WAEC Further Mathematics 2019 past questions and answers

50 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Indices and Logarithms

Solve: 8x−2 = 43x

  1. A.−2
  2. B.−1
  3. C.1
  4. D.2
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Answer: A (−2)

Explanation

In powers of 2: 2^(3x − 6) = 2^(6x), so 3x − 6 = 6x and x = −2.

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Question 2

Trigonometry

Evaluate tan 75°, leaving the answer in surd form (radicals).

  1. A.√3 + 2
  2. B.√3 + 1
  3. C.√3 − 1
  4. D.√3 − 2
Show the answer and explanation

Answer: A (√3 + 2)

Explanation

tan 75° = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1) = (4 + 2√3)/2 = 2 + √3.

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Question 3

Polynomials

Solve: p/2 + k/3 = 5 and 2p − k = 6 simultaneously.

  1. A.p = −6, k = −6
  2. B.p = −6, k = 6
  3. C.p = 6, k = 6
  4. D.p = 6, k = −6
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Answer: C (p = 6, k = 6)

Explanation

3p + 2k = 30 and k = 2p − 6, so 3p + 4p − 12 = 30, p = 6 and k = 6.

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Question 4

Surds

Rationalize: 1/(√2 + 1)

  1. A.√2 − 1
  2. B.1 − √2
  3. C.(√2 − 1)/2
  4. D.(1 − √2)/2
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Answer: A (√2 − 1)

Explanation

Multiply top and bottom by (√2 − 1): (√2 − 1)/(2 − 1) = √2 − 1.

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Question 5

Permutations and Combinations

If nC2 = 15, find the value of n.

  1. A.8
  2. B.7
  3. C.6
  4. D.5
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Answer: C (6)

Explanation

n(n − 1)/2 = 15 gives n² − n − 30 = 0, (n − 6)(n + 5) = 0, so n = 6.

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Question 6

Binary Operations

An operation (*) is defined on the set T = {−1, 0, …, 5} by x * y = x + y − xy. Which of the following operation(s) will give an image which is an element of T? I. 2(*)5 II. 3(*)2 III. 3(*)4

  1. A.I only
  2. B.II only
  3. C.I and III only
  4. D.II and III only
Show the answer and explanation

Answer: B (II only)

Explanation

2 * 5 = 7 − 10 = −3 (not in T); 3 * 2 = 5 − 6 = −1 (in T); 3 * 4 = 7 − 12 = −5 (not in T). Only II.

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Question 7

Functions

Given that g: x → 3x and f: x → cos x, find the value of g o f(20°).

  1. A.0.50
  2. B.0.94
  3. C.2.60
  4. D.2.82
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Answer: D (2.82)

Explanation

f(20°) = cos 20° ≈ 0.9397, then g gives 3 × 0.9397 ≈ 2.82.

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Question 8

Matrices and Determinants

A linear transformation is defined by T: (x, y) → (−x + y, −4y). Find the image, Q′, of Q(−3, 2) under T.

  1. A.Q′(5, −8)
  2. B.Q′(−8, 5)
  3. C.Q′(5, −3)
  4. D.Q′(−5, −8)
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Answer: A (Q′(5, −8))

Explanation

T(−3, 2) = (3 + 2, −8) = (5, −8).

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Question 9

Functions

If g: r → 5 − 2r, r is a real number, find the image of −3.

  1. A.13
  2. B.11
  3. C.−1
  4. D.−9
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Answer: B (11)

Explanation

g(−3) = 5 + 6 = 11.

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Question 10

Sets and Logic

Consider the following statements: p: Birds fly
q: The sky is blue
r: The grass is green
What is the symbolic representation of "If the grass is green and the sky is not blue, then the birds do not fly"?

  1. A.(r ∧ p) ⇒ q
  2. B.(r ∧ q) ⇒ ~p
  3. C.(r ∧ ~q) ⇒ ~p
  4. D.(r ∧ ~p) ⇒ ~q
Show the answer and explanation

Answer: C ((r ∧ ~q) ⇒ ~p)

Explanation

'Grass is green and sky is not blue' is r ∧ ~q; 'birds do not fly' is ~p. So (r ∧ ~q) ⇒ ~p.

