WAEC Further Mathematics 2019 past questions

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

Ten questions from the 2019 WAEC Further Mathematics paper, with answers and explanations. We hold 47 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2019 questions and answers

1. 1.The second and fourth terms of an exponential sequence (G.P) are 2/9 and 8/81 respectively. Find the sixth term of the sequence
   1. A
      32/729
      Correct
   2. B
      -32/729
   3. C
      16/243
   4. D
      64/2187
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   Answer: A

   Given: T₂ = 2/9 and T₄ = 8/81 of a G.P. (an exponential sequence), with T_n = ar^n-1.

   Step 1 — find the common ratio. Since T₄ = T₂ × r²:
   r² = (8/81) ÷ (2/9) = (8/81) × (9/2) = 4/9, so r = ±2/3.

   Step 2 — find the first term. From T₂ = ar = 2/9: if r = 2/3, a = 1/3; if r = -2/3, a = -1/3. Both are valid branches of the same sequence.

   Step 3 — jump straight from T₄ to T₆. The cleanest route is T₆ = T₄ × r², since moving from the 4th term to the 6th term is exactly two more multiplications by r, i.e. one multiplication by r²:
   T₆ = (8/81) × (4/9) = 32/729.

   So the sixth term of the sequence is 32/729.

2. 2.A 35 N force acts on a body of mass 5 kg for 2 seconds. Calculate the change in momentum of the body.
   1. A
      17.5 kgm/s
   2. B
      175 kgm/s
   3. C
      350 kgm/s
   4. D
      70 kgm/s
      Correct
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   Answer: D

   Change in momentum = Impulse = Force × time.

   Given: Force (F) = 35 N, time (t) = 2 seconds.

   Change in momentum = F × t = 35 × 2 = 70 kgm/s.

   You can confirm this another way: acceleration a = F/m = 35/5 = 7 m/s². Velocity gained v = at = 7 × 2 = 14 m/s. Change in momentum = mass × change in velocity = 5 × 14 = 70 kgm/s. Both routes agree, so 70 kgm/s is correct.

   The wrong options come from common slip-ups: multiplying the force by the mass instead of the time gives 175 kgm/s; dividing the force by the time instead of multiplying gives 17.5 kgm/s; and multiplying force by both time and mass together (double-counting the mass, which the impulse formula does not need) gives 350 kgm/s.

3. 3.Three soldiers, X, Y and Z have probabilities 1/3, 1/5 and 1/4 respectively of hitting a target. If each of them fires once, find, correct to two decimal places, the probability that only one of them hits the target.
   1. A
      0.43
      Correct
   2. B
      0.20
   3. C
      0.40
   4. D
      0.60
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   Answer: A

   Set up the probabilities. P(X hits) = 1/3, P(Y hits) = 1/5, P(Z hits) = 1/4. So the "miss" probabilities are P(X misses) = 2/3, P(Y misses) = 4/5, P(Z misses) = 3/4.

   "Only one hits" has three separate ways to happen: X hits while Y and Z miss, OR Y hits while X and Z miss, OR Z hits while X and Y miss. These three cases cannot happen together, so we add them.

   P(only X) = (1/3)(4/5)(3/4) = 12/60
   P(only Y) = (2/3)(1/5)(3/4) = 6/60
   P(only Z) = (2/3)(4/5)(1/4) = 8/60

   Adding: 12/60 + 6/60 + 8/60 = 26/60 = 13/30 ≈ 0.43 (correct to two decimal places).

4. 4.Given that X and Y are independent events such that P(X)=0.5, P(Y)=m and P(X union Y)=0.75, find the value of m.
   1. A
      0.6
   2. B
      0.4
   3. C
      0.25
   4. D
      0.5
      Correct
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   Answer: D

   Since X and Y are independent, P(X∩Y) = P(X) × P(Y) = 0.5m.

   Using the addition rule:
   P(X∪Y) = P(X) + P(Y) − P(X∩Y)
   0.75 = 0.5 + m − 0.5m
   0.75 = 0.5 + 0.5m
   0.5m = 0.25
   m = 0.5

   A candidate who forgets that the events are independent and simply adds the probabilities (treating them as mutually exclusive) gets 0.75 = 0.5 + m, giving the wrong value 0.25.

   Other values such as 0.6 and 0.4 come from mis-rearranging the equation or applying the wrong sign to the P(X)×P(Y) term — they do not satisfy the independence condition once checked back against P(X∪Y) = 0.75.

   Only m = 0.5 satisfies the equation correctly, since 0.5 + 0.5 − (0.5×0.5) = 0.5 + 0.5 − 0.25 = 0.75.

5. 5.Find the equation of the circle centre (2,3) which passes through the y-intercept of the line 3x-2y+6=0
   1. A
      (x − 2)² + (y − 3)² = 4
      Correct
   2. B
      (x − 2)² + (y − 3)² = 2
   3. C
      (x + 2)² + (y + 3)² = 4
   4. D
      (x − 2)² + (y − 3)² = 25
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   Answer: A

   Step 1: Find the y-intercept of the line.
   The line is 3x − 2y + 6 = 0. A y-intercept occurs where x = 0, so: −2y + 6 = 0 → y = 3. The y-intercept is the point (0, 3).

