WAEC Further Mathematics (Further Maths) past questions and answers

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

WAEC Further Mathematics: 843 objective past questions from 19 papers between 2006 and 2024, each with the answer and an explanation. Here are ten.

How the WAEC Further Maths paper is set. On WAEC's 2026 timetable for private candidates, Paper 1 (objective) takes 1 hour 30 minutes and Paper 2 (essay) 2 hours 30 minutes. If the question paper on the day gives another time, follow the question paper. The questions here are from Paper 1.

1. 1.Find the sum of all natural numbers between 403 and 603 which are divisible by 7
   1. A
      14,112
   2. B
      14,587
   3. C
      15,015
   4. D
      14,616
      Correct
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   Answer: D

   The numbers divisible by 7 that lie between 403 and 603 form an arithmetic progression (AP).

   Finding the first term: 403 ÷ 7 = 57.57…, so the first multiple of 7 above 403 is 7 × 58 = 406.

   Finding the last term: 603 ÷ 7 = 86.14…, so the last multiple of 7 below 603 is 7 × 86 = 602.

   Counting the terms: using n = (l − a)/d + 1 = (602 − 406)/7 + 1 = 28 + 1 = 29 terms.

   Summing the AP:
   Sum = n/2 × (a + l) = 29/2 × (406 + 602) = 29/2 × 1008 = 29 × 504 = 14,616.

   This is the correct total.

   The value 14,112 comes from forgetting to add the extra 1 when counting the terms (using 28 terms instead of 29), which shortchanges the sum by one full term (406).

2. 2.How many ways can 6 students be seated around a circular table?
   1. A
      120
      Correct
   2. B
      720
   3. C
      60
   4. D
      360
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   Answer: A

   Circular arrangement formula: when n distinct people are seated around a circular table, the number of ways is (n − 1)!, not n!.

   This is because, on a circle, there is no fixed "first seat" — rotating everyone round to the next chair gives the same seating arrangement. So one person's position is fixed as a reference point, and only the remaining (n − 1) people are arranged relative to that person.

   For 6 students: (6 − 1)! = 5! = 5 × 4 × 3 × 2 × 1 = 120.

   A student who forgets the circular table and treats the seats as being in a straight line gets 6! = 720 — that mistake ignores that rotated seatings are identical.

3. 3.A particle is acted upon by two forces 6N and 3N inclined at an angle of 120° to each other. Find the magnitude of the resultant force.
   1. A
      9N
   2. B
      3√3N
      Correct
   3. C
      3√7N
   4. D
      3N
   Show answerHide answer

   Answer: B

   For two forces P and Q acting at an angle θ to each other, the magnitude of the resultant is found from the parallelogram law:

   R² = P² + Q² + 2PQ cosθ

   Here P = 6N, Q = 3N and θ = 120°, so cos120° = −½.

   R² = 6² + 3² + 2(6)(3)cos120°
   R² = 36 + 9 + 36(−0.5)
   R² = 45 − 18 = 27
   R = √27 = 3√3 N (≈ 5.2N)

   So the resultant of the two forces is 3√3N.

   A candidate who simply adds the two forces together, 6N + 3N, as if they act in the same straight line gets 9N — that ignores the angle between them completely, which only works when the forces are in the same direction (0°).

4. 4.If $\log_{3}a - 2 = 3\log_{3}b$, express a in terms of b.
   1. A
      a = 27b
   2. B
      a = b³ + 9
   3. C
      a = b³/9
   4. D
      a = 9b³
      Correct
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   Answer: D

   Start with log₃a − 2 = 3log₃b and collect the logarithm terms on one side:

   log₃a = 3log₃b + 2

   Use the power law of logarithms on the right: 3log₃b = log₃(b³).

   Now the constant 2 must also be turned into a log to base 3 before you can combine it, since 2 = log₃(3²) = log₃9.

   So: log₃a = log₃(b³) + log₃9 = log₃(9b³), using the product law (log of a product = sum of logs).

   Since both sides are logarithms to the same base, the arguments are equal: a = 9b³.

   The option a³ + 9 comes from wrongly treating log₃(b³) + log₃9 as if adding logs means adding their arguments — it does not; it means multiplying the arguments.

5. 5.A linear transformation T is defined by T: (x,y) -> (3x-y, x+4y). Find the image of (2,-1) under T.
   1. A
      (5, 6)
   2. B
      (7, -2)
      Correct
   3. C
      (5, -2)
   4. D
      (-2, 7)
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   Answer: B

   We are given T(x, y) = (3x - y, x + 4y) and asked to find the image of the point (2, -1), so x = 2 and y = -1.

   First component: 3x - y = 3(2) - (-1) = 6 + 1 = 7.
   Second component: x + 4y = 2 + 4(-1) = 2 - 4 = -2.

   So the image of (2, -1) under T is (7, -2).

