math wd solns

7/27/2019 Math Wd Solns

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

Solution:

( )

( )

3 / 4

4 / 3

4

x 8

x 8

x 2

x 16

=

=

=

=

2. If 6 / 8a 0.001,− = solve for a.

Ans: 10,000

Solution:6 / 8a 0.001

6In In 0.001

8

a 10,000

− =

− α =

=

3. Solve for x if log3 x81 16=Ans: 4

Solution:

( )

( ) ( )

x3

16x

x 164

4x 16

log 81 16

81 3

3 3

3 3

4x 16

x 4

=

=

=

==

=

4. What is the value of

( ) ( )log 5 to the base 2 log 5 to the base 3+

Ans: 3.79

Solution:

2 3

2

x

3

y

log 5 log 5

Let x log 5

2 5

xlog 2 log 5

x 2.322

Let y log 5

3 5

y log3 log5y 1.465

x y 2.322 1.465

x y 3.787 say 3.79

+

=

==

=

=

=

==+ = ++ =

5. If log of 2 to the base 2 plus log of x to the base 2 is equalto 2, Find the value of x.Ans:2

Solution:

( ) ( )

( )

2

2

2

log2 log x 2

log 2 x 2

2x 2

2x 4

x 2

+ =

=

=

==

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

Solution:

( )

4 4

24 4

2

4

22

2log x log 9 2

log x log 9 2

xlog 2

9

x4

9x 12

− =

− =

= ÷

=

=

7. Solve for x : log10 8=3-3 log10 xAns: 5

Solution:

( ) ( )

( ) ( )

10 10

310 10

310

3 3

3 33

log 8 3log x 3

log 8 log x 3

log 8 x 3

8x 10

2 x 10

2x 10

x 5

+ =

+ =

=

=

=

==

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

Solution:

a

0.25

10

10

10

log 10 0.25

10 a

log10 0.25loga

a 10,000

log a x

x log 10000x 4

log a 4

=

==

==

==

=

9. What is the value of log to the base 10 of 10003.3

Ans: 9.9

Solution:

( )

3.310log 1000

3.3log1000

3.3 3 9.9=

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

Solution:

( ) ( )

( )

log 2x 7 log x 1 log5

2x 7log log5

x 1

2x 75

x 1

2x 7 5x 5

3x 12

x 4

+ − − =

+=

−+

=−+ = −=

=

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

Solution:

of 13 Mathematics Enhancers



( )

2

1 2

1 2

1 2

5x 10X 2 0

A 5

B 10

C 2

Bx x

A

10x x

5

x x 2

− + === −=

+ = −

−+ = −

+ =

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

expansion of ( )12

x 4y .+

Ans: 126720

Solution:

( )

( )

( )

( )

n r 1 r 1

n m m

48

48 4

4

k x y

k x y

m r 1

m 5 1 4

n!kn m !m!

k x 4y

k x y 4

12!4 126720

8!4!

− + −

−

= −= − =

= −

=

13. What is the sum of the coefficients of the expansion of

( )20

2x 1 .−

Ans: 0

Solution:

( )

( ) ( )

( ) ( )

20

20 20

20 20

Substitute x 1but subtract 1Sum of the coefficients

2 1 1 1

Sum of the coefficients

1 1

0

= −

− − −

= − −

=

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

( )8

x y z .+ −

Ans: 1

Solution:Substitute x = 1 y = 1 and z = 1

Sum of the coefficients = ( )8

1 1 1+ −

Sum of the coefficients = ( )8

1

Sum of the coefficients = 1

15. Find the coefficient of the expansion of (x-y)15 containingthe term x4y11

Ans: -1365

Solution:

( )

( )

( ) ( ) ( )

( ) ( ) ( )

114

4 11

n 5 ; m 11

k x y

kx y

n! 15!k

n m !m! 4!11!

15 14 13 12 11!k

4 3 2 1 11!

k 1365

= =

−

−

= − = −−

= −

= −

16. Find the term involving x8 in the expansion of ( )10

x 2y .−

Ans: 180 x8 y2

Solution:

Term involving x8 = ( )2810!x 2y

8!2!

