Some Useful Quicker Methods
Short Description
Tricks...
Description
QUICKER METHODS
.
Number System (i) (i )
Remainder Rule is applied to find the remainder for the smaller division, when the same number is divided by the two different divisors such that one divisor is a multiple of the other divisor and also the remainder for the greater divisor is known. If the remainder for the greater divisor = r and the smaller divisor = d , then the remainder rule states that, when r > d the required remainder for the smaller divisor will be the remainder found out by dividing the ‘ r ’ by ‘ d ’, d ’, and when r < d , then the required remainder is ‘ r r ’ itself.
(ii)
If two different numbers a and b , on being divided by the same divisor leave remainders r 1 and r 2 respectively, then their sum (a + b), if divided by the same divisor will leave remainder R as given below: R = (r 1 + r 2) – Divisor = (Sum of remainders) – Divisor the above equation, then then Note : If R becomes negative in the the required remainder will be the sum of the remainders. That is, the required remainder = sum of remainders.
(iii) When two numbers after being divided by the same divisor leave the same remainder, then the difference of those two numbers must be exactly divisible by the same divisor.
(iv)
If a given number is divided successively by the different factors of the divisor leaving remainders r 1, r 2 and r 3 respectively, then the true remainder (ie remainder when the number is divided by the divisor) can be obtained by using the following formula: True remainder = (First re mainder) + (Second remainder × First divisor) + (Third remainder × First divisor × Second divisor).
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M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(v) (v )
When (x + + 1)n is divided by x , then the remainder is always 1; where x and and n are are natural numbers.
(vi)
When (x – – 1)n is divided by x , then the remainder will be 1, if n is an even natural number. But the remainder will be (x – – 1), if n is is an odd natural number.
(vii) The sum of the digits of two-digit numbe r is S. If the digits are reversed, the number is decreased by N, then the number is as given below: N 1 Number = 5 S + 9 2
Sum of digi digits ts = 5 Sum
N S 2
Decrease 1 + 2 9
Decrease Sum of digi digits ts Sum 9
Note : If after reversing the digits, the number is increased by N, then the number is as given below: 1 S Number = 5 S + 9 9 2
K KUNDAN
Sum of digi digits ts = 5 Sum
Decrease 1 + 2 9
Decrease 1 Sum of digi digits ts Sum + 2 9
(viii) When the difference between two-digit number and the number obtained by interchanging the digits is given, then the difference of the two digits of the two-digit number is as given below: Differe erence in orig riginal and and Difference of two digits =
int int erch erchan anged ged number numbers s 9
Note: We cannot get the sum of two digits of the given two-digit numbers. (ix)
A number on being divided by d 1 and d 2 successively leaves the remainders r 1 and r 2 respectively. If the numbe r is divided
by d 1 × d 2, then the remainder is given by ( d 1 × r 2 + r 1).
(x) (x )
When the sum of two-digit number and the number obtained by interchanging the digits number is as given below: Sum of two digits =
Sum of orig origin ina al and intercha terchang nged ed numb number ers s 11
Highest Common Factor (i )
To find the greate st numbe r that will exactly divide x , y and z . Required number = HCF of x , y and and z
num ber that will divide x , y and z (ii) To find the gre atest number leaving remainders a , b and and c respectively. Required number = HCF of ( x – – a ), ), ( y – – b ) and (z – – c ) num ber that will divide x , y and z (iii) To find the gre atest number leaving the same remainder ‘ r r’ in each case. Required number = HCF of ( x – – r ), ), ( y – – r ) and ( z – – r ) and z living living (iv) To find the greatest number that will divide x , y and the same remainder in each case. Required number = HCF of |( x – – y )|, )|, |(y – – z )| )| and |(z – – x )| )|
(v ) To find the all possible numbers, when the product of two numbers and their HCF are given, we follow the following steps: Step I:
Find the value of
Pr oduct . (HCF)2
possible e pairs pairs of of value value got in step step I. Step II: Find the possibl the pair of prime factors factors Step III : Multiply the HCF with the obtained in step II.
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Least Common Multiple (i) (i )
To find the least number which is exactly divisible by x , y and z . Required number = LCM of x , y and and z
(ii) To find the least number which w hen divided by x , y and z leaves the remainders a , b and c respectively. It is always observed that, ( x – a ) = (y – – b ) = (z – – c c ) = K (say) Required number = (LCM of , and x y z ) – K and z (iii) To find the least num ber which, when divided by x , y and leaves the same remainder r in each case. Required number = (LCM of x, y and z) + r -digit greatest number which, when divided by (iv) To find the n -digit and z , x , y and (1) leaves no remainder (ie exactly divisible) Following step-wise methods are adopted: and z = = L Step I: LCM of x , y and -digit greatest number ( Step II: L) n -digit Remainder (R) -digit greatest number – R Step III: Required number = n -digit
(2) leaves remainder K in each case Following step-wise method is adopted: and z = = L Step I: LCM of x , y and -digit greatest number ( Step II: L) n -digit Remainder (R) -digit greatest number Step III: Required number = ( n -digit – R) + K
-digit smallest number which, when divided by (v) (v ) To find the n -digit and z . x , y and (1) leaves no remainder (ie exactly divisible) Following steps are followed: and z = = L Step I: LCM of x , y and -digit smallest number ( Step II: L) n -digit Remainder (R) -digit smallest Step III : The required number = n -digit number + (L – R) (2) leaves remainder K in each case. First two steps are the same as in the case of (1).
Step III : Required number = n -digit -digit smallest number + (L – R) + K and (vi) To find the least number which on being divided by x, y and leaves in each case a remiander R, but when divided by N z leaves leaves no remainder, following step-wise methods are adopted: say (L). Step I: Find the LCM of x , y and z say
Step II: Required number will be in the form of (LK + R); where K is a positive integer. Step III : N) L (Quotient (Q) ( Remainder (R 0) L = N × Q + R 0 Now, put the vaue of L into the expression obtained in step II. required number will be in the form of (N × Q + R 0) K + R or, (N × Q × K) + (R 0K + R) Clearly, N × Q × K is always divisible by N.
Step IV: Now make (R 0 K + R) divisible by N by putting the least value of K. Say, 1, 2, 3, 4 ...... Now, put the value of K into the expression (LK + R) which will be the required number.
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Exponents and Surds Laws of Integral Exponents For all real numbers a and b , if m and n are are positive integers, then m n m +n (i ) a × a = a For example, 2 3 × 24 = 23+4 = 27 = 128 (ii) (a m )n = a mn 1 1 For example, [(–2) 2] –3 = (–2) 2×–3 = (–2) –6 = 6 = (2) 64 (iii) (ab )m = a m b m For example, (2 × 3) 4 = 24 × 34 = 16 × 81 = 1296
a (iv) b
m
b a
m
5
2 5 10 10 3 2 3 3 4 For example, 4 4 3 4
(v )
a m a m n a n For example, 3 7 ÷ 34= 37–4 = 33 =27 a m an
K KUNDAN a m
a m m a 0 1 a m For example, 7 5 ÷ 75 = 75–5 = 70 =1
m m (vi) a a
n
a a (vii) n b b
n
4
2 16 2 For example, 4 3 3 81 4
Laws of Surds (i )
For any positive integer ‘ n n’ and a positive rational number ‘ a ’ , a’,
(ii)
n
a n
a .
If ‘ n ’, ‘ b n’ is a positive integer and ‘ a ’, b’ are rational numbers,
a n b n ab . ’, ‘ b (iii) If ‘ n n’ is a positive integer and ‘ a ’, b’ are rational numbers, then
then
n
n
a
n
b
n
a . b
(iv) If ‘ m m’ and ‘ n n’ are positive integers and ‘ a ’ is a positive rational number, then
a mn a n m a . If ‘ m m’ and ‘ n n’ are positive integers and ‘ a ’ is a positive rational number, then m n
(v )
n m
p m
a
n a p mn a pm
For example, 5 4
2 3
5
5
2 3
4
1 4
5 23 5 8
Average (i) (i )
If the average age of ‘ m ’ m’ boys is ‘ x x’ and the average age of ‘ n ’ boys out of them ( m boys) is ‘y ’ then the average age of the rest of the boys is
(ii)
mx ny ; where m > > n . m n
If the average of n quantities quantities is equal to x . When a quantity is removed or added the average becomes ‘ y ’. ’. Then the value of removed or added quantity is [ n ( ( x – – y ) + y ]. ]. In other words, it may be written as value of new entrant (or removed quantity) = Number of old members × Increase in average + New average.
