Math Wd Solns

1.

If x to the 3/4 power equals 8, Find the value of x. Ans: 16

Solution: 2log4 x − log4 9 = 2

Solution:

log4 x 2 − log4 9 = 2

x3 / 4 = 8 x = ( 8)

4/3

x = ( 2)

4

 x2  log4  ÷ = 2  9 x2 2 = ( 4) 9 x = 12

x = 16 2.

If a−6 / 8 = 0.001, solve for a. Ans: 10,000

7.

Solution:

Solution: log10 8 + 3log10 x = 3

a−6 / 8 = 0.001 6 − In α = In 0.001 8 a = 10,000 3.

log10 8 + log10 x 3 = 3 log10 ( 8) ( x ) = 3 3

8x 3 = 103

Solve for x if log3 81x = 16 Ans: 4

( 2) 3 x3 = ( 10) 3 2x = 10 x=5

Solution: log3 81x = 16 81x = ( 3)

(3 )

4 x

16

= ( 3)

8.

16

If loga 10=0.25, what is the value of log10 a=? Ans. 4 Solution: loga 10 = 0.25

34x = 316 4x = 16 x=4 4.

Solve for x : log10 8=3-3 log10 x Ans: 5

What is the value ( log5 to the base 2) + ( log5 to the base 3)

10 = a0.25 log10 = 0.25loga a = 10,000 log10 a = x

of

x = log10 10000

Ans: 3.79

x=4 log10 a = 4

Solution: log2 5 + log3 5 Let x = log2 5

9.

2x = 5 x log 2 = log 5 x = 2.322

What is the value of log to the base 10 of 10003.3 Ans: 9.9 Solution: log10 10003.3 3.3log1000

Let y = log3 5

3.3 ( 3) = 9.9

y

3 =5 y log3 = log5 y = 1.465 x + y = 2.322 + 1.465 x + y = 3.787 say 3.79 5.

10. Solve for x: log ( 2x + 7) − log ( x − 1) = log5 Ans: 4 Solution: log ( 2x + 7) − log ( x − 1) = log5

If log of 2 to the base 2 plus log of x to the base 2 is equal to 2, Find the value of x. Ans:2 Solution: log2 + log2 x = 2 log2 ( 2) ( x ) = 2 2x = ( 2)

2

( 2x + 7)

= log5 x −1 2x + 7 =5 x −1 2x + 7 = 5x − 5 3x = 12 x=4

log

2x = 4 x=2

6.

If (2 log x to the base 4) –(log 9 to the base 4) = 2, find x, Ans: 12

Page 1 of 13

11. Find the sum of the roots of 5x2 – 10x + 2 = 0 Ans: 2 Solution:

Mathematics Enhancers

n=5

5x 2 − 10X + 2 = 0 A=5 B = −10 C=2 B x1 + x 2 = − A −10) ( x1 + x 2 = − 5 x1 + x 2 = 2

k x

( x + 4y )

12

16. Find the term involving x8 in the expansion of ( x − 2y ) 8

Ans: 180 x y

.

10

.

2

Solution: Term involving x8 =

y

k x8 ( 4y )

2

5 n+2 17. Find the 8th term of the sequence −3, 4, ... 3 2n − 3 10 Ans: 13

4

Solution:

12! ( 4) 4 = 126720 8! 4! 13. What is the sum of the coefficients of the expansion of .

Ans: 0 Solution: Substitute x = 1 but subtract ( −1)

20

Sum of the coefficients 20

20

20

=0 14. What is the sum of the coefficients in the expansion of

( x + y − z) 8 .

1+ 2 = −3 2 ( 1) − 3

2+2 =4 4−3 3+2 5 when n = 3 = 6−3 3 8 + 2 10 when n = 8 = 16 − 3 13 when n = 2

Solution: 2 3 5 9 17 x 1 2 4 8 16 x = 17 + 16 x = 33 19. Find the value of x if 2 + 4 + 6 + 8 + ...x = 110 Usin g 2 + 4 + 6 + ....2n = n ( n + 1)

Ans: 1

Ans: 20

Solution: Substitute x = 1 y = 1 and z = 1 Sum of the coefficients = ( 1 + 1 − 1) Sum of the coefficients = ( 1)

when n = 1

18. What follows logically in these series of numbers 2,3,4,5,9,17….. Ans: 33

 2 ( 1) − 1 − ( −1) Sum of the coefficients − ( −1)

10! x 8 ( −2y )

8!2! Term involving x8 = 90 x8 (-2y)2 Term involving x8 = 180 x8 y2

4

k x8 y 4 ( 4)

20

15!

k = −1365

n −r +1 r −1

= ( 1)

=−

( n − m ) !m! 4!11! 15 ( 14 ) ( 13 ) ( 12 ) 11! k=− 4 ( 3 ) ( 2 ) ( 1) 11!

k x n −m y m m = r −1 m = 5 −1 = 4 n! k= ( n − m) !m!

