

A train of length 180m took 1 minute to cross a platform of length 540m. How long does it take to cross a pole? 1. 5 sec 2. 10 sec 3. 15 sec 4. 20 sec

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16.

A train moving at a speed of 72 km/ph took 55 seconds to cross a platform of length 500m. Find its length? 1. 1000mts 2. 1100mts 3. 500mts 4. None

17.

A train of length 250m took 15 seconds to cross a pole. How long does it take to cross a bridge of length 350m? 1. 12 sec 2. 24 sec 3. 36 sec 4. None

18.

A train of length 200 m took 20 seconds to cross a platform of length 300m. How long does it take to cross a pole? 1. 8 sec 2. 10 sec 3. 12 sec 4. None

19.

How many seconds will a 500m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? 1. 25 2. 30 3. 40 4. 45

20.

A 120 meter long train is traveling at a speed of 90 km per hr. It will cross a railway platform 230m long in 4 1 1. 4 sec 2. 9 sec 3. 8 sec 4. 14 sec 5 5

21.

A train passes a 50 meters long platform in 14 seconds and a man standing on the platform in 10 seconds. The speed of the train is 1. 24 km/hr 2. 26 km/hr 3. 40 km/hr 4. 45 km/hr

22.

A 120m long train takes 10 seconds to cross a man standing on a platform. What is the speed of the train? 1. 12 m/sec 2. 10 m/sec 3. 15 m/sec 4. 20 m/sec

23.

A train passes a man standing on a platform in 8 seconds and also crosses the platform which is 264 meters long in 20 seconds. The length of the train (in meters) is 1. 188 2. 176 3. 175 4. 96

24.

A train, 120 meters long, passes a telegraph post in 6 seconds. Find the speed of the train. 1. 60 km/hr 2. 72.5 km/hr 3. 80 km/hr 4. 72 km/hr

25.

A train passes a telegraph post in 8 seconds and a 264 m long bridge in 20 seconds. What is the length of the train? 1. 180m 2. 176m 3. 164m 4. 158m

26.

A train 800 meters long is running at the speed of 78 km/hr. If it crosses a tunnel is 1 minute, then the length of the tunnel (in meters) is 1. 700 2. 500 3. 400 4. 600

27.

Two trains one 160m and other 140m long are running in opposite directions on parallel rails, the first at 77 km an hour and the other at 67 km an hour. How long will they take to cross each other? 1 1. 7 sec 2. 7 sec 3. 6 sec 4. 10 sec 2 Two trains are running in opposite directions with the same speed. If the length of each train is 120 meters and they cross each other in 12 seconds, the speed of each train (in km/hr) is 1. 72 2. 10 3. 36 4. 18

28.

29.

A train 110 meters in length passes a man walking at the speed of 6 km/hr, against it in 6 seconds. The speed of the train in km per hour is 1. 60 km/hr 2. 45 km/hr 3. 50 km/hr 4. 55 km/hr

30.

A train 110 meters in length passes a man walking at the rate of 6 km/hr against it in 6 seconds, it will pass another man walking at the same speed in the same direction in time of 2 1 1 1. 9 sec 2. 10 sec 3. 8 sec 4. 7 sec 3 3 3

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31.

In how many seconds a train of 240 m length will cross a tree, when it runs at 54 km/hr? 1. 8 sec 2. 16 sec 3. 24 sec 4. 32 sec

32.

In how many seconds, a train of 160 m crosses a platform of 140 m, when it runs at the rate of 108 km/hr? 1. 10 sec 2. 20 sec 3. 30 sec 4. 40 sec

33.

A train of 240 m crosses a platform in 12 seconds. If the speed of the train is 108 km/hr, what is the length of the platform? 1. 60 m 2. 90 m 3. 120 m 4. 150 m

34.

A man is running at the speed of 6 km/hr. A train runs at 54 km/hr in the same direction. If the length of the train is 200 m, how much time will it take for the train to overtake the man? 1. 5 sec 2. 10 sec 3. 15 sec 4. 20 sec

35.

A train running at a speed of 36 m/s crosses a man running in opposite direction at 4 km/hr in 9 seconds. What is the length of the train? 1. 167 m 2. 334 m 3. 668 m 4. None

36.

A train crosses a bridge of 180 m in 27 seconds and it crosses a telegraph pole in 15 seconds. What is the length of the train? 1. 140 m 2. 280 m 3. 420 m 4. 560 m

37.

A train passes through a tunnel of 80 m in 30 seconds and it crosses a platform of 180 m in 45 seconds. What is the length of the train? 1. 30 m 2. 60 m 3. 90 m 4. 120 m

38.

Two trains of 120 m and 180 m are running on parallel lines in the same direction. If the speeds are 30 km/hr and 35 km/hr respectively, in what time will they cross each other? 1. 216 sec 2. 68 sec 3. 432 sec 4. 200 sec

39.

Two trains of 550 m and 250 m are running in opposite direction can cross each other in 24 sec. If the speed of the faster train is 78 km/hr, what is the speed of the slower train? 1. 21 km/hr 2. 42 km/hr 3. 63 km/hr 4. 84 km/hr

40.

Two trains are running on parallel lines in the same direction with the speeds 42 and 24 km/hr. The faster train crosses a man in the slower train in 52 seconds. What is the length of the faster train? 1. 130 m 2. 260 m 3. 390 m 4. 520 m

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15. BOATS AND STREAMS Still Water: If the speed of the water in the river is Zero, then the water is called Still water. (Stationary Water) Stream: If the water of the river is moving, it is called a stream. (Water in motion) Upstream: If a boat or a swimmer moves against the stream i.e. in the d irection opposite to that of the stream, it is called upstream. Downstream: If a boat or a swimmer moves with the stream i.e. along the direction of the stream, it is called downstream. Note: If the speed of a boat or a swimmer is given, it usually means the speed in still water.

Formulae: 1.

If the speed of a boat or a swimmer be x km/hr and the speed of the current be y km/hr, then

a. b.

Speed of the boat or swimmer downstream = (x + y) km/hr. Speed of the boat or swimmer downstream = (x - y) km/hr.

stream or the

2.

a. b.

Speed of the boat or swimmer in still water 1 = (Downstream speed + Upstream speed) 2 Speed of the stream 1 = (Downstream speed - Upstream speed) 2

3.

If a man is capable of rowing at the speed of x km/hr in still water, rows the same distance up and down a stream which flows at a rate of y km/hr, then his average speed throughout the journey = U ps tream Downs tream (x  y)(x  y) km/hr.  M an's rate in s tillwater x

4.

A man can row a boat in still water at x km/hr. In a stream flowing at y km/h r, if it takes t hours more in upstream than to go downstream for the same distance, then The distance =

(x2  y2 )t km. 2y

a.

A man rows a certain distance downstream in t 1 hours and returns the same distance upstream in t 2 hours. If the speed of the stream be x km/hr, then  t  t1   km/hr. The speed of the man in still water = y  2   t2  t1 

b.

A man rows a boat in still water at x km/hr. In a stream flowing at y km/hr if it takes him t hours to row to a place and come back, then

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c.

 x2  y2   km. The distance between the two places = t   2x    A boat or a swimmer takes n times as long to row upstream so as to row downstream the river. If the speed of boat or swimmer be x km/hr and the speed of stream be y km/hr, n  1 Then x  y  . n 1

SOLVED EXAMPLES 1.

The speed of a boat in still water is 10 km/hr. If the speed of the stream be 2 km/hr, then find its downstream and upstream speeds. Sol: Speed of the boat (x) = 10 km/hr. Speed of the stream (y) = 2 km/hr. Downstream speed = x + y = 10 + 2 = 12 km/hr. Upstream speed = x – y = 10 – 2 = 8 km/hr.

2.

A boat is rowed down a river 32 km in 4 hours and up a river 15 km in 3 hours. Find the speed of the boat and the river. 32 Sol: Speed of the boat downstream =  8 km/hr. 4 15 Speed of the boat upstream =  5 km/hr. 3 1 Speed of the boat = (Downstream speed + Upstream speed) 2 13 1 = (8 + 5) =  6.5 km/hr. 2 2 1 Speed of the river = (Downstream speed - Upstream speed) 2 3 1 = (8 - 5) =  1.5 km/hr. 2 2

3.

A man rows a boat at a speed of 16 km/hr in still water to a certain distance upstream and back to the starting point in a river which flows at 8 km/hr. Find his average speed for total journey. U ps tream Downs tream (x  y)(x  y) Sol: Average speed =  M an's rate in s tillwater x (1 6  8)(1 6  8) 2 4 8 =   1 2 km/hr. 16 16 A man can row 4 km/hr in still water. If the river is running at 3 km/hr, it tak es 12 hours more in upstream than to go downstream for the same distance. How far is the place?

4.

(x2  y2 )t (42  32 )  1 2   1 0 km. 2y 23 A motorboat covers a certain distance downstream in 8 hours but takes 10 hours to return upstream to the starting point. If the speed of the stream be 14 km/hr, then find the speed of the motor boat in still water.

Sol: The required distance = 5.

 t  t1   km/hr Sol: Speed of the motorboat in still water = y 2   t 2  t1 

6.

10 8  = 1 4    1 2 6km/hr. 10 8  A man can row 8 km/hr in the still water. If the river is running at 4 km/hr, it takes him 10 hours to row to a place and back. How far is the place?

115

 x2  y2   km Sol: The required distance = t   2x   

7.

 82  42  1 0 4 8   3 0km. = 1 0   28  16   A man can row at the rate of 16 km/hr in still water. If the time taken to row a certain distance upstream is 7 times as much as to row the same distance downstream, find the speed of the current.

n  1 Sol: Speed of the man =   speed of the current n  1

7  1 =   speed of the current  7  1 So, speed of the current = 12 km/hr.

 16

116

Brainstorming

1.

The speed of a boat with the current is 18 km/hr and its speed against the current is 12 km/hr. What is the speed of boat in still water? 1. 15 km/hr 2. 12 km/hr 3. 18 km/hr 4. 8 km/hr

2.

A man‟s rate against current is 11 km/hr and with the current is 15 km/hr. What is the speed of stream? 1. 8 km/hr 2. 10 km/hr 3. 2 km/hr 4. 5 km/hr

3.

A boat can row 12 km upstream in 48 minutes and 10 km with the current in 30 minutes. What is the speed of boat in still water? 1. 1.5 km/hr 2. 2.5 km/hr 3. 3.5 km/hr 4. 4 km/hr

4.

A boat can row 36 km down the stream in 6 hours. if the speed of flow is 2 km/hr, find the time required for the boat to travel the same distance against the flow. 1. 18 hrs 2. 15 hrs 3. 24 hrs 4. None

5.

The speed of a boat in still water is 17.5 km/hr and the rate of current distance traveled in down stream in 25 minutes is____? 1 1 1. 18 km 2. 8 km 3. 5 km 3 2 A man rows 3 km/hr in still water. If the river is running at 1 km/hr, it row to a place and back. How far is the place? 1. 8 km 2. 1 km 3. 5 km

6.

is 2.5 km/hr. The 4. 6 km/hr takes him 45 minutes to 4. 6 km

7.

A boat goes 80 km downstream in 8 hours and 70 km upstream in 14 hours. Find the speed of boat in still water. 1. 2.5 km/hr 2. 10 km/hr 3. 5 km/hr 4. 7.5 km/hr

8.

A man can row 80 km downstream in 5 hours and 30 km upstream in the same time. The speed of the stream is ______? 1. 6 km/hr 2. 1.5 km/hr 3. 5 km/hr 4. 2.8 km/hr

9.

A man can row 9 km/hr in still water and he finds that it takes him twice as long to row upstream as to row down the river. Find the rate of the stream. 1. 1 km/hr 2. 2 km/hr 3. 3 km/hr 4. None

10.

A man can row 50 km down stream in 5 hours and 20 km upstream in 10 hours. Find the speed of the stream. 1. 4 km/hr 2. 8 km/hr 3. 3 km/hr 4. 6 km/hr

11.

A man can row with a speed of 15 km/hr in still water. If the stream flows at 5 km/hr, then his speed in downstream is ________ 1. 10 km/hr 2. 5 km/hr 3. 20 km/hr 4. 22 km/hr

12.

A man can row upstream at 25 km/hr and down stream at 35 km/hr then find the speed of the man in still water. 1. 30 km/hr 2. 10 km/hr 3. 60 km/hr 4. 5 km/hr

13.

A man can row upstream at 21 km/hr and downstream at 25 km/hr, then the speed of current is______? 1. 4 km/hr 2. 2 km/hr 3. 23 km/hr 4. 46 km/hr

14.

A man swims downstream 30 km and upstream 18 km taking 3 hours each time. What is the speed of the man in still water? 1. 2 km/hr 2. 8 km/hr 3. 16 km/hr 4. 4 km/hr

117

15.

A man can row 6 km/hr in still water. When the river is running at 1.2 km/hr, it takes him 1 hour to row to a place and back. How far is the place? 1. 3.12 km 2. 2.88 km 3. 3 km/hr 4. 2 km/hr

16.

A man can row 25 km/hr in still water and the stream is flowing at 15 km/hr and it takes him 15 hours to row to a place and back. How far is the place? 1. 12 km 2. 60 km 3. 120 km 4. 240 km

17.

The current of a stream runs at the rate of 4 km/hr. A boat goes 6 km and back to the starting point in 2 hours, then find the speed of boat in still water. 1. 10 km/hr 2. 21 km/hr 3. 8 km/hr 4. 6 km/hr

18.

The current of a stream flows at 1 km/hr. A boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the boat in still water is ___? 1. 2 km/hr 2. 10 km/hr 3. 8 km/hr 4. 12 km/hr

19.

A man whose speed is 4.5 km/hr in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 km/hr. Find his average speed for the entire journey. 1. 8 km/hr 2. 4 km/hr 3. 2 km/hr 4. 10 km/hr

20.

Speed of a boat in still water is 7 km/hr and speed of the stream is 1.5 km/hr. A distance of 7.7 km in upstream is covered in how many minutes? 1. 2 km/hr 2. 1.5 km/hr 3. 3.5 km/hr 4. 4 km/hr

21.

A man can row downstream at 18 km/hr and upstream at 10 km/hr. Find the speed of the man in still water and the speed of stream (in km/hr). 1. 13; 3 2. 12; 6 3. 15; 3 4. 14; 4

22.

A man can swim at 6 km/hr downstream and 4 km/hr upstream. Find the speed of the man in still water and also the speed of the stream. 1. 1; 5 2. 5; 1 3. 6; 2 4. 2; 6

23.