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Question 11

Polynomials

Given that 1/(x2 − 4) ≡ P/(x + 2) + Q/(x − 2), x ≠ ±2, find the value of (P + Q)

  1. A.3/2
  2. B.1
  3. C.1/2
  4. D.0
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Answer: D (0)

Explanation

1 = P(x − 2) + Q(x + 2). x = 2 gives Q = ¼; x = −2 gives P = −¼. P + Q = 0.

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Question 12

Sequences and Series

Find the sum of the first 20 terms of the sequence −7, −3, 1, ….

  1. A.620
  2. B.660
  3. C.690
  4. D.1240
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Answer: A (620)

Explanation

a = −7, d = 4. S₂₀ = 10[2(−7) + 19(4)] = 10 × 62 = 620.

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Question 13

Surds

Find the value of x for which 6√(4x2 + 1) = 13x, where x > 0.

  1. A.6/5
  2. B.25/24
  3. C.24/25
  4. D.5/6
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Answer: A (6/5)

Explanation

Square: 36(4x² + 1) = 169x², so 25x² = 36 and x = 6/5 (x > 0).

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Question 14

Coordinate Geometry

Calculate the distance between points (−2, −5) and (−1, 3).

  1. A.√5 units
  2. B.√17 units
  3. C.√65 units
  4. D.√73 units
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Answer: C (√65 units)

Explanation

√((−1 + 2)² + (3 + 5)²) = √(1 + 64) = √65.

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Question 15

Matrices and Determinants

If P = (2 3 / −4 1), Q = (6 / 8) and PQ = k(45 / −20), find the value of k. (Matrices written row by row; / separates the rows.)

  1. A.−5/4
  2. B.−4/5
  3. C.4/5
  4. D.5/4
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Answer: C (4/5)

Explanation

PQ = (12 + 24 / −24 + 8) = (36 / −16) = (4/5)(45 / −20), so k = 4/5.

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Question 16

Sequences and Series

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.81/32
  2. B.9/8
  3. C.1/4
  4. D.32/729
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Answer: D (32/729)

Explanation

r² = (8/81) ÷ (2/9) = 4/9, so r = 2/3 and a = 1/3. T₆ = ar⁵ = (1/3)(32/243) = 32/729.

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Question 17

Trigonometry

Points X and Y are on the same horizontal base as the foot of a building such that X is 96 m due east of the building and Y is due west. If the angle of elevation of the top of the building from X is 30° and that of Y is 60°, calculate the distance of Y from the base of the building.

  1. A.30 m
  2. B.32 m
  3. C.42 m
  4. D.50 m
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Answer: B (32 m)

Explanation

Height = 96 tan 30° ≈ 55.43 m. Distance of Y = height ÷ tan 60° = 96/3 = 32 m.

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Question 18

Differentiation

Find the coordinates of the point on the curve y = 3x2 − 2x − 5 where the tangent is parallel to the line y − 5 = 8x.

  1. A.(−5/3, 0)
  2. B.(0, −5/3)
  3. C.(0, 5/3)
  4. D.(5/3, 0)
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Answer: D ((5/3, 0))

Explanation

dy/dx = 6x − 2 = 8 gives x = 5/3; y = 25/3 − 10/3 − 5 = 0. Point (5/3, 0).

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Question 21

Dynamics

A particle starts from rest and moves in a straight line such that its velocity, V ms−1, at time t second is given by V = 3t2 − 6t. Calculate the distance in 4 seconds.

  1. A.12 m
  2. B.16 m
  3. C.64 m
  4. D.96 m
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Answer: B (16 m)

Explanation

s = ∫(3t² − 6t) dt = t³ − 3t² (s = 0 at t = 0). At t = 4, s = 64 − 48 = 16 m from the start.