(iii) The average weight of ‘ n n’ persons is increased by ‘ x ’ kg when some of them [ n 1, n 2, ... n , where n 1 + n 2 + ... < n ] who weigh [y 1 + y 2 + ... where, y 1 + y 2 + ... = y kg] kg] are replaced by the same number of persons. Then the weight of the new persons is ( y + + nx ). ). Weight of new persons = Weight of removed person + Numbers of persons × Increase in average. (iv) The average age of ‘ n ’ persons is decreased by ‘ x ’ years when some of them [ n 1, n 2 ... n ; where n 1 + n 2 + ... b ) (a) a b a b ax bx and respectively (where a < b ) (b) b a b a
(vi)
If three numbers are in the ratio of a : b : c and the sum of these numbers is x , then these numbers will be ax bx cx , and respectively. a b c a b c a b c
(vii)
If the ratio between the first and the second quantities is a : b and the ratio between the second and the third quantities is c : d , then the ratio among first, second and third quantities is given by ac : : bc : : bd . The above ratio can be represented diagrammatically as
(viii) If the ratio between the first and the second quantities is a : b ; the ratio between the second and the third quantities is c : d and the ratio between the third and the fourth quantities is e : f then the ratio among the first, second, third and fourth quantities is given by
(ix)
If in x litres mixture of milk and water, the ratio of milk and water is a : b, the quantity of water to be added in order to make this ratio c : d is is
(x) (x )
x ad bc c a b
.
A mixture contains milk and water in the ratio a : b . If
x litres of water is added to the mixture, milk and water become in the ratio a : c . Then the quantity of milk in the mixture is given by
ax and that of water is given by c b
bx . c b
(xi)
If two quantities X and Y are in the ratio x : y . Then : X – Y : : : x + y : x – y X + Y :
(xii)
In any two two-dimensional figure, if the corresponding sides are in the ratio a : : b , then their areas are in the ratio a 2 : b 2.
(xiii) In any two 3-dimensional figures, if the corresponding sides or other measuring lengths are in the ratio a : : b , then their volumes are in the ratio a 3 : b 3. (xiv) The ratio betwee n two num bers is a : b . If each number be increased by x , the ratio becomes c : d . Then, the two numbers are given as
c–a (xv)
xa c d xb c d and ; where ad bc ad bc
d d – b
The incomes of two persons are in the ratio a : b and their expenditures are in the ratio c : d . If each of them saves Rs X, then their incomes are given by
Xa d c ad bc
and
Xb d c . ad bc
and their (xvi) The incomes of two persons are in the ratio a : b and expenditures are in the ratio c : d . If each of them saves Rs X , then their expenditures are given by
Xd b a . ad bc
Xc b a ad bc
and
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(xvii) Two candles of the same height are lighted at the same time. The first is consumed in T 1 hours and the second in T 2 hours. Assuming that each candle burns at a constant rate, the time after which the ratio of first candle to second
x T1T 2 1 y hours. candle becomes x : y is given by x T1 T 2 y
Partnership (i) (i )
If investments are in the ratio of a : b : c and the timing of their investments in the ratio of x : y : z then the ratio of their profits are in the ratio of ax : by : cz .
and profits in the ratio (ii) If investments are in the ratio a : b : c and : q : : r , then the ratio of time = p :
p q r : : . a b c
the ir capit als in a busi nes s. If the (iii) Three partn ers inve st their ratio of their periods of investments are t 1 : t 2 : t 3 and their profits are in the ratio of a : b : c , then the capitals will be in
a b c the ratio of t : t : t . 1 2 3
Profit and Loss (i) (i )
If certain article is bought at the rate of ‘A’ for a rupee, then to gain x %, % , the article must be sold at the rate of
100 100 x × A for a rupee (Remember the rule of fraction). (ii)
If a man purchases ‘ x x’ items for Rs ‘ y y’ and sells ‘ y y’ items for Rs ‘ x ’ , then the profit or loss [depending upon the respective x’, (+ve) or (–ve) sign in the final result] made by him is
x 2 y 2 100 % . 2 y (iii) If a man purchases ‘ a a’ items for Rs ‘ b b’ and sells ‘ c c’ items for Rs ‘ d ’, then the gain or loss [depending upon the respective d ’, (+ve) or (–ve) sign in the final result] made by him is
ad bc bc 100 % . (iv)
Problems Based on Dishonest Dealer % gain =
Error 100 True value value – Error
or, % gain =
( v) v)
True weight weight – False weight weight 100 False weight
( a) When there are two successive profits of x % and y % , then the resultant profit per cent is given by xy x y . 100 (b) When there is a profit of x % and loss of y % in a transaction, then the resultant profit or loss per cent is xy given by x y according to the + ve and the -ve 100 signs respectively. %, then (c) When there are two successive loss of x % and y %,
xy the resultant loss per cent is given by x y . 100
(vi)
If an article is sold at a profit of x % and if both the cost price and selling price are Rs A less, the profit would be y %
x y A . In other words, y
more, then the cost price is
cost price =
Initia iall Prof Profit it % Init
Incr Increas ease e in prof profit it % A
Increase in profit %
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M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(vii) If cost price of x articles is equal to the selling price of y articles, then the profit percentage =
x y 100% . y
(viii)(a) A person buys certain quantity of an article for Rs A. If he sells m th th part of the stock at a profit of x % and the remaining n th th part at y % profit, then the per cent profit
mx ny in this transaction is m n First part % profit on first part Second part % profit on sec sec ond part or Total of two parts
.
part is sold at m% profit and the rest y part part is sold (b) If x part at n % loss and Rs P is is earned as overall profit, then
P ×100 ×100 the value of the total consignment is Rs . xm – ny (ix)
If a man buys two items A and B for Rs P and sells one item A so as to lose x % and the other item B so as to gain y %, %, and on the whole he neither gains nor loses, then
Py (a) the cost of the item A is and x + y Px (b) the cost of the item B is . x + y (x )
items a rupee, ( a) By selling a certain item at the rate of ‘ X’ items a man loses x %. %. If he wants to gain y %, %, then the number number
100 x of items should be sold for a rupee is X . 100 y (b) By selling an article for Rs A , a dealer makes a profit of %. If he wants to make profit of y %, %, then he should x %.
y x increase his selling price by Rs A and the 100 x
100 y selling price is given by Rs A . 100 x (c) By selling an article for Rs A , a dealer makes a loss of %. If he wants to make a profit of y %, %, then he should x %.
x y increase his selling price by Rs A and the 100 x 100 y selling price is given by Rs A . 100 x
(xi)
When each of the two commodities is sold at the same price Rs A , and a profit of P% is made on the first and a loss of L % is made on the second, then the percentage gain or 100 P L 2PL loss is 10 0 P 100 L according to the +ve or –ve sign.
Note: (a) In the special case when P = L we have
100 0 2P 2 p 2 200 100 Since the sign is –ve, there is always loss and the value is given as
% value
2
100
.
(b) When each of the two commodities is sold at the same price Rs A, and a profit of P % is made on the first and a profit of L % is made on the second, then the percentage gain is
100 P L 2PL
100 P 100 L
.
(xii) If a merchant, by selling his goods, has a gain of x % of the selling price, then his real gain per cent on the cost price is
x 100 x 100 % . Note: Real profit per cent is always calculated on cost price and real profit per cent is always more than the % profit on selling price. %, of the (xiii) If a merchant, by selling his goods, has a loss of x %, selling price, then his real loss per cent on the cost price is
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M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
x 100 x 100 %. items for a rupee, then (xiv) If an item is bought at the rate of X items the number of items sold for a rupee in order to gain x % is
100 X 100 x .
Discount (i) (i )
If a tradesman marks his goods at x % above his cost price and allows purchasers a discount of y % for cash, then there
is x y
xy % profit or loss according to +ve or –ve sign 100
respectively.
x 2 . –ve sign 100 indicates that there will be always loss.
