WAEC 2025 Mathematics Paper 2 (Essay) Questions & Answer

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.

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WAEC 2025 Mathematics Paper 2 (Essay)

Time Allowed: 2 hours 30 minutes
Instructions:

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

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Question 1

(a) Simplify:

34÷(23−12)+5643​÷(32​−21​)+65​

(b) Solve for $x$ in the equation:

2x−13−x+24=132x−1​−4x+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.

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Answer to Question 1

(a)
Simplify:

34÷(23−12)+5643​÷(32​−21​)+65​

Step 1: Solve the expression in the brackets:

23−12=4−36=1632​−21​=64−3​=61​

Step 2: Perform the division:

34÷16=34×61=184=9243​÷61​=43​×16​=418​=29​

Step 3: Add 5665​:

92+56=27+56=326=16329​+65​=627+5​=632​=316​

Final Answer:

163316​​

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(b)
Solve for $x$:

2x−13−x+24=132x−1​−4x+2​=1

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

12⋅2x−13−12⋅x+24=12⋅112⋅32x−1​−12⋅4x+2​=12⋅1$$4(2x-1)-3(x+2)=12$$

Step 2: Expand the brackets:

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

Step 3: Solve for $x$:

$$5x=12+10$$$$5x=22$$x=225x=522​

Final Answer:

x=225x=522​​

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(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₦42,000​

(ii) Ratio of expenditure on rent to food:

Ratio=₦48,000₦30,000=4830=85Ratio=₦30,000₦48,000​=3048​=58​

Answer:

8:58:5​

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Question 2

(a) Solve the simultaneous equations:

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

(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

(c) The area of a rectangle is 48 cm248cm2. If its length is 8 cm, find its perimeter.

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Answer to Question 2

(a)
Solve the simultaneous equations:

$$2x+3y=13\text{(1)}$$$$4x-y=5\text{(2)}$$

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

$$4x-y=5⟹y=4x-5$$

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

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

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

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

Final Answer:

x=2, y=3x=2,y=3​

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

(i) Probability of drawing a red ball:

P(Red)=510=12P(Red)=105​=21​

Answer:

1221​​

(ii) Probability of not drawing a green ball:

P(Not Green)=1−P(Green)=1−210=810=45P(Not Green)=1−P(Green)=1−102​=108​=54​

Answer:

4554​​

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(c)
Area of rectangle = 48 cm248cm2
Length = 8 cm
Width = AreaLength=488=6 cmLengthArea​=848​=6cm
Perimeter = $2(\text{Length}+\text{Width})=2(8+6)=28\text{cm}$

Answer:

28 cm28cm​

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Question 3

(a) Factorize completely:

x2−5x+6x2−5x+6

(b) Solve the inequality:

$$3x-7<5$$

(c) A triangle has sides of lengths 6 cm, 8 cm, and 10 cm. Determine if it is a right-angled triangle.

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Answer to Question 3

(a)
Factorize:

x2−5x+6x2−5x+6

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

x2−5x+6=(x−2)(x−3)x2−5x+6=(x−2)(x−3)

Answer:

(x−2)(x−3)(x−2)(x−3)​

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(b)
Solve the inequality:

$$3x-7<5$$

Add 7 to both sides:

$$3x<12$$

Divide by 3:

$$x<4$$

Answer:

x<4x<4​

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

62+82=36+64=100=10262+82=36+64=100=102

Since 62+82=10262+82=102, the triangle is right-angled.

Answer:

Yes, it is a right-angled triangleYes, it is a right-angled triangle​

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Question 4

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

log⁡2(x+3)=4log2​(x+3)=4

(b) A circle has a radius of 7 cm. Calculate its circumference and area. (Take π=227π=722​)

(c) Solve the quadratic equation:

2×2−5x−3=02x2−5x−3=0

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Answer to Question 4

(a)
Solve:

log⁡2(x+3)=4log2​(x+3)=4

Convert to exponential form:

x+3=24x+3=24$$x+3=16$$$$x=13$$

Answer:

1313​

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(b)
Radius $r=7\text{cm}$
Circumference = 2πr=2×227×7=44 cm2πr=2×722​×7=44cm
Area = πr2=227×72=154 cm2πr2=722​×72=154cm2

Answer:

Circumference=44 cm, Area=154 cm2Circumference=44cm,Area=154cm2​

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(c)
Solve:

2×2−5x−3=02x2−5x−3=0

Using the quadratic formula:

x=5±(−5)2−4×2×(−3)2×2x=2×25±(−5)2−4×2×(−3)​​x=5±25+244=5±494=5±74x=45±25+24​​=45±49​​=45±7​x=124=3orx=−24=−12x=412​=3orx=4−2​=−21​

Answer:

x=3 or x=−12x=3orx=−21​​

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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=35sinθ=53​, find $\cos \theta$ and $\tan \theta$.

(c) The sum of the first $n$ terms of an arithmetic sequence is given by Sn=3n2+5nSn​=3n2+5n. Find the 10th term.

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Answer to Question 5

(a)
Total distance = $60\times 2+80\times 3=120+240=360\text{km}$
Total time = $2+3=5\text{hours}$
Average speed = 3605=72 km/h5360​=72km/h

Answer:

72 km/h72km/h​

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(b)
Given sin⁡θ=35sinθ=53​, construct a right-angled triangle with opposite side = 3 and hypotenuse = 5.
Adjacent side = 52−32=16=452−32​=16​=4

cos⁡θ=AdjacentHypotenuse=45cosθ=HypotenuseAdjacent​=54​tan⁡θ=OppositeAdjacent=34tanθ=AdjacentOpposite​=43​

Answer:

cos⁡θ=45, tan⁡θ=34cosθ=54​,tanθ=43​​

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(c)
Given Sn=3n2+5nSn​=3n2+5n, the 10th term is S10−S9S10​−S9​:

S10=3(10)2+5(10)=300+50=350S10​=3(10)2+5(10)=300+50=350S9=3(9)2+5(9)=243+45=288S9​=3(9)2+5(9)=243+45=288

10th term = $350-288=62$

Answer:

6262​

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