WAEC Further Mathematics 2015 past questions

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

Ten questions from the 2015 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 2015 questions and answers

1. 1.Four fair coins are tossed once. Calculate the probability of having equal heads and tails.
   1. A
      1/2
   2. B
      5/16
   3. C
      1/4
   4. D
      3/8
      Correct
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   Answer: D

   When 4 fair coins are tossed once, each coin can land Head (H) or Tail (T), giving a total of 2⁴ = 16 equally likely outcomes in the sample space.

   "Equal heads and tails" out of 4 tosses means exactly 2 heads and 2 tails. The number of ways to choose which 2 of the 4 coins show heads is given by the combination C(4,2):

   C(4,2) = 4!/(2!2!) = 6

   So there are 6 favourable outcomes (e.g. HHTT, HTHT, HTTH, THHT, THTH, TTHH) out of the 16 total outcomes.

   Probability = 6/16 = 3/8, which is the correct answer.

2. 2.Given that x²+4x+k = (x+r)²+1, find the value of k and r.
   1. A
      k = 17, r = 4
   2. B
      k = 4, r = 2
   3. C
      k = 5, r = −2
   4. D
      k = 5, r = 2
      Correct
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   Answer: D

   We are told x² + 4x + k = (x + r)² + 1, and we must find k and r.

   First expand the right-hand side:
   (x + r)² + 1 = x² + 2rx + r² + 1

   Now compare this with the left-hand side, x² + 4x + k, coefficient by coefficient.

   Coefficient of x: 2r = 4, so r = 2.

   Constant term: k = r² + 1 = (2)² + 1 = 4 + 1 = k = 5.

   So the answer is k = 5, r = 2. As a check: (x + 2)² + 1 = x² + 4x + 4 + 1 = x² + 4x + 5, which matches x² + 4x + k exactly when k = 5.

   The option with k = 4, r = 2 comes from forgetting to add the extra "+1" when working out the constant term — a student stops at r² instead of r² + 1.

3. 3.A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by a = (3t-2) ms⁻². Find the other time when the velocity would be zero.
   1. A
      3/2 s
   2. B
      4/3 s
      Correct
   3. C
      2/3 s
   4. D
      2 s
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   Answer: B

   Step 1: Find the velocity function. Velocity is the integral of acceleration, so:

   v(t) = ∫(3t − 2) dt = (3/2)t² − 2t + C

   Step 2: Use the starting condition. The particle starts from rest, so v(0) = 0. Substituting t = 0 gives C = 0. So:

   v(t) = (3/2)t² − 2t = t[(3/2)t − 2]

   Step 3: Find when velocity is zero. This product is zero when t = 0 (the starting instant) or when (3/2)t − 2 = 0, which gives t = 4/3 seconds. Since the question asks for the other time (not the start), the answer is t = 4/3 s (that is, 1⅓ seconds).

4. 4.Given that y = 4 - 9x and Δx = 0.1, calculate Δy.
   1. A
      Δy = -9
   2. B
      Δy = -0.09
   3. C
      Δy = -0.9
      Correct
   4. D
      Δy = 0.9
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   Answer: C

   We are given y = 4 - 9x and asked to find the small change Δy when Δx = 0.1.

   For small changes, Δy is approximated using the derivative:

   Δy ≈ (dy/dx) × Δx

   Differentiating y = 4 - 9x with respect to x:

   dy/dx = -9

   Now substitute Δx = 0.1:

   Δy = -9 × 0.1 = -0.9

   So the correct value is Δy = -0.9.

   Where the wrong answers come from:

   Δy = 0.9 comes from dropping the negative sign on the gradient, treating dy/dx as +9 instead of -9.

   ureaΔy = -9 is simply the gradient dy/dx itself, mistaking it for Δy without multiplying by Δx = 0.1.

   Δy = -0.09 comes from a decimal-point slip when multiplying -9 by 0.1 (shifting the decimal one place too far).

5. 5.In how many ways can a committee of 5 be selected from 8 students if 2 particular students are to be included?
   1. A
      35
   2. B
      15
   3. C
      20
      Correct
   4. D
      56
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   Answer: C

   Setting up the problem: The committee has 5 members, and 2 particular students must be in it. Since those 2 seats are already filled, we only need to choose the remaining 5 − 2 = 3 members. They must come from the students who are left: 8 − 2 = 6 students.

   Number of ways = ⁶C₃ = 6!/(3!×3!) = 20.

6. 6.If log3(x) = log9(3), find the value of x.
   1. A
      x = 9
   2. B
      x = 1/3
   3. C
      x = √3
      Correct
   4. D
      x = 3
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   Answer: C

   Step 1: Evaluate log9(3).
   We want the power y such that 9^y = 3. Since 9 = 3², this becomes (3²)^y = 3¹, so 2y = 1, giving y = 1/2. Therefore log9(3) = 1/2.

