WASSCE 2023 Mathematics Past Questions and Answers
The WASSCE 2023 Mathematics paper has 50 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. A point on the ground is 5 m away from the foot of a vertical wall 12 m high.
Calculate, correct to the nearest degree, the angle of depression of the point from the top of the wall.
Calculate, correct to the nearest degree, the angle of depression of the point from the top of the wall.
- A. 25$^{o}$
- B. 23$^{o}$
- C. 67$^{o}$
- D. 65$^{o}$
Answer: C. 67$^{o}$
The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.
Let θ = angle of depression.
SOHCAHTOA
SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).
SOH
sin(θ) = $\frac{Opposite}{Hypothenuse}$
CAH
cos(θ) = $\frac{Adjacent}{Hypothenuse}$
TOA
tan(θ) = $\frac{Opposite}{Adjacent}$
The longest side (side opposite/facing the right-angle) is the hypotenuse
The opposite and the adjacent sides of the angle of depression are known.
tan(θ) = $\frac{Opposite}{Adjacent}$
tan(θ) = $\frac{12 cm}{5 cm}$
θ = tan$^{-1}$$(\frac{12}{5})$
θ = 67.380135052$^{o}$ ≈ 67$^{o}$ (nearest degree)
The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.
Let θ = angle of depression.
SOHCAHTOA
SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).
SOH
sin(θ) = $\frac{Opposite}{Hypothenuse}$
CAH
cos(θ) = $\frac{Adjacent}{Hypothenuse}$
TOA
tan(θ) = $\frac{Opposite}{Adjacent}$
The longest side (side opposite/facing the right-angle) is the hypotenuse
The opposite and the adjacent sides of the angle of depression are known.
tan(θ) = $\frac{Opposite}{Adjacent}$
tan(θ) = $\frac{12 cm}{5 cm}$
θ = tan$^{-1}$$(\frac{12}{5})$
θ = 67.380135052$^{o}$ ≈ 67$^{o}$ (nearest degree)
2. The gradient of the line passing through the points (3, 6) and (x, 4) is - $\frac{2}{5}$. Find the value of x.
- A. 3
- B. 8
- C. 6
- D. 5
Answer: B. 8
Gradient = $\frac{Change in Y}{Change in X}$ = $\frac{{Y}_{2}-{Y}_{1}}{{X}_{2}-{X}_{1}}$
Y$_{2}$ = 4
Y$_{1}$ = 6
X$_{2}$ = x
X$_{1}$ = 3
Gradient = $\frac{4 - 6}{x - 3}$
Gradient = $\frac{-2}{x - 3}$
But gradient = - $\frac{2}{5}$ = $\frac{-2}{5}$
$\frac{-2}{x - 3}$ = $\frac{-2}{5}$
Since the numerator at the left and right are the same, the denominators are also the same.
x - 3 = 5
x = 5 + 3
x = 8
Gradient = $\frac{Change in Y}{Change in X}$ = $\frac{{Y}_{2}-{Y}_{1}}{{X}_{2}-{X}_{1}}$
Y$_{2}$ = 4
Y$_{1}$ = 6
X$_{2}$ = x
X$_{1}$ = 3
Gradient = $\frac{4 - 6}{x - 3}$
Gradient = $\frac{-2}{x - 3}$
But gradient = - $\frac{2}{5}$ = $\frac{-2}{5}$
$\frac{-2}{x - 3}$ = $\frac{-2}{5}$
Since the numerator at the left and right are the same, the denominators are also the same.
x - 3 = 5
x = 5 + 3
x = 8
3. A woman pours 85 litres of kerosene into a cylindrical container with radius 7 cm.
Calculate, correct to the nearest cm, the depth of the kerosene in the container.
[Take π = $\frac{22}{7}$]
Calculate, correct to the nearest cm, the depth of the kerosene in the container.
[Take π = $\frac{22}{7}$]
- A. 240 cm
- B. 552 cm
- C. 480 cm
- D. 595 cm
Answer: B. 552 cm
Volume of a cylinder = π x radius x radius x height
1 Litre = 1000 cm$^{3}$
85 litres = 85 x 1000 cm$^{3}$
85 litres = 85000 cm$^{3}$
Volume of kerosene = 85000 cm$^{3}$
π = $\frac{22}{7}$
Radius = 7 cm
Let d = depth/height of kerosene in the container
85000 cm$^{3}$ = $\frac{22}{7}$ x 7 cm x 7 cm x d
The denominator 7 cancels out one of the 7s multiplying.
85000 cm$^{3}$ = 22 x 1 cm x 7 cm x d
85000 cm$^{3}$ = 154 cm$^{2}$ x d
Divide both sides by 154 cm$^{2}$
d = $\frac{85000{cm}^{3}}{154{cm}^{2}}$
d = 551.948 cm ≈ 552 cm
Volume of a cylinder = π x radius x radius x height
1 Litre = 1000 cm$^{3}$
85 litres = 85 x 1000 cm$^{3}$
85 litres = 85000 cm$^{3}$
Volume of kerosene = 85000 cm$^{3}$
π = $\frac{22}{7}$
Radius = 7 cm
Let d = depth/height of kerosene in the container
85000 cm$^{3}$ = $\frac{22}{7}$ x 7 cm x 7 cm x d
The denominator 7 cancels out one of the 7s multiplying.
