WAEC 2025 Mathematics Paper 2 (Essay) Questions & Answer

Sky Prey — Sat, 08 Mar 2025 21:19:48 +0000

Certainly! Below is a sample of a WAEC (West African Examinations Council) Mathematics question paper for 2025, along with detailed answers. This is a simulated paper based on the WAEC format and syllabus. Please note that this is not an official WAEC paper but a practice resource.

Tips for Success in WAEC 2025 Expo

WAEC 2025 Mathematics Paper 2 (Essay)

Time Allowed: 2 hours 30 minutes
Instructions:

Answer five questions only.
All questions carry equal marks.
Show all working clearly.

Question 1

(a) Simplify:

$43 \div (32 - 21) + 65$

(b) Solve for $x$ in the equation:

$32 x - 1 - 4 x + 2 = 1$

(c) A man earns ₦120,000 per month. He spends 40% on rent, 25% on food, and saves the rest. Calculate:
(i) His monthly savings.
(ii) The ratio of his expenditure on rent to food.

Answer to Question 1

(a)
Simplify:

$43 \div (32 - 21) + 65$

Step 1: Solve the expression in the brackets:

$32 - 21 = 64 - 3 = 61$

Step 2: Perform the division:

$43 \div 61 = 43 \times 16 = 418 = 29$

Step 3: Add $6 5$ :

$29 + 65 = 627 + 5 = 632 = 316$

Final Answer:

$316$

(b)
Solve for $x$ :

$32 x - 1 - 4 x + 2 = 1$

Step 1: Find the LCM of the denominators (3 and 4), which is 12. Multiply through by 12:

$12 \cdot 32 x - 1 - 12 \cdot 4 x + 2 = 12 \cdot 1$ $4 (2 x - 1) - 3 (x + 2) = 12$

Step 2: Expand the brackets:

$8 x - 4 - 3 x - 6 = 12$ $5 x - 10 = 12$

Step 3: Solve for $x$ :

$5 x = 12 + 10$ $5 x = 22$ $x = 522$

Final Answer:

$x = 522$

(c)
(i) Monthly savings:
Total income = ₦120,000
Expenditure on rent = 40% of ₦120,000 = $0.40 \times 120, 000 = ₦48, 000$
Expenditure on food = 25% of ₦120,000 = $0.25 \times 120, 000 = ₦30, 000$
Total expenditure = ₦48,000 + ₦30,000 = ₦78,000
Savings = Total income – Total expenditure = ₦120,000 – ₦78,000 = ₦42,000

Answer:

$₦42, 000$

(ii) Ratio of expenditure on rent to food:

$Ratio = ₦30 , 000₦48 , 000 = 3048 = 58$

Answer:

$8 : 5$

Question 2

(a) Solve the simultaneous equations:

$2 x + 3 y = 13$ $4 x - y = 5$

(b) A bag contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is drawn at random, find the probability that it is:
(i) Red
(ii) Not green

Answer to Question 2

(a)
Solve the simultaneous equations:

$2 x + 3 y = 13 (1)$ $4 x - y = 5 (2)$

Step 1: From equation (2), express $y$ in terms of $x$ :

$4 x - y = 5 ⟹ y = 4 x - 5$

Step 2: Substitute $y = 4 x - 5$ into equation (1):

$2 x + 3 (4 x - 5) = 13$ $2 x + 12 x - 15 = 13$ $14 x - 15 = 13$ $14 x = 28$ $x = 2$

Step 3: Substitute $x = 2$ into $y = 4 x - 5$ :

$y = 4 (2) - 5 = 8 - 5 = 3$

Final Answer:

$x = 2, y = 3$

(b)
Total balls = 5 (red) + 3 (blue) + 2 (green) = 10

(i) Probability of drawing a red ball:

$P (Red) = 105 = 21$

Answer:

$21$

(ii) Probability of not drawing a green ball:

$P (Not Green) = 1 - P (Green) = 1 - 102 = 108 = 54$

Answer:

