WAEC Further Mathematics 2022 past questions

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

Ten questions from the 2022 WAEC Further Mathematics paper, with answers and explanations. We hold 55 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2022 questions and answers

1.Which of the following is the semi-interquartile range of a distribution?
1. A
  QU − QL, the interquartile range
2. B
  QU − Q2, the difference between the upper quartile and the median
3. C
  ½(QU − QL), half the interquartile range
  Correct
4. D
  ½(QU + QL), half the sum of the quartiles
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Answer: C
Working: For a distribution with lower quartile Q_L and upper quartile Q_U, the interquartile range is Q_U − Q_L, and the semi-interquartile range (also called the quartile deviation) is defined as half of this: ½(Q_U − Q_L).
The option giving ½(Q_U − Q_L) is therefore correct — it matches the definition exactly.
2.A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.
1. A
  3.0 ms⁻¹
2. B
  5.5 ms⁻¹
3. C
  1.0 ms⁻¹
4. D
  0.5 ms⁻¹
  Correct
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Answer: D
This is a perfectly inelastic collision (the two bodies stick together), so we use conservation of linear momentum: total momentum before collision = total momentum after collision.
Take the direction of the 18 kg body as the positive direction. Since the 6 kg body moves in the opposite direction, its velocity is taken as negative.
Momentum before collision:
(18 × 4) + (6 × -10) = 72 - 60 = 12 kg·ms⁻¹
Total mass after collision = 18 + 6 = 24 kg
Common velocity, v = 12 ÷ 24 = 0.5 ms⁻¹, in the direction the 18 kg body was originally travelling.
3.If x^2+y^2-2x-6y+5=0, evaluate dy/dx when x=3 and y=2.
1. A
  1/2
2. B
  1
3. C
  2
  Correct
4. D
  -2
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Answer: C
Differentiate x² + y² - 2x - 6y + 5 = 0 implicitly with respect to x, remembering that y is a function of x so every y-term picks up a dy/dx.
2x + 2y(dy/dx) - 2 - 6(dy/dx) = 0
Collect the dy/dx terms on one side:
dy/dx (2y - 6) = 2 - 2x
dy/dx = (2 - 2x) / (2y - 6)
Now substitute x = 3 and y = 2:
dy/dx = (2 - 6) / (4 - 6) = (-4) / (-2) = 2
4.In how many ways can six persons be paired?
1. A
  90
2. B
  15
  Correct
3. C
  360
4. D
  720
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Answer: B
Question: In how many ways can six persons be paired?
This means splitting 6 distinct people into 3 unordered pairs (couples), where it does not matter which pair is called "first" or "second" — only who is paired with whom matters.
Step-by-step count:
Pick a partner for the first person: 5 choices remain.
From the 4 people left, pick a partner for one of them: 3 choices remain.
The last 2 people must pair together: 1 way.
Total = 5 × 3 × 1 = 15 ways.
This matches the standard grouping formula for splitting 2n people into n pairs: 6!/(2³ × 3!) = 720/48 = 15.
So the straight-line counting method confirms 15 is correct.
5.A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t $^2$ + 3t. Find the magnitude of F at time t.
1. A
  9N
2. B
  6N
  Correct
3. C
  2N
4. D
  15N
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Answer: B
Given: mass m = 3 kg, distance d = t² + 3t.
Velocity is the first derivative of distance with respect to time:
v = dd/dt = 2t + 3
Acceleration is the derivative of velocity with respect to time:
a = dv/dt = 2 m/s²
Notice that the acceleration comes out as a constant (2 m/s²) with no t left in it. This means the force does not actually change with time, even though the question asks for it "at time t".
Using Newton's second law, F = ma:
F = 3 x 2 = 6N
So the magnitude of the force is 6N at every instant, since the motion has constant acceleration.
