WAEC Further Mathematics 2013 past questions

Checked by the Skolra question team · Last updated 23 October 2026

Ten questions from the 2013 WAEC Further Mathematics paper, with answers and explanations. We hold 43 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2013 questions and answers

1. 1.Given that $a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1$, solve for n.
   1. A
      n = -5/6
   2. B
      n = 6/5
      Correct
   3. C
      n = -6/5
   4. D
      n = 5/6
   Show answerHide answer

   Answer: B

   Rule used: when you multiply powers of the same base, you add the exponents: a^x × a^y = a^x+y.

   So a^5/6 × a^-1/n = a^{(5/6 - 1/n)}. Since this equals 1 = a⁰, the exponent itself must be zero:

   5/6 - 1/n = 0
   1/n = 5/6
   n = 6/5

   The value n = 6/5 is correct.

   n = 5/6 is the tempting wrong answer you get if you solve 1/n = 5/6 by just copying the fraction across instead of inverting it to find n.

   n = -6/5 comes from wrongly adding the exponents as 5/6 + 1/n = 0 (treating the second exponent as positive 1/n, forgetting the negative sign already in -1/n), which flips the sign of the final answer.

2. 2.If $\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}$, find the possible values of x.
   1. A
      x = 1
   2. B
      x = 0
      Correct
   3. C
      x = 0 or x = -1
   4. D
      x = -1
   Show answerHide answer

   Answer: B

   Square both sides of √x + √(x+1) = √(2x+1):

   x + (x+1) + 2√(x(x+1)) = 2x + 1

   This simplifies to 2x + 1 + 2√(x² + x) = 2x + 1, so 2√(x² + x) = 0, giving x² + x = 0, which factorises as x(x + 1) = 0.

   That gives two candidate roots: x = 0 and x = -1. But the original equation has square roots of x and of (x+1), so x must be greater than or equal to 0 for the terms to be real. x = -1 fails this requirement and must be discarded — it is an extraneous root introduced by squaring.

   Checking x = 0 in the original equation: √0 + √1 = 0 + 1 = 1, and √(2(0)+1) = √1 = 1. Both sides match, so x = 0 is the correct and only valid answer.

3. 3.If $2\sin^{2}\theta = 1 + \cos \theta, 0^\circ \leq \theta \leq 90^\circ$, find $\theta$.
   1. A
      45°
   2. B
      30°
   3. C
      90°
   4. D
      60°
      Correct
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   Answer: D

   Start from the identity: replace sin²θ with 1 − cos²θ, since the equation mixes sin and cos and we want a single-variable equation.

   2sin²θ = 1 + cosθ
   2(1 − cos²θ) = 1 + cosθ
   2 − 2cos²θ = 1 + cosθ
   2cos²θ + cosθ − 1 = 0

   This factorises neatly as (2cosθ − 1)(cosθ + 1) = 0, giving cosθ = 1/2 or cosθ = −1.

   Apply the given range. cosθ = −1 gives θ = 180°, which falls outside 0° ≤ θ ≤ 90°, so it is rejected. cosθ = 1/2 gives θ = 60°, which fits the range perfectly — this is the correct answer.

4. 4.The fourth term of an exponential sequence is 192 and its ninth term is 6. Find the common ratio of the sequence.
   1. A
      32
   2. B
      1/32
   3. C
      1/2
      Correct
   4. D
      2
   Show answerHide answer

   Answer: C

   In an exponential (geometric) sequence, each term is the previous term multiplied by the common ratio r. So the 4th term is ar³ and the 9th term is ar⁸, where a is the first term.

   From the question:
   ar³ = 192  (4th term)
   ar⁸ = 6  (9th term)

   Divide the 9th term by the 4th term so that a cancels out, leaving only powers of r. The gap between the 9th and 4th term is 5 steps, so you are left with r⁵:

   r⁵ = 6 ÷ 192 = 1/32

   Taking the fifth root of both sides: r = 1/2, since (1/2) × (1/2) × (1/2) × (1/2) × (1/2) = 1/32.

   A candidate who divides the terms the wrong way round (192 ÷ 6 instead of 6 ÷ 192) will wrongly get r⁵ = 32, and end up with the reciprocal answer, 2, instead of 1/2.

5. 5.Differentiate $x^{2} + xy - 5 = 0$.
   1. A
      dy/dx = -(2x + y)/(x + y)
   2. B
      dy/dx = -2x - y
   3. C
      dy/dx = -(2x + y)/x
      Correct
   4. D
      dy/dx = (2x + y)/x
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   Answer: C

   Differentiate x² + xy - 5 = 0 implicitly with respect to x.

   Take each term in turn:
   d/dx(x²) = 2x
   d/dx(xy) needs the product rule, because y is itself a function of x: d/dx(xy) = x·(dy/dx) + y
   d/dx(-5) = 0 (a constant has no derivative)

   Putting it together: 2x + x(dy/dx) + y = 0

   Now solve for dy/dx: x(dy/dx) = -2x - y, so dy/dx = -(2x + y)/x. That is the correct result.

   One tempting wrong answer drops the minus sign altogether and leaves dy/dx = (2x + y)/x — this comes from moving terms across the equals sign without flipping the sign correctly.

