WAEC Further Mathematics 2020 past questions

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

Ten questions from the 2020 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 2020 questions and answers

1. 1.Differentiate $\frac{x}{x + 1}$ with respect to x.
   1. A
      1/(x + 1)
   2. B
      -1/(x + 1)²
   3. C
      1/(x + 1)²
      Correct
   4. D
      (2x + 1)/(x + 1)²
   Show answerHide answer

   Answer: C

   Let u = x and v = x + 1, so u' = 1 and v' = 1. Using the quotient rule, (u/v)' = (vu' − uv')/v².

   Substituting: dy/dx = [(x + 1)(1) − x(1)] / (x + 1)² = (x + 1 − x) / (x + 1)² = 1/(x + 1)².

   The option 1/(x + 1)² is correct because the numerator simplifies to exactly 1 after subtraction, and the quotient rule always squares the denominator function v.

   The option 1/(x + 1) is a common slip — a student finds the correct numerator but forgets to square the denominator.

   The option −1/(x + 1)² comes from swapping the order in the quotient rule numerator (doing uv' − vu' instead of vu' − uv'), which flips the sign.

2. 2.In which of the following series can be the formula S = $\frac{a}{1 - r}$where a is the firstterm and r is the common ratio, be used to find the sum of all the terms?
   1. A
      3 + 6 + 12 + 24 + ⋯
   2. B
      1 + ½ + ¼ + ⅛ + ⋯
      Correct
   3. C
      4 + 8 + 16 + 32 + ⋯
   4. D
      2 + 4 + 6 + 8 + ⋯
   Show answerHide answer

   Answer: B

   The formula S = a/(1 − r) is the sum to infinity of a geometric series, and it only gives a finite, sensible value when the common ratio satisfies |r| < 1. When |r| ≥ 1, the terms do not shrink towards zero, so the series keeps growing and has no finite sum — plugging such an r into the formula gives a meaningless or negative result.

   Check each series for a common ratio:

   The series 1 + ½ + ¼ + ⅛ + ⋯ is geometric with first term a = 1 and common ratio r = ½. Since |½| < 1, the formula applies directly: S = 1/(1 − ½) = 1/(½) = 2. This is the series the formula can be used on.

3. 3.Find the median of the numbers 9,7, 5, 2, 12,9,9, 2, 10, 10, and 18.
   1. A
      9
      Correct
   2. B
      7
   3. C
      10
   4. D
      8
   Show answerHide answer

   Answer: A

   Step 1 — Arrange the numbers in order (smallest to largest):
   2, 2, 5, 7, 9, 9, 9, 10, 10, 12, 18

   Step 2 — Count how many numbers there are.
   There are 11 numbers, which is odd, so the median is simply the middle value — the one with an equal number of values on each side.

   Step 3 — Find the middle position.
   For n = 11 numbers, the median sits at position (n + 1) ÷ 2 = (11 + 1) ÷ 2 = 6th position.

   Counting along the ordered list: 2 (1st), 2 (2nd), 5 (3rd), 7 (4th), 9 (5th), 9 (6th), 9 (7th), 10 (8th), 10 (9th), 12 (10th), 18 (11th).

   The 6th value is 9, so the median of the data set is 9.

4. 4.The distance(s) in metres covered by a particle in motion at any time, tseconds, is given by S =120t - 16t$^2$.Find in metres, the distance covered by the body before coming to rest.
   1. A
      0 m
   2. B
      450 m
   3. C
      225 m
      Correct
   4. D
      390 m
   Show answerHide answer

   Answer: C

   The particle's distance is S = 120t − 16t². It comes to rest when its velocity (the rate of change of S) is zero.

   Velocity: v = dS/dt = 120 − 32t.

   Set v = 0: 120 − 32t = 0, so t = 120/32 = 3.75 seconds.

   Now substitute t = 3.75 back into the distance equation:
   S = 120(3.75) − 16(3.75)²
   S = 450 − 16(14.0625)
   S = 450 − 225 = 225 m.

   So the particle covers 225 metres before it comes to rest. This is the correct answer.

5. 5.Given that 2x + 3y - 10 and 3x = 2y - 11, calculate the value of (x - y).
   1. A
      -5
      Correct
   2. B
      -3
   3. C
      3
   4. D
      5
   Show answerHide answer

   Answer: A

   The two statements, read as simultaneous equations, are:

   2x + 3y = 10 ... (i)
   3x - 2y = -11 ... (ii) (since 3x = 2y - 11)

   To eliminate y, multiply (i) by 2 and (ii) by 3:

   4x + 6y = 20
   9x - 6y = -33

   Adding the two lines removes y: 13x = -13, so x = -1.

   Put x = -1 back into (i): 2(-1) + 3y = 10, so 3y = 12, giving y = 4.

   Check with (ii): 3(-1) = 2(4) - 11 → -3 = -3. The values agree, so they are correct.

   Therefore x - y = -1 - 4 = -5, which is the right answer.

6. 6.The table shows the distribution of marks scored by 64 students in a test: [TABLE] Marks | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 Frequency | 2 | 2 | 2 | 8 | 13 | 11 | 12 | 10 | 4 [/TABLE] (a) Draw a histogram for the distribution. (b) Use the histogram to estimate the modal score.
   1. A
      54.5
   2. B
      57.1
   3. C
      52.4
   4. D
      56.6
      Correct
   Show answerHide answer

   Answer: D

   The modal class is the class with the highest frequency. From the table, the highest frequency is 13, in the class 50 - 59. The class before it (40 - 49) has frequency 8, and the class after it (60 - 69) has frequency 11.