−

Term involving x8 = 90 x8 (-2y)2

Term involving x8 = 180 x8 y2

17. Find the 8th term of the sequence5 n 2

3,4, ...

3 2n 3

+−

−Ans:

10

13

Solution:

( )

1 2when n 1 3

2 1 3

2 2when n 2 4

4 3

3 2 5when n 3

6 3 3

8 2 10when n 8

16 3 13

+= = −

−

+= =

−+

= =−+

= =−

18. What follows logically in these series of numbers2,3,4,5,9,17…..Ans: 33

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+ + + + =( )Using 2 4 6 ....2n n n 1+ + + = +

Ans: 20

Solution: 20

( )

( )

( )

( ) ( )

2

2 4 6 ...2n n n 1

S n n 1

110 n n 1

n n 110 0

n 10 n 11 0

n 10

x 2n

x 20

+ + + = +

= +

= +

+ − =

− + =

===

20. Solve for x in the following equation.

x 3x 5x 7x .....49x 625.+ + + + =Ans: 1

Solution:




( )

( ) ( )

( )

( ) ( ) ( )

( )

( ) ( )

1

n 1

a x

d 2x

a a n 1 d

49x x n 1 2x

49 1 2n 2

50 2n

n 25

nS 2a n 1 d2

25625 2 x 24 2x

2

25 50 x625

2

625 25 25 x

x 1

=

=

= + −

= + −

= + −=

=

= + −

= +

=

=

=

21. Given f(x) = ( ) ( )x 4 x 3 4,− + + when f(x) is divided by

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

Ans: 4

Solution:

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

2

2

f x x 4 x 3 4

f K K 4 K 3 4 remainder

K K 4 K 3 4

K K K 12 4

K 2K 8 0

K 4 K 2 0

K 4 K 2

= − + +

= − + +

= − + +

= − − +

− − =

− + =

= = −

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

Solution:

( )

( ) ( ) ( )

( )

3 2

3 2

f x 3x 2x 32

when x 2

f 2 3 2 2 2 32

f 2 0

= + −

=

= + −

=

Note: If the remainder is zero, the number we assume is afactor. 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

Solution:

( ) ( ) ( )

( ) ( )2

3 2 2

3 2

x 1 x 2 x 4 0

x 1 x 6x 8 0

x 6x 8x x 6x 8 0

x 5x 2x 8 0

+ − − =

+ − + =

− + + − + =

− + + =

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

Solution:

12 6

6 x

36x

12

x 3

=

=

=

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

Solution:

( )

36 x

28 7

36 7x

28

x 9

=

=

=

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

Solution:

( )

( )

F kx

24 k 3

k 8

F kx 8 2

F 16N

=

=

=

= =

=

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 spherewith3.25 cm. radius of the same kind of material?Ans:

Solution:

( )

( )

( )

( )

( )

3

3

3

3

3

3

V kr

20.75 k 2.54

k 1.266

V kr

V 1.266 3.25

V 43.46cm hemisphere

V 2 43.46

V 86.92cm sphere

==

=

=

=

=

=

=

28. If x varies directly as y and inversely as z, and x=14, wheny=7 and z=2, find x when y=16 and z =4.

Ans: 16

Solution:

( )

yx k

z

714 k

2

k 4

4 16x

4

x 16

=

=

=

=

=

29. A tank is filled with 2 pipes. The first pipe can fill the tank

in 10 hours. But after it has been opened for 31

3hours,

the second pipe is opened and the tank is filled up in 4hours more. How long would it take the second pipe aloneto fill the tank? The two pipes have different diameters.Ans: 15

Solution:




( )

1 1 1 13 4 1

3 10 10 x

10 4 41

3 10 10 x

x 15 hours

+ + = ÷ ÷ ÷

+ + =

=

30. A and B working together can finish painting a house in sixdays. A working alone, can finish it in five days less than

B. How long will it take each of them to finish the workalone?Ans: 10,15

Solution:

( ) ( )

( ) ( )