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Question 23

Binomial Theorem

Find the constant term in the binomial expansion of (2x2 + 1/x2)4.

  1. A.10
  2. B.12
  3. C.24
  4. D.42
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Answer: C (24)

Explanation

The constant term has (2x²)² and (1/x²)²: ⁴C₂ × 4 × 1 = 6 × 4 = 24.

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Question 24

Coordinate Geometry

Which of these inequalities is represented by the shaded portion of the graph?

Diagram for this question
  1. A.2y + x − 3 < 0
  2. B.2y − x − 3 < 0
  3. C.2y − x + 3 < 0
  4. D.2y + x + 3 < 0
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Answer: B (2y − x − 3 < 0)

Explanation

The boundary line passes through (−3, 0) and (0, 1.5), so it is 2y = x + 3. The shaded side is below the line, where 2y − x − 3 < 0.

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Question 25

Dynamics

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.70 kg ms−1
  2. B.55 kg ms−1
  3. C.50 kg ms−1
  4. D.35 kg ms−1
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Answer: A (70 kg ms−1)

Explanation

Change in momentum = impulse = Ft = 35 × 2 = 70 kg ms⁻¹.

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Question 26

Indices and Logarithms

Solve, correct to three significant figures, (0.3)x = (0.5)8.

  1. A.4.61
  2. B.4.606
  3. C.0.461
  4. D.0.0130
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Answer: A (4.61)

Explanation

x = 8 log 0.5 / log 0.3 = 8(−0.3010)/(−0.5229) ≈ 4.606, which is 4.61 to three significant figures.

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Question 27

Sets and Logic

Given that P and Q are two non-empty subsets of the universal set, U. Find P ∩ (Q ∪ Q′)

  1. A.P
  2. B.P′
  3. C.Q
  4. D.Q′
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Answer: A (P)

Explanation

Q ∪ Q′ = U, and P ∩ U = P.

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Question 28

Binomial Theorem

Find the coefficient of the third term in the binomial expansion of (2x + 3y/4)3 in descending powers of x.

  1. A.27y2/64
  2. B.27y2/8
  3. C.8y2
  4. D.9y2
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Answer: B (27y2/8)

Explanation

The third term is ³C₂(2x)(3y/4)² = 3 × 2x × 9y²/16 = (27y²/8)x, so its coefficient (of x) is 27y²/8.

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Question 29

Coordinate Geometry

Find the coordinates of the centre of the circle 3x2 + 3y2 − 6x + 9y − 5 = 0.

  1. A.(−3, 9/2)
  2. B.(−1, 3/2)
  3. C.(1, −3/2)
  4. D.(3, −9/2)
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Answer: C ((1, −3/2))

Explanation

Divide by 3: x² + y² − 2x + 3y − 5/3 = 0. The centre is (1, −3/2).

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Question 30

Integration

Evaluate ∫09 √x dx.

  1. A.3
  2. B.9
  3. C.18
  4. D.27
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Answer: C (18)

Explanation

∫√x dx = (2/3)x^(3/2); from 0 to 9: (2/3)(27) = 18.

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Question 31

Polynomials

The function f: x → x2 + px + q has turning point when x = −3 and remainder of −6 when divided by (x + 2). Find the vale of q.

  1. A.6
  2. B.2
  3. C.−2
  4. D.−8
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Answer: B (2)

Explanation

f′(x) = 2x + p = 0 at x = −3, so p = 6. f(−2) = 4 − 12 + q = −6, so q = 2.

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Question 32

Differentiation

If y = (5 − x)−3, find dy/dx.

  1. A.−15/(5 − x)4
  2. B.−3/(5 − x)4
  3. C.3/(5 − x)4
  4. D.15/(5 − x)4
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Answer: C (3/(5 − x)4)

Explanation

dy/dx = −3(5 − x)⁻⁴ × (−1) = 3/(5 − x)⁴.