A man can swim downstream at 8 km/hr and upstream at 2 km/hr. Find man‟s rate in still water and the speed of current. 1. 5; 3 2. 3; 5 3. 15; 3 4. 3; 15

24.

A boat can cover 2 km against the stream in 20 minutes and return in 15 minutes. Find the speed of boat in still water and the speed of current. 1. 8; 3 2. 3; 6 3. 8; 1 4. None

25.

A man rows 10 km upstream and back again to the origin point in 55 minutes. If the speed of the stream is 2 km/hr, find the speed of rowing in still water. 1. 11 km/hr 2. 22 km/hr 3. 33 km/hr 4. None

26.

A boat goes down stream at U m/sec and upstream at V m/sec. Then the speed of boat in still water (in m/sec) is _______ 1 1 1. (U - V) 2. U – V 3. (U + V) 4. U + V 2 2

27.

The speed of a boat in still water is 15 km/hr and the rate of stream is 5 km/hr. Find the distance traveled downstream in 24 min. 1. 4 km 2. 8 km 3. 6 km 4. 16 km

28.

If a man‟s downstream rate is 10 km/hr and the rate of stream is 1.5 km/hr, then the man‟s upstream rate is ______ 1. 13 km/hr 2. 10 km/hr 3. 3 km/hr 4. 7 km/hr

29.

If a man rows at 8 km/hr in still water and his upstream rate is 5 km/hr, then the man‟s rate along the current is ______ 1. 21 km/hr 2. 8 km/hr 3. 16 km/hr 4. 11 km/hr

30.

The rowing speed of man in still water is 20 km/hr. Going downstream, he moves at the rate of 25 km/hr. Find the rate of stream.

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1. 45 km/hr

2. 2.5 km/hr

3. 12.5 km/hr

4. 5 km/hr

31.

If a man goes upstream at 6 km/hr and the rate of stream is 2 km/hr, then the man‟s speed in still water is ________ 1. 4 km/hr 2. 8 km/hr 3. 2 km/hr 4. 12 km/hr

32.

A boat goes 12 km upstream in 48 minutes. The speed of stream is 2 km/hr. The speed of boat in still water is _______ 1. 13 km/hr 2. 17 km/hr 3. 15 km/hr 4. None

33.

A boat takes 4 hours for traveling downstream from point A to point B and coming back to point A upstream. If the velocity of the stream is 2 km/hr and the speed of the boat in still water is 4 km/hr, what is the distance between A and B? 1. 8 km 2. 9 km 3. 4 km 4. 6 km

34.

A boat goes 20 km downstream in 1 hour and the same distance upstream in 2 hours. The speed of boat in still water is ______. 1. 15 km/hr 2. 10 km/hr 3. 5 km/hr 4. 7.5 km/hr

35.

A boat takes 9 hours to travel upstream and 3 hours to travel the same distance downstream. If its speed in still water is 4 km/hr, what is the velocity of the stream? 1. 4 km/hr 2. 3 km/hr 3. 6 km/hr 4. None

36.

A person can swim at 7.5 km/hr in still water. In a river with 1.5 km/hr current, the person swims to a certain distance and comes back with in 50 minutes. What is the distance between the two points? 1. 3 km 2. 4 km 3. 2 km 4. 1 km

37.

A man rows a boat in 18 km in 4 hours downstream and returns upstream in 12 hours. The speed of the stream is ____ km/hr. 1. 1 2. 1.5 3. 1.75 4. 2

38.

A boat goes 8 km in 1 hour along the stream and 2 km in 1 hour against the stream. The speed of the stream in km/hr is _____. 1. 2 2. 3 3. 4 4. 5

39.

In one hour a boat goes 11 km along the stream and 5 km against the stream. The speed of the boat in still water in km/hr is _____. 1. 6 2. 8 3. 9 4. 5

40.

The speed of a boat is 15 km/hr in still water. If the speed of the stream is 3 km/hr, and it takes 15 hours for the boat to reach to a place and return. What distant is the place from stating point? 1. 96 km 2. 54 km 3. 72 km 4. 108 km

16. SIMPLE INTEREST Read this please: Raju borrowed Rs.1000 from Ramesh. After some days Raju paid back Rs.1200 to Ramesh. Means Raju paid Rs.200 excess than the amount he borrowed from Ramesh.  This excess amount is called interest.  The total amount of money borrowed by Raju from Ramesh is called the Principal or Sum.  The money paid back to Ramesh, which is the combination of both Principal and interest is called the Amount.  So, Amount = Principal + Interest

119

Rate of interest: The interest is usually charged according to a specified term, which is expressed as some per cent of the principal and is called the rate of interest for the fixed period of time.  If the fixed period is a year, the rate of interest is charged annually.  If the fixed period is six months, the rate of interest is charged semi -annually.  If the fixed period is three months, the rate of interest is charged quarterly.  If the fixed period is a month, the rate of interest is charged monthly. Example: If the rate of interest is 10% per annum, then the interest payable on Rs.100 for one year is Rs.10. Simple Interest: When the interest is payable on the principal only, it is called the simple interest. It is the interest calculated on the principal for the entire period it is borrowed. It is denoted by S.I. Example: S.I. on Rs.100 at 10% per annum will be Rs.10 each year. At the end of one year, the total amount will be Rs.100+10 = Rs.110. At the end of second year, the total amount will be Rs.100+10+10 = Rs.120 and so on. Formulae: 1. If P stands for Principal, R the rate per cent per annum, T the number of years, A the amount and S.I. the simple interest then P T R S.I. = 100 1 0 0 S.I. 2. P RT 1 0 0 S.I. 3. R  % PT 1 0 0× S.I. 4. years T= P ×R RT   5. Amount, A  P 1  1 0 0  6.

7.

8.

9.

10.

11.

12.

13.

If a certain sum in T years at R% per annum amounts to Rs.A, then the sum will 1 0 0 A be P  . 1 0 0 (R  T) The annual payment that will discharge a debt of Rs.A due in T years at R% per ann um 1 0 0 A = . RT(T  1) 1 0 0T  2 If a certain sum is invested in n types of investments in such a manner that equal amount is obtained on each investment where interest rates are R 1 ,R 2 ,…….,R n respectively and time periods and time periods are T 1 ,T 2 ,…., T n respectively, then the ratio in which the amounts are invested is 1 1 1 . : : .......... .......... .... 1 0 0 R1T1 1 0 0 R2T2 1 0 0 RnTn If a certain sum of money becomes n times itself in T years at S.I., then the rate of 1 0 0(n  1) interest per annum is, R  %. T If a certain sum of money becomes n times itself at R% per annum S.I. in T years, then 100(n  1) years. T R If a certain sum of money becomes n times itself in T years at a simple interest, then m  1 the time T ' in which it will become m times itself is given by T' =    T years.  n 1  If the rate of interest (R) changes from R 1 to R 2 and P, T are constant, then PT Change in S.I. =  (R 1  R 2 ) . 100 If Principal (P) changes from P 1 to P 2 and R, T are constant, then

120

RT  (P1  P2 ) . 100 14. If Rate (R) Changes from R 1 to R 2 and time (T) changes from T 1 to T 2 but principal (P) is constant, then P Change in S.I. =  (R1T1  R2T2 ) . 100 15. If a debt of Rs.X is paid in n number of installments and if the value of each installment n(n  1) Ra is a, then the borrowed (debt) amount is given by X  na  .  1 0 0 b 2 Where, R is the rate of interest per annum b is the no. of installments per year Note: b = 1, when each installment is paid yearly. b = 2, when each installment is paid half-yearly. b = 4, when each installment is paid quarterly. b = 12, when each installment is paid monthly. Change in S.I. =

16.

If a certain sum of money P lent out at S.I. amounts to A 1 in T 1 years and to A 2 in T 2 A T ~ A2 T1 A1 ~ A2 × 1 0 0% years, then, P = 1 2 and R = T2 ~ T1 A1T2 ~ A2 T1

17.

If a certain sum of money P lent out for a certain time T amounts to A 1 at R 1 % per annum and to A 2 at R 2 % per annum, then A1 ~ A2 A R ~ A1R 2 P= 2 1 × 1 0 0years. and T = R1 ~ R 2 A2R1 ~ A1R 2

18.

If an amount P 1 lent at S.I. rate of interest R 1 % p.a. and another amount P 2 at S.I. rate  P R  P2R 2  of R 2 % p.a., then the rate of interest for the whole sum is R   1 1 .  P1  P2 

19.

If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained as S.I. on each part where interest rates are R 1 ,R 2 ,……,R n respectively and time periods are T 1 ,T 2 ,…..,T n respectively, then the ratio in which the sum will be divided in n parts is given by 1 1 1 : : .......... .......... .... . R1T1 R2T2 RnTn

20.

If there is a change in principal (P), rate of interest (R) and time (T), then the va lue of S.I also changes and is given by S.I1 P  R 1  T1  1 S.I2 P2  R 2  T2 

21.

A 1  P1 P  R 1  T1  1 A 2  P2 P2  R 2  T2

Out of a certain sum P,

1 1 part is invested at R 1 %, part is invested at R 2 % and the y x

 1 1 1 remainder 1    say, part at R 3 %. If the annual income from all these z x y  investments is Rs.A, the original sum is given by     A  1 0 0 . P    R1 R 2 R3      y z   x

SOLVED EXAMPLES 1.

Find the S.I. on Rs.3500 for 3 years at 5% per annum.

121

Sol: Here, P = Rs.3500, T = 3 years and R = 5%  S.I. = 3 5 0 0 3  5  Rs.5 2 5 . 100 2.

Find the Principal, if S.I. = 624 for 2 years at 3% per annum. Sol: P 

3.

1 0 0 S.I. 1 0 0 6 2 4   Rs.10 4 0 0 . RT 23

At what rate per annum will a sum of Rs.4000 amount to Rs. Rs.6000 in 5 years? Sol: S.I. = 6000 – 4000 = 2000

R 4.

1 0 0 S.I. 1 0 0 2 0 0 0   1 0% . PT 4 0 0 0 5

In what time will Rs.1500 earn an interest of Rs.150 at 2% per annum? Sol: T 

5.

1 0 0 S.I. 1 0 0 1 5 0   5years. PR 1 5 0 0 2

Sunny borrowed Rs.2000 from his friend Praveen at 10% per annum for 5 years. Find the interest and money returned by Sunny to Praveen. 2000  5  10 Sol: S.I. =  1000 100 A = P + S.I. = 2000 + 1000 = Rs.3000

1 years? 2 1 0 0 A 1 0 0 4 2 0 1 0 0 4 2 0 1 0 0 4 2 0     Rs.4 0 0 Sol: P  1 5 1 0 0 (R  T) 105 1 0 0 (2  2 ) 1 0 0 (2  ) 2 2

6.

What sum will amount to Rs.420 at 2% per annum in 2

7.

Find the annual installment that will discharge a debt of Rs.16200 in 5 years at 4% per annum. 1 0 0 A Sol: Annual installment = RT(T  1) 1 0 0T  2 1 0 0 1 6 2 0 0 1 0 0 1 6 2 0 0 1 0 0 1 6 2 0 0    Rs.3 0 0 0. = 4  5  (5  1) 80 540 1 0 0(5)  5 0 0 2 2

8.

A sum of Rs.3454 is divided among three such parts that amount obtained on these three parts of money after 3, 4 and 5 years, respectively at rate of 5% per annum remains equal. Find such three parts of the sum. Sol: The three parts will be in the ratio 1 1 1 : : .......... .......... .... 1 0 0 R1T1 1 0 0 R2T2 1 0 0 RnTn

1 1 1 1 1 1 : :  : : 1 0 0 (5  3) 1 0 0 (5  4) 1 0 0 (5  5) 1 1 5 1 2 0 1 2 5 1 69000 1 69000 1 69000 = : :  6 0 0: 5 7 5: 5 5 2 115 120 125 [Since the L.C.M. of 115, 120 and 125 is 69000.] =



Ratio = 600: 575: 552 Sum of proportional = 600 + 575 + 552 = 1727

122

600  3 4 5 4 Rs.1 2 0 0 1727 575 part =  3 4 5 4 Rs.1 1 5 0 1727 552 part =  3 4 5 4 Rs.1 1 0 4. 1727



1 st part =



2 nd



3 rd

9.

A certain sum of money quadruples itself in 6 years S.I. Find the rate percent per annum. 1 0 0(n 1) 1 0 0(4 1) Sol: R = %= = 5 0% . T 6

10.

In what time a sum of money will double itself at a rate of S.I. of 6% per annum?

(2  1) 4 2 n  1   1 0 0years   1 0 0  1 6  1 6 years Sol: T   6 6 3  R  11.

A sum of money put on S.I. doubles itself in 10 years. In how many years would it quadruple itself?  4  1 Sol: Required Time, T ' =    10 = 30 years. 2  1

12.

If simple interest on Rs.500 increases by Rs.20, when the rate % increases by 8% per annum, find the time. PT Sol: Change in S.I. =  (R 1  R 2 ) 100 5 0 0T 1  20   8  2 0  4 0T  T  years. 100 2

13.

If the S.I. on Rs.3500 be more than the interest on Rs.2000 by Rs.60 in 4 years, then find the rate per cent per annum. RT Change in S.I. =  (P1  P2 ) 100 4R 4R  60   (3 5 0 0 2 0 0 0)  6 0   1 5 0 0 R  1% . 100 100

14.

If the S.I. on a certain sum at 3% per annum for 5 years is Rs.90 more than the interest on the same sum for 2 years at 6% per annum. Find the sum. P Sol: Change in S.I. =  (R1T1  R2T2 ) 100

 90  15.

P P  (3  5  2  6)  90   3  P  Rs.3000. 100 100

A sum of Rs.4 is lent to be paid back in 3 equal monthly installments of Re.1 each. Find the rate percent. Sol: Here, X = Rs.4, a = Re.1, n = 3, b = 12, R = ? n(n  1) Ra Requiredformula, X  na   1 0 0 b 2

123

R 1 3(3  1)  1 0 0 1 2 2 R  4 3 3 1200 1 2 0 0 R 4 400  1 2 0 0 R  1 6 0 0  R  4 0 0% .  4  3(1) 

16.

Anitha deposits Rs.8000 in S.B.I at 2% per annum and Rs.6000 at 5% per annum in ICICI bank. Find the rate of interest for the whole sum.  P R  P2R 2  Sol: Required rate, R   1 1   P1  P2  8 0 0 0 2  6 0 0 0 5 8 0 0 0 6 0 0 0 19000 5  1 %. 14000 14 

17.