( 2x − 1)

n!

k=−

Solution:

20

( −y )

−kx y

Ans: 126720

k x

m = 11

; 11

4 11

12. Compute the numerical coefficient of the 5th term of the expansion of

4

Solution: 20 2 + 4 + 6 + ...2n = n ( n + 1)

8

S = n ( n + 1)

110 = n ( n + 1)

8

n2 + n − 110 = 0

( n − 10 ) ( n + 11) = 0

Sum of the coefficients = 1

n = 10 x = 2n x = 20

15. Find the coefficient of the expansion of (x-y) 15 containing the term x4y11 Ans: -1365

20. Solve for x in the x + 3x + 5x + 7x + .....49x = 625. Ans: 1 Solution:

Solution:

Page 2 of 13

Mathematics Enhancers

following

equation.

a1 = x

12 6 = 6 x 36 x= 12 x=3

d = 2x an = a1 + ( n − 1) d

49x = x + ( n − 1) ( 2x )

49 = 1 + 2n − 2 50 = 2n n = 25 n S =  2a + ( n − 1) d 2 25  2 ( x ) + ( 24) ( 2x )  625 = 2  25 ( 50) x 625 = 2 625 = ( 25) ( 25) x

25. Find x if 7 is the fourth proportional to 36 and 28, and x. Ans: 9 Solution: 36 x = 28 7 36 ( 7) x= 28 x=9

x =1

( x − 4) ( x + 3) + 4, when

21. Given f(x) =

f(x) is divided by

( x − k ) , the remainder is K. Find the value of K.

Solution: F = kx

Ans: 4

24 = k ( 3)

Solution: f ( x ) = ( x − 4) ( x + 3) + 4

k=8 F = kx = 8 ( 2)

f ( K ) = ( K − 4) ( K + 3) + 4 ( remainder )

K = ( K − 4) ( K + 3) + 4

F = 16N

K = K 2 − K − 12 + 4

27. The volume of a hemisphere varies directly as the cube of it’s radius. The volume of the hemisphere with 2.54 cm. radius is 20.75 cm3. What is the volume of a sphere with3.25 cm. radius of the same kind of material? Ans:

K 2 − 2K − 8 = 0

( K − 4 ) ( K + 2) K=4

=0

K = −2

Solution: V = kr 3

22. Which of the following is a factor of 3x3 + 2x2 -32? Ans: x-2

20.75 = k ( 2.54)

Solution: f ( x ) = 3x3 + 2x 2 − 32 3

3

k = 1.266 V = kr 3

when x = 2

V = 1.266 ( 3.25)

2

f ( 2) = 3 ( 2) + 2 ( 2) − 32 f ( 2) = 0

3

V = 43.46cm3 ( hemisphere)

Note: If the remainder is zero, the number we assume is a factor. Therefore x-2 is a factor of 3x3+2x2 -32 23. Write a cubic equation whose roots are (-1,2,4) Ans: x3 – 5x2 + 2x + 8 = 0

2

V = 2 ( 43.46)

V = 86.92cm3 ( sphere) 28. If x varies directly as y and inversely as z, and x=14, when y=7 and z=2, find x when y=16 and z =4. Ans: 16

Solution: ( x + 1) ( x − 2) ( x − 4) = 0

( x + 1) ( x

26. The force required to stretched a spring is proportional to the elongation. If 24 N stretches a spring 3 mm, find the force required to stretch the spring 2 mm. Ans: 16

Solution:

)

y z 7 14 = k 2 k=4

− 6x + 8 = 0

x=k

x 3 − 6x 2 + 8x + x 2 − 6x + 8 = 0 x 3 − 5x 2 + 2x + 8 = 0

x=

4 ( 16) 4

x = 16

24. The mean proportion between 12 and x is equal to 6. Find the value of x. Ans: 3 Solution:

29. A tank is filled with 2 pipes. The first pipe can fill the tank 1 in 10 hours. But after it has been opened for 3 hours, 3 the second pipe is opened and the tank is filled up in 4 hours more. How long would it take the second pipe alone to fill the tank? The two pipes have different diameters. Ans: 15 Solution:

Page 3 of 13

Mathematics Enhancers

Then,

 1  1   1 1  + + 4 =1  3 3 ÷   10 ÷   10 x ÷  10 4 4 + + =1 3 ( 10) 10 x

80 ( 40 ) + 50 ( x ) = 72 ( 100) x = 80

x = 15 hours 30. A and B working together can finish painting a house in six days. A working alone, can finish it in five days less than B. How long will it take each of them to finish the work alone? Ans: 10,15