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Question 34

Vectors

Find correct to the nearest degree, the angle between p = 12i − 5j and q = 4i + 3j.

  1. A.59°
  2. B.60°
  3. C.75°
  4. D.76°
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Answer: A (59°)

Explanation

p·q = 48 − 15 = 33; |p| = 13, |q| = 5. cos θ = 33/65 ≈ 0.5077, so θ ≈ 59°.

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Question 35

Integration

Find, the area between line y = x + 1 and the x − axis from x = −2 to x = 0.

  1. A.5 square units
  2. B.4 square units
  3. C.2 square units
  4. D.1 square unit
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Answer: D (1 square unit)

Explanation

The line crosses the x-axis at x = −1. From −2 to −1 the region is a triangle of area ½ (below the axis) and from −1 to 0 another of area ½. Total area = 1 square unit.

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Question 36

Permutations and Combinations

How many numbers greater than 200 can be formed from the digits 1, 2, 3, 4, 5 if no digit is to be repeated in any particular number?

  1. A.50
  2. B.60
  3. C.288
  4. D.300
Show the answer and explanation

Answer: C (288)

Explanation

3-digit numbers above 200: 4 × 4 × 3 = 48. 4-digit: 5 × 4 × 3 × 2 = 120. 5-digit: 5! = 120. Total 288.

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Question 37

Probability

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.0.28
  2. B.0.67
  3. C.0.72
  4. D.0.97
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Answer: D (0.97)

Explanation

P(neither) = 0.1 × 0.3 = 0.03, so P(at least one) = 0.97.

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Question 38

Probability

Given that X and Y are independent events such that P(X) = 0.5, P(Y) = m and P(X ∪ Y) = 0.75, find the value of m.

  1. A.0.6
  2. B.0.5
  3. C.0.4
  4. D.0.3
Show the answer and explanation

Answer: B (0.5)

Explanation

P(X ∪ Y) = P(X) + P(Y) − P(X)P(Y): 0.75 = 0.5 + m − 0.5m, so 0.5m = 0.25 and m = 0.5.

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Question 39

Statics

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.69
  2. B.60
  3. C.35
  4. D.30
Show the answer and explanation

Answer: D (30)

Explanation

Moments about the support: 54 × 40 = 35 × 10 + F × 60, so 60F = 1810 and F ≈ 30.

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Theory question 101

Indices and Logarithms

Simplify: (625(3x/4 − 1) + 125(x−1))/5(3x−2).

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Model answer

625^(3x/4 − 1) = 5^(3x − 4); 125^(x − 1) = 5^(3x − 3).
Divide by 5^(3x − 2): 5^(−2) + 5^(−1) = 1/25 + 1/5 = 6/25.

Explanation

Write everything as a power of 5.

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Theory question 102

Coordinate Geometry

(a) Find the coordinates of the point which divides the line joining (7, −5) and (−2, 7) externally in the ratio 3 : 2.
(b) Without using calculators or mathematical tables, evaluate 2/(1 − √2) − 2/(2 + √2), leaving the answer in the form p + q√n, where p, q and n are integers.

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Model answer

(a) External division: ((3(−2) − 2(7))/(3 − 2), (3(7) − 2(−5))/(3 − 2)) = (−20, 31).
(b) 2/(1 − √2) = −2 − 2√2; 2/(2 + √2) = 2 − √2.
Difference = −2 − 2√2 − 2 + √2 = −4 − √2 (p = −4, q = −1, n = 2).

Explanation

For external division use (mx₂ − nx₁)/(m − n).

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Theory question 103

Sequences and Series

How many terms of the series −3 − 1 + 1 + … add up to 165?

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Model answer

a = −3, d = 2. Sₙ = (n/2)[−6 + 2(n − 1)] = n(n − 4) = 165.
n² − 4n − 165 = 0, (n − 15)(n + 11) = 0, so n = 15.

Explanation

Set Sₙ = 165 and solve the quadratic.

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Theory question 104

Trigonometry

If sin X = (p − q)/(p + q), where 0° ≤ X ≤ 90°, find 1 − tan2X.