85000 cm$^{3}$ = 22 x 1 cm x 7 cm x d
85000 cm$^{3}$ = 154 cm$^{2}$ x d
Divide both sides by 154 cm$^{2}$
d = $\frac{85000{cm}^{3}}{154{cm}^{2}}$
d = 551.948 cm ≈ 552 cm
4. Given that x$^{2}$ - 11*x* + m is a perfect square, find the value of m.
- A. $\frac{11}{2}$
- B. $\frac{121}{8}$
- C. $\frac{121}{4}$
- D. 121
Answer: C. $\frac{121}{4}$
If a polynomial ax$^{2}$ + bx + c is a perfect square if b$^{2}$ = 4*ac*
For the polynomial x$^{2}$ - 11*x* + m
a = 1
b = -11
c = m
Since the polynomial x$^{2}$ - 11*x* + m is a perfect square, b$^{2}$ = 4*ac*
(-11)$^{2}$ = 4 x 1 x m
121 = 4*m*
Divide both sides by 4
m = $\frac{121}{4}$
If a polynomial ax$^{2}$ + bx + c is a perfect square if b$^{2}$ = 4*ac*
For the polynomial x$^{2}$ - 11*x* + m
a = 1
b = -11
c = m
Since the polynomial x$^{2}$ - 11*x* + m is a perfect square, b$^{2}$ = 4*ac*
(-11)$^{2}$ = 4 x 1 x m
121 = 4*m*
Divide both sides by 4
m = $\frac{121}{4}$
5. The sides of a scalene triangle are 4 cm, 9 cm and 11 cm. Calculate, correct to the nearest whole number, the area of the triangle.
- A. 13 cm$^{2}$
- B. 17 cm$^{2}$
- C. 21 cm$^{2}$
- D. 19 cm$^{2}$
Answer: B. 17 cm$^{2}$
Heron's formula
If you know the length of the three sides of the triangle (a, b, c), then Heron's formula can be applied to calculate for the area of the triangle.
Using Heron's formula
Area = $\sqrt{S(S - a)(S - b)(S - c)}$
S = $\frac{a + b + c}{2}$
S = $\frac{4 + 9 + 11}{2}$
S = $\frac{24}{2}$ = 12
Area = $\sqrt{12(12 - 4)(12 - 9)(12 - 11)}$
Area = $\sqrt{12(8)(3)(1)}$
Area = $\sqrt{12 x 8 x 3 x 1}$
Area = $\sqrt{288}$
Area = 16.970562748 ≈ 17 cm$^{2}$
Heron's formula
If you know the length of the three sides of the triangle (a, b, c), then Heron's formula can be applied to calculate for the area of the triangle.
Using Heron's formula
Area = $\sqrt{S(S - a)(S - b)(S - c)}$
S = $\frac{a + b + c}{2}$
S = $\frac{4 + 9 + 11}{2}$
S = $\frac{24}{2}$ = 12
Area = $\sqrt{12(12 - 4)(12 - 9)(12 - 11)}$
Area = $\sqrt{12(8)(3)(1)}$
Area = $\sqrt{12 x 8 x 3 x 1}$
Area = $\sqrt{288}$
Area = 16.970562748 ≈ 17 cm$^{2}$
6. In a hall, there are 175 persons. 12% are children, 56 are men and the rest are women. If one person is selected at random from the hall, find the probability that a woman is selected.
- A. $\frac{3}{25}$
- B. $\frac{11}{25}$
- C. $\frac{8}{25}$
- D. $\frac{14}{25}$
Answer and explanation: practise this paper on ePrep.
7. A student measured the length of a classroom and obtained 3.99 m which is less than the actual length. If the percentage error was 5%, what was the actual length?
- A. 3.80 m
- B. 3.78 m
- C. 4.18 m
- D. 4.20 m
Answer and explanation: practise this paper on ePrep.
8. A car covers the first 80 km of a journey in 2 hours and completes the journey by travelling further 2.5 hours at 50 km/h. What is the average speed of the entire journey?
- A. 44$\frac{4}{9}$ km/h
- B. 45$\frac{5}{9}$ km/h
- C. 47$\frac{1}{4}$ km/h
- D. 63$\frac{1}{4}$ km/h
Answer and explanation: practise this paper on ePrep.
9. If log x = 2 - 3log 2, find the value of x.
- A. $\frac{5}{2}$
- B. $\frac{2}{5}$
- C. $\frac{2}{25}$
- D. $\frac{25}{2}$
Answer and explanation: practise this paper on ePrep.
10. The graph of y = x$^{2}$ - 5*x* + k passes through the point (3, 1). Find the value of k.
- A. 2
- B. 3
- C. 7
- D. 5
Answer and explanation: practise this paper on ePrep.
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