2

1 1 1

A B 6

A B 5

1 16 1

B 5 B

6 B B 5 B B 5

B 17B 30 0

B 15 B 2 0B 15 B 2

Use B 15

A 15 5

A 10

+ = ÷

= −

+ = ÷− + − = −

− + =

− − == =

== −=

31. Crew no.1 can finish the installation of an antenna tower in200 man-hours while Crew No.2 can finish the same job in300 man-hours. How long will it take both crews to finishthe same job, working together?Ans: 120

Solution:

( )

( )

1 1 1200 300 x

300 200 1

200 300 x

x 120 man hours

+ = ÷

+=

= −

32. An experienced statistical clerk submitted the followingstatistics to his manager on the average rate of productionof transistorized radios in an assembly line, 1.5 workersproduced 3 radios in 2 hours. How many workers areemployed 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( )1.5 2

3=

No. of man-hours to produce 480 radios( )x 40

480=

By proportion:

( )1.5 240x

480 3

x 12 wor ker s

=

=

33. A certain job can be done by 72 men in 100 days. Therewere 80 men at the start of the project but after 40 days,

30 of them had to be transferred to another project. Howlong will it take the remaining workforce to complete the job?Ans: 80

Solution:Let x =no. of days it will take for the remainingworkforce to complete the job

Then,

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

x 80

+ =

=

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

= + −

=

Total number of days to finish the job

20 7

27 days

= +=

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

Solution:

x 1st no.

y 2nd no.

2x y 12

x y 36

3x 48

x 16

y 20

==− =

+ ====

36. The product of 1 1

and4 5

of a number is 500. What is the

number.

Ans: 100

Solution:

( )2

1 1x x 500

4 5

x 500 20

x 100

= ÷ ÷

=

=

37. The square of the number increased by 16 is the same as10 times the number. Find the numbers.Ans: 2,8

Solution:

( ) ( )

2

2

x 16 10x

x 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 areaof the new triangle.Ans: 240

Solution:

( )16 30

A 2

A 240 sq.m.

==

39. Compute the median of the following set of numbers4,5,7,10,14,22,25,30.Ans: 12




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

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

Solution:

84 88 93 x90

4

x 95

+ + +=

=

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

Solution:

( )

75 83 78

Average 3

Average 78.67

x final score

x 178.67 80

2 2

x 81.33 say 82

+ +

==

=

+ =

=

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

Solution:Factors:

437 19 and 23

483 3 and 161

417 3 and 139

487 1 and 487

=

===

Note: A prime number is a positive integer that hasexactly two factors, the number itself and 1.

43. The boat travels downstream in 2/3 the time as it doesgoing 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

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. Thevelocity of the wind due east is 25 kph. If the plane travelseast and returns back to its base again after 4 hours. Atwhat distance does the plane travel due east?Ans: 240 km.

Solution:

x x4125 25 125 25

x x4

150 100

x 240 km.

+ =+ −

+ =

=

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

Solution:x = age of Roberto’s father x – 25 = age of Robertox + 10= age of Roberto’s father 10 years from now

( ) ( )

( )

x 10 2 x 25 10

x 10 2x 30

x 40 father

x 25 15 Roberto

+ = − +

+ = −=

− =

46. The sum of the ages of Maria and Anna is 35. When Mariawas two thirds her present age and Anna was 3/4 of her present age, the sum of their ages was 25. How old isMaria now?Ans: 15

Solution:

( )

M A 35

2

M 253

8M 9A 300

M A 35

9M 9A 315

8M 9A 300

M 15 age of Maria

3

A4

+ =

+ =+ =

+ =+ =

+ ==

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

Solution:Past(6 yrs. Ago) Present(10 yrs. Later)

A-6 A A + 10B-6 B B + 10

( ) ( )