1 2 years at the rate of 2% per annum, equals the S.I. on the second part for 4 years at the rate of 4% per annum, then find two such divisions of the sum. 1 1 Sol: Required Ratio = : R1T1 R2T2 If a sum of Rs.3800 is divided into two such parts that the S.I. on the first part for 1

1 1 1 :  :  1 6: 3 1 4  4 3 16 2 1 2 Sum of proportionals = 16 + 3 = 19 16 So, 1 st part =  3 8 0 0 Rs.3 2 0 0 19 3 2 nd part =  3 8 0 0 Rs.6 0 0. 19



1 st part: 2 nd part =

18.

1

If Rs.50 amounts to Rs.85 in 2 years, what Rs. 340 amount to in 15 years at the same rate? A P P  R1  T1 Sol: By using, 1 1  1 we get A2  P2 P2  R2  T2

85  50 85  R  2  A2  340 340  R  15 35 170 35 1    A2  340 340  15 A2  340 2  15  A2  340  30  35  1050 

 A2  1050  340  1390 19.

1 1 th at 2%, th at 4% and the rest at 6%. If the 4 5 annual income of Krishna is Rs.1840 from all these investments, then find the original sum.     A  100  Sol: Required formula is, P    R1 R 2 R3      y z   x 1 1 1 1 1 1 1 1 1 2 0 5  4 1 1 1  1     ,  and z x y 4 5 20 20 x 4 y 5 From a certain sum Krishna invested,

124

Original sum =

1 8 4 0 1 0 0 2 4 6   4 5 20 11

1 8 4 0 1 0 0 1 8 4 0 1 0 0  2 4 6 6 1 0 1 6 6 6   4 5 20 20 1 8 4 0 1 0 0   2 0  Rs.4 0,0 0 0. 92 

125

Brainstorming 1.

Find the S.I. on Rs.500 at 6% p.a. from 3rd May to July 15th on the same year. 1. Rs.9 2. Rs.6 3. Rs.4 4. None

2.

Uday borrowed a Sum of Rs.10000 from a bank for 6 years at 8% p.a. Find the amount returned by him. 1. Rs.14800 2. Rs.12600 3. Rs.13300 4. None

3.

The Principal that will yield Rs.60 as S.I. at 6% p.a. in 5 years is ________. 1. Rs.175 2. Rs.350 3. Rs.200 4. None

4.

Calculate the Sum of money that will produce Rs.1700 interest in 7 ½ years at 8% S.I. p.a. 1. Rs.2950 2. Rs.3120 3. Rs.2800 4. None If the S.I. on a certain Sum of money after 6 ¼ years is 3/8 of the Principal, then find the rate% p.a. 1. 5% 2. 6% 3. 4% 4. None

5.

6.

Ramesh borrowed Rs.5000 from Ganesh at S.I. If Ganesh got Rs.500 more than his capital after 5 years, then find the rate% p.a. 1. 2% 2. 3% 3. 4% 4. None

7.

The rate% at which Rs.1200 amounts to Rs.1400 in 4 years is _________. 1. 5% 2. 4% 3. 6% 4. None

8.

The S.I. on a certain Sum of money is Rs.256 and the rate of interest p.a. equals the number of years, then find the rate% p.a. 1. 13% 2. 14% 3. 16% 4. Data insufficient

9.

If S.I. on a certain Sum of money for 2 years is 1/5th of the Sum then find the rate% p.a. 1. 9% 2. 10% 3. 8% 4. Data insufficient

10.

If S.I. on a certain Sum of money for 2 years is 4/25th of the Sum and the rate% is same as the time period, then find the rate% p.a. 1. 2% 2. 3% 3. 4% 4. None

11.

If a certain Sum of money borrowed at 5% p.a. S.I. amounts to Rs.1020 in 4 years, then find the Sum of money borrowed. 1. Rs.850 2. Rs.925 3. Rs.750 4. None

12.

In what time will Rs.1200 amounts to Rs.1344 at 6% p.a? 1. 2 ½ years 2. 3 years 3. 2 years

4. None

13.

In what time will Rs.8100 produce the same income at 3% as Rs.225 in 4 years at 3%? 1. 1/7 years 2. 1/9 years 3. 1/6 years 4. None

14.

If Rs.1000 be invested at 5% and the interest be added to the Principal every 10 years then find the time in which it will amount to Rs.2000. 1. 16 2/3 years 2. 16 ¼ years 3. 1/6 years 4. None

15.

If Rs.500 amounts to Rs.725 at 9% S.I. in some time, what will be Rs.600 amount to at 11% in the same time? 1. Rs.870 2. Rs.930 3. Rs.910 4. None

16.

Sanjay lends Rs.10000 for 2 years at 20% p.a. S.I. After 1 year, he gets Rs.6000. How much will he get in the next year? 1. Rs.5900 2. Rs.6400 3. Rs.7200 4. Data insufficient

17.

What Principal will amount to Rs.15000 at 10% p.a. in 5 years? 1. Rs.10,000 2. Rs.8700 3. Rs.10,500

4. Data insufficient

126

18.

The annual payment that will discharge a debt of Rs.47,250 due in 3 years at 5% S.I. is _____ 1. Rs.8000 2. Rs.10000 3. Rs.15000 4. None

19.

Find the annual installment that will discharge a debt of Rs.4200 due in 5 years at 10% S.I. 1. Rs.700 2. Rs.750 3. Rs.800 4. None

20.

If the amount obtained by Mahesh by investing Rs.1500 for 2 ½ years at 8% p.a. is equal to the amount obtained by Suresh by investing a certain Sum for 2 years at 5% p.a. S.I., then the find the Sum invested by Suresh. 1. Rs.1636 4/11 2. Rs.1636 3. Rs.1636 1/2 4. None

21.

A man invests Rs.3965 in the names of his three daughters Nitu, Lekha and Ratna in such a way that they get the same amount after 2, 3 and 4 years respectively. If the rate of interest is 5% p.a. then find the amounts invested for Nitu, Lekha and Ratna. 1. Rs.1380, Rs.1320, Rs.1265 2. Rs.1330, Rs.1360, Rs.1380 3. Rs.1265, Rs.1320, Rs.1340 4. None of these

22.

A Sum of money at S.I. becomes 4 times in 24 years. Find the rate% p.a. 1. 13 ¾ % 2. 12 ½ % 3. 11 ¾ % 4. None

23.

In how many years will a Sum of money triples itself at 10% p.a. S.I.? 1. 15 years 2. 19 years 3. 20 years 4. None

24.

A Sum of money doubles itself in 8 years. In how many years will it triple? 1. 16 years 2. 15 years 3. 14 years 4. None

25.

A Sum of money put at S.I. at a certain rate for 4 years. Had it been put at 2% higher rate, it would have fetched Rs.56 more. Find the Sum. 1. Rs.680 2. Rs.700 3. Rs.720 4. None

26.

If the interest on Rs.800 be more than the interest on Rs.400 by Rs.40 in 2 years then find the rate% p.a. 1. 5% 2. 5 ½ % 3. 6% 4. None

27.

If the difference between S.I. on a certain Sum for 4 years at 2 ½ % p.a. and the S.I. on the same Sum for the same period at 3% p.a. is Rs.60 then find the Sum. 1. Rs.3000 2. Rs.2900 3. Rs.3100 4. None

28.

If a certain Sum of money at S.I. amounts to Rs.2800 in two years and Rs.3250 in 5 years, then find rate % p.a. 1. 4% 2. 6% 3. 5% 4. None

29.

If a certain Sum of money amounts to Rs.1760 in two years and Rs.2000 in 5 years at S.I. then find the Sum. 1. Rs.1960 2. Rs.1590 3. Rs.1600 4. None

30.

If a certain Sum is invested for a certain time, it amounts to Rs.450 at 7% p.a. but when invested at 5% p.a. it amounts to Rs.350 then find the Sum. 1. Rs.60 2. Rs.100 3. Rs.120 4. None

31.

A certain Sum is invested for 40 years, it amounts to Rs.400 at 10% p.a. but when invested at 4% p.a. it amounts to Rs.200. Find the time. 1. 36 years 2. 37 years 3. 37½ years 4. 36½ years

32.

If a Sum of Rs.20 is lent to be paid back in 10 equal month installments of Re.1 each then find the rate%. 1. 266 2/3 % 2. 265 ¾ % 3. 266% 4. None

33.

A Sum of Rs.7700 is to be divided among three brothers Veeru, Vinay and Ram in such a way that S.I. on each part at 5% p.a. after 1, 2 and 3 years respectively remains equal. The share of Veeru is more than that of Vinay by ____. 1. Rs.2800 2. Rs.2500 3. Rs.3000 4. None

127

34.

If S.I. on a certain Sum of money for 4 years at 5% p.a. is same as the S.I. on Rs.560 for 10 years at the rate of 4% p.a. then find the Sum. 1. Rs.1190 2. Rs.1120 3. Rs.1210 4. None

35.

Mr. Hari invested an amount of Rs.12000 at S.I. rate of 10% p.a. and another at 20%. The total interest at the end of one year on the total amount invested became 14% p.a. Find the total amount invested. 1. Rs.20,000 2. Rs.20,800 3. Rs.21,000 4. None

36.

Mr. Gupta deposits Rs.3000 in a bank at 10% p.a. and Rs.5000 in another bank at 8% p.a. Find the rate% for the whole Sum. 1. 8 ½ % 2. 8 ¾ % 3. 8% 4. None

37.

A person invested 2/3rd of his capital at 3%, 1/6th at 6% and the remaining at 12%. If his annual income is Rs.25 then find the capital. 1. Rs.490 2. Rs.510 3. Rs.500 4. None

38.

The S.I. on a Sum of money will be Rs.600 after 10 years. If the Principal is trebled after 5 years, what will be the total interest at the end of 10th year? 1. Rs.1200 2. Rs.1190 3. Rs.1210 4. None

39.

Rs.1500 is invested at 10% S.I. and interest is added to the Principal after every 5 years. In how many years will it amount to Rs.2500? 1. 6 1/9 years 2. 6 ¼ years 3. 7 years 4. None

40.

Anitha lends some money to Navitha at 5% p.a. S.I. Navitha lend the amount to Radhika on the same day at 8 ½% p.a. In this transaction after a year Navitha earned a profit of Rs.350. Find the Sum of money lent by Anitha to Navitha. 1. Rs.9000 2. Rs.10,000 3. Rs.10,200 4. None

41.

Poornima borrowed Rs.1000 to build a hut. She pays 5% S.I. She lets the hut to Srikanth and receives the rent of Rs.12 ½ per month from him. In how many years, Poornima ought to clear off the debt? 1. 10 years 2. 10 ¼ years 3. 10 ½ years 4. None

42.

The rate of interest on a Sum of money is 4% p.a. for the first two years and 6% p.a. for the next 4 years, 8% p.a. for the period beyond 6 years. If the S.I. occurred by the Sum for a total period of 9 years is Rs.1120 then find the Sum. 1. Rs.2400 2. Rs.2200 3. Rs.2000 4. None

43.

Find the simple Interest on Rs.5000 for 219 days at 10% p.a. 1. Rs.300 2. Rs.250 3. Rs.350

4. None

44.

Mr. Shyam borrowed Rs.25000 from a bank at the rate of 6% p.a. What is the interest paid by him at the end of the 5th year if simple interest is calculated? 1. Rs.7000 2. Rs.7500 3. Rs.7250 4. None

45.

The simple interest on a sum of money is 1/9 of the principal, and the number of years is equal to the rate % p.a. Find the rate % p.a. 1 1. 2% 2. 3 % 3. 2.5% 4. 3% 3

17.COMPOUND INTEREST Compound Interest: In this method, the interest for each period is added to the principal before; interest is calculated for the next period. So, the principal grows as the interest is added to it. It is denoted by C.I. Formulae:

128

1.

If a principal P is given on C.I. at the rate of interest R% p.a. , then the Amount A after t years is t

2.

R   A  P 1  . 1 0 0  C.I. = A – P t t   R  R   = P 1   P  P 1    1  1 0 0 1 0 0     

1   A t  1%p.a. Rate of interest (R) = 1 0 0    P     Note: S.I. and C.I. for 1 year at a given rate of interest per annum are always equal.

3.

4.

If the interest is compounded half-yearly, then 2t

R   a) Amount, A  P 1   . 1 0 0  2   2t   R  b) C.I. = P 1    1 . 1 0 0 2     

5.

1   2  A  t   1%p.a. c) Rate (R) = 2  1 0 0   P      If the interest is compounded quarterly, then

R   a) Amount, A  P 1  1 0 0 4  

4t

.

4t   R  b) C.I. = P 1    1 . 1 0 0 4     

6.

1   4  A  t   1%p.a. c) Rate (R) = 4  1 0 0   P      If the interest is compounded n times a year, then

R   a) Amount, A  P 1  1 0 0 n  

n t

.

n t   R   1 . b) C.I. = P 1   1 0 0 n     

7.

1   n  A  t   1%p.a. c) Rate (R) = n  1 0 0   P       If the rate of interest is different for different years, say R 1 %, R 2 %, R 3 % for first, second and third years respectively, then R  R  R   Amount, A  P1  1 1  2 1  3 . 1 0 0 1 0 0 1 0 0 

8.

If the time is in the form of fraction, say

9.

x

y z

years, then

y   x R  R   Amount, A  P1    1  z .  1 0 0 1 0 0      The difference between the C.I. and the S.I. on a certain sum of money for 2 years at R% p.a., is

129

2

10.

 R  a) C.I. – S.I. = P  if P and R are given  1 0 0 R  S.I. b) C.I. – S.I. = if S.I. and R are given 2  100 The difference between the C.I. and the S.I. on a certain sum of money for 3 years at R% p.a., is 2  R  3  R   a) C.I. – S.I. = P    3   if P and R are given  1 0 0  1 0 0   2 S.I.  R   R     3  if S.I. and R are given 3  1 0 0  1 0 0  If a certain sum becomes n times in t years at C.I., a) Then the same sum becomes n t in mt years.  1  b) Rate of C.I., R  1 0 0(n) t  1% .     If a certain sum of money at C.I. amounts to Rs.x in A years and to Rs.y in B years, then the rate of interest p.a. is 1    y  B  A  R     1  1 0 0% .  x     If a loan of Rs.P at R% C.I. p.a., is to be repaid in n e qual yearly installments, then the P value of each installment = Rs. . 2 n  100   100   100        ........     1 0 0 R   1 0 0 R   1 0 0 R 

b) C.I. – S.I. =

11.

12.

13.

SOLVED EXAMPLES 1.