Note : When x = y, then formula becomes
(ii)
A person marks his goods x % above the cost price but allows y % discount for cash payment. If he sells the article for
100 100 . 100 x 100 y
Rs X, then the cost price is X
(iii) If a trader buys an article at x % discount on its original price and sells it at y % increase on the price he buys it, then the percentage of profit he makes on the original price
xy is y x . 100
(iv)
A dealer buys an item at x % discount on its original price. If he sells it at a y % increase on the original price, then the
y x per cent profit he gets is 100 . 100 x
(v) (v )
A businessman marks an article at Rs A and allows x % discount (on the marked price). He gains y %. % . If the cost price of the article is Rs B, then the selling price of the article can be calculatd from the equation given below A 100 x B 100 y = selling price. 100 100
(vi)
If a person buys an article with x per per cent discount on the marked price and sells the article with y per cent profit on the marked price, then his per cent profit on the price he
x y 100 per cent. 100 x
buys the article is given by
each after giving x % discount (vii) A person sells articles at Rs A each on marked price. Had he not given the discount, he would have earned a profit of y % on the cost price. Then the cost
1002 A . price of each article is given by Rs 100 x 1 00 y per cent discount for wholesale (viii) A certain company declares x per buyers. If a person buys articles from the company for Rs A after getting discount. He fixed up the selling price of the articles in such a way that he earned a profit y % on original company price. Then the total selling price is given by
100 y . 100 x
Rs A
(ix)
A shopkeeper sold an article for Rs A after giving x % discount on the labelled price and made y % profit on the cost price. Had he not given the discount, the percentage profit would
x y 100 per cent. have been 100 x
(x )
( a) Equivalent discount of two successive discounts x % and xy y % = x y %. 100 (b) Equivalent discount of three successive discounts x % ,
y % and z % = x y z
xy yz zx xyz %. 100 (100)2
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M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
Simple Interest (i) (i )
If a person deposits Rs x 1 in a bank at r 1% per annum and Rs x 2 in another bank at r 2% per annum, then the rate of
x1r1 x 2r2 interest for the whole sum is x x . 2 1
(ii)
If the simple interest on a sum of money is
1 of the n
principal, and the number of years is equal to the rate per
cent per annum, then the rate per cent is 100
(iii)
1 %. n
If the simple interest on certain sum ‘ P ’ is ‘ I I’ and the number of years is equal to the rate per cent per annum,
100 I . P
then the rate per cent or time is given by
(iv)
The annual payment that will discharge a debt of Rs A due in t years at the rate of interest r % per annum is
100A 100t rt t 1 . 2 (v) (v )
If a sum of money becomes ‘ x ’ times in ‘ t t’ years at SI, the rate of interest is given by
(vi)
100 x 1 t
%.
A certain sum is invested for certain time. It amounts to Rs A 1 at r 1% per annum. But when invested at r 2% per annum, it amounts to Rs A 2, then the time is given by
A1 A 2 100 yea rs. A2r1 A1r2
(vii)
A certain sum is invested for certain time. It amounts to
Rs A 1 at r 1% per annum. But when invested at r 2% per annum, it amounts to Rs A 2, then the sum is given by Rs
A 2r1 A1r2 . r1 r2
y e ar s. (viii) A sum was put at SI at a certain rate for t ye Had it been put at x % higher rate, it would have
A 100 fetched Rs ‘A’ more, then the sum is Rs or t x More Interest 100 . Time More Rate (ix)
If a certain sum of money amounts to Rs A 1 in t 1 years and
A 2t1 A1t 2 t1 t2 .
to Rs A 2 in t2 years, then the sum is given by
(x) (x )
The simple interest on a sum of mo ney will be Rs x after after ‘ t ’ t’ years. If in the next ‘ t t’ years principal becomes n times, then the total interest at the end of the ‘2 t ’th ’th year is given by Rs [( n + + 1) x ]. ].
(xi)
The simple interest on a sum of money will be Rs x after after t 1 years. If in the next t 2 years principal becomes n times, then the total interest at the end of ( t 1+ t 2)th year is given
t 2 n . t 1
by Rs x 1
(xii)
A sum of Rs X is lent out in n parts in such a way that the interest on first part at r 1% for t 1 years, the interest on second part at r 2% for t 2 years the interest on third part at r 3% for t 3 years, and so on, are equal, the ratio in which the sum was divided in n parts is given by
1 1 1 1 : : : .... r1t1 r2t2 r3t 3 rn tn . (xiii) If a sum of money becomes ‘ n n’ times at the simple interest rate of r % per annum, then it will become ‘ m ’ times at the
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M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
m 1 simple interest rate of r per cent. n 1
(xiv)
When different amounts mature to the same amount at simple rate of interest, the ratio of the amounts invested are in inverse ratio of (100 + time × rate). That is, the ratio in which the amounts are investe d is
1 1 1 1 : : : .... : 1 00 r1t1 100 r2t 2 100 r3t 3 1 00 rnt n .
(xv)
There is a direct relationship betwee n the principal and the amount and is given by sum =
(xvi)
100 Amount . 100 rt
A person lent a certain sum of money at r % simple interest and in ‘ t t’ years the interest amounted to Rs A less than
100 A the sum lent, then the sum lent is given by Rs 100 rt
(xvii) If a sum of money becomes ‘ n n’ times in ‘ t t’ years at a simple interest, then the time in which it will amount to ‘ m ’ times
m 1 itself is given by t years. n 1
(xviii) If the simple interest on Rs P 1 is less than the interest on Rs P 2 at r % simple interest by Rs A , then the time is
A 100 years. r P2 P 1
given by
(xix) Two equal amounts of money are deposited at r 1% and r 2% for t 1 and t 2 years respectively. If the difference between their interests is I d then the sum =
I d 100 . r1 t1 r2 t 2
(xx)
If the difference between the interest received from two different banks on Rs X for t years is Rs I d , then the difference between their rates is given by
I d 100 X t per cent.
(xxi)
If a sum amounts to Rs A 1 in t 1 years and Rs A 2 in t 2 years at simple rate of interest, then rate per annum
100 A2 A 1
A1t2 A2t1 .
1 1 (xxii) A person invested n of his capital at x 1%, n at x 2% and 1 2 1 the remainder n at x 3%. If his annual income is Rs A , 3
A 100 . the capital is given by Rs x1 x 2 x 3 n1 n 2 n 3 whi ch a sum of mon ey becom be com es n times (xxiii) The tim e in which itself at r % per annum simple interest is given by
n 1 100 years. r
Compound Interest (i) (i )
A sum of money, placed at compound interest, becomes n times in t years and m times in x years. We calculate the value of x from from the equation given below:
n
1t
m 1 x .
(ii) If the compound interest on a certain sum for 2 years is Rs ‘C’ and simple interest is Rs ‘S’, then
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M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
2 C S 100 % S
the rate of interest per annum is or
2 Di fference 100 per cent. SI
(iii) When difference between the compound interest and simple interest on a certain sum of money for 2 years at r % rate is Rs x , then the sum is given by 2
x 100 Difference 100 1 00 100 x Sum = = 2 r Rate Rate r
2
(iv) If the difference between CI and SI on a certain sum for 3 years at r % is Rs x , then the sum will be
Difference 100 r 2 300 r
3
.
(v) (v ) If an amount of money grows upto Rs A 1 in n years and upto Rs A 2 in (n + 1) years on compound interest, then the rate
A2 A 1 100 or A 1 Differe Difference nce of amount amount after n years years and (n 1) years years 100 . Amount after n years
per cent is given by
(vi) If the compound rate of interest for the first t 1 ye ars is r %, for the next t years is r %, for the next t 3 ye y e ar s is r 3%, ... and the last t n years is r n % , t h e n compound interest on Rs x for ( t 1 + t 2 + t 3..... t n ) years is t r x 1 1 100
1
t t r 2 r n 1 . . . . 1 x . 100 100 2
n
(vii) If a sum of money, say Rs x, is divided among n parts in such a manner that when placed at compound interest, amount obtained in each case remains equal while the rate of interest on each part is r 1, r 2, r 3, ..., r n respectively and time period for each part is t 1 , t 2, t 3, ..., t n respectively, then the divided parts of the sum will in the ratio of 1 r1 1 10 0
t1
:
1 r2 1 100
t2
:
1 r3 1 100
t3
: .. . :
1 rn 1 10 0
tn
.
