WAEC Mathematics 2025: Top Likely Questions

Hello guys! Welcome to today’s tutorial by AllCBTs. In this lesson, we’re solving WAEC Mathematics past questions with step-by-step solutions.
Please get your writing materials and calculators ready — let’s dive in!

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You can watch the full class in the video below:

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Question 1: Find the Sum of the First Four Terms of a Geometric Progression (GP)

Given:
The second and fifth terms of a geometric progression (GP) are 6 and 48, respectively.
Task: Find the sum of the first four terms.

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Step-by-Step Solution

Let the first term be a and the common ratio be r.
From GP formula:

• T2=arT_2 = arT2​=ar
• T5=ar4T_5 = ar^4T5​=ar4

Given:

• ar=6ar = 6ar=6
• ar4=48ar^4 = 48ar4=48

Divide both equations: ar4ar=486⇒r3=8⇒r=83=2\frac{ar^4}{ar} = \frac{48}{6} \Rightarrow r^3 = 8 \Rightarrow r = \sqrt[3]{8} = 2arar4​=648​⇒r3=8⇒r=38​=2

Now, plug back r=2r = 2r=2 into ar=6ar = 6ar=6: a×2=6⇒a=62=3a \times 2 = 6 \Rightarrow a = \frac{6}{2} = 3a×2=6⇒a=26​=3

First Four Terms:

• a=3a = 3a=3
• ar=3×2=6ar = 3 \times 2 = 6ar=3×2=6
• ar2=3×4=12ar^2 = 3 \times 4 = 12ar2=3×4=12
• ar3=3×8=24ar^3 = 3 \times 8 = 24ar3=3×8=24

Sum = 3 + 6 + 12 + 24 = 45

Correct Answer: Option E

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Question 2: Trigonometry – Find tan⁡θ\tan \thetatanθ Given sin⁡θ=k\sin \theta = ksinθ=k

Given: sin⁡θ=k\sin \theta = ksinθ=k
Find: tan⁡θ\tan \thetatanθ

Recall:

• sin⁡θ=OppositeHypotenuse=k1\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{k}{1}sinθ=HypotenuseOpposite​=1k​

Using Pythagoras theorem: Adjacent2=12−k2⇒Adjacent=1−k2\text{Adjacent}^2 = 1^2 – k^2 \Rightarrow \text{Adjacent} = \sqrt{1 – k^2}Adjacent2=12−k2⇒Adjacent=1−k2​ tan⁡θ=OppositeAdjacent=k1−k2\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{k}{\sqrt{1 – k^2}}tanθ=AdjacentOpposite​=1−k2​k​

Correct Answer: Option C

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Question 3: Simplify Using Difference of Squares

Question:
Evaluate 101.52−100.52101.5^2 – 100.5^2101.52−100.52

Trick:

Use difference of squares identity: a2−b2=(a+b)(a−b)a^2 – b^2 = (a + b)(a – b)a2−b2=(a+b)(a−b) =(101.5+100.5)(101.5−100.5)=202×1=202= (101.5 + 100.5)(101.5 – 100.5) = 202 \times 1 = 202=(101.5+100.5)(101.5−100.5)=202×1=202

Answer: 202

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Question 4: Express Product in Standard Form

Question:
What is the standard form of 0.06×0.090.06 \times 0.090.06×0.09 ?

Convert to standard form:

• 0.06=6.0×10−20.06 = 6.0 \times 10^{-2}0.06=6.0×10−2
• 0.09=9.0×10−20.09 = 9.0 \times 10^{-2}0.09=9.0×10−2

=(6.0×10−2)(9.0×10−2)=54.0×10−4=5.4×10−3= (6.0 \times 10^{-2})(9.0 \times 10^{-2}) = 54.0 \times 10^{-4} = 5.4 \times 10^{-3}=(6.0×10−2)(9.0×10−2)=54.0×10−4=5.4×10−3

Answer: 5.4×10−35.4 \times 10^{-3}5.4×10−3

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Watch more educational videos and past questions: https://youtube.com/@allcbts

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Quick Exam Tips for WAEC Maths

• Use tricks like difference of squares, factorization, and recall of trigonometric identities.
• Convert decimals to standard form carefully.
• Apply Pythagorean theorem in trigonometry problems.
• Memorize key GP and AP formulas.

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