Ravi invested Rs.25000 at C.I. rate 4% p.a., for a period of 3 years. What amount will he receive at the end of 2 years? Sol: P = Rs.25000, R = 4%, t = 3 years, A =?

R   A  P 1  1 0 0 

t

3

2.

4  26 26 26   25000 1  1 0 0  2 5 0 0 0 2 5  2 5  2 5  Rs.2 8,1 2 1.6 0.   Find the C.I. on Rs.2000 for 2 years at 5% p.a. Sol: P = Rs.2000, R = 5%, t = 2 years, C.I. =? t   R  C.I. = P 1    1  1 0 0     2    2 1 2  5  = 2 0 0 01    1  2 0 0 0   1   1 0 0    2 0       2000 (1.025 1)  2000 (0.1025 )  Rs.205.

3.

Ram invested Rs.5548 for 3 years at C.I. and received an amount of Rs.6750 on maturity. What is rate percent?

130

Sol: P = Rs.5548, A = Rs.6750, t= 3 years, R =? 1   A t  Rate of interest (R) = 1 0 0    1%p.a.  P     1 1      6 7 5 0 3   3 3 7 5 3   1 0 0   1 %  1 0 0  1   2 7 7 4 %  5 5 4 8          

4.

1     1 53  3  1 5   1 0 0    1%    1%  1 0 0    1 4 1 4            100 50 1    7 % .p.a. 14 7 7 1 Find the amount of Rs.16000 in 1 years at 10% p.a., C.I. payable half-yearly. 2 1 3 Sol: P = Rs.16000, R = 10%, t = 1 = years, A =? 2 2

R   Amount, A  P 1  1 0 0 2  

2t

10    16000 1  1 0 0 2   

2

3 2

3

 2 1  16000    2 0

(2 1)3  2  9 2 6 1 Rs.1 8,5 2 2. 8000 Find the C.I. on Rs.8192 at 50% p.a., compounded quarterly for 1 year.  1 6 0 0 0

5.

Sol: P = Rs.8192, R = 50%, t = 1 year, C.I. =? 4t   R  C.I. = P 1    1 1 0 0 4      4 1    9 4  50   8 1 9 21   1  8 1 9 2   1  1 0 0 4     8       2465  6 5 6 1 4 0 9 6  8 1 9 2  Rs.4 9 3 0   8 1 9 2 4096 4096  

6.

Find the C.I. on Rs.2000 at 12% p.a. for 2 months compounded monthly. 2 1 Sol: P = Rs.2000, R = 12%, t = 2 months =  years, C.I. =? 12 6 12  t   R  C.I. = P 1   1  1 0 0 1 2     1   12   1 0 12  12  6    2 0 0 01   1  2 0 0 0    1 1 0 0 1 2  1 0 0         2 0 0 01.0 2 0 1 1  2 0 0 0 0.0 2 0 1 Rs.4 0 .2 .

7.

Lakshmi invests Rs.50000 in a mutual fund which gives interest at 2% p .a., during first year, 5% during the second year and 8% during the third year. How much does she get at the end of the third year? Sol: P = Rs.50000, R 1 = 2%, R 2 = 5%, R 3 = 8% R  R  R   Amount at the third year, A  P1  1 1  2 1  3  1 0 0 1 0 0 1 0 0 

2  5  8    50000 1  1  1   1 0 0 1 0 0 1 0 0 

131

8.

 5 1 2 1 2 7 5 0 0 0 0 2 8 9 1 7  50000  Rs.5 7 8 3 4 .  5 0 2 0 2 5  25000     1 What will be the C.I. on Rs.93750 for 2 years at 4% p.a.? 2 1 Sol: P = Rs.93750, t= 2 years, R = 4%. 2   y   x R    R  z    1 Required C.I. = P 1    1    1 0 0 100             1   2  4   4  2    9 3 7 5 0 1   1   1  1 0 0 100         2 6 2 6 5 1   9 3 7 5 0     1 2 5 2 5 5 0  9 3 7 5 0 3 2 2 6  Rs.9 6 7 8 31250 Find the difference between C.I. and S.I. on a sum of Rs.6250 put for 2 years at 4% p.a. Sol: P = Rs.6250, t = 2 years, R = 4% 

9.

 R  For 2 years, C.I. – S.I. = P   1 0 0

2

if P and R are given 2

10.

1 1  4   6 2 5 0    6 2 5 0 2 5  2 5  Rs.1 0 1 0 0   The difference between C.I. and S.I. on a certain sum of money for 3 years, at 4% p.a. is Rs.76. Find the sum. Sol: C.I. – S.I. = Rs.76, R = 3% 2  R  3  R   For 3 years, C.I. – S.I. = P    3   if P and R are given  1 0 0  1 0 0   2  4  3  4   76 = P    3    1 0 0  1 0 0   2  1  3  1   76 = P    3    2 5  2 5   2  1  3  1   76 = P    3    2 5  2 5  

 1  (3  2 5)  76 = P    2 5  2 5  2 5

 76  76 = P   1 5 6 2 5 P  Rs.15625. 11.

Certain sum of money placed at C.I. doubles itself in 3 yea rs. In how many years will it amount to eight times itself? Sol: Here, n = 2, t = 3 years and m = 3 Then the same sum becomes nm in mt years = 3 x 3 = 9 years.

12.

At what percent C.I. does a sum of money become eight -fold in 3 years.

132

 1   1  Sol: Required Rate percent is, R  1 0 0(n) t  1  1 0 0(8)3  1  1 0 0% .         13. A certain sum of money at C.I. amounts to Rs.800 in two years and to Rs.1152 in four years. Find the rate of interest per annum. Sol: x = Rs.800, y = Rs.1152, A = 2 and B = 4 1    y  B  A   1  1 0 0% Required rate of interest, R    x      1 1      1 1 5 2 4  2   5 7 6 2     1  1 0 0%      1  1 0 0% 8 0 0 4 0 0             1   2 2    2 4   2 4  2 0  =     1  1 0 0%     1 0 0  2 0% .  2 0    20        1 14. If a sum of Rs.18120 is to be paid back in two equal installments at 1 % per annum, 3 what is the amount of each installment? 1 4 Sol: P = Rs.18120, R = 1 % = % 3 3 P Each installment = Rs. 2  100   100        1 0 0 R   1 0 0 R 

=

18120 2



18120

             100    100   1 0 0    3 0 4 4  4  1 0 0   1 0 0    3  3   3  18120 18120 =  2 3 0 0 3 0 0   3 0 0  3 0 0 1   3 0 4   3 0 4 3 0 4 3 0 4     =

     1 0 0  3 0 4    3 

2

18120 1 8 1 2 0 3 0 4 3 0 4 3 0 4 3 0 4  =  Rs.9 2 4 1 .6 .0 300 604 3 0 0 6 0 4 10  304 304

133

Brainstorming

1. 2. 3. 4. 5.

Find the C.I. on Rs.12,000 at 10% for 2 ½ years. 1. Rs.3000 2. Rs.3246 3. Rs.2000

4. None

On what sum of money will be C.I. be Rs.328 in 2 years at 5% p.a.? 1. Rs.3200 2. Rs.4000 3. Rs.3500

4. Rs.4500

At what rate% p.a. will Rs.400 amount to Rs.441 in 2 years? 1. 7% 2. 9% 3. 11%

4. 5%

The C.I. on Rs.10,000 for 4 years is Rs.641. Find the rate% p.a. 1. 10% 2. 11% 3. 12%

4. 19%

In what time will Rs.1200 amount to Rs.1323 at 5% p.a. C.I.? 1. 3 years 2. 2 years 3. 6 years

4. 9 years

6.

Find the C.I. on Rs.24,000 at 10% p.a. for 1 year 6 months the interest is being compounded half-yearly? 1. Rs.12,000 2. Rs.15,000 3. Rs.14,000 4. None

7.

Find the rate% p.a. if Rs.2000 amount to Rs.2315.25 in 1 ½ years interest being compounded six monthly? 1. 10% 2. 11% 3. 15% 4. 20%

8.

The difference between S.I. and C.I. on a certain sum of money for 2 years at 4% p.a. is 20. Find the sum. 1. Rs.12,500 2. Rs.13,000 3. Rs.14,000 4. None

9.

The difference between S.I. and C.I. is Rs.1395 for 3 years is at 10%. Find the sum. 1. Rs.45,000 2. Rs.90,000 3. Rs.4500 4. None

10.

Find the difference between S.I. and C.I. on Rs.3125 for 3 years at 4% p.a. 1. Rs.15.20 2. Rs.60.00 3. Rs.20 4. Rs.22

11.

The C.I. and S.I. on a certain sum of money for 2 years are Rs.816 and Rs.800 at the same rate. Find the sum and the rate%. 1. Rs.10,000, 6% 2. Rs.4000, 4% 3. Rs.10,000, 4% 4. None

12.

The C.I. on a sum of money for 2 years at 5% p.a. is Rs.246. Find the corresponding S.I. 1. Rs.240 2. Rs.900 3. Rs.680 4. None

13.

At a C.I. a sum becomes double itself in 3 years. In 12 years, it will become how many times? 1. 15 times 2. 16 times 3. 18 times 4. 20 times

14.

At a C.I. a sum becomes double itself in 7 years. In how many years it will become 32 times? 1. 30 years 2. 40 years 3. 35 years 4. 39 years

15.

The annual increase in the price of a TV is 10% if the present value is Rs.15215. What was it 2 years ago? 1. Rs.12000 2. Rs.15000 3. Rs.12500 4. None

16.

A building of worth Rs.81,000 is constructed on a land worth Rs.1,21,000. After how many years will the value of both the same if the building appreciates at 10% p.a. and the land depreciates at 10% p.a.? 1. 2 years 2. 3 years 3. 5 years 4. 7 years

17.

A sum of Rs.13,360 is borrowed at 8 ¾ % p.a. C.I. and paid back in equal installments. What is the value of each installment?

134

1. Rs.7579

2. Rs.7569

3. Rs.7000

4. Rs.6500

18.

What is the nominal % p.a. when interest is payable half-yearly that would give an effective rate of 8% p.a.? 1. 7.8% 2. 8.7% 3. 9% 4. None

19.

A loan was repaid in two annual installments of Rs.121 each. If the rate of interest be 10% p.a. compounded annually, the sum borrowed was ____. 1. Rs.200 2. Rs.210 3. Rs.300 4. Rs.310

20.

A sum of Rs.8448 is to be divided between A and B who are respectively 18 and 19 years old in such a way that if their share be invested at 6.25% p.a. C.I. they shall receive equal amounts on attaining the age of 21 years. Find the present share of each and how much each will receive at the age of 21? 1. Rs.4096, Rs.4352 2. Rs.5000, Rs.3448 3. Rs.6000, Rs.2448 4. None

21.

A sum of money is borrowed and paid in two equal installments of Rs.729 allowing 8% C.I. What was the sum borrowed? 1. Rs.13,640 2. Rs.12,000 3. Rs.1300 4. None

22.

A sum of money put at C.I. for 2 years at 20%, it would fetch more Rs.482 more, if the interest were payable half-yearly, then if it were payable half-yearly, than if it were payable yearly. Find the sum? 1. Rs.25,000 2. Rs.20,000 3. Rs.14,000 4. None

23.

At what rate% C.I. does a sum of money becomes four-fold in 2 years? 1. 150% 2. 100% 3. 200% 4. 400%

24.

If the C.I. on a certain sum for 2 years is Rs.60.60 and the S.I. is Rs.60 then find the rate of interest p.a. 1. 2% 2. 3% 3. 4% 4. None

25.

If the C.I. on a certain sum for 2 years is Rs.105 and S.I. is Rs.100 then find the sum. 1. Rs.300 2. Rs.400 3. Rs.500 4. None

26.

Find the difference between S.I. and C.I. on Rs.1250 for 2 years at 4% p.a. 1. Rs.3 2. Rs.4 3. Rs.2 4. None

27.

On a certain sum of money, the S.I. for 2 years is Rs.200at 7% p.a. Find the difference in C.I. and S.I. 1. Rs.3 2. Rs.4 3. Rs.7 4. None

28.

The difference of C.I. and S.I. on a certain sum at 5% for 2 years is Rs.1.50. Find the sum. 1. Rs.700 2. Rs.600 3. Rs.500 4. None

29.

Find the difference between the S.I. and C.I. on a certain sum at 3% p.a. for 3 years is Rs.27.27. Find the sum. 1. Rs.12000 2. Rs.15000 3. Rs.10000 4. None

30.

Find the difference between the C.I. and S.I. on Rs.8000 for 3 years at 5% p.a. 1. Rs.61 2. Rs.63 3. Rs.65 4. None

31.

If a sum of money at C.I. amounts to thrice itself in 3 years, then in how many years will it be 9 times itself? 1. 9 years 2. 6 years 3. 7 years 4. None

32.

At what rate% C.I. does a sum of money becomes 16 times in 4 years? 1. 75% 2. 100% 3. 50% 4. None

33.

A certain sum of money at C.I. grows up to Rs.12,960 in 2 years and up to Rs.13,176 in 3 years. Find the rate% p.a. 1. 1 1/3% 2. 2 1/3% 3. 1 2/3% 4. None

135

34.

What sum of money at C.I. will amount to Rs.650 at the end of 1st year and to Rs.676 at the end of the second year? 1. Rs.825 2. Rs.925 3. Rs.625 4. None

35.

A sum of Rs.1260 is borrowed from a money lender at 10% p.a. compounded annually. If the amount is paid back in two equal installments, find the annual installment. 1. Rs.726 2. Rs.626 3. Rs.526 4. None

36.

A tree increases annually by 1/8th of its height. By how much will it increase after 2 years, if it stands today 64 cm high? 1. 72 cm 2. 74 cm 3. 78 cm 4. 81 cm

37.

Find the least number of completed years in which a sum of money put out at 20% C.I. will be more than doubled. 1. 3 2. 4 3. 5 4. 6

38.

A man borrows Rs.4000 from a bank at 7 ½ % C.I. At the end of every year, he pays Rs.1500 as part repayment of loan and interest. How much does he still owe to the bank after three such installments? 1. Rs.123.25 2. Rs.125 3. Rs.400 4. Rs.469.18

39.

If in a certain no. of years Rs.3000 amount to Rs.4320 at C.I., in half that time Rs.3000 will amount to _____. 1. Rs.3400 2. Rs.3600 3. Rs.3800 4. Rs.3520

40.