Mixture and Alligation (i) (i ) The proportion in which rice at Rs x per kg must be mixed with rice at Rs y per per kg, so that the mixture be worth Rs z a
y z kg, is given by . z x (ii) A mixture of a certain quantity of milk with ‘l’ litres of water is worth Rs x per litre. If pure milk be worth Rs y per litre,
x then the quantity of milk is given by l litres. y x (iii) n gm of sugar solution has x % sugar in it. The quantity of sugar should be added to make it y % in the solution is given
y x gm. 100 y
by n
K KUNDAN requir ired ed % value alue requ presen entt % valu value e pres
Solution
or Quantity of sugar added =
100 required %
value
(iv) In a group, there are some 4-legged creatures and some 2-legged creatures. If heads are counted, there are x and if leggs are counted there are y , then the number of 4-legged creatures are given by
y 2x Total legs 2 Total heads 2 or and the number 2 of
2-legged
creatures
are
given
by
4x y 2
or
4 Total heads Total legs . 2 (v) (v ) If x glasses of equal size are filled with a mixture of spirit and water. The ratio of spirit and water in each glass are as follows: a 1 : b 1, a 2 : b 2 , ... a x : b x . If the contents of
30
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e all the x glasses are emptied into a single vessel, then proportion of spirit and water in it is given by
a1 a 2 a x a b a b ... a b : 1 1 2 2 x x b1 b 2 b x a b a b ... a b 1 1 2 2 x x (vi) If x glasses of different sizes, say S 1 , S 2 , S 3 , ... S x , are filled with a mixture of spirit and water. The ratio of spirit and water in each glass are as follows, a 1 : b 1, a 2 : b 2, a 3 : b 3, ...., : b . If the contents of all the glasses are emptied into a a x x single vessel, then proportion of spirit and water in it is given by
a1S1 a S aS a S 2 2 3 3 ... x x : a x bx a1 b1 a 2 b 2 a 3 b 3 b1S1 b S bS bS 2 2 3 3 ... x x a x bx a1 b1 a 2 b 2 a 3 b 3
K KUNDAN
Time and Work (i) (i )
If M 1 persons can do W 1 works in D 1 days and M 2 persons can do W 2 works in D 2 days, then we have a very general formula in the relationship of M 1D1W2 = M2D2W1.
(ii)
If M 1 persons can do W 1 works in D 1 days working T 1 hours a day and M 2 persons can do W 2 works in D 2 days working T 2 hours a day, then we have a very general formula in the relationship of M 1 D1 T 1 W2 = M2 D2 T 2 W1.
(iii)
If A and B can do a piece of work in x days, days, B and C in y days, C and A in z days, days, then (A + B + C) working together
2xyz days . xy yz xz
will do the same work in
2xyz Let be ‘r’, then xy yz xz ‘A’
yr alone will will do the same work work in days or y r
2xyz xy yz zx days, ‘B’ alone will do the same work in
zr days or z r
2xyz yz zx xy days and ‘C’ alone will do the same work in
xr x r days or
2xyz xz xy yz days.
(iv)
If A can do a piece of work in x days days and B can do it in y days, then A and B working together will do the same work
xy in days. x y (v) (v )
If A, B and C can do a work in x , y and z days days respectively, then all of them working together can finish the work in
xyz xy yz xz days. (vi)
If A and B together can do a piece of work in x days and A alone can do it in y days, then B alone can do the work in xy days. y x
(vii)
If x 1 men or y 1 women can reap a field in ‘D’ days, then x 2
32
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
D x1y 1 men and y 2 women take to reap it x y x y days. 1 2 2 1
days and (viii) If a 1 men and b 1 boys can do a piece of work in x days days, then the following a 2 men and b 2 boys can do it in y days,
yb2 xb1 relationship is obtained: 1 man = xa ya boys.. 2 1
(ix)
A certain number of men can do a work in ‘D’ days. If the re were ‘ x’ x’ men less it could be finished in ‘ d ’ days more,
x D d . d
then the number of men originally are
(x) (x )
If x 1 men or x 2 women or x 3 boys can do a work in ‘D’ days, then the number of days in which 1 man, 1 woman and 1 boy do the same work is given by the following formula:
D x1 x 2 x 3 days. x1x 2 x1x3 x2 x3
Number of required days =
(xi)
A certain number of men can do a work in ‘D’ days. If the re were ‘ x x’ men more it could be finished in ‘ d ’ days less,
x D d . d
then the number of men originally are
Or Number of more workers × Number of days
taken by the second group Number Number of less less days days
(xii)
If A working alone takes ‘x ’ days more than A and B, and B working alone takes ‘ y ’ days more than A and B together, then the number of days taken by A and B working together is given by xy days.
and y days days respectively and A (xiii) If A and B can do a work in x and leaves the work after doing for ‘a ’ days, then B does the
x a y days. x
remaining work in
(xiv)
If A and B can do a work in x and and y days days respectively, and B leaves the work after doing for ‘ a’ days, days, then A does the
y a x days. y
remaining work in
(xv)
A and B can do a piece of work in x and and y days days respectively and both of them starts the work together. If B leaves the work ‘ a a’ days before the completion of work, then the total
y a x x y
time in which the whole work is completed days.
(xvi)
A and B can do a piece of work in x and and y days days respectively and both of them starts the work together. If A leaves the work ‘ a a’ days before the completion of the work, then the
K KUNDAN total time in which the whole work is complete d
x a y x y
days. days. If A does the work only (xvii) A can do a piece of work in x days. for ‘ a a’ days and the remaining work is done by B in ‘ b ’ days,
xb the B alone can do the work in days. x a
days respectively. (xviii) A and B can do a piece of work in x and y days Both starts the work together. But due to some problems A leaves the work after some time, and B does the remaining work in ‘ a a’ days, then the time after which A leaves
y a x days. x y
the work is given by
(xix)
A completes a work in ‘ x ’ days. B completes the same work in ‘y ’ days. A started working alone and after ‘ a’ days B
34
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e joined him. Then the time in which they will take together
x a y . x y
to complete the remaining work is given by
Work and Wages (i ) A can do a work in x days and B can do the same work in y days. If the contract for the work is Rs X and both of them work togehter, then the share of A and B is given by
X X y and x respectively. Rs x y x y
(ii) A, B and C can do a work in x , y and z days respectively. If doing that work together they get an amount of Rs X , then the
Xyz Xxz Share of A = Rs , Share of B = Rs xy xz yz xy xz yz
K KUNDAN
Xxy and Share of C = Rs and ratio of their shares is xy xz yz
given by A : B : C = yz : : xz : : xy
(iii) A, B and C contract a work for Rs X . If together A and B are supposed to do
Rs X 1
x of the work, then the share of C is given by y
x y .
Pipes and Cisterns (i) (i )
If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, hours, then the net part filled in 1
1 1 hour, when both the pipes are opened . x y
time
(T) taken to fill the tank, when both the pipes are
opened =
xy y x
Note: If T is (+ve), then cistern gets filled up and if T is (-ve), then cistern gets emptied. (ii)
If a pipe fills a tank in x hours and another fills the same tank in y hours, but a third one empties the full tank in z hours, and all of them are opened together, the net part
1
filled in 1 hour
x
time
1 y
1
z
xyz hours. yz xz xy
taken to fill the tank
minutes (iii) Two pipes A and B can fill a tank in x minutes and y minutes respectively. If both the pipes are opened simultaneously, then the time after which pipe B should be closed, so that
t the tank is full in t minutes, is y 1 minutes. x
K KUNDAN
hours and y hours (iv) Two pipes P and Q will fill a cistern in x hours respectively. If both pipes are opened together, then the time after which the first pipe must be turned off, so that
t the cistern may be just filled in t hours, hours, is x 1 hours. y (v) (v )
A cistern is normally filled in x hours but takes t hours longer to fill because of a leak in its bottom. If the cistern is
x x t hours. t
full, the leak will empty it in
Time and Distance (i) (i )
If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, then the average speed 2xy during the whole journey is km/hr. x y
(ii)
A person is walking at a speed of x km/hr. After every
36
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e kilometre, if he takes rest for t hours, hours, then the time he will
y take to cover a distance of y km km is y 1 t hour. Or, x in other words, required time =
Dist Distan ance ce to be cover covered ed + Speed
Number of rest Time for each rest
(iii) A person covers a certain distance between two points. Having an average speed of x km/hr, km/hr, he is late by x 1 hours. However, with a speed of y km/hr km/hr he reaches his destination y 1 hours earlier. The distance between the two points is
xy x1 y1 km. km. Or, Or, Required Required distance distance = y x
given by
Product of two speeds × Difference Difference between arrival times Difference of two speeds
(iv)
A person goes to a destination at a speed of x km/hr and returns to his place at a speed of y km/hr. If he takes T hours in all, the distance between his place and
K KUNDAN xy T km. In other words, destination is x y Required distance =
Total time taken×
(v) (v )
Product of the two speeds Addition of the two speeds
If a person does a journey in T hours and the first half at S1 km/hr and the second half at S 2 km/hr, then the distance =
2 Time S1 S2 S1 S2
Where, S1 = Speed during first half and S 2 = Speed during second half of journey
(vi)
If the new speed of a person is
a of the usual speed, b
then the change in time taken to cover the same
b distance is 1 × usual time or, usual time is given by a Change in time b hours. – 1 a
km/hr (vii) If two persons A and B start from a place walking at x km/hr and y km/hr km/hr respectively, at the end of t hours, hours, when they are moving in same direction and x < y , they will be (y – – x km apart. x )t km km/hr (viii) If two persons A and B start from a place walking at x km/hr and y km/hr km/hr respectively, at the end of t hours, hours, when they are moving in opposite directions, they will be ( x + y ) t km apart.