34. Twenty (20) men can finish the job in 30 days. Twenty five (25) men were hired at the start and 10 quit after 20 days. How many days will it take to finish the job? Ans: 27 Solution: No. of man-days: 20 ( 30) = 25 ( 20) + ( 25 − 10) x x = 6.67 say 7 days

Solution:  1 1 1  A + B ÷= 6   A =B−5

Total number of days to finish the job = 20 + 7 = 27 days

1  1  B − 5 + B ÷6 = 1   6 ( B + B − 5) = B ( B − 5 )

35. Given two numbers such that the difference of twice the first and the second number is 12. If the sum of the first and the second number is 36, find the number. Ans: 16,20

B2 − 17B + 30 = 0

( B − 15 ) ( B − 2 )

=0

B = 15 B=2 Use B = 15 A = 15 − 5 A = 10 31. Crew no.1 can finish the installation of an antenna tower in 200 man-hours while Crew No.2 can finish the same job in 300 man-hours. How long will it take both crews to finish the same job, working together? Ans: 120

Solution: x = 1st no. y = 2nd no. 2x − y = 12 x + y = 36 3x = 48 x = 16 y = 20 36. The product of

Solution: 1  1  1 =  200 + 300 ÷  x

( 300 + 200) 200 ( 300)

=

number. Ans: 100 Solution: 1 1  x x = 500  4 ÷ 5 ÷ 

1 x

x = 120 man − hours 32. An experienced statistical clerk submitted the following statistics to his manager on the average rate of production of transistorized radios in an assembly line, 1.5 workers produced 3 radios in 2 hours. How many workers are employed in the assembly line working 40 hrs each per week if weekly production is 480 radios? Ans: 12 Solution: No. of man-hours to produced 3 radios =

x 2 = 500 ( 20 ) x = 100 37. The square of the number increased by 16 is the same as 10 times the number. Find the numbers. Ans: 2,8 Solution: x 2 + 16 = 10x

1.5 ( 2)

No. of man-hours to produce 480 radios =

3 x ( 40) 480

By proportion: 40x 1.5 ( 2) = 480 3 x = 12 wor ker s

x 2 − 10x + 16 = 0

( x − 2) ( x − 8 )

=0

x=2 x=8 38. The sides of the right triangle are 8, 15 and 17 units. If each side is doubled, how many square units will the area of the new triangle. Ans: 240

33. A certain job can be done by 72 men in 100 days. There were 80 men at the start of the project but after 40 days, 30 of them had to be transferred to another project. How long will it take the remaining workforce to complete the job? Ans: 80 Solution: Let x =no. of days it will take for the remaining workforce to complete the job

Page 4 of 13

1 1 and of a number is 500. What is the 4 5

Solution: A=

16 ( 30 )

2 A = 240 sq.m. 39. Compute the median of the following set of numbers 4,5,7,10,14,22,25,30. Ans: 12

Mathematics Enhancers

Solution: The middle number is 10 and 14 Therefore the median is the average of 10 and 14 = 12

45. Roberto is 25 years younger than his father, However his father will be twice his age in 10 years. Find the ages of Roberto and his father. Ans: 15,40

40. To get an A in a course a student’s average must be at least 90(and at most 100). If a particular student’s grade so far are 84,88 and 93. What scores must that student get on the last test to earn an A if all tests court equally? Ans: 95

Solution: x = age of Roberto’s father x – 25 = age of Roberto x + 10= age of Roberto’s father 10 years from now ( x + 10) = 2 ( x − 25 + 10) x + 10 = 2x − 30 x = 40 father

Solution: 84 + 88 + 93 + x = 90 4 x = 95

x − 25 = 15

41. A students has test scores of 75, 83 and 78. The final tests counts half the total grade. What must be the minimum (integer) score on the final so that the average will be 80? Ans: 82

( Roberto)

46. The sum of the ages of Maria and Anna is 35. When Maria was two thirds her present age and Anna was 3/4 of her present age, the sum of their ages was 25. How old is Maria now? Ans: 15 Solution: M + A = 35 2 3 M + A = 25 3 4 8M + 9A = 300 M + A = 35 9M + 9A = 315 8M + 9A = 300 M = 15 ( age of Maria )

Solution: 75 + 83 + 78 3 Average = 78.67 x = final score x 1 + ( 78.67 ) = 80 2 2 x = 81.33 say 82 Average =