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Model answer

Opposite p − q, hypotenuse p + q, adjacent √((p + q)² − (p − q)²) = 2√(pq).
tan²X = (p − q)²/(4pq).
1 − tan²X = (4pq − (p − q)²)/(4pq) = (6pq − p² − q²)/(4pq).

Explanation

Draw a right triangle and use Pythagoras for the third side.

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Theory question 105

Statics

Three forces 10 N, 14 N and 16 N acting on a particle keep it in equilibrium. Find the angle between the forces 10 N and 16 N.

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Model answer

The 14 N force balances the resultant of the 10 N and 16 N forces.
14² = 10² + 16² + 2(10)(16) cos θ, so 196 = 356 + 320 cos θ and cos θ = −½.
θ = 120°.

Explanation

The resultant of two forces equals the third force reversed.

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Theory question 106

Statics

A uniform beam, WX, of length 90 cm and weight 50 N is suspended on a pivot, 35 cm from W. It is kept in equilibrium by means of forces T and 20 N applied at Y and Z respectively. |WY| = 10 cm and |XZ| = 10 cm. Find the value of T.

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Model answer

Positions from W: Y at 10 cm, pivot at 35 cm, centre of the beam at 45 cm, Z at 80 cm.
Taking both forces downwards, moments about the pivot: T × 25 = 50 × 10 + 20 × 45 = 500 + 900.
T = 1400/25 = 56 N.

Explanation

Don't forget the beam's own weight acting at its midpoint.

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Theory question 107

Statistics

The table shows the age distribution in years of a group of people.
Age (in years): 1 − 5, 6 − 10, 11 − 15, 16 − 20, 21 − 25, 26 − 30
Number of People: 18, 12, 25, 15, 20, 10
Using an assumed mean of 13 years, find the mean age of the people.

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Model answer

Class marks 3, 8, 13, 18, 23, 28; d = x − 13 = −10, −5, 0, 5, 10, 15.
Σf = 100; Σfd = −180 − 60 + 0 + 75 + 200 + 150 = 185.
Mean = 13 + 185/100 = 14.85 years.

Explanation

Mean = A + Σfd/Σf.

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Theory question 108

Probability

Two fair dice are thrown together two times. Find the probability of obtaining a sum of seven in the first throw and a sum of four in the second throw.

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Model answer

P(sum 7) = 6/36 = 1/6; P(sum 4) = 3/36 = 1/12.
The throws are independent: 1/6 × 1/12 = 1/72.

Explanation

List the outcome pairs for each sum.

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Theory question 109

Coordinate Geometry

The curve y = 7 − 6/x and the line y + 2x − 3 = 0 intersect at two points. Find the: (a) coordinates of the two points; (b) equation of the perpendicular bisector of the line joining the two points.

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Model answer

(a) 7 − 6/x = 3 − 2x gives 2x² + 4x − 6 = 0, i.e. x² + 2x − 3 = 0, x = 1 or −3. Points (1, 1) and (−3, 9).
(b) Midpoint (−1, 5); the joining line has gradient −2, so the bisector has gradient ½.
y − 5 = ½(x + 1), i.e. 2y − x − 11 = 0.

Explanation

The perpendicular bisector passes through the midpoint with gradient −1/m.

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Theory question 110

Polynomials

(a) Find the range of values of p for which 4x2 − px + 1 = 0 has real roots. (b)(i) Expand (1 + 3xy)6 in assending powers of x. (ii) Using the expansion in 10(b)(i), find, correct to four significant figures, the value of (1.03)6.

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Model answer

(a) p² − 16 ≥ 0, so p ≤ −4 or p ≥ 4.
(b)(i) 1 + 18xy + 135x²y² + 540x³y³ + 1215x⁴y⁴ + 1458x⁵y⁵ + 729x⁶y⁶.
(ii) 3xy = 0.03, so xy = 0.01: 1 + 0.18 + 0.0135 + 0.00054 + 0.0000122 + … ≈ 1.194.