6. 6.A linear transformation is defined by T: (x,y) maps to (-x+y,-4y). Find the image, Q, of Q(-3,2) under T
   1. A
      (5, -8)
      Correct
   2. B
      (5, 8)
   3. C
      (-1, -8)
   4. D
      (-1, 8)
   Show answerHide answer

   Answer: A

   The transformation is T: (x, y) → (-x + y, -4y). To find the image of Q(-3, 2), substitute x = -3 and y = 2 directly into both rules.

   For the first coordinate: -x + y = -(-3) + 2 = 3 + 2 = 5.
   For the second coordinate: -4y = -4(2) = -8.

   So the image of Q under T is (5, -8).

   A candidate who drops the negative sign in front of x gets x + y = -3 + 2 = -1, landing on (-1, -8) by mistake. A candidate who forgets the negative sign in -4y gets 4y = 8, landing on (5, 8). Making both sign slips at once gives (-1, 8). Each of these comes from mishandling just one of the two signs in the transformation rule, so it is important to apply -x and -4y carefully, keeping track of the signs throughout.

7. 7.A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by $v = (3t^{2} - 2t) ms^{-1}$. Calculate the distance covered in the first 2 seconds.
   1. A
      4 8/27 m
      Correct
   2. B
      3 19/27 m
   3. C
      4 m
   4. D
      4 4/27 m
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   Answer: A

   Step 1 — check for a direction reversal. v = 3t² − 2t = t(3t − 2), so v = 0 at t = 0 and at t = 2/3 s. Since t = 2/3 s lies inside the given 0–2 s interval, the particle slows to rest, reverses, and comes back — so "distance covered" is not simply the net displacement; it must be worked out in two stages.

   Step 2 — find the displacement function. Integrating, x(t) = t³ − t² (the constant is zero since the particle starts from the origin).

   x(2/3) = 8/27 − 4/9 = 8/27 − 12/27 = −4/27 m
   x(2) = 8 − 4 = 4 m

8. 8.If y = 4x-1, list the range of the domain {-2≤x≤2}, where x is an integer.
   1. A
      {-9, -5, -1, 3, 7}
      Correct
   2. B
      {-8, -4, 0, 4, 8}
   3. C
      {-9, -5, -1, 3}
   4. D
      {-7, -3, 1, 5, 9}
   Show answerHide answer

   Answer: A

   The domain is the set of integer values of x satisfying -2 ≤ x ≤ 2, which gives x = -2, -1, 0, 1, 2 (five values, since the endpoints are included).

   Substitute each one into y = 4x - 1:

   x = -2: y = 4(-2) - 1 = -8 - 1 = -9
   x = -1: y = 4(-1) - 1 = -4 - 1 = -5
   x = 0: y = 4(0) - 1 = -1
   x = 1: y = 4(1) - 1 = 4 - 1 = 3
   x = 2: y = 4(2) - 1 = 8 - 1 = 7

   So the range is {-9, -5, -1, 3, 7}, which is the correct answer.

9. 9.If (x-5) is a factor of x³-4x²-11x+30, find the remaining factors.
   1. A
      (x − 3) and (x + 2)
   2. B
      (x + 3) and (x − 2)
      Correct
   3. C
      (x − 6) and (x + 5)
   4. D
      (x + 6) and (x − 5)
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   Answer: B

   Since (x − 5) is given as a factor of x³ − 4x² − 11x + 30, we divide the cubic by (x − 5) using synthetic division with the root x = 5, taking the coefficients 1, −4, −11, 30.

   1 brought down → 1 × 5 = 5, so −4 + 5 = 1 → 1 × 5 = 5, so −11 + 5 = −6 → −6 × 5 = −30, so 30 + (−30) = 0.

   The zero remainder confirms (x − 5) truly divides the expression, and the quotient is x² + x − 6.

   Factorising x² + x − 6: we need two numbers that multiply to −6 and add to +1, which are +3 and −2. This gives (x + 3)(x − 2), so the remaining factors are (x + 3) and (x − 2).

   Checking: (x − 5)(x + 3)(x − 2) expands back to x³ − 4x² − 11x + 30, confirming the result.

10. 10.A and B are two independent events such that $P(A) = \frac{2}{5}$ and $P(A \cap B) = \frac{1}{15}$. Find $P(B)$.
    1. A
       1/3
    2. B
       2/75
    3. C
       1/6
       Correct
    4. D
       7/15
    Show answerHide answer

    Answer: C

    Key fact: when two events A and B are independent, the multiplication law for probability says P(A ∩ B) = P(A) × P(D).

    We are given P(A) = 2/5 and P(A ∩ B) = 1/15. Making P(D) the subject:

    P(D) = P(A ∩ B) ÷ P(A) = (1/15) ÷ (2/5)

    Dividing by a fraction means multiplying by its reciprocal, so:

    P(D) = (1/15) × (5/2) = 5/30 = 1/6

    That is the correct value.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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