Year : 
1990
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 47 Questions

# Question Ans
21.

Solve the equation 7y\(^2\) = 3y

A. y = 3 or 7

B. y = 0 or 7

C. y = 0 or 3/7

D. y = 0 or 9

E. y = 0 or 10

Detailed Solution

7y\(^2\) = 3y
7y\(^2\) - 3y = 0
y(7y - 3) = 0
y = 0 or y = \(\frac{3}{7}\)
22.

Find the value of m which makes x\(^2\) + 8 + m a perfect square

A. 2

B. 4

C. 8

D. 12

E. 16

Detailed Solution

x\(^2\) + 8 + m
To make it a perfect square, we add \((\frac{b}{2})^2\)
= \((\frac{8}{2})^2\)
= 16
23.

Solve the equation 2a\(^2\) - 3a - 27 = 0

A. 3/2, 9

B. -2/3, 9

C. 3, 9/2

D. -3, -9/2

E. -3, 9/2

Detailed Solution

2a\(^2\) - 3a - 27 = 0
2a\(^2\) - 9a + 6a - 27 = 0
a(2a - 9) + 3(2a - 9) = 0
(a + 3)(2a - 9) = 0
a = -3 or a = \(\frac{9}{2}\)
24.

A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7]

A. 6o

B. 26o

C. 52o

D. 103o

E. 206o

Detailed Solution

πr2 = 360o
44cm2 = θ; ∴ θ =
44/πr2
x 360o = 102.9o = 103o
25.

Calculate the surface area of a sphere of radius 7cm [Take π = 22/7]

A. 86cm2

B. 154cm2

C. 616cm2

D. 1434cm2

E. 4312cm2

Detailed Solution

Surface area of a sphere = \(4\pi r^2\)
= \(4 \times \frac{22}{7} \times 7 \times 7\)
= \(616 cm^2\)
26.

If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ?

A. 75o

B. 60o

C. 45o

D. 30o

E. 0o

Detailed Solution


\(\theta_{1} = \theta_{2} = 30°\)
27.

A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]
Find the total surface area of the container

A. 35cm2

B. 154cm2

C. 220cm2

D. 528cm2

E. 770cm2

Detailed Solution

TSA of a cylinder = \(2\pi r^2 + 2\pi rh\)
= \(2\pi r (r + h)\)
= \(2 \times \frac{22}{7} \times 7 (7 + 5)\)
= \(44 \times 12\)
= \(528 cm^2\)
28.

A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]. What is the volume of the container?

A. 35cm3

B. 154cm3

C. 220cm3

D. 528cm3

E. 770cm3

Detailed Solution

Volume = \(\pi r^2 h\)
= \(\frac{22}{7} \times 7^2 \times 5\)
= \(770 cm^3\)
29.

In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR

A. 72o

B. 64o

C. 52o

D. 32o

E. 26o

Detailed Solution

From the figure, < PSR = 32° (base angles of an isos. triangle)
\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle)
< QRP = 180° - 116° = 64° (angle on a straight line)
< PQR = 64° (base angles of an isos. triangle)
< QPR = 180° - (64° + 64°) = 52°
30.

In the diagram above, WXYZ is a rhombus and ∠WYX = 20°. What is the value of ∠XZY

A. 20o

B. 30o

C. 45o

D. 60o

E. 70o

Detailed Solution

Diagonals bisect at 90°; < YXZ = 90° - 20° = 70°
But ZY = XY (sides of a rhombus)
\(\therefore\) < XYZ = 70° (base angle of an isos. triangle)
21.

Solve the equation 7y\(^2\) = 3y

A. y = 3 or 7

B. y = 0 or 7

C. y = 0 or 3/7

D. y = 0 or 9

E. y = 0 or 10

Detailed Solution

7y\(^2\) = 3y
7y\(^2\) - 3y = 0
y(7y - 3) = 0
y = 0 or y = \(\frac{3}{7}\)
22.

Find the value of m which makes x\(^2\) + 8 + m a perfect square

A. 2

B. 4

C. 8

D. 12

E. 16

Detailed Solution

x\(^2\) + 8 + m
To make it a perfect square, we add \((\frac{b}{2})^2\)
= \((\frac{8}{2})^2\)
= 16
23.

Solve the equation 2a\(^2\) - 3a - 27 = 0

A. 3/2, 9

B. -2/3, 9

C. 3, 9/2

D. -3, -9/2

E. -3, 9/2

Detailed Solution

2a\(^2\) - 3a - 27 = 0
2a\(^2\) - 9a + 6a - 27 = 0
a(2a - 9) + 3(2a - 9) = 0
(a + 3)(2a - 9) = 0
a = -3 or a = \(\frac{9}{2}\)
24.

A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7]

A. 6o

B. 26o

C. 52o

D. 103o

E. 206o

Detailed Solution

πr2 = 360o
44cm2 = θ; ∴ θ =
44/πr2
x 360o = 102.9o = 103o
25.

Calculate the surface area of a sphere of radius 7cm [Take π = 22/7]

A. 86cm2

B. 154cm2

C. 616cm2

D. 1434cm2

E. 4312cm2

Detailed Solution

Surface area of a sphere = \(4\pi r^2\)
= \(4 \times \frac{22}{7} \times 7 \times 7\)
= \(616 cm^2\)
26.

If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ?

A. 75o

B. 60o

C. 45o

D. 30o

E. 0o

Detailed Solution


\(\theta_{1} = \theta_{2} = 30°\)
27.

A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]
Find the total surface area of the container

A. 35cm2

B. 154cm2

C. 220cm2

D. 528cm2

E. 770cm2

Detailed Solution

TSA of a cylinder = \(2\pi r^2 + 2\pi rh\)
= \(2\pi r (r + h)\)
= \(2 \times \frac{22}{7} \times 7 (7 + 5)\)
= \(44 \times 12\)
= \(528 cm^2\)
28.

A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]. What is the volume of the container?

A. 35cm3

B. 154cm3

C. 220cm3

D. 528cm3

E. 770cm3

Detailed Solution

Volume = \(\pi r^2 h\)
= \(\frac{22}{7} \times 7^2 \times 5\)
= \(770 cm^3\)
29.

In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR

A. 72o

B. 64o

C. 52o

D. 32o

E. 26o

Detailed Solution

From the figure, < PSR = 32° (base angles of an isos. triangle)
\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle)
< QRP = 180° - 116° = 64° (angle on a straight line)
< PQR = 64° (base angles of an isos. triangle)
< QPR = 180° - (64° + 64°) = 52°
30.

In the diagram above, WXYZ is a rhombus and ∠WYX = 20°. What is the value of ∠XZY

A. 20o

B. 30o

C. 45o

D. 60o

E. 70o

Detailed Solution

Diagonals bisect at 90°; < YXZ = 90° - 20° = 70°
But ZY = XY (sides of a rhombus)
\(\therefore\) < XYZ = 70° (base angle of an isos. triangle)