A 10 B 10 50

A B 30

A 6 B 6 6

A B 6

A B 30

2A 36

A 18

B 30 18 12

+ + + =+ =

− − − =

− =+ =

=== − =

u

v

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

Solution:20% 50% x%

=


4

62



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

2.4 6x

x 0.40

x 40%

+ =

===

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

Solution:35% 50% 40%

( ) ( ) ( ) ( )40 0.35 x 0.50 40 x 0.40

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 willyou mark a small camera for sale the cost P600 in order that you may offer 20% discount on the marked price andstill makes a profit of 25 % on the selling price?Ans: P1000

Solution:

( )

x marked price

x 0.20x 600 0.25 0.80x

0.80x 600 0.20x

0.60x 600

x 1000

=

− = +

= +=

=

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 andscrew C costs P50 per box. How many boxes of screw Adid 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

( )

( ) ( ) ( )

x y z 24

300x 150y 50z 22006x 3y z 44

x y z 24

5x 2y 20

Try y 5

5x 2 5 20

x 2

2 5 z 24

z 17

Check :

300x 150y 50z 2200

300 2 150 5 17 50 22002200 2200

+ + =

+ + =+ + =

+ + =+ ==

+ =

=+ + ==

+ + =

+ + ==

u

v

52. Dalisay Corporation’s gross margin is 45% of sales.Operating expenses such as sales and administration are15% of sales. Dalisay is in 40% tax bracket. What percentof sales is their profit after taxes?Ans: 18%

Solution:Gross margin=45% of sales

Operating expenses = 15% of sales

Net profit = 45-15

Net profit =30% of sales

Tax =40%

Profit = 0.60 of 30% of salesProfit = 18% of sales

53. What is the sum of the progression 4,9,14,19… up to the20th term?Ans: 1030

Solution:

( )

( ) ( ) ( )

a 4

d 9 4 5

n 20

nS 2a n 1 d

220

S 2 4 19 52

S 1030

== − ==

= + −

= +

=

54. The are 9 arithmetic mean between 11 and 51. Computethe sum of the progression.Ans: 341

Solution:

[ ]

[ ]

1 nn

S a a2

11S 11 51

2

S 341

= +

= +

=

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

terms.Ans: 189

Solution:

1st 2nd 3rd 4th 5th 6th 7th 8th 9th

a =1 a+d a+2d a+3d a+4d a+5d a+6d a+7d a+8d

Middle term = a + 4d

( )

( ) ( ) ( )

21 1 4d

d 5

nS 2a n 1 d

2

9S 2 1 9 1 5

2S 189

= +=

= + −

= + −

=

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

Solution:


x40 40+x



( )

( )

( )

( )

1 n

n 1

nS a a

2

n220 10 30

2

n 11

a a n 1 d

30 10 11 1 d

d 2

= +

= +

=

= + −

= + −

=

57. Determine the sum S of the following series

S 2 5 8 11.....with 100 terms= + + +Ans: 15050

Solution:

( )

( ) ( ) ( )

nS 2a n 1 d

2

100S 2 2 100 1 3

2

S 15050

= + −

= + −

=

58. Find the sum of the first n positive multiples of 4.

Ans: 2n(n+1)

Solution:

( )

( ) ( )

[ ]

( )

1

1

a 4 d 4

nS 2a n 1 d

2

nS 2 4 n 1 4

2

nS 4n 4

2

S 2n n 1

= =

= + −

= + −

= +

= +

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:

( ) ( ) ( )2 2 2

2 2

2

a d 1st no.

a 2nd no.

a d 3rd no.

a d a a d 33

a 11

11 d 11 11 d 461

121 22d d 121 121 22d d 461

2d 98

d 7

11 7 4

11 11

11 7 18

The numbers are 4,11,18

− ==+ =− + + + ==

− + + + =

− + + + + + =

==− =

=+ =

60. Find the value of x if the following forms a harmonic

progression.1 1 1

, ,2 x 1 6

−

Ans: 7

Solution:x 2 16 x

2x 14

x 7

+ = −=

=

61. Determine the positive value of x so that x,x2-5 and 2x willbe a harmonic progression.

Ans: 3

Solution:

( )

( ) ( )