42. Which of the following is a prime number? Ans: 487

47. Ten (10) yrs from now the sum of the ages of A and B is equal to 50. Six (6) yrs ago, the difference of their ages is equal to 6. How old is A and B? Ans: A=18, B=12

Solution: Factors: 437 = 19 and 23 483 = 3 and 161 417 = 3 and 139 487 = 1 and 487

Solution: Past(6 yrs. Ago) Present(10 yrs. Later) A-6 A A + 10 B-6 B B + 10

Note: A prime number is a positive integer that has exactly two factors, the number itself and 1. 43. The boat travels downstream in 2/3 the time as it does going upstream, if the velocity of the river current is 8 kph, determine the velocity of the boat in the still water. Ans: 40 kph Solution: D 2 D = V + 8 3 ( V − 8)

u A + 10 + B + 10 = 50 A + B = 30 v ( A − 6) − ( B − 6) = 6 A −B = 6 A + B = 30 2A = 36 A = 18 B = 30 − 18 = 12

2 ( V + 8) = 3 ( V − 8) 2V + 16 = 3V − 24 V = 40 kph

44. The velocity of an airplane in still air is 125 kph. The velocity of the wind due east is 25 kph. If the plane travels east and returns back to its base again after 4 hours. At what distance does the plane travel due east? Ans: 240 km. Solution: x x + =4 125 + 25 125 − 25 x x + =4 150 100 x = 240 km.

Page 5 of 13

48. Two gallons of 20% salt solution is mixed with 4 gallons of 50% salt solution. Determine the percentage of salt solution in the new mixture. Ans: 40% Solution: 20%

2

Mathematics Enhancers

50%

4

x%

6=

2 ( 0.2) + 4 ( 0.5 ) = 6 ( x )

Solution: Gross margin=45% of sales

2.4 = 6x x = 0.40 x = 40%

Operating expenses = 15% of sales Net profit = 45-15

49. Two alcohol solution consists of a 40 gallons of 35% alcohol and other solution containing 50% alcohol. If the two solutions are combined together, they will have a mixture of 40% alcohol. How many gallons of solutions containing 50% alcohol? Ans: 20 Solution: 35%

40

50%

x

40% 40+x

Net profit =30% of sales Tax =40% Profit = 0.60 of 30% of sales Profit = 18% of sales 53. What is the sum of the progression 4,9,14,19… up to the 20th term? Ans: 1030 Solution:

40 ( 0.35 ) + x ( 0.50) = ( 40 + x ) ( 0.40 )

a=4 d= 9−4 = 5 n = 20 n S =  2a + ( n − 1) d 2 20  2 ( 4 ) + ( 19) ( 5)  S= 2  S = 1030

14 + 0.50x = 16 + 0.40x 0.10x = 2 x = 20 gallons 50. If you owned a sari-sari store in Kuwait, at what price will you mark a small camera for sale the cost P600 in order that you may offer 20% discount on the marked price and still makes a profit of 25 % on the selling price? Ans: P1000 Solution: x = marked price x − 0.20x = 600 + 0.25 ( 0.80x )

54. The are 9 arithmetic mean between 11 and 51. Compute the sum of the progression. Ans: 341 Solution: n [ a1 + an ] 2 11 S = [ 11 + 51] 2 S = 341

0.80x = 600 + 0.20x 0.60x = 600 x = 1000

S=

51. A mechanical Engineer bought 24 boxes of screws for P2200.00. There were three types of screws bought. Screw A costs P300 per box, screw B costs P150 and screw C costs P50 per box. How many boxes of screw A did he buy? Ans: 2 Solution: x=no. of boxes of screw A y=no. of boxes of screw B z= no. of boxes of screw C

55. An arithmetic progression starts with 1, has 9 terms and the middle term is 21. Determine the sum of the first 9 terms. Ans: 189

u x + y + z = 24 v 300x + 150y + 50z = 2200 6x + 3y + z = 44 x + y + z = 24 5x + 2y = 20 Try y = 5

Solution:

5x + 2 ( 5) = 20

2nd

3rd

4th

a =1

a+d

a+2d

a+3d

5th a+4d

6th

7th

a+5d a+6d

8th

9th

a+7d a+8d

Middle term = a + 4d 21 = 1 + 4d d=5 n S =  2a + ( n − 1) d 2 9 S =  2 ( 1) + ( 9 − 1) ( 5 )  2

x=2 2 + 5 + z = 24 z = 17 Check : 300x + 150y + 50z = 2200 300 ( 2) + 150 ( 5) + 17 ( 50) = 2200 2200 = 2200

