Year : 
2011
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 49 Questions

# Question Ans
31.

John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container

A. 11cm

B. 18cm

C. 21cm

D. 34cm

Detailed Solution

volume of rectangular container = L x B x H

1 litre = 1000cm3

96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres

volume = 96000cm3

96000 = 220 x 40 x 11

96--- = 8800H

H = \(\frac{96000}{8800}\)

= 10.97cm

= 11cm
32.

John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container

A. 11cm

B. 18cm

C. 21cm

D. 34cm

Detailed Solution

volume of rectangular container = L x B x H

1 litre = 1000cm3

96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres

volume = 96000cm3

96000 = 220 x 40 x 11

96--- = 8800H

H = \(\frac{96000}{8800}\)

= 10.97cm

= 11cm
33.

In a quiz competition, a student answers n questions correctly and was given D(n + 50) for each question correctly answered. If he gets D600.00 altogether, how many questions did he answer correctly?

A. 18

B. 15

C. 12

D. 10

Detailed Solution

let the number of question = n

n(n + 50) = 600

n2 + 50n - 600 - 0 by quadratic formular

x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2}\)

n = \(\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)}\)

= \(\frac{-50 \pm \sqrt{4900}}{2}\)

= \(\frac{-50 + 70}{2}\)

= \(\frac{20}{2}\)

= 10
34.

If 27x = 9y. Find the value of \(\frac{x}{y}\)

A. \(\frac{1}{3}\)

B. \(\frac{2}{3}\)

C. 1\(\frac{1}{2}\)

D. 3

Detailed Solution

27x = 9y

33x = 32y

3x = 2y
35.

A boy looks through a window of a building and sees a mango fruit on the ground 50m away from the foot of the building. If the window is 9m from the ground, calculate, correct to the nearest degree, the angle of depression of the mango from the window

A. 9o

B. 10o

C. 11o

D. 12o

Detailed Solution

tan \(\theta\) = \(\frac{9}{50}\) = 0.18

\(\theta = tan^{-1} 0.18\)

\(\theta\) = 10.20
36.

The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?

A. \(\bigtriangleup\) PQK

B. \(\bigtriangleup\) PKS

C. \(\bigtriangleup\) SKR

D. \(\bigtriangleup\) PSR

C
37.

In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x

A. 60o

B. 55o

C. 50o

D. 45o

Detailed Solution

< Q = \(\frac{240}{2}\) (angle at centre twice that at the circumference)

< Q = 120o

Also < POR = 360 - 240

= 120o

( < s at centre) since /PQ/ = /QR/, < x = < R

Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)

x + R + 120 = 360o

x + R = 360 - 240 = 120; Since x = R

x + x = 120

2x = 120

Since x = R

x + x = 120

2x = 120

x = \(\frac{120}{2}\)
<
38.

Find the size of the angle marked x in the diagram.

A. 108o

B. 112o

C. 128o

D. 142o

Detailed Solution

x + 52o = 90

x = 90 - 52

x = 38o

k = opposite angle Z

k = 38o

y + k = 90o

y + 38o = 90o

y = 90o - 38o

y = 52o

y = x = 180o(sum of angles on straight line)

52 + x = 180o

x = 180 - 52

x = 128
39.

The histogram shows the age distribution of members of a club. How many members are in the club?

A. 52

B. 50

C. 48

D. 40

Detailed Solution

2 + 2 + 5 + 8 + 12 + H + 7 + 3 + 2 = 52
40.

The histogram shows the age distribution of members of a club. What is their modal age?

A. 44.5

B. 42.5

C. 41.5

D. 40.5

A
31.

John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container

A. 11cm

B. 18cm

C. 21cm

D. 34cm

Detailed Solution

volume of rectangular container = L x B x H

1 litre = 1000cm3

96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres

volume = 96000cm3

96000 = 220 x 40 x 11

96--- = 8800H

H = \(\frac{96000}{8800}\)

= 10.97cm

= 11cm
32.

John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container

A. 11cm

B. 18cm

C. 21cm

D. 34cm

Detailed Solution

volume of rectangular container = L x B x H

1 litre = 1000cm3

96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres

volume = 96000cm3

96000 = 220 x 40 x 11

96--- = 8800H

H = \(\frac{96000}{8800}\)

= 10.97cm

= 11cm
33.

In a quiz competition, a student answers n questions correctly and was given D(n + 50) for each question correctly answered. If he gets D600.00 altogether, how many questions did he answer correctly?

A. 18

B. 15

C. 12

D. 10

Detailed Solution

let the number of question = n

n(n + 50) = 600

n2 + 50n - 600 - 0 by quadratic formular

x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2}\)

n = \(\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)}\)

= \(\frac{-50 \pm \sqrt{4900}}{2}\)

= \(\frac{-50 + 70}{2}\)

= \(\frac{20}{2}\)

= 10
34.

If 27x = 9y. Find the value of \(\frac{x}{y}\)

A. \(\frac{1}{3}\)

B. \(\frac{2}{3}\)

C. 1\(\frac{1}{2}\)

D. 3

Detailed Solution

27x = 9y

33x = 32y

3x = 2y
35.

A boy looks through a window of a building and sees a mango fruit on the ground 50m away from the foot of the building. If the window is 9m from the ground, calculate, correct to the nearest degree, the angle of depression of the mango from the window

A. 9o

B. 10o

C. 11o

D. 12o

Detailed Solution

tan \(\theta\) = \(\frac{9}{50}\) = 0.18

\(\theta = tan^{-1} 0.18\)

\(\theta\) = 10.20
36.

The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?

A. \(\bigtriangleup\) PQK

B. \(\bigtriangleup\) PKS

C. \(\bigtriangleup\) SKR

D. \(\bigtriangleup\) PSR

C
37.

In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x

A. 60o

B. 55o

C. 50o

D. 45o

Detailed Solution

< Q = \(\frac{240}{2}\) (angle at centre twice that at the circumference)

< Q = 120o

Also < POR = 360 - 240

= 120o

( < s at centre) since /PQ/ = /QR/, < x = < R

Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)

x + R + 120 = 360o

x + R = 360 - 240 = 120; Since x = R

x + x = 120

2x = 120

Since x = R

x + x = 120

2x = 120

x = \(\frac{120}{2}\)
<
38.

Find the size of the angle marked x in the diagram.

A. 108o

B. 112o

C. 128o

D. 142o

Detailed Solution

x + 52o = 90

x = 90 - 52

x = 38o

k = opposite angle Z

k = 38o

y + k = 90o

y + 38o = 90o

y = 90o - 38o

y = 52o

y = x = 180o(sum of angles on straight line)

52 + x = 180o

x = 180 - 52

x = 128
39.

The histogram shows the age distribution of members of a club. How many members are in the club?

A. 52

B. 50

C. 48

D. 40

Detailed Solution

2 + 2 + 5 + 8 + 12 + H + 7 + 3 + 2 = 52
40.

The histogram shows the age distribution of members of a club. What is their modal age?

A. 44.5

B. 42.5

C. 41.5

D. 40.5

A