2 2

2 2

2

1 1 1 1

x 2xx 5 x 5

2x 2 x 5 x 5 2x

3x 4x 15 0

x 3 3x 5

x 3

− = −− −

− − = − −

− − =

− +

=

62. There are 4 geometric mean between 3 and 729, Find thesum of the G.P.

Ans: 1092

Solution:

( )

( ) ( )

n 1

5

5

5

n

6

n 6 a 3

a r 729

a r 729

3r 729

r 243

r 3

a r 1S

r 1

3 3 1 3 728S

3 1 2

S 1092

−

= =

=

=

=

==

−=

−

− = =−

=

63. The number x, 2x + 7, 10x – 7 form a G.P. Find the valueof x:Ans: 7

Solution:

( )

( ) ( )

2 2

2

2x 7 10x 7

x 2x 72x 7 10x 7x

6x 35x 49 0

x 7 6x 7 0

x 7

+ −=

++ = −

− − =

− + =

=

64. Find the value of x from the given Geometric Progression

1 2 4, , ...

5 x 4 5

Ans: 15

Solution:




( ) ( )

2

2

2 / x 4 / 45

1/ 5 2 / x

2 4 1

x 45 5

45 5 4x

4

x 15

=

= ÷ ÷

=

=

65. The arithmetic mean of 6 numbers is 17. If two numbersare added to the progression, the new set of number willhave an arithmetic mean of 19. What are the two numbersis their difference is 4?Ans: 23 and 27

Solution:let x = one number x + 4 = other number

S17

6

S 102

S x x 419

8

102 2x 4 152

x 23

x 4 27

=

=+ + +

=

+ + ==+ =

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

Solution:

1

1

n 1n 1

5

n

n

a amount of remaining air

1 2a 1

3 3

2r n 63

a a r

2 2a

3 3

a 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 fromthe container, how many strokes are required to removed95%of the air?

Ans: 19

Solution:

( )

( )( )

( ) ( )

1

2

n 1n

n

n 1n

n 1

n 1

a 85% amount of air left after 1st stroke

a 0.85 85 72.25%

amount of air left after 2nd strokes

72.25r 0.85

85

a a r

a 5% amount of air left after n strokes

a a r

5 85 0.85

0.85 0.05882

n 1 In 0.85 In 0.05882

n 18.43 say 1

−

−

−

−

=

= =

= =

=

=

=

==

− =

= 9 strokes

68. The sum of a geometric series is as follows:

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

Solution:

1

n1

50

1.1 1.21 1.331a 1 r

1 1.1 1.21

r 1.1

a r 1S

r 1

1 1.1 1S

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

Solution:

( ) ( )

2

3 y

x 3

xy 9

For A.P.

y 3 x y

2y 3 x

92 3 x

x

x 3x 18 0

x 6 x 3 0

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:

1

50

a 1

1.1 1.21 1.331r 1.1

1 1.1 1.21

1 1.1 1S

1.1 1

S 1163.91

=

= = = =

− =−

=

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 theprevious swing.Ans: 300

Solution:

a 6cm

r 0.98

aS

1 r

6S

1 0.98

S 300cm

==

=−

=−

=

72. Suppose a ball rebounds one half the distance if falls. If itis dropped from a height of 40 fet, how far does it travel

before coming to stop?Ans: 120 feet

Solution:




( ) ( )

1

1

1

1

aS

a r

1a 40

2

a 40

aS

1 r

40S

2

80

112

=−

=

=

=−

= =−

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

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

Solution:Let:

x = additional leaps of the thief 5

x6

= number of leaps of the policeman

By proportion:

x 72 4

5 3x6

x 648 leaps of the thief

5x 540 leaps of the policeman

6

+=

=

=

74. Find the angle whose supplement exceeds 5 times itscomplement.

Ans: 67.5o

Solution:Let:

x angle

180 x sup plement

90 x complement

=− =− =

Then,180 x 5(90 x)

x 67.5

− = −= °

75. Find the angle whose supplement exceeds 6 times its

complement by 20o

.

Ans: 76o

Solution:Let:

x angle

180 x sup plement

90 x complement

=− =− =

Then,(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?