S = 189

52. Dalisay Corporation’s gross margin is 45% of sales. Operating expenses such as sales and administration are 15% of sales. Dalisay is in 40% tax bracket. What percent of sales is their profit after taxes? Ans: 18%

Page 6 of 13

1st

56. If the sum is 220 and the first term is 10, find the common difference if the last term is 30. Ans: 2 Solution:

Mathematics Enhancers

Ans: 7

n ( a1 + an ) 2 n 220 = ( 10 + 30) 2 n = 11 S=

Solution: x + 2 = 16 − x 2x = 14 x=7

an = a1 + ( n − 1) d

30 = 10 + ( 11 − 1) d

61. Determine the positive value of x so that x,x2-5 and 2x will be a harmonic progression. Ans: 3

d=2 57. Determine the sum S of the S = 2 + 5 + 8 + 11.....with 100 terms Ans: 15050

following

series

Solution: 1 x2 − 5

(

1 1 1 = − x 2x x 2 − 5

)

2x − 2 x 2 − 5 = x 2 − 5 − 2x

Solution: n S =  2a + ( n − 1) d 2 100  2 ( 2) + ( 100 − 1) ( 3 )  S= 2  S = 15050

3x 2 − 4x − 15 = 0

( x − 3) ( 3x + 5) x=3 62. There are 4 geometric mean between 3 and 729, Find the sum of the G.P. Ans: 1092

58. Find the sum of the first n positive multiples of 4. Ans: 2n(n+1) Solution: a1 = 4

−

Solution: n=6

d=4

n S =  2a1 + ( n − 1) d 2 n S =  2 ( 4) + ( n − 1) 4 2 n S = [ 4n + 4] 2 S = 2n ( n + 1)

ar

a=3

n−1

= 729

5

a r = 729 3r 5 = 729 r 5 = 243 r =3 S=

)

(

a rn − 1

r −1 6 3  ( 3) − 1 3 ( 728)   S= = 3 −1 2 S = 1092

59. The sum of the three numbers in AP is 33, if the sum of their squares is 461, find the numbers. Ans: 4,11,18 Solution:

63. The number x, 2x + 7, 10x – 7 form a G.P. Find the value of x: Ans: 7 Solution: 2x + 7 10x − 7 = x 2x + 7

( 2x + 7) 2 = 10x 2 − 7x

a − d = 1st no. a = 2nd no. a + d = 3rd no. a − d + a + a + d = 33 a = 11

6x 2 − 35x − 49 = 0

( x − 7) ( 6x + 7)

64. Find the value of x from the given Geometric Progression 1 2 4 , , ... 5 x 45

( 11 − d) 2 + ( 11) 2 + ( 11 + d) 2 = 461 121 − 22d + d2 + 121 + 121 + 22d + d2 = 461 2d2 = 98 d=7 11 − 7 = 4 11 = 11 11 + 7 = 18 The numbers are 4,11,18

Ans: 15 Solution:

60. Find the value of x if the following forms a harmonic 1 1 1 progression. − , , 2 x 16

Page 7 of 13

=0

x=7

Mathematics Enhancers

S = 1.01 + 1.1 + 1.21 + 1.331 + ....up to the 50th term Ans: 1163.91

2 / x 4 / 45 = 1/ 5 2/ x 2

4  1  2 =  x ÷   45  5 ÷ 2

x =

Solution:

65. The arithmetic mean of 6 numbers is 17. If two numbers are added to the progression, the new set of number will have an arithmetic mean of 19. What are the two numbers is their difference is 4? Ans: 23 and 27 Solution: let x = one number x + 4 = other number S = 17 6 S = 102 S+x+x+4 = 19 8 102 + 2x + 4 = 152 x = 23 x + 4 = 27

Solution: −3 y = x −3 xy = 9 For A.P. y+3 = x−y 2y + 3 = x

1 2 = 3 3

an = a1 r n−1 2  2  3  3 ÷ an = 0.088(remaining air after 6 strokes) Air removed after 6 strokes : = 1 − 0.088 = 0.9122 67. If a stroke of a vacuum pump removed 15% of the air from the container, how many strokes are required to removed 95%of the air? Ans: 19 Solution: a1 = 85% amount of air left after 1st stroke

an = 5% amount of air left after n strokes an = a r n −1 5 = 85 ( 0.85)

n −1

( 0.85) n−1 = 0.05882 ( n − 1) In 0.85 = In ( 0.05882)

71. Find the total distance traveled by the tip of a pendulum if the distance of the first swing is 6 cm. and the distance of each succeeding swing is 0.98 of the distance of the previous swing. Ans: 300 Solution: a = 6cm r = 0.98 a S= 1− r 6 S= 1 − 0.98 S = 300cm 72. Suppose a ball rebounds one half the distance if falls. If it is dropped from a height of 40 fet, how far does it travel before coming to stop? Ans: 120 feet Solution:

n = 18.43 say 19 strokes 68. The sum of a geometric series is as follows:

Page 8 of 13

=0

1.1 1.21 1.331 = = = 1.1 1 1.1 1.21 11.150 − 1  S=  1.1 − 1 S = 1163.91

an =

an = a r n −1

( x − 6) ( x + 3)

r=

5

amount of air left after 2nd strokes 72.25 r= = 0.85 85

x 2 − 3x − 18 = 0

Solution: a1 = 1

n=6

a2 = 0.85 ( 85) = 72.25%

 9 2 ÷+ 3 = x  x

x=6 x = −3 70. The sum of a geometric series is as follows: S = 1.0 + 1.1 + 1.21 + 1.331 + ...up to the 50th term Ans: 1163.91

Solution: a1 = amount of remaining air

2 3

a1 r n − 1  S=  r −1 11.150 − 1  S=  1.1 − 1 S = 1163.91 69. x and y are positive numbers. If x,-3,y forms a G.P and -3,y,x forms an A>P. Find the value of x. Ans: 6,-3

66. If one third of the air in a tank is removed by each stroke of an air pump, what fractional part of the total air is removed in 6 strokes? Ans: 0.9122

r=

r=

4

x = 15

a1 = 1 −

1.1 1.21 1.331 = = 1 1.1 1.21 r = 1.1

a1 = 1

45 ( 5) ( 4)

Mathematics Enhancers

(n − 2)180 = 165n n = 24

a1 a −r 1 a1 = ( 40 ) ( 2 ) 2 a1 = 40 S=

77. Find the largest angle of a triangle if the sum and difference of two angles are 100 and 20 degrees, respectively.

a1 1− r 40 S= = 80 1 1− 2 S=

Ans: 80 degrees Solution: Let:

x = one angle y = second angle

Total distance the ball has traveled = 80 + 40 = 120 feet 73.

Then,

A policeman is pursuing a thief who is ahead by 72 of his own leaps. The thief takes 6 leaps while the policeman is taking 5 leaps, but 4 leaps of the thief are as long as 3 leaps of the policeman. How many leaps will each make before the thief is caught? Ans: 648 leaps & 540 leaps

x=

180 − x = 5(90 − x) x = 67.5°

Ans: 76o

56 x

56 ( 5)

2

( 4) 2

80. The sides of a pentagon measures 6,5,2,8 and 4 m. respectively. The shortest side of a similar pentagon is 1 meter, find the perimeter of this pentagon. Ans: 12.5 m. Solution: Perimeter of given pentagon = 6+5+2+8+4 Perimeter of given pentagon = 25 m.

Solution: Let:

x = angle 180 − x = supplement 90 − x = complement

Ratio of corresponding smaller sides = ratio of perimeter

(180 − x) − 6(90 − x) = 20 x = 76°

76. How many sides has an equiangular polygon if each of its interior angles is 165 degrees?

Page 9 of 13

=

x = 87.5 sq.cm.

75. Find the angle whose supplement exceeds 6 times its complement by 20o.

Then,

79. The corresponding sides of two similar polygon is 4:5. If the area of the smaller polygon is 56 sq. cm., what is the area of the bigger polygon? Ans: 87.5 sq.cm.

( 4) 2 ( 5) 2

x = angle 180 − x = supplement 90 − x = complement

n = number of sides = number of interior angles

= 180 − ( 60 + 40 )

Solution:

Solution: Let:

Let:

Thus, the largest angle is the third angle:

Solution: 4 x = 6 48 4 ( 48) x= 6 x = 32cm.

Ans: 67.5o

Ans: 24 sides

x = 60 y = 40

78. The two corresponding sides of two similar polygon is 4:6 If the perimeter of the larger polygon is 48 cm, what is the perimeter if the smaller polygon? Ans: 32 cm.

74. Find the angle whose supplement exceeds 5 times its complement.

Then,

Solving the two equations simultaneously, we get:

= 80°

Solution: Let: x = additional leaps of the thief 5 x = number of leaps of the policeman 6 By proportion: x + 72 4 = 5 3 x 6 x = 648 leaps of the thief 5 x = 540 leaps of the policeman 6

Then,

x + y = 100   x − y = 20 

1 P = 2 25 P = 12.5 m. (perimeter of the smaller pentagon) 81. The corresponding sides of two similar polygons is 2:4. Determine the ratio of the area to the perimeter of the biggest polygon if the smallest polygon has a perimeter of 24 cm. and an area of 36 sq. cm. Ans: 3 Solution:

Mathematics Enhancers

2 24 = 4 x x = 48 cm. ( perimeter of biggest polygon)

( 2) 2 ( 4) 2

=

86. An engineer places his transit along the line tangent to the circle at point A such that PA = 200 m. He locates another point B on the circle and finds PB = 80 m. If a third point C, on the circle lies along PB, how far from point B will it be? Ans: 420 m.