Rs.3757 is to be divided between A and b such that A‟s share at the end of 7 years be equal to B‟s share at the end of 9 years. If rate% be 10% p.a. C.I. then find the B‟s share. 1. Rs.1700 2. Rs.1400 3. Rs.1500 4. Rs.2057

136

18. CLOCKS Minute Spaces: The circumference of a dial of a clock or watch is divided into 60 equal parts. These parts are called minute spaces. Hands: The clock has two hands. 1. Hour hand

2. Minute hand

Hour hand: The hour hand or short hand indicates time in hours. In an hour, the hour hand covers 5 minute spaces. Minute hand: The minute hand or long hand indicates time in minutes. In an hour, the minute hand covers 60 minute spaces. Note:    

In every hour, both the hands coincide once. In every hour, the hands are straight (points in opposite directions) once. In this position, the hands are 30 minutes apart. In every hour, the hands are twice at right angles. In this position, the hands are 15 minute spaces apart. The minute hand moves through 6 o in each minute whereas the hour hand moves through o

o 1 1 in each minute. Thus, in one minute, the minute hand gains 5 than the hour hand. 2 2

0o .



When the hands coincide, the angle between them is

 

When the hands point in opposite directions, the angle between them is 180 . The hands are in the same straight line, when they are coincident or opposite to each other.



So, the angle between the two hands is 0 or 180 . The minute hand moves 12 times as fast as the hour hand.

o

o

o

Formulae:

1. 2. 3.

4. 5.

 60 H The two hands of the clock will be together between H and (H + 1) O‟ clock at   O‟  11  clock. The two hands of the clock will be at right angles between H and (H + 1) O‟ clock at (5H ± 12 15) minutes past H O‟ clock. 11 The two hands of the clock will be in the same straight line but not together between H and (H + 1) O‟ clock at 12 (5H – 30) minutes past H when H > 6, 11 12 (5H + 30) minutes past H when H < 6 11 Between H and (H + 1) O‟ clock, the two hands of a clock are M m inutes apart at (5H ± M) 12 minutes past H O‟ clock. 11 Angle between hands of a clock: a. When the minute hand is behind the hour hand, the angle between the two hands at M M  M minutes past H O‟ clock = 30  H   + degrees. 5 2 

b. When the minute hand is ahead of the hour hand, the angle between the two hands, M  M at M minutes past H O‟ clock = 30   H - degrees. 5  2

6.

The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct 720  6 0  2 4  M time. The clock gains or loses in a day by   minutes. M  11  

137

Solved Examples 1.

At what time between 6 and 7 O‟ clock are hands of a clock together? Sol: Here, H = 6   6 0H  = 6 0  6  3 6 0  3 2 8 11 11 11  11 

8 minutes past 6 O‟ clock. 11 At what time between 7 and 8 O‟ clock will the hands of a clock be at right angle?



2.

Hands of a clock are together at 3 2

Sol: Here, H = 7  (5H ± 15) 1 2 11

 3.

12 11 6 9 = 54 and 2 1 11 11 6 9 Hands of a clock are at right angle at 5 4 minutes past 7 and 2 1 minutes past 7. 11 11 = (5 x 7 ± 15)

Find at what time between 3 and 4 O‟ clock will the hands of a clock be in the same straight line but not together. Sol: Here, H = 3 < 6  (5H + 30) 1 2 = (5 x 3 + 30) 1 2 = 5 4 0  4 9 1 . 11 11 11 11



The hands will be in the same straight line but not together at 4 9

1 minutes past 3 11

O‟ clock. 4.

Find the time between 5 and 6 O‟ clock when the two hands of a clock are 5 minutes apart. Sol: Here, H = 5 and M = 5  (5H ± M) 1 2 = (5 x 5 ± 5) 1 2 = 3 2 8 and 2 1 9 11 11 11 11 8  The hands will be 5 minutes apart at 3 2 past and 2 1 9 past 5 O‟ clock. 11 11

5.

Find the angle between two hands of a clock at 25 minutes past 8 O‟ clock. Sol: Here, H = 8 and M = 25  The required angle = 30 H  M  + M degrees 5 2 

6.

2 5 5  = 30  8   + degrees 5  2  205 = 1 0 2.5o . = 2 The minute hand of a clock overtakes the hour hand at intervals of 45 minutes. How much a day does the clock gain or lose? Sol: Here, M = 45



720   M The clock gains or loses in a day by  1 1  

 6 0  2 4   minutes  M 

720   4 5 =  1 1  

 6 0  2 4    45 

138

720   6 0  2 4 =   4 5   1 1    45  =

225 7200 6 minutes. 32   654 11 11 11

139

Brainstorming

1.

At what time between 3 and 4 „O‟ clock will the hands of a clock at right angles? 1. 3hr, 32 8/11 hrs. min 2. 3.15hr, 32 11/8 hrs. min 3. 3.10hr, 31 8/11 hrs. min 4. None

2.

What is the angle between the two hands of a clock when the clock shows 3hrs 25mins? 1. 45 ½o 2. 46o 3. 46 ½o 4. 47 ½o

3.

By how many degrees does an hour hand move in one quarter of an hour? 1. 5o 2. 7.5o 3. 10o 4. 12.5o

4.

At what time between 6 and 7 o‟clock, are the hands of a clock together? 1. 6 hours, 32 8/11 minutes 2. 6 hours, 33 6/11 minutes 5 3. 6 hours, 34 /11 minutes 4. 6 hours, 29 7/11 minutes

5.

At what time between 5 and 6 are the hands of a clock co-incident? 1. 22 minutes past 5 2. 30 minutes past 5 3. 278/11 minutes past 5 4. 27 3/11 minutes past 5

6.

In accurate clock shows 8 o‟clock in the morning. Though how many degrees will the hour hand rotate when the clock shows 2 o‟clock in the afternoon? 1. 144o 2. 150o 3. 168o 4. 180o

7.

How many times do the hands of a clock point towards each other in a day? 1. 24 2. 20 3. 21 4. 22

8.

At what time between 9 to 10 will the hands of a watch be together? 1. 45 minutes past 9 2. 50 minutes past 9 3. 49 1/11 minutes past 9 4. 48 2/11 minutes past 9

9.

The angle between the minute hand and the hour hand of a clock when the time is 8.30 is ? 1. 80o 2. 75o 3. 60o 4. 105o

10.

At what time between 3 and 4o o‟clock are the hands of a clock in the opposite direction? 1. 3 hours, 48 6/11 minutes 2. 3 hours, 49 1/11 minutes 4 3. 3 hours, 50 /11 minutes 4. 3 hours, 47 2/11 minutes

11.

When the clock shows 3 hours 14 minutes, what is the angle between the hands of the clock? 1. 10o 2. 12o 3. 13o 4. 14o

12.

At what angle the hands of a clock are inclined at 15minutes past 5? 1. 72 ½o 2. 67½o 3. 58 ½o

13.

How many times in a day, the hands of a clock are on a straight line? 1. 22 2. 24 3. 44 4. 48

14.

The angle between the two hands of a clock is 70o , when the hour hand is between 7 and 8 what time does the watch show? 1. 7 hours, 50 10/11 minutes 2. 7 hours, 25 5/11 minutes 8 3. 7 hours, 42 /11 minutes 4. Both (1) & (2)

15.

How many times, the minute hand of clock overlaps with the hour hand from 9.00 am to 4.00 pm in a day? 1. 5 2. 6 3. 7 4. 8

16.

At what time between 5 and 6 o‟clock will the hands of a clock be at an angle of 62o? 1. 5 hours, 17 2/11 minutes 2. 5 hours, 38 6/11 minutes

4. 64o

140

3. 5 hours, 16 minutes

4. Both (2) & (3)

17.

At what time between 7 o‟clock and 8 o‟clock will both the hands of a clock in same line? 1. 7 Past, 5 5/11 minutes 2. 7 Past, 5 11/5 minutes 5 3. 7 Past, 11 /11 minutes 4. None

18.

At what time, in minutes between 3 o‟clock and 4 o‟clock, both the needles will coincide each other? 1. 51/11 2. 124/11 3. 134/11 4. 164/11

19.

At what time between 4 & 5 o‟clock are the hands of a clock in the opposite directions? 1. 52 3/11 minutes, Past 4 o‟clock 2. 54 6/11 minutes, Past 4 o‟clock 7 3. 51 /11 minutes, Past 4 o‟clock 4. 53 9/11 minutes, Past 4 o‟clock

20.

A watch, which gains uniformly, was observed to be 4 minutes slow at 6am on a Monday. On the subsequent Thursday at 7pm it was noticed that the was 6 minutes fast. When did the watch show the correct time? 1. 5pm Tuesday 2. 4pm Tuesday 3. 6pm Tuesday 4. 3pm Tuesday

21.

The angle between the hands of a clock is 20o and the hour hand is in between 2 and 3. What is the time shown by the clock? 1. 2hrs, 7 3/11 minutes 2. 2hrs, 15 5/11 minutes 6 3. 2hrs, 14 /11 minutes 4. Both (1) & (3)

22.

At what time between 9 and 10 o‟clock will be hands of a watch be together? 1. 45 minutes past 9 2. 50 minutes past 9 3. 49 1/11 minutes past 9 4. 49 2/11 minutes past 9

23.

Which of the following can be the time shown and 5 and the angle between the two hands of 1. 16 4/11 minutes past 4 3. 32 8/11 minutes past 4

24.

The minute hand of a clock overtakes the hour hand at intervals of 66 minutes of the correct time. How much in a day does the clock gain or lose? 1. 10 113/121 minutes loss 2. 11 115/121 minutes gain 109 3. 11 /121 minutes loss 4. 10 104/121 minutes gain

25.

At what time between 5.30 and 6 will the hands of a clock be at right angles? 1. 43 5/11 minutes past 5 2. 43 7/11 minutes past 5 3. 40 minutes past 5 4. 45 minutes past 5

26.

A clock which gains uniformly is 2 minutes slow at noon on Monday and is 4minutes 48 seconds fast at 2pm on the following Monday when was it correct? 1. 2 pm on Tuesday 2. 2 pm on Wednesday 3. 3 pm on Thursday 4. 1 pm on Friday

27.

The minute hand of a clock overtakes the hour hand at intervals of 62 minutes of the correct time. How much in a day does the clock gain or lose? 1. 80 80/341 minutes 2. 80 70/341 minutes 90 3. 80 /341 minutes 4. 80 60/341 minutes

28.

A watch which gains uniformly was observed to be 6 minutes slow at 9.00am on a Tuesday and 3 minutes fast at 12.00noon on the subsequent Wednesday. When did the watch show the correct time? 1. 9.00 am on Tuesday 2. 12.00 am on Wednesday 3. 3.00 am on Wednesday 4. 6.00 am on Wednesday

29.

A watch which gains uniformly was observed to be 5minutes slow at 12noon on Monday. It was noticed 10 minutes fast at 6.00 p.m. on the next day. When did the watch show the correct time? 1. 9pm the same day 2. 9.30 pm the same day 3. 10.30 pm, the same day 4. 10 pm the same day

by the clock when the hour hand is in between 4 the clock is 60o? 2. 18 9/11 minutes past 4 4. 36 5/11 minutes past 4

141

30.

At what time between 4 and 4:30 will the hands of a clock be at right angle? 1. 5 5/11 minutes past 4 2. 5 11/5 minutes past 4 11 3. 5 /5 minutes past 5 4. 5 11/5 minutes past 3

31.

A watch showed 5 minutes past 3 o‟clock on Sunday evening when the correct time was 3 o‟clock. It loses uniformly and was observed to be 10minutes slow on the subsequent Tuesday at 9pm when did the watch show the correct time? 1. 8am Monday 2. 10am Monday 3. 7am Monday 4. 9am Monday

32.

A watch showed 10minutes past 6 o‟clock on Thursday morning when the correct time was 6 o‟clock. It loses uniformly and was observed to be 15minutes slow at 8 o‟clock on Saturday morning. When did the watch show the correct time? 1. 1 o‟clock on Friday afternoon 2. 12 o‟clock noon on Friday 3. 4 o‟clock on Friday evening 4. 2 o‟clock on Friday morning

33.

Find the time between 2 and 3 o‟clock at which the minute hand and the hour hand are coincide each other? 1. 10 10/11 minutes past 2 2. 12 10/11 minutes past 2 11 3. 10 /10 minutes past 2 4. 27 3/11 minutes past 2

34.

Find the time between 2 & 3 o‟clock at which the minute hand and hour hand are on the same straight line but are facing opposite directions? 1. 43 7/11 minutes past 2 o‟clock 2. 42 7/11 minutes past 2 o‟clock 3. 43 11/7 minutes past 2 o‟clock 4. None

35.

There are 2 clocks on a wall, both set to show the correct time at 8am. One clock loses two minutes in an hour while the other gains one minute in one hour. By how many minutes do the two clocks differ at 12noon on the same day? 1. 6minutes 2. 9minutes 3. 12minutes 4. 15minutes

36.

A watch which loses uniformly was observed to be 12 minutes fast at 4am on 6th of a month. It showed 20minutes less than the correct time at 6pm on the 10th of the same month when did the watch show the correct time? 1. 9:15am on the 7th 2. 9:05am on the 8th 3. 9:35am on the 9th 4. 9:20am on the 7th

37.

The minute hand of a clock overtakes the hour hand at intervals of 65minnutes of the correct time. How much a day does the clock gain or lose? 1. 10 10/43 minutes in 24hours 2. 10 43/10 minutes in 24hours 10 3. 11 /43 minutes in 24hours 4. None

38.

The hands of a correct clock coincide after every ___________ 6 1. 60 min 2. 61 min 3. 64 min 11

4. 65

5 min 11

39.

If a clock takes 33 seconds to strike 12, how much time will it take to strike 6 ______ 1. 15 2. 33/2 3. 18 4. 6

40.

A man go out in between 5 p.m. and 6 p.m. When he comes back in between 5 p.m. and 7 p.m. he observes that the two hands of a clock have interchanged their position. Find when the man did go out? 7 5 4 2 1. 34 min past 5 2. 36 min past 5 3. 32 min past 5 4. 37 min past 5 13 13 13 13

19. CALENDARS   

Here you mainly deal in finding the day of the week on a particular given date. The process of finding this depends on the number of odd days. Odd days are quite different from the odd numbers.