Ans: 24 sides

Let:n number of sides

number of interior angles

==

Then,

(n 2)180 165n

n 24

− ==

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

Ans: 80 degrees

Solution:

Let:x one angle

y second angle

==

Then,

x y 100

x y 20

+ = − =

Solving the two equations simultaneously, we get:x 60

y 40

==

Thus, the largest angle is the third angle:

( )180 60 40

80

= − +

= °78. The two corresponding sides of two similar polygon is 4:6

If the perimeter of the larger polygon is 48 cm, what is theperimeter if the smaller polygon?Ans: 32 cm.

Solution:

( )

4 x

6 48

4 48x

6

x 32cm.

=

=

=

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

Ans: 87.5 sq.cm.

Solution:

( )

( )

( )

( )

2

2

2

2

4 56

x5

56 5x

4

x 87.5 sq.cm.

=

=

=

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.

Ratio of corresponding smaller sides= ratio of perimeter

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 thebiggest polygon if the smallest polygon has a perimeter of 24 cm. and an area of 36 sq. cm.Ans: 3

Solution:




( )

( )

( )

2

2

2 24

4 x

x 48 cm. perimeter of biggest polygon

2 36

A4

A 144 area of biggest polygon

144Ratio of area to perimeter

48Ratio of area to perimeter 3

=

=

=

=

=

=

82. Seven regular hexagons, each with 6 cm. sides arearranged so that they share the same sides and thecenters of the six hexagons are equidistant from theseventh central hexagon. Determine the ratio of the totalarea of the hexagons to the total outer perimeter enclosingthe hexagons.Ans: 6.06

Solution:

( ) ( ) ( )

( )

2

2

6 Sin60 6 7 A

2

A 654.72 cm

P 6 18

P 108

ARatio

P

654.72Ratio

108

Ratio 6.06

=

=

=

=

=

=

=

—

83. A chord is 36 cm. long and its mid point is 36 cm. from themidpoint of the longer arc. Find the radius of the circle.

Ans: 22.5

Solution:

( ) ( )x 36 18 18

x 9

2r 9 36

r 22.5 cm.

=

== +

=

84. A circular play area is being laid out in a field. The pointP,R and Q have been found such that PS=24 m., PQ=66m., 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

x 84

TR 84 12

TR 96 m.

=

== +=

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

Solution:

( ) ( ) ( )

( ) ( )

AB AE

AD AC

AB AC AD AE

40 80 AD 100

AD 32 m.

=

=

=

=

86. An engineer places his transit along the line tangent to thecircle 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.

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. Atriangle ABC is inscribe in a circle. If the angle BCA = 30o

and BAC = 50o compute the area of the triangle.Ans: 301.73 m2

Solution:

( ) ( ) ( ) ( ) ( )

( )

2 2 2 o

2 o o

o

AC 20 20 2 20 20 Cos 160

AC 39.39 cm.

39.39 Sin 30 Sin 50 Area of triangle

2 Sin 100

Area of triangle 301.73 sq.m.

= + −

=

=

=

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

Solution:

2 60 1802 120

60

10 r 2r

r 10

θ + =θ =

θ =+ =

=

89. The distance between the center of the circles which aremutually tangent to each other externally are 10,12 and 14units, The area of the largest circle is:Ans: 64 π

Solution:

1 31 2

2 3

3 2

3 2

3

3

2

1

r r 10

r r 12

r r 14

and

r r 2

r r 14

2r 12

r 6

r 14 6 8

r 10 6 4

+ =

+ =

+ =

− = −

+ ==

=

= − =

= − =

u

v

w

u v

Area of largest circle = ( )2

8π

Area of largest circle = 64 π

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


x



( )

( )

2

o

2

2

2

S R

A R

360

R A

2

R A

2

168002

6.25

S R 16 6.25

S 100 m.

= θ

π=

θ

πθ=

π

θ=

θ=

θ =

= θ =

=

91. A sector is bent to form a cone. If the angle of a sector is30 degrees and radius of 6 cm., what is the altitude of thecone?Ans: 5.98 cm.