(ix)
If two runners A and B cover the same distance at the rate of x km/hr and y km/hr respectively, then the distance
K KUNDAN
xy t y x
travelled, when A takes t hours hours longer than B, is
km. Or Distance =
Multiplication of speeds Difference in time to Difference of Speeds cover the distance
(x) (x )
A person takes x hours to walk to a certain place and ride back. However, he could have gained t hours, if he had covered both ways by riding, then the time taken by him to walk both ways is (x + t ) hours. Or, Both ways walking = One way walking and one way riding time + gain in time
(xi)
A man takes x hours to walk to a certain place and ride back. However, if he walks both ways he nee ds t hours more, then the time taken by him to ride both ways is ( x – – t ) hours.
38
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(xii) A person A leaves a point P and reaches Q in x hours. If another person B leaves the point Q, t hours later than A and reaches the point P in y hours, hours, then the time in which
x A meets to B is ( y + t ) x y hours.
(xiii) A person A leaves a point P and reaches Q in x hours. If another person B leaves the point Q, t hours hours earlier than A and reaches the point P in y hours, hours, then the time in which
x A meets to B is (y – t ) x y hours.
(xiv) Speed and time taken are inversely proportional. Therefore, S 1T 1 = S 2 T 2 = S 3 T 3 ..... Where, S 1, S 2 , S 3 ... are the speeds and T 1, T 2 , T 3 .... are the time taken to travel the same distance.
(xv) A thief is spotted by a policeman from a distance of d km. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief x kilometres an hour, and that of the policeman y kilometres an hour, then
x the thief will run before he is overtaken = d km. y x Or, The dista nce cove red by the thief before bef ore he get s caught
=
Lead of distance × Speed of thief Relative speed
Trains (i) (i )
Two trains are are moving in the same direction at x km/hr km/hr and > y ). ). If the faster train crosses a man in y km/hr (where x > the slower train in ‘ t t’ seconds, then the length of the faster
5 train is given by x y t metres. 18
(ii)
Two trains are moving in opposite directions at x km/hr and y km/hr (where x > > y ), ), if the faster train crosses a man in the slower train in t seconds, then the length of the
5 x y t metres. faster train is given by 18 (iii)
A train running at x km/hr takes t 1 seconds to pass a platform. Next it takes t 2 seconds to pass a man walking at y km/hr in the opposite direction, then the length of the
5 train is x y t 2 metres and that of the platform is 18 5 x t1 t 2 yt 2 metres. 18 (iv)
A train running at x km/hr takes t 1 seconds to pass a platform. Next it takes t 2 seconds to pass a man walking at y km/hr in the same direction, then the length
5 x y t 2 metres and that of the of the train is 18 5 x t1 t 2 yt 2 metres. platform is 18 (v) (v )
L metres long train crosses a bridge of length L metres in seconds. Time taken by the train to cross a platform of L 2 T seconds.
L L 2 T seconds. L L 1
metres is given by
(vi)
If L metres metres long train crosses a bridge or a platform of length L 1 metres in T seconds, then the time taken by train to
L T cross a pole is given by seconds. L L 1 start at the same time from A and B and proceed (vii) Two trains start towards each other at the rate of x km/hr and y km/hr respectively. When they meet it is found that one train has travelled d km more than the other. Then the distance
x y between A and B is d km. Or x y
40
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e Distance = Difference in distance ×
Sum Sum of spee speeds ds . Diff Differe erenc nce e in speeds speeds
(viii) A train passes by a stationary man standing on the platform or a pole in t 1 seconds and passes by the platform completely in t 2 seconds. If the length of the platform is ‘ p ’ metres, then the length of the train is
t1 p t t metres and the speed of the 2
1
p train is t t m/sec. Or Length of the train 2 1
Length of the platform Difference in time
Time taken to cross a stationary pole or man
an d
Lengt Length h of the the plat platfo form rm . Diffe Differen rence ce in time time
speed of the train =
(ix) Two trains of the same leng th but with diffe rent spee ds pass a static pole in t 1 seconds and t 2 seconds respectively. They are moving in the same direction. The time they will
K KUNDAN 2t1t 2 take to cross each other is given by t t seconds. 2 1
(x) (x )
Two trains of the same leng th but with diffe rent spee ds pass a static pole in t 1 seconds and t 2 seconds respectively. They are moving in the opposite directions. The time they
2t1t 2 will take to cross each other is given by t t seconds. 1 2
trains of the length l 1 and l 2 m respectively with different (xi) Two trains speeds pass a static pole in t 1 seconds and t 2 seconds respectively. When they are moving in the same direction,
l1 l 2 t 1t 2 seconds. t 2l1 t1l 2
they will cross each other in
(xii) Two trains of the lengt h l 1 m and l 2 m respectively with different speeds pass a static pole in t 1 seconds and t 2
seconds respectively. When they are moving in the opposite
l1 l 2 t 1t 2 seconds. t 2l1 t1l 2
direction they will cross each other in
le ngth l 1 m and l 2 m respectively run on (xiii) Two trai ns of length parallel lines of rails. 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 the speed of each train is given as the following. Speed of the faster
l1 l 2 t1 t 2 2 t1t 2 m/sec and the speed of the slower
train =
l1 l 2 t1 t 2 m/sec. Thus a general formula 2 t1t 2
train =
for the speed is given as Average length of two trains
1 1 Opposite directio direction's n's time Same directio direction's n's time time Opposite
×
(xiv) If a train crosses L m and L m long bridge or platform or tunnel in T seconds seconds and T seconds seconds respectively, respectively, then the the
L1T2 L2T1 T1 T 2 m and the speed of the
length of the train is
L1 L 2 train is T T m/sec. 1 2 (xv)
A goods train and a passenger train are running on parallel tracks in the same or in the opposite direction. The driver of the goods train observes that the passenger train coming from behind overtakes and crosses his train completely in T 1 seconds. Whereas a passenger on the passenger train marks that he crosses the goods train in T 2 seconds. If the speeds of the trains be in the ratio of a : b , then the ratio of
T 2 their lengths is given by T T . 1 2 (xvi) Two trains A and B start from P and Q towards Q and P respectively. After passing each other they take T 1 hours and T 2 hours to reach Q and P respectively. If the train
42
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e from P is moving x km/hr, km/hr, then the speed of the other train
is x
T 1 km/hr. T 2
Or
Spee d of the first train ×
Time taken by first train after meeting Time taken by sec ond train after meeting
Boats and Streams (i) (i )
If x km be the rate of stream and a man takes n times as long to row up as to row down the river, then the rate of the
n 1 man in still water is given by x km/hr. n 1 (ii) If the speed of the boat in still water is x km/hr and the rate of current is y km/hr, km/hr, then the distance travelled downstream in ‘ T + y )T )T km, ie Distance travelled downstream T ’ hours is ( x + = Downstream Rate × Time. And the distance travelled upstream in ‘T’ hours is ( x – y )T )T km, ie Distance travelled upstream = Upstream Rate × Time. km/hr in still waters. If in a stream which (iii) A man can row x km/hr is flowing at y km/hr, km/hr, it takes him z hours hours to row to a place and back, the distance between the two places is
z x 2 y 2 2x
.