142

Odd Days: The days more than the complete number of weeks in a given period are called odd days. Ordinary Year: An year that has 365 days is called Ordinary Year. Leap Year: The year which is exactly divisible by 4 (except century) is called a leap year. E.g. 1968, 1972, 1984, 1988 and so on are the examples of Leap Years. 1986, 1990, 1994, 1998, and so on are the examples of no n leap years. Note: The Centuries divisible by 400 are leap years. Important Points: An ordinary year has 365 days = 52 weeks and 1 odd day. A leap year has 366 days = 52 weeks and 2 odd days. Century = 76 Ordinary years + 24 Leap years. Century contain 5 odd days. 200 years contain 3 odd days. 300 years contain 1 odd day. 400 years contain 0 odd days. Last day of a century cannot be Tuesday, Thursday or Saturday. First day of a century must be Monday, Tuesday, Thursday or Saturday. Explanation: 100 years



= 76 ordinary years + 24 leap years = 76 odd days + 24 x 2 odd days = 124 odd days = 17 weeks + 5 days 100 years contain 5 odd days.

No. of odd days in first century = 5  Last day of first century is Friday. No. of odd days in two centuries = 3  Wednesday is the last day. No. of odd days in three centuries = 1  Monday is the last day. No. of odd days in four centuries = 0  Sunday is the last day. Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday. So, the last day of a century should be Sunday, Monday, Wednesday or Friday. Therefore, the first day of a century must be Monday, Tuesday, Thursday or Saturday.

Working Rules: Working rule to find the day of the week on a particular date when ref erence day is given: Step 1: Find the net number of odd days for the period between the reference date and the given date (exclude the reference day but count the given date for counting the number of net odd days). Step 2: The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day). Working rule to find the day of the week on a particular d ate when no reference day is given

143

Step 1: Count the net number of odd days on the given date Step 2: Write: For For For . . . For

0 odd days – Sunday 1 odd day – Monday 2 odd days – Tuesday . . . . . . . . . 6 odd days - Saturday

SOLVED EXAMPLES 1. If 11 th January 1997 was a Sunday then what day of the week was on 10 th January 2000? Sol: Total number of days between 11 th January 1997 and 10 th January 2000 = (365 – 11) in 1997 + 365 in 1998 + 365 in 1999 + 10 days in 2000 = (50 weeks + 4 odd days) + (52 weeks + 1 odd day) + (52 weeks + 1 odd day) + (1 week + 3 odd days) Total number of odd days = 4 + 1 + 1 + 3 = 9 days = 1 week + 2 days Hence, 10 th January, 2000 would be 2 days ahead of Sunday i.e. it was on Tuesday. 2. What day of the week was on 10 th June 2008? Sol: 10 th June 2008 = 2007 years + First 5 months up to May 2008 + 10 days of June 2000 years have 0 odd days. Remaining 7 years has 1 leap year and 6 ordinary years  2 + 6 = 8 odd days So, 2007 years have 8 odd days. No. of odd days from 1 st January 2008 to 31 st May 2008 = 3+1+3+2+3 = 12 10 days of June has 3 odd days. Total number of odd days = 8+12+3 = 23 23 odd days = 3 weeks + 2 odd days. Hence, 10 th June, 2008 was Tuesday.

Brainstorming

1.

The no. of odd days in 400 years is _______. 1. 1 2. 0

3. 3

4. 4

2.

Smt. Indira Gandhi died on 31st October, 1984. The day of the week is _____. 1. Monday 2. Tuesday 3. Wednesday 4. Friday

3.

What will be the day of the week on April 29, 2005? 1. Friday 2. Tuesday 3. Saturday

4. Sunday

4.

The first day of the year 1998 was Wednesday. If the birthday of Raju falls on 25th June, then on which day of 1998 was his birthday? 1. Friday 2. Tuesday 3. Wednesday 4. Sunday

5.

If you are born on 13th April, 1992 which is a Saturday, then which day of the week is your birthday in 1993?

144

1. Sunday 6. 7.

2. Monday

3. Tuesday

4. Wednesday

Find the day of the week on 15th July, 1776? 1. Sunday 2. Monday

3. Tuesday

4. Wednesday

How many odd days are there in 352 days? 1. 1 2. 2

3. 3

4. 4

8.

If 9th March, 1988 is Tuesday, on what day of the week will 9th March, 1989 falls? 1. Sunday 2. Monday 3. Tuesday 4. Wednesday

9.

What will be the day of the week on April 29, 2005? 1. Saturday 2. Monday 3. Tuesday

4. Friday

On which day of the week does 5th June, 2001 fall? 1. Thursday 2. Monday 3. Tuesday

4. Wednesday

10. 11.

If Ram was born on 29th June, 1988 which is Monday, on what day was your birthday in 1989? 1. Friday 2. Monday 3. Tuesday 4. Wednesday

12.

Find the day of the week of April 16th, 1976, if April 16th 1974 was Tuesday? 1. Friday 2. Monday 3. Tuesday 4. Saturday

13.

Find the day of the week on 15th January, 1979? 1. Sunday 2. Monday 3. Friday

4. Wednesday

14.

If 23rd May, 2003 is a Friday, what day of the week will be 23rd December? 1. Tuesday 2. Monday 3. Tuesday 4. Friday

15.

January 3rd, 1992 was a Friday. What day of the week was January 3rd, 1993? 1. Sunday 2. Thursday 3. Friday 4. Saturday

16.

On which day of the week does 28th May, 2003 fall? 1. Monday 2. Tuesday 3. Wednesday

4. Thursday

If your birthday is on 28th May, 1991, then what was the day it was? 1. Tuesday 2. Wednesday 3. Friday

4. Sunday

What day of the week is 1st March, 1990? 1. Sunday 2. Thursday

4. Friday

17. 18. 19. 20.

3. Tuesday

Which year will have the same calendar that of 2007? 1. 2008 2. 2013 3. 2010

4. 2018

Which year will have same calendar that of 2001? 1. 2002 2. 2005 3. 2006

4. 2007

21.

You entered the college first time on 4th June, 2001. What was the day? 1. Monday 2. Tuesday 3. Wednesday 4. Thursday

22.

The no. of odd days in an ordinary year is _______. 1. 1 2. 2 3. 3

4. 4

23.

If 3rd April, 2003 was Thursday then which day of the week is 28th December of the same year? 1. Sunday 2. Monday 3. Tuesday 4. Wednesday

24.

Which among the following is a leap year? 1. 2600 2. 2700

3. 2800

4. 3000

What day of the week was 25th April, 1901? 1. Monday 2. Tuesday

3. Wednesday

4. Thursday

25.

145

26.

Raju‟s brother was born on 7th November, 1984. The day of the week was _____. 1. Monday 2. Tuesday 3. Wednesday 4. Thursday

27.

What was the day of the week on 23rd July, 1970? 1. Monday 2. Tuesday 3. Wednesday

4. Thursday

28.

If today is Monday, then what day of the week will be 427th day from today? 1. Sunday 2. Monday 3. Tuesday 4. Wednesday

29.

Which day of the week was January 29th, 1601? 1. Monday 2. Tuesday 3. Wednesday

4. Thursday

30.

If Sundays are holidays, and in a particular year 2nd April is Sunday. Is 30th September in that year a holiday? 1. Yes 2. No 3. Insufficient Data 4. None

31.

On what dates of September 1972 did Wednesday 1. 6, 13, 20, 27 2. 4, 11, 18, 25 3. 5, 12, 19, 26

4. 3, 10, 17, 24

32.

If 15th February, 1995 was a Wednesday then 15th February, 1994 was on which day? 1. Monday 2. Tuesday 3. Wednesday 4. Thursday

33.

Today is Saturday. After 100 days it will be _______. 1. Monday 2. Tuesday 3. Wednesday

4. Thursday

34.

On which day this year (2008) you celebrated the Independence Day? 1. Monday 2. Friday 3. Wednesday 4. Thursday

35.

4th April, 2002 was Thursday. The day of the week on 4th April, 2003 was ______. 1. Monday 2. Friday 3. Wednesday 4. Thursday

36.

31st March, 2003 was Monday. The day of the week on 31st March, 2004 was _____. 1. Monday 2. Friday 3. Wednesday 4. Thursday

37.

15th August, 1947 was Friday. Find the day of the week on 15th August, 1960. 1. Monday 2. Tuesday 3. Wednesday 4. Thursday

38.

17th May, 1989 was Wednesday. The day of the week on 9th November, 1989 was _______. 1. Monday 2. Tuesday 3. Wednesday 4. Thursday

39.

27th October, 1992 was Tuesday. The day of the week on 29th March, 1993 was ________. 1. Monday 2. Tuesday 3. Wednesday 4. Thursday

40.

The day of your birthday was __________

146

20. MENSURATION: 2D Types of Plane Figures 1. Triangle 2. Quadrilateral 3. Polygon 6. Rectangular Paths 7. Circular paths

4. Circle

5. Sector of a circle

I. Triangle B

(a). Any triangle a, b and c are three sides of the triangle; h is the altitude and AC is the base.

1 1  any side   base  altitude = 2 2 length of  r dropped on that side =

a

c

Perimeter (P) : P = a + b + c = 2s

h

Area (A) : A =

A

C

b

s (s  a) (s  b) (s  c)

B

(b). Equilateral a is the length of each side

a

a

Perimeter (P) : P = 3a Area (A) : A =

3 2 a 4

A

(c). Right-angled

C

a

B

b, c are the lengths of the two legs Perimeter (P) : P = a + b + c = 2s Area (A) : A =

a

c

1  product of two legs 2

90 A

o

b

(d). Isosceles a is the length of two equal sides b is the base BD is the perpendicular dropped on base such that it divides the b base equally AD = CD = 2

C B

a

a

Perimeter (P) : P = 2a+b Area (A) : A =

A

b 4a2  b2 4

b/2

D

C b/2

(e). Right-angled Isosceles Perimeter (P) : P =

2  a  ( 2 +1)

Hypotenuse - h a

a

147

Area (A) : A =

1  (a)2 2

II. Quadrilateral (a). Any Quadrilateral

C

AC is the diagonal = d, DE and BF are two perpendicular drawn on the diagonal (AC) P1, and P2 are the lengths of the two perpendiculars

D

d P1

Perimeter (P) : P = sum of the four sides.

F

1 1 Area (A) : A =  d  (p1+p2) =  any diagonal  2 2 (sum of  rs drawn on that diagonal)

(b). Rectangle

A

B

l

D

l = length b = breadth d = diagonal

P2

E

b

d

C

b

O

Perimeter (P) : P = 2(l + b) = 2(l+ d2  l2 ) = 2(b+Error! Objects cannot be created from editing field codes.) = 2

A

d2  2A

B

l

Area (A): A = l  b = l  d2  l2 = b  Error! Objects cannot be created from editing field codes., when p and A are known and l and b are unknown. The two values of x will give l and b.

D A

(c). Square a = length of side d = diagonal

a O B A

Perimeter (P) : P = 4a = 2d 2 Area (A) : A = a2 =

d2 p2 = 2 16

d C B A

A

D C B A

(d). Rhombus a = each side d1 = one diagonal d2 = another diagonal h = height Perimeter (P) : P = 4a = 2 d12  d2 2

A

a D C B A

d1

a D C B A

a O B A 90o aO BAO

C B A

a O B A

B A o

90 aO BA

B A

a O B A

d2

90o Oa DCO o a BA 90 Oa D DC C BA B A

a D C B Ah

B A

C B A a D C B A

148

d 1 d   d1  d2 = 1  a2   1  2 2 2

Area (A) : A =

2

=

d2 d   a2   2  2 2

2

=a  h

D C B

(e). Trapezium a and b are two parallel sides, h is the height

a A

C B h b a A

1 1  (sum of parallel sides) (a + b)  h = 2 2  (perpendicular distance between parallel sides)

Area (A) : A =

A

(f). Parallelogram

b a A

D C B

b is the base h is the perpendicular distance between the base and its opposite side

Polygon is a n-sided closed figure bounded only by line segments.

C B

h b a A

Area (A) : A = b  h = base  (perpendicular distance between the base and its opposite sides) = 2  area of Δ ABD (or Δ BCD)

III. Polygon

B

A

B

b a A

In a polygon if the internal angle at each vertex is less than 180o then the polygon is a convex polygon, else a concave polygon.

Convex Polygon: i.

Area of a regular polygon =

1  perimeter  r distance from the center of the polygon to 2

any side.

ii. iii. iv. v. vi.

n(n  3) 2 Sum of all interior angles of a polygon = (2n-4)  90o n  2  o Each interior angle of n-sided regular polygon =    180  n  Number of diagonals in a polygon =

Sum of all exterior angles of n-sided regular polygon = 360o 360 Each exterior angle of n-sided regular polygon = n

IV. Circle

B A

O is the center of the circle OA = OC = OB = OD = radius of circle = r AC = BD = diameter of circle = d = 2r

A

O

Circumference (or Perimeter) C = 2 π r = π d

d2 4 If C = circumference, A = area then Area of circle A = π r2 = π

A=

C2 A r and  4π C 2

D C B A

C B A

149

V. Sector of a circle θ

 πr2 3 6 0o θ  2πr Length of the Arc AB = 3 6 0o Area of sector AOB =

O

θ B

A

VI. Rectangular Paths Case - I K

A D N D K N K N K Case - I

L Rectangle C B A D N K

B A D C N B K A D N K

L M N K Path way

M N K

A D N K

K N K

D N K

Case - II L C B A D WN wK M W N K

Pathway is outside the rectangle The length of rectangle AB = l, Breadth BC = b and , Width of path way = W, then Area of Pathway = 2W (l+b+2w) (shaded portion) Case – II Path way is inside the rectangle Area of Pathway = 2W(l+b-2W) (shaded portion)

VII. Circular Pathway A

W

r R W

O

A r W W O

W

C A W

C OAC is a circle of radius = r, there is pathway, outside the circle of width = W Area of circular pathway = π  W (2r+W) When, the pathway is inside the circle, Area of circular pathway = π  W (2r - W)

L M N M K N K

B A D N K

C B A D N K

150

SOLVED EXAMPLES 1.

If three sides of a triangle are 5, 6 and 7 cm respectively, find the area of triangle. Sol: Area of  = Now, s =

 Area

s(s  a)(s b)(s c) abc 56 7 = 9  2 2 = =

2.

9  (9  5)(9  6)(9 7)

9 4 3 2

2

216  6 6 cm .

ABC is an equilateral triangle of side 24 cm. Find the in radius of the triangle. Sol: In a equilateral triangle, the altitude, median and perpendicular are equal.

 AD =

3 /2 x 24 = 12 3

GD (in radius) = 1/3 x 12 3 = 4 3 cm 3.

The base and other side of an isosceles triangle is 10 and 13 cm respectively. Find its area.

b 4a2  b2 4 base b = 10 Other side a = 13

Sol: Area of Isosceles  = Given,

Area (A)

4.