36 A

A = 144 area of biggest polygon 144 Ratio of area to perimeter = 48 Ratio of area to perimeter = 3 82. Seven regular hexagons, each with 6 cm. sides are arranged so that they share the same sides and the centers of the six hexagons are equidistant from the seventh central hexagon. Determine the ratio of the total area of the hexagons to the total outer perimeter enclosing the hexagons. Ans: 6.06

Solution: 200 80 = PC 200 PC = 500 BC = 500 − 80 BC = 420 m. 87. A circle having a center at point O has a radius of 20 m. A triangle ABC is inscribe in a circle. If the angle BCA = 30 o and BAC = 50o compute the area of the triangle. Ans: 301.73 m2 Solution:

Solution:

( AC) 2 = ( 20) 2 + ( 20) 2 − 2 ( 20) ( 20) Cos 160o

( 6) 2 Sin60—( 6) ( 7) A=

AC = 39.39 cm.

2

A = 654.72 cm2 P = 6 ( 18)

Area of triangle =

2 Sin 100o Area of triangle = 301.73 sq.m.

P = 108 A P 654.72 Ratio = 108 Ratio = 6.06 Ratio =

88. The central circle has 10 cm. radius. Six equal smaller circles are to be arranged so that they are externally tangent to the central circle and each tangent to the adjacent small circle. What should be the radius in cm. of each small circle? Ans: 10

83. A chord is 36 cm. long and its mid point is 36 cm. from the midpoint of the longer arc. Find the radius of the circle. Ans: 22.5 Solution: x ( 36) = 18 ( 18 )

x

x=9 2r = 9 + 36 r = 22.5 cm. 84. A circular play area is being laid out in a field. The point P,R and Q have been found such that PS=24 m., PQ=66 m., and RS=12 m. If T is to be another point on the circle, how far from R will it lie? Ans: 96 m. Solution: 12x = 42 ( 24)

AD = 32 m.

Page 10 of 13

89. The distance between the center of the circles which are mutually tangent to each other externally are 10,12 and 14 units, The area of the largest circle is: Ans: 64 π Solution: u r1 + r3 = 10 wr2 + r3 = 14 u and v r3 − r2 = −2

85. Two secants AC and AE have lengths of 80 m. and 100 m. respectively are drawn from point A outside a circle and intersects the circle at points C,B, D and E. If the angle between the two secants is 25o, and AB is equal to 40 m. compute the length of AD. Ans: 32 m.

40 ( 80 ) = AD ( 100)

Solution: 2θ + 60 = 180 2θ = 120 θ = 60 10 + r = 2r r = 10

v r1 + r2 = 12

x = 84 TR = 84 + 12 TR = 96 m.

Solution: AB AE = AD AC ( AB) ( AC) = AD ( AE)

( 39.39) 2 Sin 30o Sin 50o

r3 + r2 = 14 2r3 = 12 r3 = 6 r2 = 14 − 6 = 8 r1 = 10 − 6 = 4 Area of largest circle = π ( 8) Area of largest circle = 64 π

2

90. The area of a circular sector is 800 m2. If the radius is equal to 16 m, compute for the length of arc Ans: 100 m. Solution:

Mathematics Enhancers

S = Rθ

94. The ratio of the volume to the lateral area of a right circular cone is 2:1. if the altitude is 15 cm., what is the ratio of the slant height to the radius. Ans: 5:2

2

A πR = θ 360o A=

πθR2 2π

A=

θR2 2

800 =

Solution: S = πrL πr 2h 3 V 2 = S 1 V = 2S V=

θ ( 16)

2

2

θ = 6.25 S = Rθ = 16 ( 6.25 )

91. A sector is bent to form a cone. If the angle of a sector is 30 degrees and radius of 6 cm., what is the altitude of the cone? Ans: 5.98 cm. Solution: S = rθ S=

6 ( 30) π

180 S=π S = 2πR π = 2πR 1 R= 2 h = ( 6) 2

Ratio of slant height to radius = 5:2

2

 1 − ÷  2

2

πr 2h 3 6S πr = rh S πr = L 6S S = rh L 6L = rh L h = r 6 L 15 = r 6 L 5 = r 2 2S =

S = 100 m.

h = 5.98 cm.