(iv) A man rows a certain distance downstream in x hours and returns the same distance in y hours. hours. If the stream flows at the rate of z km/hr, km/hr, then the speed of the man in still water is given by
z x y y x
km/hr. Or, Speed in still water
Rate of stream Sum of upstream up stream and downstream time km/hr Differen Difference ce of upstrea upstream m anddownstr anddownstreamtime eamtime
=
(v) (v )
. If a man can row at a speed of x km/hr in still water to a certain upstream point and back to the starting point in a river which flows at y km/hr, km/hr, then the averge speed for total
x y x y km/hr. x
journey (up + down) is given by
Plane Mensuration (1) To find the area of an equilat eral triangl e if its heig ht is given. Area of the equilateral triangle =
Height
2
3
(2) To find the area of a rectangle when its perimeter and diagonal are given.
Perimeter 2 Diagonal 2 sq units. Area of a rectangle = 8 2 (3) To find the perimeter of a square if its diagonal is given. Perimeter of the square =
2
2 Diag onal
K KUNDAN
(4) If the diagonal of a square becomes x times, then the area of the square becomes x 2 times. (5) If the ratio of the areas of square A and square B is a : b, then (i) (i ) the ratio of their sides = a : b , (ii) the ratio of their perimeters = a : b and (iii) the ratio of their diagonals =
a : b .
(6) If the perimeter of a square is equal to the perimeter of a
circle, then the side of the square is
r and radius of 2
2x the circle is is the side of the square and r . Where, x is is the radius of the circle. (7) To find the area of a parallelogram, if the lengths o f the two adjacent sides and the length of the diagonal connecting the ends of the two sides are given. (see the figure).
44
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
Where, ‘ a a’ and ‘ b b’ are the two adjacent sides and ‘D’ is the diagonal connecting the ends of the two sides. Area of a parallelogram = S=
2 s s a s b s D an d
a b D 2
(8) In a parallelogram, the sum of the squares of the diagonals = 2 × (the sum of the squares of the two adjacent sides) Or,
D12 D2 2 2 a 2 b2 Where, D 1 and D 2 are the diagonals and a and b are the adjacent sides. (9) To find the sides of a parallelogram if the distance be tween its opposite sides and the area of the parallelogram is given.
Here, ABCD is a parallelogram, h 1 and h 2 are the distance between opposite sides, ‘l ’ and ‘ b b ’ are the sides of the parallelogram. ‘A’ is area of the parallelogram. A = lh1 bh2
A
A
= h and b = = h l = 1 2
(10) To find the area of a trapezium, when the lengths of parallel sides and non-parallel sides are given. a b s s k s c s d where, Area of a trapezium = k ), ie the difference between the parallel sides and c k = (a – b ), and d are the two non-parallel sides of the trapezium. And
k c d . 2 (11) To find the perpendicular distance be tween the two parallel sides of the trapezium. s
Perpendicular distance =
2 s s k s c s d where, k
k = (a – b ), ie the difference between the parallel sides and c and d are the two non-parallel sides of the trapezium. And s =
k c d 2
(12) There are two concentric circles of radii R and r respectively. Now consider the following cases. larger ger cir circle cle makes makes ‘ n Case I: If lar n ’ revolutions to cover a certain distance, then the smaller circle makes n revolutions to cover the same distance.
R smaller circle circle makes n revolutions to cover Case II: If smaller r a certain distance, then the larger circle makes
r n revolutions to cover the same distance. R (13) Length of a carpet ‘ d d ’ m wide, required to cover the floor of a
xy room which is x m m long and y m broad is given by m. d Or Length required =
Length of room Breadth of room Widt Width h of carp carpet et
(14) A ‘ d d’ m wide carpet is used to cover the floor of a room which is x m long and y m broad. If the carpet is available at Rs A per metre, then the total amount required to xy cover the floor of the room is given by Rs A . Or d Amount required
Rate per per metr metre e = Rs Rat
length of room breadth of room widt width h of carp carpet et
46
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e carpet = Note: Length of the carpet
xy or, d
length of room breadth of room widt width h of carp carpet et (15) Number of tiles, each measuring d 1 m × d 2 m, required to pave a rectangular courtyard x m long and y m wide are
x y given by d d . 1 2 Or Number of tiles required =
leng length th × brea breadt dth h of cour courty tyar ard d len length gth × brea breadt dth h of each each tile tile
(16) Certain number of tiles, each measuring d 1 m × d 2 m, are required to pave a rectangular courtyard x m m long and y m wide. If the tiles are available at Rs A per piece, then the
x y amount needs to be spent is given by Rs A d d . 1
2
K KUNDAN Or, Amount required
= price per tile ×
len length gth brea bread dth of cou courty rtyard ard length bread eadth of eac each tile
(17) A room x m long and y m broad is to be paved with square tiles of equal sizes. The largest possible tile so that the tiles exactly fit is given by “HCF of length and breadth of the room” and the number of tiles required are
x y 2 . HCF of x and y metres long is surrounded by a verandah (18) If a square hall x metres (on the outside of the square hall) d metres wide, then the area of the verandah is given by 4 d (x + d ) sq metres. m long. It has a gravel path d m m wide all (19) If a square plot is x m round it on the inside, then the area of the path is given by 4d (x – – d ) sq m. m long and y m m broad, is surrounded (20) If a rectangular hall x m
by a verandah (on the outside of the rectangular hall) d m wide, then the area of the verandah is given by 2d [( [(x + + y ) + 2 d ] m 2. Or Area of verandah = 2(width of verandah) × [length + breadth of room + 2 (width of verandah)] m by ‘y ’ m. It has a gravel path ‘ d ’ (21) If a rectangular plot is ‘x’ m d’ m wide all round it on the inside, then the area of the path is given by 2 d (x + + y – – 2d) sq m.
(22) A rectangular garden is ‘ x x’ metres long and ‘ y y’ metres broad. It is to be provided with pavements ‘d ’ metres wide all round it both on its outside as well as inside. Then the total area of the pavement is given by 4 d (x + + y ) sq m. (23) A square garden is ‘ x x’ metres long. It is to be provided with pavements ‘ d d’ metres wide all round it both on its outside as well as inside. Then the total area of the pavement is given by (8 dx ) sq metres. m by y m. m. From the (24) An oblong piece of ground measures x m centre of each side a path ‘ d ’ m wide goes across to the centre of the opposite side.
I.
Are a of the path = d (x + + y – – d ) = (width of path) (length + breadth of park – width of
path)
II .
Area Area of the the pa park min minus the the pa path = (x – – d ) ( y – – d ) = (length of park – width of path) × (breadth of park – width of the path)
(25) There is a square garden of side ‘ x x’ metres. From the centre of each side a path ‘ d d ’ metres wide goes across to the centre of opposite side. I.
Are a of the path = d (2 (2x – d ) sq metres.
Area Area of the gard garden en – min minus us the the pa path = ( x – d )2 sq II . metres.
(26) To find the area of a rhombus if one side and one diagonal are given. Area of a rhombus = diagonl ×
2
diagonal 2
side
2
48
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(27) To find the other diagona l of a rhombu s, if perim ete r of rhombus and one of its diagonals are given. Other diagonal 2
=2×
Perimeter diagonal side ; where side = 4 2 2
rad ius of a circu ci rcu lar whe el is r m. The number of (28) The radius
d . 2r
revolutions it will make in travelling ‘ d ’ km is given by Or Number of revolutions =
Distance 2r
(29) The circumference of a circular garden is ‘c ’ metres. Inside the garden a road of ‘ d’ metres width runs round it. The area of the ring-shaped road is given by d (c – d ) or d (2r – – d ) [ c = 2 r ], ], where r = = radius of the circle. Area of ring-shaped road = width of ring (circumference of the circle – × width of ring). Or
× width of ring (2 × radius of the circle – width of ring)
OAC is a circle of radius = r , there is pathway, inside the circle of width = d .
(30) The circumference of a circular garden is ‘ c ’ metres. Outside the garden, a road of ‘ d ’ m width runs round it. The area of the ring-shaped road is given by d (c + d ) sq metres. Or d (2 (2r + + d ) [ c = 2 r ] where r = = radius of the circle Area of ring-shaped road = width of ring (circumference +
× width of ring)
OAC is a circle of radius = r , there is pathway, outside the circle of width d .