6. 6.Two bodies of masses 3kg and 5kg moving with velocities 2 m/s and V m/s respectively in opposite directions collide. If they move together after collision with velocity 3.5 m/s in the direction of the 5kg mass, find the value of V.
   1. A
      6.8 m/s
      Correct
   2. B
      12.7 m/s
   3. C
      4.4 m/s
   4. D
      2.6 m/s
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   Answer: A

   Take the direction of motion of the 5 kg mass as the positive direction. Since the question says the two bodies move in opposite directions, the 3 kg mass's velocity must be taken as -2 m/s, while the 5 kg mass's velocity is +V.

   By conservation of linear momentum:
   m₁u₁ + m₂u₂ = (m₁ + m₂)v

   3(-2) + 5(V) = (3 + 5)(3.5)
   -6 + 5V = 28
   5V = 34
   V = 6.8 m/s

   So 6.8 m/s is correct. Check: total momentum before = 3(-2) + 5(6.8) = -6 + 34 = 28 kgm/s, and total momentum after = 8(3.5) = 28 kgm/s. They match, confirming the answer.

7. 7.If $\alpha$ and $\beta$ are the roots of the equation $2x^{2} - 6x + 5 = 0$, evaluate $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$.
   1. A
      8/5
      Correct
   2. B
      4/5
   3. C
      6/5
   4. D
      18/5
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   Answer: A

   For 2x² − 6x + 5 = 0, using sum and product of roots:

   Sum: α + β = −(−6)/2 = 3
   Product: αβ = 5/2

   We need β/α + α/β. Combine over a common denominator:

   β/α + α/β = (α² + β²)/(αβ)

   Now express α² + β² using the sum and product: α² + β² = (α + β)² − 2αβ = 3² − 2(5/2) = 9 − 5 = 4.

   So β/α + α/β = 4 ÷ (5/2) = 4 × 2/5 = 8/5.

   The option 18/5 comes from forgetting to subtract the 2αβ term and simply dividing (α+β)² straight by αβ. The option 4/5 comes from using αβ = 5 instead of correctly dividing the constant term by the leading coefficient to get 5/2. The option 6/5 comes from mixing up the identity and dividing the sum (α+β) by the product (αβ) instead of using α² + β².

8. 8.A circular ink blot on a piece of paper increases its area at the rate $4mm^{2}/s$. Find the rate of the radius of the blot when the radius is 8mm. $[\pi = \frac{22}{7}]$.
   1. A
      7/44 mm/s
   2. B
      7/176 mm/s
   3. C
      7/11 mm/s
   4. D
      7/88 mm/s
      Correct
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   Answer: D

   The area of a circle is A = πr². Differentiating both sides with respect to time using the chain rule gives dA/dt = 2πr · dr/dt.

   We are given dA/dt = 4 mm²/s and r = 8 mm, so:
   4 = 2π(8) × dr/dt
   4 = 16π × dr/dt
   dr/dt = 4/(16π) = 1/(4π)

   Using π = 22/7: dr/dt = 1 ÷ (4 × 22/7) = 7/88 mm/s (about 0.08 mm/s). This is the correct rate at which the radius is increasing.

   A candidate who forgets the factor of 2 that comes from differentiating r² (treating the rate as dA/dt = πr · dr/dt instead of 2πr · dr/dt) gets 7/44 mm/s — exactly double the correct value.

9. 9.Given that $P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}$ and PQ = R, find the value of x.
   1. A
      x = -4
   2. B
      x = 5
      Correct
   3. C
      x = -5
   4. D
      x = 3
   Show answerHide answer

   Answer: B

   Multiply the matrices first. With $P=\begin{pmatrix}3&4\\2&x\end{pmatrix}$ and $Q=\begin{pmatrix}1&3\\-2&4\end{pmatrix}$, the product $PQ$ is:

   Row 1: $(3\times1)+(4\times-2)=3-8=-5$, and $(3\times3)+(4\times4)=9+16=25$. This already matches the top row of R, $(-5\ \ 25)$, which confirms the arrangement is correct.

   Row 2: $(2\times1)+(x\times-2)=2-2x$, and $(2\times3)+(x\times4)=6+4x$.

   Compare with R's second row, $(-8\ \ 26)$:
   $2-2x=-8 \Rightarrow -2x=-10 \Rightarrow x=5$
   $6+4x=26 \Rightarrow 4x=20 \Rightarrow x=5$

   Both entries agree, so x = 5 is confirmed from two independent equations — a strong check that the value is correct.

10. 10.Find the equation of the line that is perpendicular to $2y + 5x - 6 = 0$ and bisects the line joining the points P(4, 3) and Q(-6, 1).
    1. A
       2x - 5y + 12 = 0
       Correct
    2. B
       5x + 2y + 1 = 0
    3. C
       2x - 5y + 8 = 0
    4. D
       5x - 2y + 9 = 0
    Show answerHide answer

    Answer: A

    Step 1: Find the slope of the given line.
    2y + 5x - 6 = 0 → 2y = -5x + 6 → y = -5/2 x + 3, so its slope is -5/2.

    Step 2: Find the slope of the perpendicular line.
    For perpendicular lines, the product of the slopes is -1. So the required slope m satisfies (-5/2)(m) = -1, giving m = 2/5.

    Step 3: Find the midpoint of PQ.
    P(4, 3) and Q(-6, 1): midpoint = ((4 + (-6))/2, (3 + 1)/2) = (-1, 2).

    Step 4: Form the equation through the midpoint with slope 2/5.
    y - 2 = (2/5)(x + 1)
    5(y - 2) = 2(x + 1)
    5y - 10 = 2x + 2
    2x - 5y + 12 = 0

    This is the correct equation of the required line.

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