   Step 2: Solve log3(x) = 1/2.
   This means x = 3^(1/2) = √3, which is the correct value.

   Getting x = 3 happens when a student assumes the "3" on the right side of the original equation must be the answer itself, without actually working out what log9(3) equals.

   Getting x = 9 happens when log9(3) is confused with log3(9), which equals 2 (since 3² = 9); using that wrong value gives x = 3² = 9 instead of the correct 3^(1/2).

7. 7.Given the statements: p: the subject is difficult; q: I will do my best. Which of the following is equivalent to 'Although the subject is difficult, I will do my best'?
   1. A
      p ∧ q
      Correct
   2. B
      q → p
   3. C
      p → q
   4. D
      p ∨ q
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   Answer: A

   Let p: the subject is difficult. Let q: I will do my best.

   The statement "Although the subject is difficult, I will do my best" is really saying two things at the same time: the subject IS difficult, and the speaker WILL do their best. In logic, connecting words like although, but, yet, even though behave exactly like and — they have the same truth table, even though in ordinary English they add a sense of contrast. So the correct symbolic translation is p ∧ q (conjunction).

   So the statement is a straightforward conjunction of the two given statements: p ∧ q.

8. 8.Simplify: nCr ÷ nC(r-1)
   1. A
      (n − r + 1)/r
      Correct
   2. B
      (n + r − 1)/r
   3. C
      (n − r)/r
   4. D
      r/(n − r + 1)
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   Answer: A

   Write out both combinations using the factorial formula.

   nCr = n! ÷ [r!(n − r)!]
   nC(r − 1) = n! ÷ [(r − 1)!(n − r + 1)!]

   Dividing nCr by nC(r − 1) means multiplying nCr by the reciprocal of nC(r − 1):

   nCr ÷ nC(r − 1) = [n!/(r!(n − r)!)] × [(r − 1)!(n − r + 1)!/n!]

   The n! cancels top and bottom, leaving:

   [(r − 1)!(n − r + 1)!] ÷ [r!(n − r)!]

   Now split this into two simple fractions. Since r! = r × (r − 1)!, the term (r − 1)!/r! reduces to 1/r. Since (n − r + 1)! = (n − r + 1) × (n − r)!, the term (n − r + 1)!/(n − r)! reduces to (n − r + 1).

   Multiplying these two results together gives (n − r + 1)/r, which is the simplified answer.

9. 9.Find the 3rd term of (x/2 - 1)⁸ in descending order of x.
   1. A
      7x⁶/8
   2. B
      x⁶/4
   3. C
      7x⁶/16
      Correct
   4. D
      -7x⁶/16
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   Answer: C

   Using the binomial expansion, the general term of (x/2 − 1)⁸ in descending powers of x is:

   T(r+1) = C(8,r) × (x/2)^(8−r) × (−1)^r

   The 3rd term corresponds to r = 2 (the 1st term is r = 0, the 2nd term is r = 1, the 3rd term is r = 2).

   So T3 = C(8,2) × (x/2)^6 × (−1)²
   C(8,2) = 28, (x/2)^6 = x⁶/64, and (−1)² = 1.

   T3 = 28 × x⁶/64 = 7x⁶/16, which is the correct 3rd term.

   The option −7x⁶/16 comes from wrongly treating (−1)² as −1 instead of +1 — a squared negative sign always turns positive, so the sign should not carry over.

   The option 7x⁶/8 comes from simplifying (x/2)^6 incorrectly as x⁶/32 instead of x⁶/64, doubling the final coefficient.

10. 10.If the determinant |1+2x, -1; 6, 3-x| = -3, find the values of x.
    1. A
       x = 0 or x = 5/2
    2. B
       x = 4 or x = -3/4
    3. C
       x = 4 or x = -3/2
       Correct
    4. D
       x = -4 or x = 3/2
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    Answer: C

    The determinant of a 2×2 matrix |a b; c d| is calculated as ad − bc.

    Here, a = 1+2x, b = −1, c = 6, d = 3−x. So the determinant is (1+2x)(3−x) − (−1 × 6).

    Expanding (1+2x)(3−x) gives 3 − x + 6x − 2x² = 3 + 5x − 2x².

    Subtracting (−1 × 6), which is subtracting −6, actually means adding 6. So the full determinant is 3 + 5x − 2x² + 6 = −2x² + 5x + 9.

    Setting this equal to −3: −2x² + 5x + 9 = −3, which simplifies to −2x² + 5x + 12 = 0. Multiplying through by −1 gives the cleaner form 2x² − 5x − 12 = 0.

    Applying the quadratic formula: x = [5 ± √(25 + 96)] / 4 = [5 ± 11] / 4.

    This gives x = 4 or x = −3/2, which is the correct pair of values.

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