92. Find the volume of a cone to be constructed from a sector having a diameter of 72 cm. and a central angle of 150o. Ans: 7711.82 Solution: S = Rθ S=

36 ( 150) π

Solution: V1 = 4V2

180 S = 94.25 94.25 = 2πr r = 15 2

h2 = ( 36) − ( 15)

V1 =4 V2 V1 ( 8) = 3 V2 h

3

2

h = 32.73

4=

2

πr 2h π ( 15) ( 32.73) = 3 3 V = 7711.83 93. The lateral area of a right circular cone is 40 π sq.m. The base radius is 4 m. What is the slant height? Ans: 10

40 π = π ( 40) L L = 10

Page 11 of 13

( 8) 3

h3 h = 5.04 cm.

V=

Solution: S = πrL

95. Given a solid right circular cone having a height of 8 cm. has a volume equal to 4 times the volume of the smaller cone that could be cut from the same cone having the same axis. Compute the height of the smaller cone. Ans: 5.04 cm.

96. A circular cone having an altitude of 9 m. is divided into 2 segments having the same vertex. If the smaller altitude is 6 m., find the ratio of the volume of small cone to the big cone. Ans: 0.296 Solution: V1 ( 6 ) = = 0.296 V2 ( 9 ) 3 3

Mathematics Enhancers

97. The height of a right circular cone is “h”. If it contains oil and water at equal depth of h/2, what is the ratio of the volume of the volume of oil to that of the water. Ans: 7 Solution: V1 ( h / 2) = V2 h3 V V1 = 8 V2 = V − V1 ( vol. of oil) 3

V2 = V −

Vol. of cone = Vol. of Cyl. π ( 32)

2

( 54)

3 h = 32 in.

= π ( 24 ) h 2

101.A pyramid has a triangular base whose sides are 10 cm., 14 cm. and 18 cm. respectively. Compute the volume of the inscribed cone if it has an altitude of 28 cm. Ans: 323.19 Solution: A = S ( s − a ) ( s − b) ( s − c )

V 8

10 + 14 + 18 = 21 2 S − a = 11, S − b = 7, S − c = 3 S=

7 V2 = V 8 V2 7 / 8V = V1 V/8

A = 2 ( 11) ( 7) ( 3) A = 69.65 sq.m. A = rS

V2 =7 V1

69.65 = r ( 21)

98. A plane is passed parallel to the base of a triangular pyramid of altitude of 9 m. such that the area of the base is 9 times the area of the triangle of intersection. How far from the vertex does the plane intersect the altitude? Ans: 3 m.

r = 3.32 cm. πr 2h π ( 3.32) ( 28 ) = 3 3 V = 323.19 cu.m. V=

Solution: A1 ( h ) = A 2 ( 9) 2 2

A1 h2 = 9A1 81 h= 3 m.

99. The volume of a cone having an inclined axis at an angle of 60o with the base is equal to 1884.96 cu.m., find the length of the axis of the cone, if the radius at the base is 10 m. Ans: 20.78 m. Solution: V=

Ah 3

1884.96 =

π ( 10) 3

2

Solution:

h

The radius of the base of a cone of revolution is 32 inches and its altitude is 54 inches. What is the altitude of a cylinder of the same volume whose diameter of the base is 48 inches? Ans: 32 in.

103.The bases of a right prism is a hexagon with one of each side equal to 6 cm. If the volume of the right prism is 500 cu. cm., find the distance between the bases. Ans: 5.35 cm. Solution: V = Ah 500 =

6 ( 6) Sin 60o

h = 5.35 cm

Page 12 of 13

)

V = 917.13 cu.cm. 4 3 πr ( 9 ) = 917.13 3 r = 2.90 cm.

x = 20.78m. ( axis length)

Solution:

(

h b + B + bB 3 24  V= 20 + 60 20 ( 60 )   3  V=

h = 18m Length of axis : h Sin 60o = x 18 x= Sin 60o

100.

102.A solid gold in the form of a frustum of a pyramid has smaller base area of 20cm2 and a bigger base area of 60 cm2 . It has an altitude of 24 cm. If the gold is melted to form a 9 spherical balls, find the radius of each balls. Ans: 2.90 cm.

Mathematics Enhancers

2

( 6) h

104.The bases of a right prism is a hexagon with one of each side equal to 6 cm. The bases are 12 cm. apart. What is the volume of the right pris. Ans: 1122.4 cu. cm. Solution: V = Base x height 360 θ= = 60 6 1  Base =  ( 6) ( 6 ) Sin 60 60 2   Base = 93.53 V = 93.53 ( 12) V = 1122.4 cu. cm.

Page 13 of 13

Mathematics Enhancers