10 10 4  (13)2  102  676  100 4 4 10 =  24 = 60 cm 2 . 4

=

In a right-angled triangle, the length of two legs are 12 and 5 cm. Find the length of hypotenuse and its area. Sol: In a right angled triangle, (Hypotenuse) 2 = (one leg) 2 + (other leg) 2 = 12 2 + 5 2

 Hypotenuse

=

122  52 =

169 = 13 cm.

In a right angled triangle, 1 1 Area =  (leg)1  (leg)2   12  5 = 30 cm 2 . 2 2 5.

If the perimeter and diagonal of a rectangle and 14 and 15 cm respectively. Find its area. Sol: In a rectangle,

(Perimeter)2 (14)2 = (diagonal)2 + 2 x Area ; = (5) 2 + 2 x Area 4 4

 2 x Area =

196 49  25 - 25  Area = = 12 cm 2 . 4 2

151

6.

Find the length of the diagonal and the perimeter of a square plot if its area is 900 square metres. Sol: In a square, A =

d2 p2  2 16

 (Diagonal) 2 = 2 x Area = 900  Diagonal (d) =

2  900  30  2 = 42.42 metres

(Perimeter) 2 = 16 x Area = 16 x 900

 Perimeter (P) = 7.

16  900 = 120 metres.

A field in the shape of a rhombus has the distances between pairs of opposite vertices as 14 m and 48 m. What is the cost (in rupees) of fencing the field at Rs.20 per metre? Sol: The diagonals are 14 m and 48 m 2

Sides of rhombus =

2

 14   48       625 = 25  2   2 

Perimeter of rhombus = 4 x 25 = 100 m. Cost of fencing the field = 100 x 20 = Rs.2000 8.

In a trapezium, the length of parallel sides are 20 and 25 metres respectively and the perpendicular distance between the parallel sides is 12 metres. Find the area of trapezium. Sol: One parallel side a = 20 metres. Second parallel side b = 25 metres. Height (perpendicular distance between a and b) = 12 metres. Area =

9.

1 1 (a  b)  h  (20  25)  12 = 270 m 2 . 2 2

The distance between a pair of opposite vertices of a quadrilateral is 32 units. The lengths of the perpendiculars drawn on to this diagonal from the other two vertices are 4 1/3 units and 6 2/3 units respectively. Find the area (in sq units) of the quadrilateral?

 13 20   Sol: Area of quadrilateral = 1/2 x 32 x   = 178 sq units. 3   3 A

10.

C

B

D

In the above parallelogram ABCD, A = x + 30 o and D = x – 40 o , what is the measure of DCB ? Sol: In a parallelogram, sum of adjacent angles is equal to 180 o

 x + 30 + x – 40 = 180  x = 95 o DAB = x + 30 = 95 + 30 = 125 o  DCB = DAB = 125 o (opposite angles of a parallelogram are equal)

152

11.

In a circle of radius 49 cm, an arc subtends an angle of 36 o at the centre. Find the length of the arc and the area of the sector. Sol:

12.

Length of the arc

=

2 rθ 2  22  49  36 = 30.8 cm  360 7  360

Area of the sector

=

r2θ 22  49  49  36 = 754.6 cm 2  360 7  360

A rectangular plot of dimensions 13 m x 17 m is surrounded by a garden of w idth 5 m. What is the area (in sq m) the garden? Sol: Let ABCD be the rectangular plot of given dimension. The shaded part is the surrounding garden. Now, the plot ABCD together with the garden forms another rectangular form PQRS. Dimensions of PQRS, as can be seen from the diagram, are: Length PQ = width of garden + AB + width of garden = 5 + 17 + 5 = 27 m Similarly, breadth = PS = 5 + 13 + 5 = 23 m Area of garden = Area of PQRS – Area of ABCD = (27 x 23) – (17 x 13) = 621 – 221 = 440 sq m.

13.

There is a rectangular field of length 100 m and breadth 40 m. A carpet of 2 m width is to be spread from the centre of each side to the opposite side. What is the area of the carpet? Sol: Area of the carpet ABCD Area of the carpet EFGH But the common area of two carpets So, area of the carpet

14.

= = = =

40 m x 2 m = 80 m 2 100 m x 2 m = 200 m 2 2 x 2 = 4m 2 200 + 80 – 4 = 276 m 2

There is an equilateral triangle of which each side is 3 m. With all the three vertices as centres, circles with radius 1.5 cm are described (i) Calculate the area common to all the circles and the triangle. (ii) Find the area of the remaining portion of the triangle. Sol: (i) Area of each sector =

1    r2 6

So area common to the all the circles and triangle = 3  =

1 1    r 2   r2 6 2

1 22   1.5  1.5 = 3.53 m 2 2 7

(ii) Area of the shaded portion = Area of the triangle – Area common to the triangle and the circles 3 2 3 9 3 2 But area of the triangle = m a  (3)2  4 4 4 So area of the shaded portion =

9 3 2 m – 3.53 m 2 = 3.89 m 2 – 3.53 m 2 = 0.36 m 2 4

Brainstorming 1.

Find the area of a triangle having sides 3 m, 4 m and 5 m.

153

1. 60 sq m 2. 3.

4. 6 sq m 4. None

Find the area of an equilateral triangle each of whose sides measures 6 cm. 2. 3 3 sq cm

Length of the side of an equilateral triangle is 1. 2 cm

5.

3. 12 sq m

Find the area of a triangle whose base is 4.6m and height is 67 cm. 1. 154.10 sq m 2. 15410 sq m 3. 15.410 sq m

1. 36 sq cm 4.

2. 10 sq m

2. 4 cm

3. 9 3 sq cm

4 3

4. 12 sq cm

cm. Find its height.

3. 6 cm

4. None

Height of an equilateral triangle is 4 3 cm. Find its area. 1. 4 3 sq cm

2. 2 3 sq cm

3. 16 3 sq cm

4. 8 3 sq cm

6.

An isosceles right-angled triangle has two equal sides of length 6 m each. Find its area 1. 8 sq m 2. 36 sq m 3. 18 sq m 4. None

7.

The perimeter of an isosceles triangle is 80 cm. If the length of the equal sides is given by 0.15 m, find the length of the base. 1. 40 m 2. 50 m 3. 12 m 4. 90.5 m

8.

The perimeter of an isosceles triangle is 42 cm. If the base is 16 cm, find the length of equal sides. 1. 13 cm 2. 8 cm 3. 21 cm 4. 29 cm

9.

The two adjacent sides of a parallelogram are 5 m and 6 m respectively, and if the diagonal connecting the ends is 9 m, find the area of the parallelogram (approximately). 1. 29 sq m 2. 28 sq m 3. 58 sq m 4. 50 sq m

10.

Find the area of a quadrilateral of whose diagonal is 38 cm long and the lengths of perpendiculars from the other two vertices are 31 cm and 19 cm, respectively. 1. 950 sq cm 2. 475 sq cm 3. 138 sq cm 4. 276 sq cm

11.

Find the area of a parallelogram whose two adjacent sides are 130 m and 140 m and one of the diagonals is 150 m long. 1. 8400 sq cm 2. 16,800 sq cm 3. 2100 sq cm 4. None

12.

Find the diagonal of a rectangle whose sides are 8 cm and 6 cm. 1. 14 cm 2. 5 cm 3. 20 cm

4. 10 cm

Find the perimeter of a rectangle of length 12 m and breadth 6 m. 1. 18 m 2. 72 m 3. 36 m

4. 144 m

13. 14.

Calculate the area of a rectangular field whose length is 12.5 cm and breadth is 8 cm. 1. 10 sq cm 2. 100 sq cm 3. 200 sq cm 4. 1 sq cm

15.

Calculate the area of a rectangular field whose one side is 16 cm and the diagonal is 20 cm. 1. 192 sq cm 2. 96 sq cm 3. 294 sq cm 4. 72 sq cm

16.

A rectangular carpet has an area of 120 sq m and perimeter of 46 m. Find the length of its diagonal. 1. 34 m 2. 51 m 3. 93 m 4. 17 m

17.

The perimeter of a rectangle is 82 cm and its area is 400 sq m. Find the length of the rectangle. 1. 8 m 2. 16 m 3. 32 m 4. 64 m

18.

If the area of a square field be 6050 sq m, find the length of its diagonal. 1. 220 m 2. 110 m 3. 55 m 4. None

19.

Find the area of a square with perimeter 48 m.

154

1. 288 sq m 20.

2. 72 sq m

3. 144 sq m

4. 96 sq m

Find the diagonal of a square field whose side is of 6 m length. 1. 12 2 m

2. 6 2 m

3.

2m

4. 3 2 m

21.

Perimeter of a square field is 16 2 cm. Find the length of its diagonal. 1. 16 cm 2. 4 cm 3. 8 cm 4. 64 cm

22.

The area of a rhombus is 156 sq m. If one of its diagonals is 13 m, find the length of the other diagonal. 1. 12 m 2. 6 m 3. 48 m 4. 24 m

23.

Find the area of a rhombus whose one side is 13 cm and one diagonal is 24 cm. 1. 60 sq cm 2. 120 sq cm 3. 240 sq cm 4. 74 sq cm

24.

If the perimeter of a rhombus is 73 cm and one of its diagonals is 27.5 cm, find the other diagonal and the area of the rhombus. 1. 24 cm, 330 sq cm 2. 20 cm, 115 sq cm 3. 30 cm, 660.8 sq cm 4. 40 cm, 100.5 sq cm

25.

In a rhombus, the lengths of two diagonals are 18 m and 24 m. Find its perimeter. 1. 15 m 2. 30 m 3. 60 m 4. 120 m

26.

The diagonally of Rhombus are 12 cm and 5 cm respectively. Find the side of the Rhombus. 1. 5 cm 2. 6.5 cm 3. 6 cm 4. 8.5 cm

27.

What is the radius of a circular plot whose circumference is 176 m? 1. 14 m 2. 56 m 3. 88 m

4. 28 m

28.

A circular plot covers an area of 154 sq m. How much wire is required for fencing the plot? 1. 44 m 2. 22 m 3. 88 m 4. 77 m

29.

Find the area of sector of a circle whose radius is 10 cm and the angle at the center is 36o. 3 7 7 3 1. 30 sq cm 2. 31 sq cm 3. 30 sq cm 4. 31 sq cm 7 3 3 7 Find the area of sector of a circle whose radius is 12 cm and the length of the arc is 20 cm. 1. 60 sq cm 2. 240 sq cm 3. 120 sq cm 4. 64 sq cm

30.

31.

Find the side of a regular hexagon whole area is 30 3 sq cm. 1.

32.

5 cm

2. 2 5 cm

3. 3 5 cm

Find the area of a regular octagon whole side measures 1. 4( 2 +1) sq cm

2. 8( 2 +1) sq cm

4. None

8 cm.

3. 16 ( 2 +1) sq cm

4. ( 2 +1) sq cm

33.

Find the sum of interior angles of a regular polygon of 12 sides. Also, find the value of each interior angle. 5Π 5Π 6Π 1. 10 Π , 2. ∏, 3. 8 Π , 4. None 6 6 5

34.

Find the sum of all the exterior angles of a regular polygon of 10 sides. Also, find the value of each exterior angle. Π Π Π 1. Π , 2. 2 Π , 3. 3 Π , 4. None 5 5 5 The length and breadth of a rectangle are increased by 20% and 5%, respectively. Find the percentage increase in its area. 1. 25% 2. 26% 3. 13% 4. 15%

35.

36.

Two poles 15 m and 30 m high stand up right in a play ground if their feet be 36 cm a part find the distance between their tops. 1. 41 cm 2. 39 cm 3. 29 cm 4. 42 cm

155

37.

If all the sides and the diagonals of a square are increased by 8% each, then find the percentage increase in its perimeter? 1. 8% 2. 6% 3. 1% 4. None

38.

Ratio of the areas of two squares is 16 : 9. Find the ratio of their diagonals. 1. 2 : 4 2. 9 : 16 3. 4 : 3 4. 64 : 9

39.

The diagonal of a square is doubled. How many times will the area of the new square become? 1. 2 times 2. 4 times 3. 6 times 4. 8 times

40.

How many meters of a carpet 12 cm wide will be required to cover the floor of a room which is 600 cm long and 420 cm broad? Also, calculate the amount required in carpeting the floor if the cost of carpet is Rs.15 per meter. 1. Rs.3150 2. Rs.9000 3. Rs.1800 4. Rs.10,800

41.

A hall of length 24 cm and breadth 20 cm is to be paved with equal square tiles. What will be the size of the largest tile so that the tiles exactly fit and also find the number of tiles required? 1. 60 2. 30 3. 480 4. 120

42.

A rectangular park 18 m x 12 m, is surrounded by a path 4 m wide. Find the area of the path. 1. 304 sq m 2. 152 sq m 3. 608 sq m 4. 864 sq m

43.

A park is square in shape with side 18 m. Find the area of the pavement 3 m wide to be laid all around it on its inside. 1. 360 sq m 2. 180 sq m 3. 90 sq m 4. None

44.

A playground measures 27 m x 13 m. From the center of each side a path 2 m wide goes across to the center of the opposite side. Calculate the area of the path and the cost of constructing it at Rs.4 per sq m. 1. Rs.101 2. Rs.404 3. Rs.202 4. Rs.304

45.

A square field is surrounded by a path 2 m wide on its outside. The area of the path is 72 sq m. What is the area of the field? 1. 15 sq cm 2. 121 sq m 3. 36 sq m 4. 18 sq m

46.

A circular park of radius 22 m has a path of width 1.4 m around it on its inside. Find the area of the path. 1. 178.45 sq m 2. 187.45 sq m 3. 187.54 sq m 4. None

47.

If the area of a square is 33 sq cm, then find the area of the circle formed by the same perimeter. 1. 21 sq cm 2. 66 sq cm 3. 33 sq cm 4. 42 sq cm

48.

Find the area of largest circle inscribed in a square of side 112 cm. 1. 4928 sq cm 2. 8856 sq cm 3. 9856 sq cm

49.

Find the side of the square inscribed in a circle whose circumference is 308 cm. 1. 25 2 cm

50.