(31) A circular garden has ring-shaped road around it both on its inside and outside, each of width ‘ d ’ units. If ‘ r’ is the radius r’ is of the garden, then the total area of the path is (4 dr ) sq units or 2 Cd [where [where C = perimeter = 2 r] (32) To find the area of the shaded portion of the fo llowing figure:
Area of the shaded portion ABCD =
360
r12 r 22
(33) There is an equilateral triangle of which each side is x m. With all the three corners as centres, circles are described x each of radius m. The area area common to all the the circles and and 2 the triangle is
1 1 x 2 or (radius) 2 and the area of the 8 2
remaining portion (shaded portion) of the triangle is
3
2 radius or (0.162) (radius) 2 or, (0.0405) x 2. 2
(34) The diamete r of a coin is x cm. If four of these coins be placed on a table so that the rim of each touches that of the other two, then the area of the unoccupied space between
50
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
4 2 3 2 x or (0.215) x 2 sq cm and area of x or 4 14
them is
1 x 2 sq cm. each sector is given by 16
% , then the (35) If the length of a rectangle is increased by x %, percentage decrease in width, to maintain the same area,
x is given by 100 . 100 x
% , then the (36) If the length of a rectangle is increased by x %, percentage decrease in width, to reduce the area by y %, %, is
x y given by 100 . 100 x
% , then the (37) If the length of a rectangle is increased by x %, percentage decrease in width, to increase the area by y %, %,
K KUNDAN Difference in x and y 100 . 100 x
is given by
%, then the (38) If the length of a rectangle is decreased by x %, percentage increase in width, to increase the area by y %, %, is
x y given by 100 . 100 x
%, then the (39) If the length of a rectangle is decreased by x %, percentage increase in width, to maintain the same area, is
given by
x 100 . 100 x
and y per per (40) If length and breadth of a rectangle is increased x and
xy cent respectively, then area is increased by x y % . 100 Note :
If any of the two measuring sides of rectangle is decreased then put negative value for that in the given formula.
(41) If all the measuring sides of any two dimensional figure is
x 2
%. changed by x %, %, then its area changes by 2x 100
(42) If all the measuring sides of any two-dimensional figure are changed (increased or decreased) by x % then its perimeter also changes by the same, ie x %. %. %, then its (43) If all sides of a quadrilateral are increased by x %, corresponding diagonals also increased by x %. %.
(44) The area of the largest triangle inscribed in a semi-circle of radius r is r 2. (45) The area of the largest circle that can be drawn in a square of side x is
x
2
2
(46) To find the area of the quadrilateral when its any diagonal and the perpendiculars drawn on this diagonal from other two vertices are given. 1 any diagonal × (sum of 2 perpendiculars drawn on diagonal from two vertices) Area of the quadrilateral =
(47) The area of a circle circumscribing an equilateral triangle of side x is
3
x 2 . (See figure)
52
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(48) The area of a circle inscribed in an equilateral triangle of
side x is
12
x 2 . (See figure)
(49) An equilateral triangle is circumscribed by a circle and another circle is inscribed in that triangle, then the ratio of the areas of the two circles is 4 : 1. (See figure)
(50) There is a relation betwee n the number of sides and the number of diagonals in a polygon. The relationship is given below. Number of diagonals =
3 n n 2
; where, n = number
of sides in the polygon.
(51) The area of a rectangular plot is ‘ x x’ times its breadth. If the
difference between the length and breadth is ‘y’ metres, then the breadth is given by ( x – y y ) metres.
Solid Mensuration (1) To find volume of a cuboid if its area o f base or top, area of side face and area of other side face are given. Volume of the cuboid =
=
A1 A 2 A 3
area of b ase o r top area of one face area rea of the oth other face ace
Where, A1 = area of base or top, A 2 = area of one side face and A 3 = area of other side face.
(2) To find total surface area of a cuboid if the sum of all three sides and diagonal are given. Total surface area = (Sum of all three sides) 2 – (Diagonal) 2
K KUNDAN
(3) To find numbe r of bricks when the dimensions of brick and wall are given. Volume of wall Required number of bricks = Volume of o ne brick (4) To find capacity, volume of material and weight of material of a closed box, when external dimensions (ie length, breadth and height) and thickness of material of which box is made, are given. (i) (i ) Capacity of box = (External length – 2 × thickness) × (External breadth – 2 × thickness) × (External height – 2 × thickness) (ii) Volume of material = External Volume – Capacity (iii) Weight of wood = Volume of wood × Density of wood. (5) To find the volume of a cube if the surface area of the cube is given.
Surfac Surface e area Volume of cube = 6
3
54
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(6) To find the volume of rain water at a place if the annual rainfall of that place is given. Volume of rain water = Height (or level) of water (ie Annual rainfall) × Base area (ie area of the place) (7) Total volume of a solid does not change eve n when its shape changes. Old volume = New volume (8) To find the number of possible cubes when disintegration of a cube into identical cubes.
Orig Origin inal al len length gth of side side Number of cubes = ength h of side ide New lengt (9)
3
A hollow cylindrical tube open at both ends is made of a thick metal. If the internal diameter or radius and length of the tube are given, then the volume of metal is given by [ × height × (2 × Internal radius + thickness) × thickness] cu units. Note: In the given formula, we can write 2 × intenal radius = internal diameter
(10) A hollow cylindrical tube open at both ends is made of a thick metal. If the internal and external diameter or radius of the tube are given, then the volume of metal is given by 2 2 ius cu cm. × height × External radius Internal rad iu
(11) A hollow cylindrical tube open at both ends is made of a thick metal. If the external diameter or radius and length of the tube are given, then the volume of metal is given by
height 2 outer radius thickness thickness cu units. Note: In the given formula, we can write 2 × outer radius = outer diameter.
(12) If a rectangular sheet is rolled into a cylinder so that the one side becomes the height of the cylinder then the volume of the cylinder so formed is given by
height other side of the sheet
2
4
.
(13) If a sphere of certain diameter or radius is drawn into a cylinder of certain diameter or radius, then the length or
4 (radius of sphere )3 height of the cylinder is given by
3 radius of cylinder
2
.
%, (14) If length, breadth and height of a cuboid is increased by x %, y % and z % respectively, then its volume is increased by
xy xz yz xyz %. x y z 2 100 100
% , then its volume (15) If side of a cube is increased by x %, 3 x 3x 2 x 3 1 1 1 00% . 10 3 x % increases by or 100 100 1002
%, then its surface area (16) If side of a cube is increased by x %,
x 2 2 x increases by per cent. 100 (17) If the radius (or diameter) of a sphere or a hemisphere is changed by x % then its volume changes by 3 x 3x 2 x 3 1 10 1 00% . 1 3x 2 % or 100 100 100
(18) If the radius (or diameter) of a sphere or a hemisphere is changed by x % then its curved surface area changes by
x 2 2 x per cent. 100
(19) If height of a right circular cylinder is changed by x % and radius remains the same then its volume changes by x %. %.
56
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e
(20) If radius of a right circular cylinder is changed by x % and height remains the same the volume changes by 2 x x 2 1 1 00% . 2x % Or, 100 1 10 100
(21) If radius of a right circular cylinder is changed by x % and height is changed by y% then volume changes by
x 2 2xy x 2y 2 x y % . 100 1002
(22) If height and radius of a right circular cylinder both changes
3x 2 x 3 by x % then volume changes by 3x % . 100 100 2
(23) If the radius of a right circular cylinder is changed by x % and height is changed by y% then curved surface area xy changes by x y per cent. 100
K KUNDAN
(24) If height and radius of a right-circular cone both change by
3x 2 x 3 3 x x % then volume changes by % . 100 1002
(25) If the ratio of surface areas of the two spheres are given, then the ratio of their volumes will be obtained from the following result: (Ratio of the surface areas) 3 = (Ratio of volumes) 2 (26) If the ratio of the radii of two spheres are given, then the ratio of their surface areas will be obtained from the following result: (Ratio of radii) 2 = Ratio of surface areas. (27) If the ratio of the radii of two spheres are given, then the ratio of their volumes will be obtained from the following
result: (Ratio of radii) 3 = Ratio of volumes
(28) If the ratio of the heights of two circular cylinders of equal volume are given, then the ratio of their radii is given by the following result: Ratio of radii =
inve invers rse e rati ratio o of heig height hts s
(29) If the ratio of curved surface areas of two circular cylinders of equal volume are given, then the ratio of their heights is given by the following result: Ratio of curved surface areas =
rati ratio o of heig height hts s.