   In a histogram, the lower class boundary of the modal class (not the class limit) is used as the starting point. Since there is a gap of 1 unit between consecutive class limits (e.g. 49 and 50), the lower boundary of the modal class is 49.5, and the class width (h) is 10.

   Estimating the mode from the histogram (drawing lines from the top corners of the modal bar to the corners of the two neighbouring bars and reading off where they cross) is equivalent to:

   Mode = L + [(f_m − f₁) / ((f_m − f₁) + (f_m − f₂))] × h
   = 49.5 + [(13 − 8) / ((13 − 8) + (13 − 11))] × 10
   = 49.5 + (5/7) × 10
   = 49.5 + 7.1
   = 56.6

7. 7.A stone was dropped from the top of a building 40m high. Find, correct to one decimal place, the time it took the stone to reach the ground. [Take g = 9.8ms$^{-2}$]
   1. A
      2.0 seconds
   2. B
      2.9 seconds
      Correct
   3. C
      4.1 seconds
   4. D
      2.8 seconds
   Show answerHide answer

   Answer: B

   For a body dropped from rest and falling freely under gravity, the distance fallen is given by h = ½gt², since the initial velocity is zero.

   Here h = 40 m and g = 9.8 ms²⁻¹, so:

   40 = ½ × 9.8 × t²
   t² = 80 ÷ 9.8 = 8.1633
   t = √8.1633 = 2.857 s

   Correct to one decimal place, t = 2.9 seconds is correct.

   A candidate who carelessly uses g = 10 ms⁻² (a common rounding habit) instead of the given 9.8 would get t = √8 = 2.8 s, which looks right but ignores the value stated in the question.

   A candidate who forgets the ½ in the formula and solves t = √(h/g) instead would get t = √(40/9.8) = 2.0 s, missing the factor of 2 entirely.

8. 8.A binary operation * is defined on the set of real numbers R, by p*q = p + q - $\frac{pq}{2}$, where p, q $\in$ R. Find the: (a) inverse of -1 under * given that the identity clement is zero. (b) truth set of m* 7 = m* 5,
   1. A
      Inverse of −1 is −2/3; truth set of m*7 = m*5 is {2}
   2. B
      Inverse of −1 is 2/3; truth set of m*7 = m*5 is {4}
   3. C
      Inverse of −1 is 2; truth set of m*7 = m*5 is {−2}
   4. D
      Inverse of −1 is 2/3; truth set of m*7 = m*5 is {2}
      Correct
   Show answerHide answer

   Answer: D

   Finding the inverse of −1: For p*q = p + q − pq/2, the inverse p′ of an element p satisfies p*p′ = e = 0 (the given identity).

   With p = −1: −1 + p′ − (−1)(p′)/2 = 0, which simplifies to −1 + p′ + p′/2 = 0, so (3/2)p′ = 1, giving p′ = 2/3.

   Check: −1 * (2/3) = −1 + 2/3 − (−1)(2/3)/2 = −1 + 2/3 + 1/3 = 0. Correct.

   Finding the truth set of m*7 = m*5:

   m*7 = m + 7 − 7m/2
   m*5 = m + 5 − 5m/2

   Setting them equal: m + 7 − 7m/2 = m + 5 − 5m/2. The m terms cancel, leaving 7 − 7m/2 = 5 − 5m/2, so 2 = 7m/2 − 5m/2 = m. Hence m = 2, and the truth set is {2}.

9. 9.If sin X = $\frac{3}{5}$ and cos Y = $\frac{24}{25}$, where X and Y are acute, find the value of cos (X + Y).
   1. A
      7/25
   2. B
      4/5
   3. C
      117/125
   4. D
      3/5
      Correct
   Show answerHide answer

   Answer: D

   Step 1: Find the missing sides using Pythagoras' theorem.

   Since sin X = 3/5 and X is acute, X belongs to a right triangle with opposite = 3, hypotenuse = 5, so the adjacent side = 4 (the familiar 3-4-5 triangle). Therefore cos X = 4/5.

   Since cos Y = 24/25 and Y is acute, Y belongs to a right triangle with adjacent = 24, hypotenuse = 25, so the opposite side = 7 (the 7-24-25 triangle). Therefore sin Y = 7/25.

   Step 2: Apply the cosine addition formula.

   cos(X + Y) = cos X cos Y − sin X sin Y
   = (4/5 × 24/25) − (3/5 × 7/25)
   = 96/125 − 21/125
   = 75/125
   = 3/5

   This is the correct value of cos(X + Y).

10. 10.Evaluate; $\int^3_1(3x - 2)^5 dx$
    1. A
       -6536
    2. B
       58824
    3. C
       19608
    4. D
       6536
       Correct
    Show answerHide answer

    Answer: D

    Let u = 3x - 2, so du = 3 dx, which gives dx = du/3. The integral becomes:

    (1/3)∫u⁵ du = (1/3) × u⁶/6 = u⁶/18 = (3x - 2)⁶/18

    Now substitute the limits. At x = 3, u = 3(3) - 2 = 7, so u⁶ = 117649. At x = 1, u = 3(1) - 2 = 1, so u⁶ = 1.

    Result = (117649 - 1)/18 = 117648/18 = 6536.

    A common slip is forgetting the 1/3 that comes from the substitution and dividing only by 6 (the power rule denominator) — that gives 19608. Multiplying by 3 instead of dividing by it gives 58824. Putting the lower limit in first and the upper limit second flips the sign and gives -6536, but the upper limit (x = 3) must always be substituted first and the lower limit subtracted from it.

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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