1) 4 2) 2 3) 3 4) 1 5) 3 6) 3 7) 2 8) 1 9) 2 10) 1

4. None

2. 49 2 cm

3. 36 2 cm

4. 16 2 cm

The diameter of a wheel is 2 cm. If it rolls forward covering 10 revolutions, find the distance traveled by it. 1. 62.8 cm 2. 31.4 cm 3. 15.7 cm 4. 58.2 cm 11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

2 4 3 2 1 4 2 2 3 2

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

3 4 1 1 3 2 4 1 4 3

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

2 3 1 2 2 2 1 3 2 1

41) 42) 43) 44) 45) 46) 47) 48) 49) 50)

4 1 2 4 2 2 4 3 2 1

156

21. MENSURATION : 3D Cuboid : b

A right prism with a rectangular base is called a Cuboid. The sides of the base are length (l) and breadth (b). The height is h.

h

Lateral Surface Area = 2h(l + b) Total Surface Area = 2h(l + b) + 2lb = 2(lb + bh + hl) Longest diagonal Volume

l

= l2  b2  h2 = lbh

Cube:

a

If the length, breadth and height of a cuboid are all equal, it is called a cube. Then, if edge of the cube = a Longest diagonal Lateral Surface Area Total surface Area Volume

a

a

= 3a = 6a 2 = 6a 2 = a3

Cylinder :

r

A cylinder can be considered to be a right prism except that polygons a cylinder has identical circles for its top and base lateral surface also called curved surface, instead of several The basic measurements are the radius of the base (or top)

instead of identical and it has a single rectangular surfaces. r and the height h.

h

Curved Surface (Lateral Surface Area) = 2 π rh Total surface Area = 2 π rh+2 π r 2 = 2 π r(h + r) Volume = π r2h

Hollow Cylinder: The cross section of a hollow cylinder is a ring. Volume of the material of a hollow cylinder = π h(R 2 -r 2 ) Here R is outer radius and r is inner radius of the hollow cylinder.

h r R

Cone:

A cone can be formed from the sector of a circle by rolling it and joining together its two straight edges. If r is the radius of the cone, and R is the radius of the sector of angle θ , then θ 1. r = R 360 2. Relation between r, l and h. (the radius, the slant height and height) is l 2 = h 2 +r 2 3. Curved Surface area of Cone = prl 4. Total Surface Area = π rl + π r 2 = π r(l + r) 1 π r2h 5. Volume = 3

h

r

Sphere: r

157

All points on the surface of a sphere are at the same distance from the center of the sphere. This distance is called the radius, r. Surface Area of Sphere = 4 π r 2 4 Volume of a Sphere = π r3 3 The sphere has only one surface and hence only one surface area.

Hemisphere: The radius is r. r

Curved Surface Area = 2 π r 2 Total Surface Area = 2 π r 2 + π r 2 = 3 π r 2 1 4 3 2 Volume =  πr  π r 3 2 3 3

SOLVED EXAMPLES 1.

A cuboid is 20 m x 10 m x 8 m. Find the length of diagonal, surface area and volume. Sol: In a cuboid , Diagonal d Surface are S Volume

2.

= l2  b2  h2 = 202  102  82 = 23.75 = 2 (20 x 10 + 10 x 8 + 8 x 20) = 880 m2 = l x b x h = 20 x 10 x 8 = 1600 m3.

A cube has edge 12 m. Find its length of diagonal, surface area and volume. Sol: In a cube Diagonal d Surface area S Volume V

3.

= Edge x 3 = 12 x 3 = 20.78 m = 6 x (Edge)2 = 6 x (12)2 = 864 m2 = Edge3 = (12)3 = 1728 m3.

The base of a right prism is a regular pentagon of side 18 cm. If the height of the prism is 2/3rd of the side of the base, how much is the lateral surface area (in sq cm) of the prism? Sol: Perimeter of the base of the prism Lateral surface area of a right prism

4.

= number of sides x length of each side = 5 x 18 = 90 cm. = (Base perimeter) x (height) 2  = (90)   18  = 1080 sq cm 3  

If the radius of a sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area. Sol: Let original radius = R. Then new radius =

150 3R . R 100 2 3

Original volume =

4  3R  4 3 R3 , New volume     9R . 3  2  3

 19  3  R3   100  % = 237.5% Increase % in volume =  4R3  6 

158

2

 3R  2 Original surface area = 4  R2. New surface area = 4    9R .  2   5R2  Increase % in surface area =   100  % = 125%.  4R2    5.

A cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm. Find the radius of the cone? Sol: Let the radius of the cone be r cm. Then,

6.

1 8  8  2  3    r2  6    8  8  2  r2    = 64  r = 8 cm. 3 6  

A brick measures 10 cm x 5 cm x 3 cm. How many bricks will be required for a wall of 100 metre long 6 metre high and 1.5 metre thick? Sol: Volume of the wall = 100 m x 6 m x 1.5 m = 900 m3 Volume of one brick =

1 1 3 3 m m m m3 10 20 100 20000

900 m3 = 60,00,000 3 3 m3 20000

 No. of bricks required =

7.

What is the maximum length of a pencil which can be inscribed in a box of length 24 units, breadth 3 units and height 4 units? Sol: Maximum length in a cuboid is its diagonal

 Length of main diagonal is = 8.

length2  breadth2  height2

242  32  22 

576  9  4  589 units

The height and base-radius of a right circular cone are 10 cm and 24 cm respectively. What is the area of the curved surface area (in sq cm) if the cone? Sol: Curved surface area of a cone =  rl, R and l being radius and slant height. It is given that height h = 10 cm and radius = 24 cm.

L2 = h2 + r2 = 102 +242  l = 26 (10 and 24 are in the ratio of 5 : 12; hence l will be the 2 x 13 = 26 Hence, curved surface area =  rl =  x 24 x 26 624  sq cm.

Brainstorming 1.

For any regular solid, No. of faces + No. of vertices = No. of _______ + 2 1. Faces 2. Vertices 3. Edges 4. Sides

2.

Find the volume of a cuboid whose dimensions are 20 m, 15 m and 3 m. 1. 900 m3 2. 800 m3 3. 700 m3 4. 450 m3

159

3.

Find the length of the longest bamboo stick that can be placed in a room of 16 m long, 11 m broad and 8 m high. 1. 441 m 2. 21 m 3. 210 m 4. 35 m

4.

The area of a side of a box is 150 sq cm. The area of the other side of the box is 20 sq cm. If the area of the upper surface of the box is 30 sq cm, then find the volume of the box. 1. 300 m3 2. 200 m3 3. 900 m3 4. 900 m3

5.

Cube is a special type of cuboid in which each face is a ________. 1. Rectangle 2. Rhombus 3. Square

4. Circle

6.

Find the volume of a cube whose sides measure 3 cm. 1. 12 cm3 2. 6 cm3 3. 9 cm3

4. 27 cm3

7.

Find the volume of a cube whose surface area is 384 sq cm. 1. 512 cm3 2. 192 cm3 3. 128 cm3

4. 64 cm3

8.

The diagonal of a cube is 4 3 cm. Find its total surface area and volume. 1. 16 sq cm, 64 cm3 2. 96 sq cm, 16 cm3 3. 6 sq cm, 4 cm3 4. 96 sq cm, 64 cm3

9.

If the volumes of two cubical blocks are in the ratio of 27 : 8, what will be the ratio of their sides? 1. 2 : 3 2. 3 : 2 3. 3 : 8 4. 27 : 2

10.

The volumes of two cubes are in the ratio of 1 : 27. Find the ratio of their surface areas. 1. 9 : 1 2. 1 : 3 3. 2 : 3 4. 1 : 9

11.

The sides of two cubes are in the ratio of 3 : 4. Find the ratio of their surface areas. 1. 9 : 16 2. 9 : 4 3. 3 : 16 4. 16 : 9

12.

A right circular cylinder is a solid with circular ends of equal radius and the line joining their centres _____________ to them. 1. Parallel 2. Perpendicular 3. Both 1 and 2 4. None

13.

The diameter of the base of a right circular cylinder is 42 cm and the height is 10 cm. Find the volume of the cylinder. 1. 1386 cm3 2. 13860 cm3 3. 13680 cm3 4. 8610 cm3

14.

Find the area of the curved surface of the right circular cylinder if the diameter of the base is 28 cm and the height is 20 cm. 1. 1760 cm2 2. 176 cm2 3. 560 cm2 4. 880 cm2

15.

A cylinder of height 20 cm has base of radius 8 cm. Find the total surface area of the cylinder. 1. 1480 sq cm 2. 1408 sq cm 3. 160 sq cm 4. 704 sq cm

16.

Two circular cylinders of equal heights have their radii in the ratio of 3 : 4. Find the ratio of their volumes. 1. 16 : 9 2. 4 : 3 3. 9 : 16 4. 27 : 64

17.

Two circular cylinders of equal heights have their curved surface areas in the ratio of 4 : 5. Find the ratio of their volumes. 1. 16 : 25 2. 5 : 4 3. 25 : 16 4. 4 : 25

18.

Two circular cylinders of equal curved surface areas have their heights in the ratio of 5 : 7. Find the ratio of their volumes. 1. 25 : 49 2. 7 : 25 3. 5 : 49 4. 7 : 5

19.

The heights and curved surface areas of two right circular cylinders are in the ratio 3 : 5 and 7 : 10 respectively. Find the ratio of their radii. 1. 6 : 7 2. 7 : 6 3. 21 : 50 4. 50 : 21

160

20.

The radii of two right circular cylinders are in the ratio of 2 : 3 and their curved surface areas are in the ratio of 4 : 5. Find the ratio of their heights. 1. 15 : 8 2. 6 : 5 3. 8 : 15 4. 5 : 6

21.

A right circular cone is a solid obtained by rotating a __________ triangle around its height. 1. Right-angled 2. Acute-angled 3. Obtuse-angled 4. All the three

22.

Find the slant height of a cone whose volume is 8624 cm3 and radius of the base is 14 cm. 1. 6 cm

23.

2.

1 9 6 cm

3. 42 cm

4.

1 9 6 0cm

A tent is of diameter 6 m at the base and its height is 4 m. Find the slant height. 1. 5 m

2. 6 m

3. 10 m

4.

42m

24.

A tent is of radius 3 m and its slant height is 5 m. Find the canvas required in sq m. 1. 15 sq m 2. 8 sq m 3. 47.10 sq m 4. 42.70 sq m

25.

How many persons can the tent accommodate, at the most, if each person requires 18m3 of air, if diameter is 12 m at the base and its height is 8 m? 1. 15 2. 16 3. 17 4. 18

26.

Two right circular cones of equal volumes have their heights in the ratio of 25 : 16. Find the ratio of their radii. 1. 5 : 4 2. 4 : 5 3. 125 : 64 4. None

27.

The volumes of two cones are in the ratio of 4 : 5 and their heights are in the ratio of 16 : 125. Find the ratio of their radii. 1. 25 : 8 2. 5 : 2 3. 25 : 4 4. 8 : 25

28.

The heights of two cones are in the ratio of 3 : 2 and their radii are in the ratio of 5 : 4. Find the ratio of their volumes. 1. 75 : 32 2. 45 : 16 3. 15 : 8 4. 225 : 64

29.

The volumes of two cones are in the ratio of 2 : 7 and their diameters are in the ratio of 3 : 5. Find the ratio of their heights. 1. 63 : 50 2. 50 : 63 3. 6 : 35 4. 10 : 21

30.

The radius of a sphere is 21 cm. Find its volume. 1. 38808 cm3 2. 38880 cm3 3. 38088 cm3

31.

Find the curved surface area of the hemisphere, if its diameter is 56 cm. 1. 2464 sq cm 2. 9856 sq cm 3. 4829 sq cm 4. 4928 sq cm

32.

The length, breadth and height of a Cuboid are increased by 25%, 20% and 33 1/3%. Find the percentage increase in the volume. 1. 78 1/3% 2. 100% 3. 78% 4. None

33.

The sides of a cube are decreased by 20% each. Find the percentage decrease in the volume. 1. 48.8% 2. 40% 3. 22% 4. None

34.

The diameter of a sphere is increased by 20%. Find the percentage increase in its volume. 1. 172.8% 2. 800% 3. 72.8% 4. None

35.

The radius of a right circular cylinder is decreased by 5% and the height is increased by 10%. What is the percentage change in its volume? 1. 0.725% decrease 2. 0.725% increase 3. 7.25% decrease 4. None

36.

The radius of a hemisphere is decreased by 10%. Find the percentage decrease in the surface area. 1. 18% 2. 19% 3. 81% 4. None

37.

Each edge of a cube is increased by 30%. What is the percentage increase in its surface area?

4. 30808 cm3

161

1. 96%

2. 31%

3. 169%

4. 69%

38.

The radius and height of a cylinder are increased by 20% and 15%. Find the percentage increase in the surface area. 1. 38% 2. 35% 3. 5% 4. 45%

39.

Find the number of lead balls of radius 1 cm each that can be made from a sphere of radius 5 cm. 1. 25 2. 5 3. 125 4. None

40.

A sphere of maximum possible volume is carved out of a cubical wooden block. Approximately what percentage of the wood is unused? 1. 48% 2. 50% 3. 96% 4. None

1) 3 2) 1 3) 2 4) 4 5) 3 6) 4 7) 1 8) 4 9) 2 10) 4

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

1 2 2 1 2 3 1 4 2 2

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

1 4 1 3 2 2 2 2 2 1

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

4 2 1 3 1 2 4 1 3 1

162

13. Time & Distance

14. Problems on Trains

15. Boats & streams

1) 1 2) 3 3) 2 4) 1 5) 2 6) 2 7) 4 8) 3 9) 3 10) 1

11) 3 12) 1 13)2 14) 2 15) 2 16) 3 17) 3 18)1 19) 2 20) 2

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

16. Simple Interest

4 2 1 4 2 3 2 4 4 4

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

2 2 4 1 4 1 2 2 2 4

1) 2 2) 1 3) 3 4) 4 5) 2 6) 1 7) 4 8) 4 9) 2 10) 4

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

17. Compound Interest 1) 2 2) 1 3) 4 4) 1 5) 2 6) 1 7) 1 8) 1 9) 1 10) 1

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

3 1 2 3 3 1 2 1 2 1

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

3 2 2 1 3 3 3 2 3 1

1 3 2 1 2 3 1 3 1 1

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

1 2 3 1 2 1 1 2 3 2

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

3 1 4 2 1 2 3 1 1 2

41) 42) 43) 44) 45)

18. Clocks 31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

2 2 3 3 1 4 2 4 2 1

1) 1 2) 4 3) 2 4) 1 5) 4 6) 4 7) 4 8) 3 9) 2 10) 2

11) 12) 13) 14) 15) 16) 17) 18) 19) 20)

3 2 3 4 2 4 1 4 2 2

21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

4 3 3 3 2 2 1 3 4 1

31) 32) 33) 34) 35) 36) 37) 38) 39) 40)

4 4 1 1 3 1 1 4 1 3

1 3 1 2 2

163

19. Calendar

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