Skolra home
Download
Get the app

WAEC Further Mathematics 2021 past questions

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

Ten questions from the 2021 WAEC Further Mathematics paper, with answers and explanations. We hold 44 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2021 questions and answers

  1. 1.The polynomial f(x) =2x3^3 + px+ qx - 5 has (x-1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.
    1. A
      p = 0, q = 3
    2. B
      p = 9, q = 6
    3. C
      p = 9, q = -6
      Correct
    4. D
      p = 5, q = -2
    Show answerHide answer

    Answer: C

    Setting up the two conditions. Since (x−1) is a factor of f(x) = 2x³ + px² + qx − 5, the Factor Theorem tells us f(1) = 0:

    f(1) = 2 + p + q − 5 = 0, so p + q = 3.

    Since dividing by (x+2) leaves a remainder of 27, the Remainder Theorem tells us f(−2) = 27:

    f(−2) = 2(−8) + 4p − 2q − 5 = 4p − 2q − 21 = 27, so 4p − 2q = 48, giving 2p − q = 24.

    Solving the pair. Adding p + q = 3 and 2p − q = 24 cancels q directly: 3p = 27, so p = 9. Then q = 3 − 9 = −6. Checking both conditions confirms it: f(1) = 2+9−6−5 = 0, and f(−2) = −16+36+12−5 = 27. Both hold, so p = 9 and q = −6 is correct.

  2. 2.The table shows the distribution of marks obtained by some students in a test [TABLE] Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 Frequency | 4 | 12 | 16 | 6 | 2 [/TABLE] What is the upper class boundary of the upper quartile class?
    1. A
      29.5
      Correct
    2. B
      39.5
    3. C
      29
    4. D
      19.5
    Show answerHide answer

    Answer: A

    Finding the quartile class: Total frequency N = 4 + 12 + 16 + 6 + 2 = 40.

    The upper quartile sits at the 3N/4 = 30th position. Building the cumulative frequencies: 0-9 gives 4, 10-19 gives 16, 20-29 gives 32, 30-39 gives 38, 40-49 gives 40.

    The 30th value falls between the cumulative totals 16 and 32, so it lies inside the class 20-29 — that is the upper quartile class.

    Since these class limits (0-9, 10-19, 20-29, ...) leave a gap of one unit between consecutive classes, the actual class boundaries are found by adjusting each limit by 0.5. So 20-29 really runs from 19.5 to 29.5.

  3. 3.If f(x) = 4x3^3 + px2^2 + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p
    1. A
      p = 8
    2. B
      p = -4
    3. C
      p = -8
      Correct
    4. D
      p = 4
    Show answerHide answer

    Answer: C

    Setting up with the remainder theorem: when a polynomial f(x) is divided by (2x - 5), the remainder equals f(x) evaluated at the root of the divisor. Setting 2x - 5 = 0 gives x = 5/2.

    So f(5/2) must equal the given remainder, 7:

    f(5/2) = 4(5/2)3 + p(5/2)2 + 7(5/2) - 23

    (5/2)3 = 125/8, so 4 × 125/8 = 500/8 = 62.5
    (5/2)2 = 25/4, so the p-term is 6.25p
    7 × 5/2 = 17.5

    Adding: 62.5 + 6.25p + 17.5 - 23 = 7, which simplifies to 57 + 6.25p = 7, so 6.25p = -50, giving p = -8. This is correct.

    A candidate who squares (5/2) as 25/2 instead of 25/4 (forgetting to square the denominator properly) gets a coefficient of 12.5 instead of 6.25 for the p-term, and that slip lands on p = -4.

  4. 4.Given that P = { x: 0≤ x≤ 36, x is a factor of 36 divisible by 3} and Q= { x: 0≤ x≤ 36, x is an even number and a perfect square}, find P n Q.
    1. A
      {4, 16, 36}
    2. B
      {9, 36}
    3. C
      {4, 36}
    4. D
      {36}
      Correct
    Show answerHide answer

    Answer: D

    Finding set P: the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. Picking out only those divisible by 3 gives P = {3, 6, 9, 12, 18, 36}.

    Finding set Q: the perfect squares between 0 and 36 are 1, 4, 9, 16, 25, 36. Picking out only the even ones gives Q = {4, 16, 36}.

    P ∩ Q means the elements that appear in both lists. Checking each member of Q against P: 4 is not in P, 16 is not in P, but 36 is in P (it is a factor of 36 divisible by 3) and it is also in Q (it is an even perfect square, since 6² = 36). So P ∩ Q = {36}.

  5. 5.If 2y2^2 + 7 = 3y - xy, finddydx\frac{dy}{dx}
    1. A
      dy/dx = -y/(x + 4y - 3)
      Correct
    2. B
      dy/dx = y/(x + 4y - 3)
    3. C
      dy/dx = y/(4y - x - 3)
    4. D
      dy/dx = -y/(x + 2y - 3)
    Show answerHide answer

    Answer: A

    Differentiate every term of 2y² + 7 = 3y − xy with respect to x, remembering that y is a function of x, so every y needs the chain rule, and xy needs the product rule.

    d/dx(2y²) = 4y(dy/dx)
    d/dx(7) = 0
    d/dx(3y) = 3(dy/dx)
    d/dx(xy) = x(dy/dx) + y

    Putting this together:

    4y(dy/dx) = 3(dy/dx) − [x(dy/dx) + y]
    4y(dy/dx) = 3(dy/dx) − x(dy/dx) − y

    Now collect all the dy/dx terms on one side:

    4y(dy/dx) − 3(dy/dx) + x(dy/dx) = −y
    (dy/dx)(4y + x − 3) = −y

    Dividing through gives dy/dx = −y/(x + 4y − 3), which is the correct result.