$54$

(c)
Area of rectangle = $48 cm^{2}$
Length = 8 cm
Width = $Length Area = 8 48 = 6 cm$
Perimeter = $2 (Length + Width) = 2 (8 + 6) = 28 cm$

Answer:

$28 cm$

Question 3

(a) Factorize completely:

$x^{2} - 5 x + 6$

(b) Solve the inequality:

$3 x - 7 < 5$

Answer to Question 3

(a)
Factorize:

$x^{2} - 5 x + 6$

Find two numbers whose product is 6 and sum is -5. The numbers are -2 and -3.

$x^{2} - 5 x + 6 = (x - 2) (x - 3)$

Answer:

$(x - 2) (x - 3)$

(b)
Solve the inequality:

$3 x - 7 < 5$

Add 7 to both sides:

$3 x < 12$

Divide by 3:

$x < 4$

Answer:

$x < 4$

(c)
Check if the triangle is right-angled using Pythagoras’ theorem:

$6^{2} + 8^{2} = 36 + 64 = 100 = 1 0^{2}$

Since $6^{2} + 8^{2} = 1 0^{2}$ , the triangle is right-angled.

Answer:

$Yes, it is a right-angled triangle$

Question 4

(a) Find the value of $x$ in the equation:

$log_{2} (x + 3) = 4$

(b) A circle has a radius of 7 cm. Calculate its circumference and area. (Take $π = 7 22$ )

$2 x^{2} - 5 x - 3 = 0$

Answer to Question 4

(a)
Solve:

$log_{2} (x + 3) = 4$

Convert to exponential form:

$x + 3 = 2^{4}$ $x + 3 = 16$ $x = 13$

Answer:

$13$

(b)
Radius $r = 7 cm$
Circumference = $2 π r = 2 \times 7 22 \times 7 = 44 cm$
Area = $π r^{2} = 7 22 \times 7^{2} = 154 cm^{2}$

Answer:

$Circumference = 44 cm, Area = 154 cm^{2}$

(c)
Solve:

$2 x^{2} - 5 x - 3 = 0$

Using the quadratic formula:

$x = 2 \times 25 \pm ( -5 ) ^{2} - 4 \times 2 \times ( -3 )$ $x = 45 \pm 25 + 24 = 45 \pm 49 = 45 \pm 7$ $x = 412 = 3 or x = 4-2 = - 21$

Answer:

$x = 3 or x = - 21$

Question 5

(a) A car travels at a speed of 60 km/h for 2 hours and then at 80 km/h for 3 hours. Calculate the average speed for the entire journey.

(b) If $sin θ = 5 3$ , find $cos θ$ and $tan θ$ .

Answer to Question 5

(a)
Total distance = $60 \times 2 + 80 \times 3 = 120 + 240 = 360 km$
Total time = $2 + 3 = 5 hours$
Average speed = $5 360 = 72 km/h$

Answer:

$72 km/h$

(b)
Given $sin θ = 5 3$ , construct a right-angled triangle with opposite side = 3 and hypotenuse = 5.
Adjacent side = $5^{2} - 3^{2} = 16 = 4$

$cos θ = Hypotenuse Adjacent = 54$ $tan θ = Adjacent Opposite = 43$

Answer:

$cos θ = 54, tan θ = 43$

(c)
Given $S_{n} = 3 n^{2} + 5 n$ , the 10th term is $S_{10} - S_{9}$ :

$S_{10} = 3 (10)^{2} + 5 (10) = 300 + 50 = 350$ $S_{9} = 3 (9)^{2} + 5 (9) = 243 + 45 = 288$

10th term = $350 - 288 = 62$

Answer:

$62$

This concludes the sample WAEC 2025 Mathematics paper. Let me know if you need further assistance!

WAEC 2025 Mathematics Paper 2 (Essay) – SkeyPrey.com

WAEC 2025 Mathematics Paper 2 (Essay) Questions & Answer