WASSCE 2022 Mathematics Past Questions and Answers
The WASSCE 2022 Mathematics paper has 49 objective and 13 theory questions. Here are sample objective questions with the correct answers and explanations. Sit the full paper with the real time limit on ePrep.
Sample questions with answers
1. Find the number of even integers between 11 and 97.
- A. 43
- B. 44
- C. 45
- D. 53
Answer: A. 43
Even numbers are number divisible by 2 without a remainder.
Set of even integers between 11 and 97 = {12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,80,92,94,96}
The total number of elements in the set = 43
Alternate Solution
Find the interval between 97 and 11 and divide by 2
97 - 11 = 86
Even integers between 11 and 97 = $\frac{86}{2}$ = 43
Even numbers are number divisible by 2 without a remainder.
Set of even integers between 11 and 97 = {12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,80,92,94,96}
The total number of elements in the set = 43
Alternate Solution
Find the interval between 97 and 11 and divide by 2
97 - 11 = 86
Even integers between 11 and 97 = $\frac{86}{2}$ = 43
2. Find the values of x for which $\frac{1}{2{x}^{2} -13x + 15}$ is not defined.
- A. 5, $\frac{3}{2}$
- B. 1, $\frac{13}{15}$
- C. 2, 15
- D. 13, 15
Answer: A. 5, $\frac{3}{2}$
For the equation to be undefined, the denominator must be 0.
2*x*$^{2}$ - 13*x* + 15 = 0
Solve for the value of x
2 x 15 = 30
Two numbers when multiplied the result will be 30 and when added the result will be - 13.
The numbers are -10 and - 3.
Express - 13*x* as -10*x* - 3*x*
2*x*$^{2}$ - 10*x* - 3*x* + 15 = 0
Factorize out
2*x*(x - 5)-3(x-5) = 0
(x - 5)(2*x*-3) = 0
(x - 5) = 0 or (2*x*-3) = 0
x - 5 = 0
x = 0 + 5
x = 5
2*x*-3 = 0
2*x* = 3
Divide both sides of the equation by 2
x = $\frac{3}{2}$
For the equation to be undefined, the denominator must be 0.
2*x*$^{2}$ - 13*x* + 15 = 0
Solve for the value of x
2 x 15 = 30
Two numbers when multiplied the result will be 30 and when added the result will be - 13.
The numbers are -10 and - 3.
Express - 13*x* as -10*x* - 3*x*
2*x*$^{2}$ - 10*x* - 3*x* + 15 = 0
Factorize out
2*x*(x - 5)-3(x-5) = 0
(x - 5)(2*x*-3) = 0
(x - 5) = 0 or (2*x*-3) = 0
x - 5 = 0
x = 0 + 5
x = 5
2*x*-3 = 0
2*x* = 3
Divide both sides of the equation by 2
x = $\frac{3}{2}$
3. 
In the diagram P,Q,R,S are points on the circle. If ∠PQS = 32$^{o}$ and ∠PRQ = 61$^{o}$, find ∠QPS.
In the diagram P,Q,R,S are points on the circle. If ∠PQS = 32$^{o}$ and ∠PRQ = 61$^{o}$, find ∠QPS.
- A. 32$^{o}$
- B. 61$^{o}$
- C. 87$^{o}$
- D. 93$^{o}$
Answer: C. 87$^{o}$
Inscribed angles subtended by the same are equal.
∠PSQ = ∠PRQ = 61$^{o}$
Sum of angles in are triangle is 180$^{o}$
∠QPS + ∠PSQ + ∠PQS = 180$^{o}$
∠QPS + 61 + 32 = 180
∠QPS + 93 = 180
∠QPS = 180 - 93
∠QPS = 87$^{o}$
Inscribed angles subtended by the same are equal.
