Mathematics Past Questions | West African Examinations Council (WAEC)

Question 1

The sum of the interior angles of regular polygon is ${1800}^{\circ }$

. How many sides has the polygon?

Options

A) 16

B) 12

C) 10

D) 8

The correct answer is B.

Explanation:

Sum $=(n-2)180$

$1800=(n-2)180$
Divide both sides by ${180}^{\circ }$
$\frac{1800}{180}=(n-2)\frac{180}{180}$
$10=n-2$
$10+2=n$
$n=12$

Question 2

Find the coefficient of m in the expression of $(\frac{m}{2}-1\frac{1}{2})(m+\frac{2}{3})$

Options

A) $-\frac{1}{6}$

B) $-\frac{1}{2}$

C) $-1$

D) $-1\frac{1}{6}$

The correct answer is D.

Explanation:

$(\frac{m}{2}-1\frac{1}{4})(m+\frac{2}{3})$

$(\frac{m}{2}-\frac{3}{2})(\frac{m}{1}+\frac{2}{3})$
$=\frac{m^2}{3}+\frac{3m}{6}-\frac{6}{6}$
$=\frac{2m-9m}{6}$
$=\frac{-7m}{6}$
$=1\frac{1}{6}$

Question 3

The volume of a cuboid is 54 cm³. If the length, width and height of the cuboid are in the ratio 2:1:2 respectively, find its total surface area

Options

A) 108 cm²

B) 90 cm²

C) 80 cm²

D) 72 cm²

The correct answer is B.

Explanation:

V = L × B × H
V = 2x × x × x = 2x³
= 54 cm³
i.e. 2x³ = 54
x³ = 54 ÷ 2 = 27
x = 3
L = 2x i.e. 2 × 3 = 6 cm
B = x i.e. x = 3 cm
H = x i.e. x = 3 cm
But A = 2(LB + LH + BH) = 2(6 × 3) + (6 × 3) + (3 × 3)
= 2(18 + 18 + 9) = 2(45)
= 90 cm²

Question 4

A side and a diagonal of a rhombus are 10cm and 12cm respectively, Find its area

Options

A) 20 cm²

B) 24 cm²

C) 48 cm²

D) 96 cm²

The correct answer is D.

Question 5

Factorise completely: 32x²y - ³y³

Options

A) 16x²y(2 - 3xy²)

B) 8xy(4x - 6x²y²)

C) 8x²y(4 - 6xy²)

D) 16xy(2x - 3x²y²)

The correct answer is A.

Question 6

If $x$

and $y$ are variables and $k$ is a constant, which of the following describes an inverse relationship between $x$ and $y$

?

Options

A) $y=kx$

B) $y=\frac{k}{x}$

C) $y=k\sqrt{x}$

D) $y=x+k$

The correct answer is B.

Question 7

In the diagram, |QR| = 10 m, |SR| = 8 m, < QPS = 30°, < QRP = 90° and |PS| = x. Find x.

Options

A) 1.32 m

B) 6.32 m

C) 9.32 m

D) 17.32 m

The correct answer is C.

Explanation:

In right angled ΔQPR
tan 30° = $\frac{10}{x+8}$

(x + 8) tan 30° = 10
x + 8 = $\frac{10}{0.5773}$

x + 8 = 17.3
x = 17.3 - 8
x = 9.32 m

Question 8

In the diagram, O is a circle centre of the circle PQRS and < PSR = 86°. If < PQR = x°, find x

Options

A) 108°

B) 172°

C) 130°

D) 50°

The correct answer is B.

Explanation:

x° = 86 × 2(angle at centre = 2 × angle at circumference)
= 172°

Question 9

The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n

Options

A) m = $\frac{n}{2}$

B) m = 2n

C) m = n - 2

D) m = n + 2

The correct answer is A.

Explanation:

If n = 2m
m = $\frac{n}{2}$

Question 10

In the diagram, MQ//RS, < TUV = 70° and < RLV = 30°. Find the value of x

Options

A) 150°

B) 110°

C) 100°

D) 95°

The correct answer is C.

Explanation:

L + 30° - 180°(Sum of
L = 180° - 30° = 150°
L = q = 150° (opposite angles)
y = b = 30° (alternating angles)
b + c = 180° (sum of angles on a straight line)
30° + c = 180°
c = 180 - 30
c = 150°
b = a = 30° (opposite angles)
c = d = 150° (opposite angles)
a + k + 70° = 180° (sum of angles on a straight line)
30° + k + 70° = 180°
k + 100° = 180°
k = 180 - 100
k = 80°
x + 80° = 180(sum of angles on a straight line)
x = 180° - 80°
x = 100°

Question 11

In the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true?
i. t = y
ii. x + y + z + m = 180°
iii. x + m + n = 180°
iv. x + n = m + z

Options

A) i and iv

B) ii

C) iii

D) iv

The correct answer is C.