Solution:

( )

( )2

22

S r

6 30S180

S

S 2 R

2 R

1R

2

1h 6

2

h 5.98 cm.

= θ

π=

= π= π

π = π

=

= − ÷

=

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:

( )

( ) ( )

( ) ( )

2 22

22

S R

36 150S

180

S 94.25

94.25 2 r

r 15

h 36 15

h 32.73

15 32.73r hV

3 3

V 7711.83

= θ

π=

=

= π=

= −

=

ππ= =

=

93. The lateral area of a right circular cone is 40 sq.m.π The

base radius is 4 m. What is the slant height?Ans: 10

Solution:

( )

S rL

40 40 L

L 10

= π

π = π

=

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 theslant height to the radius.Ans: 5:2

Solution:

2

2

S rL

r hV

3

V 2S 1

V 2S

r h2S

3

6Sr

rh

Sr

L

6S S

rh L

6L rh

L hr 6

L 15

r 6

L 5

r 2

= π

π=

=

=

π=

π =

π =

=

=

=

=

=

Ratio of slant height to radius = 5:2

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 thesame axis. Compute the height of the smaller cone.Ans: 5.04 cm.

Solution:

( )

( )

1 2

1

2

3

13

2

3

3

V 4V

V4

V

8V

V h

84

h

h 5.04 cm.

=

=

=

=

=

96. A circular cone having an altitude of 9 m. is divided into 2segments having the same vertex. If the smaller altitude is6 m., find the ratio of the volume of small cone to the bigcone.

Ans: 0.296

Solution:

( )

( )

31

32

6V0.296

V 9= =




97. The height of a right circular cone is “h”. If it contains oiland water at equal depth of h/2, what is the ratio of thevolume of the volume of oil to that of the water.Ans: 7

Solution:

( )

( )

3

13

2

1

2 1

2

2

2

1

2

1

h / 2V

V h

V

V 8

V V V vol. of oil

VV V

8

7V V

8

V 7/8V

V V / 8

V7

V

=

== −

= −

=

=

=

98. A plane is passed parallel to the base of a triangular

pyramid of altitude of 9 m. such that the area of the baseis 9 times the area of the triangle of intersection. How far from the vertex does the plane intersect the altitude?Ans: 3 m.

Solution:

( )

( )2

1

1

2

12

2

A h

9A 81h 3 m.

h A

A 9

=

=

=

99. The volume of a cone having an inclined axis at an angleof 60o with the base is equal to 1884.96 cu.m., find thelength of the axis of the cone, if the radius at the base is10 m.Ans: 20.78 m.

Solution:

( )

( )

2

o

o

AhV

3

10

1884.96 h3

h 18m

Length of axis:

hSin 60

x

18x

Sin 60

x 20.78m. axis length

=

π

==

=

=

=

100. The radius of the base of a cone of revolution is 32inches 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.

Solution:

( ) ( )( )

22

Vol. of cone Vol. of Cyl.

32 5424 h

3

h 32 in.

=

π= π

=

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:

( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( )2

A S s a s b s c

10 14 18S 21

2

S a 11, S b 7, S c 3

A 2 11 7 3

A 69.65 sq.m.

A rS

69.65 r 21

r 3.32 cm.

3.32 28r hV

3 3

V 323.19 cu.m.

= − − −

+ += =

− = − = − =

=

==

=

=

ππ= =

=

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

Solution:

( )

( )

( )3

hV b B bB3

24V 20 60 20 60

3

V 917.13 cu.cm.

4r 9 917.13

3

r 2.90 cm.

= + +

= + =

π =

=

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

Solution:

( )( )

o

V Ah

6 6 Sin 60500 6 h

2

h 5.35 cm

=

=

=




104.The bases of a right prism is a hexagon with one of eachside equal to 6 cm. The bases are 12 cm. apart. What isthe volume of the right pris.Ans: 1122.4 cu. cm.

Solution:

( ) ( )

( )

V Base x height

36060

6

1Base 6 6 Sin 60 602

Base 93.53

V 93.53 12

V 1122.4 cu. cm.

=

θ = =

= =

=

=


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