∠PSQ = ∠PRQ = 61$^{o}$
Sum of angles in are triangle is 180$^{o}$
∠QPS + ∠PSQ + ∠PQS = 180$^{o}$
∠QPS + 61 + 32 = 180
∠QPS + 93 = 180
∠QPS = 180 - 93
∠QPS = 87$^{o}$
4. The bearing of a point P from a point Q is 065$^{o}$. What is the bearing of Q from P?
- A. 115$^{o}$
- B. 155$^{o}$
- C. 245$^{o}$
- D. 295$^{o}$
Answer: C. 245$^{o}$
Bearing of Q from P = 90$^{o}$ + 90$^{o}$ + 65$^{o}$
Bearing of Q from P = 245$^{o}$
Bearing of Q from P = 90$^{o}$ + 90$^{o}$ + 65$^{o}$
Bearing of Q from P = 245$^{o}$
5. The arc of a circle 50 cm long subtends an angle 75$^{o}$ at the centre of the circle.
Find, correct to three significant figures, the radius of the circle.
[Take π = $\frac{22}{7}$]
Find, correct to three significant figures, the radius of the circle.
[Take π = $\frac{22}{7}$]
- A. 76.4 cm
- B. 61.2 cm
- C. 38.2 cm
- D. 8.74 cm
Answer: C. 38.2 cm
Length of arc = $\frac{\theta }{360}$ x 2π*r*
Where r = radius of the circle.
Length of the arc = 50 cm
θ = 75
π = $\frac{22}{7}$
$\frac{75}{360}$ x 2 x $\frac{22}{7}$ x r = 50
$\frac{75 x 2 x 22}{360 x 7}$ x r = 50
$\frac{3300}{2520}$ x r = 50
Get rid of the fraction by multiplying both sides of the equation by the denominator (2520)
3300 x r = 50 x 2520
3300 x r = 126000
Divide both sides of the equation by 3300
r = $\frac{126000}{3300}$
r = 38.1818 ≈ 38.2 cm (three significant figures)
Length of arc = $\frac{\theta }{360}$ x 2π*r*
Where r = radius of the circle.
Length of the arc = 50 cm
θ = 75
π = $\frac{22}{7}$
$\frac{75}{360}$ x 2 x $\frac{22}{7}$ x r = 50
$\frac{75 x 2 x 22}{360 x 7}$ x r = 50
$\frac{3300}{2520}$ x r = 50
Get rid of the fraction by multiplying both sides of the equation by the denominator (2520)
3300 x r = 50 x 2520
3300 x r = 126000
Divide both sides of the equation by 3300
r = $\frac{126000}{3300}$
r = 38.1818 ≈ 38.2 cm (three significant figures)
6. Find the truth set of (3 + x)(1 - x) > 9 - x$^{2}$.
- A. {x:x > 3}
- B. {x:x > -3}
- C. {x:x < 3}
- D. {x:x < -3}
Answer and explanation: practise this paper on ePrep.
7. A bag contains 32 blue, 28 black and 20 red identical balls of same size. If a ball is picked at random, what is the probability that it is not red?
- A. $\frac{1}{4}$
- B. $\frac{2}{5}$
- C. $\frac{3}{5}$
- D. $\frac{3}{4}$
Answer and explanation: practise this paper on ePrep.
8. Simplify: $\frac{1}{2}$(x-1)-$\frac{1}{3}$($\frac{1}{2}$x-1).
- A. $\frac{1}{6}$(x-1)
- B. $\frac{1}{6}$(2*x*+1)
- C. $\frac{1}{6}$(2*x*-1)
- D. $\frac{1}{6}$(1-2*x*)
Answer and explanation: practise this paper on ePrep.
9. A woman buys 3 exercise books at GH₵ 0.4 each and 9 exercise books at GH₵ 0.2 each. What is the average cost of an exercise book?
- A. GH₵ 0.35
- B. GH₵ 0.30
- C. GH₵ 0.22
- D. GH₵ 0.25
Answer and explanation: practise this paper on ePrep.
10. Given that 4*x* + 6*y* = 5 and 2*x* + 4*y* = 3, find the value of (x + 2*y*)
- A. -1$\frac{1}{2}$
- B. -1
- C. 1
- D. 1$\frac{1}{2}$
Answer and explanation: practise this paper on ePrep.
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