Explanation:

m + y + x = 180°(sum of angles on a straight line)
y = n(vertically opp. angle)
m + n + x = 180°

Question 12

The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050° and < PQR = 72°. Calculate the bearing of R and Q

Options

A) 155°

B) 80°

C) 158°

D) 91°

The correct answer is C.

Explanation:

< NPQ = < PQB = 50°(alternating angles)
< PQB = 50°
< PQR = < PQR < PQB + < QBR = 72°
< QBR = < PQR - < PQB = 72° - 50°
= 22°
< NQR = 180 - < QBR = 180° - 22 = 158°, the bearing of R from Q = 158°

Question 13

The bar chart shows the frequency distribution of marks scored by students in a class test. How many students are in the class?

Options

A) 10

B) 24

C) 25

D) 30

The correct answer is C.

Explanation:

6 + 8 + 6 + 5 = 25

Question 14

The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.

Options

A) 6.0

B) 3.0

C) 2.4

D) 1.8

The correct answer is C.

Explanation:

x	f	fx
1	6	6
2	8	16
3	8	18
4	5	20

Mean $\overline{x}=\frac{\sum fx}{\sum f}$

 $=\frac{60}{25}$

$\overline{x}=2.4$

Question 15

The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?

Options

A) 2

B) 4

C) 6

D) 8

The correct answer is A.

Explanation:

Median = ($\frac{N+1}{2}$

)th = ($\frac{25+1}{2}$)th
= $\frac{26th}{2}$

= 13th
The 13th is 2

Question 16

 

The graph is that of y = 2x² - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x² - 5x - 3 at the point x = 4?

Options

A) 11.1

B) 10.5

C) 10.3

D) 9.9

The correct answer is A.

Explanation:

dy/dx = gradient
y = 2x${}^2$

- 5x - 3
dy/dx = 4x - 5
but x = 4
dy/dx = (4 × 4) - 5 = 11
Gradient = 11

Explanation provided by Johnny Ifiok Eunice

Question 17

In the diagram, MN//PO, < PMN = 112°, < PNO = 129° and < MPN = y°. Find the value of y.

Options

A) 51°

B) 54°

C) 56°

D) 68°

The correct answer is B.

Explanation:

In Δ NPO + PNO + PNO + < NOP = 180° (sum of interior angles of a Δ)
i.e. NPO + 129 + 37 = 180°
< NOP = 180 - (129 + 37) = 14°
< MNP = < NOP = 14° (alternating angles)
In Δ MPN
< PMN + < MNP + y = 180° (sum of interior angles of a Δ)
i.e. 112 + 14 + y = 180°
y = 180 - (112 + 14) = 180 - 126 = 54°

Question 18

The diagram is a circle with centre O. PRST are points on the circle. Find the value of < PRS.

Options

A) 144°

B) 72°

C) 40°

D) 36°

The correct answer is A.

Explanation:

Reflex < POS = 2x (angle at centre is twice that at circumference)
Reflex < POS + < POS = 350° (angles at a point)
i.e. 2x + 8x = 360°
10x = 360°
x = $\frac{360°}{10}$

= 36°
< PRS = 1212
< POS (angle at centre twice that circumference)
= 1212 $\times$ 8x = 4x
4 $\times$

36°
< PRS = 144

Question 19

The diagram is a circle of radius |QR| = 4cm. $\overline{R}$

is a tangent to the circle at $R$

. If TPO = 120°, find |PQ|.

Options

A) 2.32cm

B) 1.84cm

C) 0.62cm

D) 0.26cm

The correct answer is C.

Question 20

In the diagram, |SR| = |QR|. < SRP = 65° and < RPQ = 48°, find < PRQ

Options

A) 65°

B) 45°

C) 25°

D) 19°

The correct answer is D.

Explanation:

< RSQ = < RPQ = 48° (angle in the same segment)
< SQR < RSQ (Base angle of an isosceles Δ)
< SQR = 48°
< QRS + < RSQ + < RSQ = 180°(sum of interior angles of a Δ)
i.e. < QRS + 48° + 48° = 180°
< QRS = 180 - (48 + 48) = 180 - 96 = 84°
but < PRQ + < PRS = < QRS
< PRQ = < QRS - < PRS - 84 - 65
= 19°

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