  6. 6.If (1−2x)4=1+px+qx2−32x3+16x4(1-2x)^4 = 1 + px + qx^2 - 32x^3 + 16x^4, find the value of (q−p)(q-p)
    1. A
      16
    2. B
      -32
    3. C
      -16
    4. D
      32
      Correct
    Show answerHide answer

    Answer: D

    Expand (1 − 2x)⁴ using the binomial theorem:

    (1−2x)⁴ = C(4,0)(1)⁴ + C(4,1)(1)³(−2x) + C(4,2)(1)²(−2x)² + C(4,3)(1)(−2x)³ + C(4,4)(−2x)⁴
    = 1 + 4(−2x) + 6(4x²) + 4(−8x³) + 16x⁴
    = 1 − 8x + 24x² − 32x³ + 16x⁴

    This matches the given form 1 + px + qx² − 32x³ + 16x⁴ term for term (notice the −32x³ and 16x⁴ already agree, which confirms the expansion is correct).

    So p = −8 and q = 24.

    Therefore q − p = 24 − (−8) = 24 + 8 = 32, which is the correct answer.

  7. 7.A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement. Calculate the probability of picking: (a) same color of bulbs; (b) different color of bulbs (c) at least one red bulb.
    1. A
      (a) 37/120 (b) 83/120 (c) 11/24
    2. B
      (a) 37/120 (b) 83/120 (c) 13/24
      Correct
    3. C
      (a) 37/60 (b) 83/120 (c) 13/24
    4. D
      (a) 45/128 (b) 83/128 (c) 135/256
    Show answerHide answer

    Answer: B

    Total bulbs in the box = 5 red + 7 blue + 4 green = 16. Picking 2 bulbs without replacement, the total number of ways to do this is C(16,2) = 120.

    (a) Same colour: add the ways of picking 2 of the same colour from each group: C(5,2) + C(7,2) + C(4,2) = 10 + 21 + 6 = 37. So P(same colour) = 37/120.

    (c) Different colours: "different colours" is the complement of "same colour", so P(different colours) = 1 − 37/120 = 83/120.

    (b) At least one red: the easiest route is through the complement "no red bulb at all". The non-red bulbs number 7 + 4 = 11, so P(no red) = C(11,2)/C(16,2) = 55/120 = 11/24. Therefore P(at least one red) = 1 − 11/24 = 13/24.

  8. 8.Given that f: x --> x2^2 - x + 1 is defined on theSet Q = { x : 0≤ x < 20, x is a multiple of 5}. find the set of range of F.
    1. A
      {0, 21, 91, 211}
    2. B
      {1, 21, 91, 211}
      Correct
    3. C
      {1, 21, 101, 211}
    4. D
      {1, 31, 111, 241}
    Show answerHide answer

    Answer: B

    Step 1: Write out the domain. Q = { x : 0 ≤ x < 20, x is a multiple of 5 }. The multiples of 5 from 0 up to (but not including) 20 are 0, 5, 10 and 15. Notice 20 itself is left out because the inequality is strict (x < 20), so there are exactly four values to map.

    Step 2: Apply the rule f(x) = x² − x + 1 to each value.
    f(0) = 0² − 0 + 1 = 1
    f(5) = 5² − 5 + 1 = 25 − 5 + 1 = 21
    f(10) = 10² − 10 + 1 = 100 − 10 + 1 = 91
    f(15) = 15² − 15 + 1 = 225 − 15 + 1 = 211

    Step 3: Collect the images. The range is the set {1, 21, 91, 211}, and this is the correct answer.

  9. 9.If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.
    1. A
      x = 30° or x = 210°
    2. B
      x = 120° or x = 300°
    3. C
      x = 150° or x = 330°
      Correct
    4. D
      x = 60° or x = 240°
    Show answerHide answer

    Answer: C

    Start by collecting the equation so that tan x stands alone.

    √5 cos x + √15 sin x = 0 → √15 sin x = −√5 cos x → tan x = −√5/√15 = −1/√3.

    Since tan 30° = 1/√3, the reference angle is 30°. Tangent is negative here, and tangent is only negative in the second and fourth quadrants (where sine and cosine have opposite signs).

    Second quadrant: 180° − 30° = 150°.
    Fourth quadrant: 360° − 30° = 330°.

    So the correct values are 150° and 330°.

  10. 10.Simplify 13\frac{1}{3} log8 +13\frac{1}{3} log 64 - 2 log6
    1. A
      log(2/9)
      Correct
    2. B
      log(9/2)
    3. C
      log 8
    4. D
      log(2/3)
    Show answerHide answer

    Answer: A

    We simplify term by term using two log rules: (1/n) log a = log(a^(1/n)) (taking a root) and n log a = log(a^n) (taking a power).

    First term: (1/3) log 8 = log(8^(1/3)) = log 2, since the cube root of 8 is 2.

    Second term: (1/3) log 64 = log(64^(1/3)) = log 4, since the cube root of 64 is 4.

    Third term: 2 log 6 = log(6²) = log 36.

    Now combine using the addition and subtraction rules for logs (log A + log B = log AB, and log A − log B = log(A/B)):
    log 2 + log 4 − log 36 = log(2 × 4) − log 36 = log 8 − log 36 = log(8/36)

    Reducing 8/36 to lowest terms gives 2/9, so the expression simplifies to log(2/9).

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

Spotted a wrong answer? Tell us from the app or on the contact page and we check it.

Practise all 44 questions we hold from the 2021 paper, on a timed CBT screen

Every question has the answer and an explanation, in the Skolra app.

Get it on Google PlayDownload for WindowsDownload for Mac

About the app · Skolra for Windows · Skolra for Mac

More Further Mathematics past questions