(30) If the ratio of radii of two circular cylinders of equal volume are given, then the ratio of their curved surface areas are given by the following result: Ratio of curved surface areas = inverse ratio of radii.
(31) If the ratio of heights of two circular cylinders of equal ra dii are given then the ratio of their volumes are given by the following result: Ratio of volumes = Ratio of heights.
(32) If the ratio of heights of two circular cylinders of equal radii are given then the ratio of their curved surface areas are given by the following result: Ratio of curved surface areas = Ratio of heights.
(33) If the ratio of volumes of two circular cylinders of equal radii are given then the ratio of their curved surface areas are given by the following result: Ratio of volumes = Ratio of curved surface areas.
(34) If the ratio of radii of two circular cylinders of equal heights are given, then the ratio of their volumes is given by the following result: Ratio of volumes = (Ratio of radii) 2
(35) If the ratio of radii of two circular cylinders of equal heights
58
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e are given, then the ratio of their curved surface areas is given by the following results: Ratio of curved surface areas = ratio of radii
(36) If the ratio of curved surface areas of two circular cylinders of equal heights are given, then the ratio of their volumes is given by the following results: Ratio of volumes = (Ratio of curved surface areas) 2
(37) If the ratio of radii of two circular cylinders of equal curved surface areas are given, then the ratio of volumes is calculated from the following result: Ratio of volumes = Ratio of radii
(38) If the ratio of heights of two circular cylinders of equal curved surface areas are given, then the ratio of their volumes is calculated from the following result: Ratio of volumes = inverse ratio of heights.
(39) If the ratio of heights of two circular cylinders of equal curved surface areas is given, then the ratio of their radii is given by the following result: Ratio of radii = Inverse ratio of heights.
(40) If the ratio of slant heights of two right circular cones of equal curves surface areas is given, then the ratio of their radii is given by the following result: Ratio of radii = inverse ratio of slant heights.
(41) If the ratio of sides of two cubes is given, then the ratio of their volumes is calculated from the following result: Ratio of volumes = (Ratio of sides) 3
(42) If the ratio of sides of two cubes is given, then the ratio of their surface areas is given by the following result: Ratio of surface areas = (Ratio of sides) 2
(43) If the ratio of volumes of two cubes is given, then the ratio of their surface areas is given from the following result:
(ratio of surface areas) 3 = (ratio of volumes) 2
(44) If the ratio of heights (not slant height) and the ratio of diameters or radii of two right circular cones are given, then the ratio of their volumes can be calculated by the given formula: Ratio of volumes = (Ratio of radii) 2 × (Ratio of heights)
(45) If the ratio of radii and the ratio of volumes of two right circular cones are given, then the ratio of their heights can be calculated by the following result: Ratio of heights = (inverse ratio of radii) 2 (ratio of volumes)
(46) If the ratio of volumes and the ratio of heights of two right circular cones are given, then the ratio of their radii is given by the following result: ratio of radii =
ratio o of rati
volu volume mes s inv invers erseratioof eratioof heig height hts s
(47) If the ratio of heights and the ratio of radii of two circular cylinders are given, then the ratio of their curved surface areas is given by (ratio of radii) (ratio of heights). (48) If the ratio of radii and the ratio of curved surface areas of two circular cylinders are given then the ratio of their heights are given by (Ratio of curved surface areas) (Inverse ratio of radii). (49) If the ratio of heights and the ratio of curved surface areas of two circular cylinders are given, then the ratio of their radii is given by (Ratio of curved surface areas) (Inverse ratio of heights) (50) If a cylinder, a hemisphere and a cone stand on the same base and have the same heights, then (a) The ratio of their volumes = 3 : 2 : 1 and (b) The ratio of their curve surface areas =
2 :
2 : 1.
(51) The ratio of the volum es of a cube to that of the sphere
60
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e which will fit inside the cube is (6 : ).
(52) If a cube of maximum volume (each corner touching the surface from inside is cut from a sphere, then the ratio of
3 ).
the volumes of the cube and the sphere is (2 :
(53) The curved surface area of a sphere and that of a cylinder which circumscribes the sphere is the same.
Problems Based on Ages (i) (i )
If t years earlier the father’s age was x times that of his son. At present the father’s age is y times that of his son. Then the present ages of the son and the father are
t x 1 x y
and
t x 1 respectively. x y
y
K KUNDAN (ii)
If the present age of the father is x times times the age of his son. t years hence, the father’s age becomes y times the age of his son. Then the present ages of the father and his son are
y 1 t y 1 t and years respectively. x y x y
x
(iii) If t 1 years earlier the age of the father was x times the age of his son. t 2 years hence, the age of the father becomes y times the age of his son. Then the present ages of the son and the father are
t2 y 1 t1 x 1
x y
and
t Son's age t Son's x + y – 1 x – 1 + 2 y – 1 years respectively. 2 2 2 Note : When t 1 = t 2 = t , then the formulae become the following:
t x y 2 years and x y
(a) present age of the son =
(b) present age of the father t Son's age x y x y years 2 2
=
(iv) If t years earlier, the father’s age was x times that of his son. At present the father’s age is y times that of his son. Then the sum total of the age of the father and the son is
t x 1 y 1 years. x y
(v) (v )
If the present age of the father is x times that the age of his son. t years hence, the father’s age becomes y times the age of his son. Then the sum of the present ages of father and
t y 1 x 1 years. x y
his son is
(vi)
If the sum of the present ages of A and B is x years. ‘t ’ years ago, the age of A was ‘ y’ times the age of the B. Then the present ages of A and B are as follows:
K KUNDAN (a) Age of B =
x t y 1 y 1
(b) Age of A =
xy t y 1 y 1
(vii) If the sum of the present ages of A and B is ‘ x ’ years. After ‘ t t’ years, the age of A will be ‘ y ’ times that of B. Then the ages of A and B are as given below: (a) The age of A =
(b) The age of B =
xy t y 1 y 1 x t y 1 y 1
and
years.
(viii) If the ratio of the ages of A and B at present is a : b . After ‘T’ years the ratio will become c : d. Then the present ages of A and B are as follows: Age of A = a ×
T c – d
T c – d =a ad – bc difference of cross products
62
M a g i c a l Bo B o ok o k o n Ar Ar i t h m e t i c a l F o r m u l a e Age of B = b ×
(ix)
(x) (x )
T c – d
T c – d =b ad – bc difference of cross products
If the ratio of the ages of A and B at present is a : : b . ‘T’ years earlier, the ratio was c : d . Then the present
(a) Age of A = a ×
T c – d T(c – d) =a bc – ad difference of cross products
(b) Age of B = b ×
T c – d T(c – d) =b bc – ad difference of cross products
If the ratio of the ages of A and B at present is a : : b . After ‘T’ years the ratio will become c : d . Then the sum of present ages of A and B is
T c – d T c d × a + b . a b or differ ence ce of cros cross s produ product cts s differen ad bc
(xi)
If the product of the present ages of A and B is ‘ x ’ years and the ratio of the present ages of A and B is a : b . Then the present
K KUNDAN (a) Age of A = a
x years and ab
(b) Age of B = b
x years ab
: b . ‘T’ years (xii) If the ratio of the ages of A and B at present is a : earlier, the ratio was c : : d , then the sum of the present ages of A and B is
T c – d T c – d or a + b . differen ence ce of cross cross produ product cts s bc – ad differ
a + b
(xiii) If a man’s age is x % of what it was t 1 years ago, but y % of what it will be after t 2 years. Then his present age is
xt1 yt 2 years. x y
x y t years. Note: If t 1 = t 2 = t , then formula will become x y
: b . If the difference between (xiv) The ratio of A’s and B’s ages is a : the present ages of A and B ‘ t ’ years hence is ‘ x x’ then
t x years (a) the present age of A (younger) = b 1 a b t x a (b) the present age of B (older) = years and b 1 a (c) the sum of the present ages of A and B =
t x b 1 years. b a 1 a
K KUNDAN
Note :Here A is always younger than B and a : b i s younger : older. Hence
b older . a younger
’ (xv) The differe nce of present ages of A and B is ‘ x ’ years. If ‘ t t’ years back the ir ages were in the ratio a : b , then
x (a) the age of A t x years a 1 b x years and (b) the age of B t a 1 b x x years. (c) the total Age of A and B = 2 t a 1 b > b Note: Here A > B, ie A is older than B. Hence a >
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