   Step 2: Find the radius.
   The circle has centre (2, 3) and passes through (0, 3), so the radius is the distance between these two points: r = √[(2−0)² + (3−3)²] = √4 = 2.

   Step 3: Write the equation.
   A circle with centre (h, k) and radius r has equation (x−h)² + (y−k)² = r². With centre (2,3) and r = 2, r² = 4, giving (x − 2)² + (y − 3)² = 4. This is the correct equation.

6. 6.The probabilities that John and Jane will pass an examination are 0.9 and 0.7 respectively. Find the probability that at least one of them will pass the examination.
   1. A
      0.63
   2. B
      0.60
   3. C
      0.37
   4. D
      0.97
      Correct
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   Answer: D

   P(John passes) = 0.9, so P(John fails) = 1 − 0.9 = 0.1.
   P(Jane passes) = 0.7, so P(Jane fails) = 1 − 0.7 = 0.3.

   "At least one passes" is the opposite of "both fail". Since the two events are independent, the probability that both fail is found by multiplying their individual failure probabilities:

   P(both fail) = 0.1 × 0.3 = 0.03

   P(at least one passes) = 1 − P(both fail) = 1 − 0.03 = 0.97. This is the correct answer.

   A candidate who multiplies the two pass probabilities together (0.9 × 0.7 = 0.63) has actually found the probability that both John and Jane pass, not "at least one" — that mix-up produces the 0.63 distractor.

7. 7.Find the constant term in the binomial expansion of (2x^2+1/x^2)^4
   1. A
      96
   2. B
      16
   3. C
      24
      Correct
   4. D
      32
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   Answer: C

   Expand (2x² + 1/x²)⁴ using the binomial theorem. The general term is:

   T_k+1 = C(4,k)·(2x²)^4−k·(1/x²)^k = C(4,k)·2^4−k·x^8−4k

   The constant term is the one where the power of x is zero, so set 8 − 4k = 0, giving k = 2.

   Substituting k = 2: C(4,2)·2² = 6 × 4 = 24. That is the correct constant term.

   The distractor 96 comes from forgetting to reduce the power on the 2 (using 2⁴ = 16 instead of 2² = 4, then 6 × 16 = 96). The distractor 32 comes from picking the wrong term (k = 1, giving C(4,1)·2³ = 4 × 8 = 32, which is actually the coefficient of x⁴, not the constant term). The distractor 16 comes from using only 2⁴ and ignoring the binomial coefficient altogether.

8. 8.Rationalize: 1/sqrt(2+1)
   1. A
      1/3
   2. B
      3
   3. C
      √3
   4. D
      √3/3
      Correct
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   Answer: D

   First simplify what is under the root: 2 + 1 = 3, so the expression to rationalize is 1/√3.

   To clear the surd from the denominator, multiply both the numerator and the denominator by √3:

   (1 × √3) / (√3 × √3) = √3/3.

   So the rationalized form is √3/3.

   A student who multiplies the denominator by √3 but forgets to also multiply the numerator ends up cancelling the surd away entirely, landing on 1/3 — the surd never moved to the top where it belongs.

   A student who multiplies correctly but forgets to divide by the resulting 3 is left with just √3, stopping one step short.

9. 9.A uniform beam, PQ, is 100 cm long and weighs 35 N. It is placed on a support at a point 40 cm from P. If weights of 54 N and F N are attached at P and Q respectively in order to keep it in a horizontal position, calculate, correct to the nearest whole number, the value of F.
   1. A
      36 N
   2. B
      30 N
      Correct
   3. C
      42 N
   4. D
      72 N
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   Answer: B

   The support is 40 cm from P, so it is 100 − 40 = 60 cm from Q. The beam's own weight (35 N) acts at its midpoint, 50 cm from P, which is 50 − 40 = 10 cm from the support, on the Q side.

   Take moments about the support itself — this is the smart move because it cancels out the unknown reaction force at the support, leaving only the three known forces.

   Clockwise moments = Anticlockwise moments
   54 × 40 = (35 × 10) + (F × 60)
   2160 = 350 + 60F
   60F = 1810
   F = 30.17 N

   Correct to the nearest whole number, 30 N is correct.

10. 10.If the mean of 2, 5, (x+1), (x+2), 7 and 9 is 6, find the median.
    1. A
       6
    2. B
       6.5
       Correct
    3. C
       7.5
    4. D
       7
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    Answer: B

    First find x using the mean. The six values are 2, 5, (x+1), (x+2), 7 and 9, and their sum is:

    2 + 5 + (x+1) + (x+2) + 7 + 9 = 2x + 26

    Since the mean of 6 numbers is 6:

    (2x + 26) ÷ 6 = 6
    2x + 26 = 36
    2x = 10
    x = 5

    So (x+1) = 6 and (x+2) = 7. The data set is now: 2, 5, 6, 7, 7, 9, already arranged in ascending order.

    With an even number of values (6 of them), the median is the average of the two middle values — the 3rd and 4th. These are 6 and 7, so:

    Median = (6 + 7) ÷ 2 = 6.5

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