6.A binary operation delta is defined on the set of real numbers R, by x delta y = sqrt(x+y-xy/4), where x,y are real. Find the value of 4 delta 3
1. A
  3
2. B
  1
3. C
  2
  Correct
4. D
  4
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Answer: C
x Δ y = √(x + y − xy/4)
Put x = 4 and y = 3 into the rule.
First find xy/4: (4 × 3)/4 = 12/4 = 3.
Then x + y − xy/4 = 4 + 3 − 3 = 4.
Finally, take the square root: √4 = 2.
So 4 Δ 3 = 2, which is correct.
7.If α and β are roots of x $^2$ + mx - n = 0, where m and n are constants, form the equation whose roots are 1/α and 1/β.
1. A
  mx² − nx − 1 = 0
2. B
  nx² − mx − 1 = 0
  Correct
3. C
  nx² − mx + 1 = 0
4. D
  nx² + mx − 1 = 0
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Answer: B
For x² + mx − n = 0 with roots α and β, compare with x² + (sum)x... carefully: sum of roots α + β = −m, and product of roots αβ = −n.
We need the equation whose roots are 1/α and 1/β, so we find their sum and product in terms of m and n.
Sum: 1/α + 1/β = (α + β)/(αβ) = (−m)/(−n) = m/n.
Product: (1/α)(1/β) = 1/(αβ) = 1/(−n) = −1/n.
A quadratic with a given sum S and product P can be written as x² − Sx + P = 0. Substituting:
x² − (m/n)x + (−1/n) = 0, which simplifies to x² − (m/n)x − 1/n = 0.
Clearing the fraction by multiplying every term by n gives nx² − mx − 1 = 0, which is the required equation.
8.Evaluate 4P2 + 4C2 - 4P3
1. A
  42
2. B
  18
3. C
  -6
  Correct
4. D
  6
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Answer: C
Work out each term separately before combining them.
4P2 = 4!/(4-2)! = 4 × 3 = 12
4C2 = 4!/(2! × 2!) = (4 × 3)/(2 × 1) = 6
4P3 = 4!/(4-3)! = 4 × 3 × 2 = 24
Now substitute: 12 + 6 - 24 = -6.
A candidate who forgets to carry out the subtraction and only adds the first two terms gets 12 + 6 = 18, which leaves out the 4P3 term completely. Another common slip is reading the final sign wrongly and adding all three values (12 + 6 + 24 = 42) instead of subtracting the last one. A third mistake comes from working out 4P3 the same way as 4P2 (stopping at 4 × 3 = 12 instead of continuing to 4 × 3 × 2), which gives 12 + 6 - 12 = 6.
9.If PQ = -2i + 5j and RQ = -i - 7j, find PR
1. A
  -i + 12j
  Correct
2. B
  -3i - 2j
3. C
  i - 12j
4. D
  -i - 2j
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Answer: A
PR can be found by going from P to Q, then from Q to R, so PR = PQ + QR. Since QR is the reverse of RQ, QR = -RQ.
Given PQ = -2i + 5j and RQ = -i - 7j, so QR = -RQ = i + 7j.
PR = PQ + QR = (-2i + 5j) + (i + 7j) = (-2+1)i + (5+7)j = -i + 12j.
(This is the same as computing PQ - RQ directly: (-2-(-1))i + (5-(-7))j = -i + 12j.)
The result -3i - 2j comes from wrongly adding PQ and RQ together instead of subtracting (going P→Q→Q→R as if both arrows point the same way).
The result i - 12j is the vector RP, not PR — it is the correct answer with every sign flipped, a mix-up of direction.
10.If Un = kn $^2$ + pn, U $_1$ = -1, U $_5$ = 15, find the values of k and p.
1. A
  k = 2, p = -3
2. B
  k = -2, p = 1
3. C
  k = 1, p = -2
  Correct
4. D
  k = -1, p = 2
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Answer: C
We are given U_n = kn² + pn, with U₁ = -1 and U₅ = 15.
Put n = 1: k(1) + p(1) = -1, so k + p = -1 ... (i)
Put n = 5: k(25) + p(5) = 15, that is 25k + 5p = 15. Dividing through by 5 gives 5k + p = 3 ... (ii)
Subtract (i) from (ii): (5k + p) - (k + p) = 3 - (-1), which gives 4k = 4, so k = 1.
Put k = 1 back into (i): 1 + p = -1, so p = -2.
Check: U₁ = 1(1) + (-2)(1) = -1 ✓, and U₅ = 1(25) + (-2)(5) = 25 - 10 = 15 ✓. So k = 1 and p = -2 is correct.

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WAEC Further Mathematics 2022 past questions

WAEC Further Mathematics 2022 questions and answers

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