Explanation

Choose xy so that 1 + 3xy = 1.03.

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Theory question 111

Matrices and Determinants

Given that M: (x, y) → (7y, 3x − y) and N: (x, y) → (2x − y, 5x + 3y), (a) write down matrices M and N of the linear transformations; (b) find the image of P(2, −3) under the linear transformation N followed by M; (c) find the coordinates of the point Q whose image is Q′(2, 4) under the linear transformation N.

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Model answer

Matrices written row by row:
(a) M = (0 7 / 3 −1), N = (2 −1 / 5 3).
(b) N(2, −3) = (7, 1); then M(7, 1) = (7, 20). (MN = (35 21 / 1 −6).)
(c) 2x − y = 2 and 5x + 3y = 4 give x = 10/11, y = −2/11. Q(10/11, −2/11).

Explanation

'N followed by M' means apply N first.

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Theory question 112

Probability

In an examination, 60% of the candidates passed. If 10 candidates are selected at random, find the probability that: (a) at least two of them failed; (b) exactly half of them passed; (c) at most two of them failed.

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Model answer

Binomial: n = 10, P(fail) = 0.4, P(pass) = 0.6.
(a) 1 − 0.6¹⁰ − 10(0.4)(0.6)⁹ ≈ 1 − 0.0060 − 0.0403 = 0.9536.
(b) ¹⁰C₅(0.6)⁵(0.4)⁵ ≈ 0.2007.
(c) P(0) + P(1) + P(2) = 0.0060 + 0.0403 + 0.1209 ≈ 0.1673.

Explanation

Use P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ with p = probability of failing.

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Theory question 113

Statistics

In a research to determine the relationship between performance of students in an entrance examination and subsequent school performance, the results of ten randomly selected students were obtained as follows:
Students: A, B, C, D, E, F, G, H, I, J
Performance in Entrance Examination: 11, 12, 8, 13, 6, 15, 10, 14, 17, 16
School Performance: 5, 10, 9, 7, 4, 8, 6, 14, 11, 12
(a) Calculate the Spearman's rank correlation coefficient. (b) What would be the researcher's conclusion from the result in 13(a)?

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Model answer

(a) Ranks (highest = 1): Entrance 7, 6, 9, 5, 10, 3, 8, 4, 1, 2; School 9, 4, 5, 7, 10, 6, 8, 1, 3, 2.
Σd² = 4 + 4 + 16 + 4 + 0 + 9 + 0 + 9 + 4 + 0 = 50.
ρ = 1 − 6(50)/990 ≈ 0.70.
(b) There is a fairly strong positive relationship: students who do well in the entrance examination tend to do well in school.

Explanation

Rank each set, find the differences, then use ρ = 1 − 6Σd²/(n(n² − 1)).

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Theory question 114

Dynamics

A body, moving at 20 ms−1 accelerates uniformly at 2½ ms−2 for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest t seconds after with uniform retardation. If the ratio of the acceleration to retardation is 3 : 4, (a) sketch the velocity − time graph of the journey; (b) find t; (c) find the total distance of the journey.

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Model answer

Speed after 4 s: 20 + 2.5 × 4 = 30 ms⁻¹.
(a) Sketch: a line from (0, 20) up to (4, 30), horizontal to (12, 30), then down to (12 + t, 0); label axes.
(b) Retardation = (4/3) × 2.5 = 10/3 ms⁻², so t = 30 ÷ (10/3) = 9 s.
(c) Distance = area = ½(20 + 30)(4) + 30 × 8 + ½ × 30 × 9 = 100 + 240 + 135 = 475 m.

Explanation

Distance is the area under the velocity–time graph.

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Theory question 115

Vectors

(a) Given that m = i − j, n = 2i + 3j and 2m + n − r = 0, find |r|. (b) The distance, S metres of a moving particle at any time t seconds is given by S = 3t − t3/3 + 9. Find the: (i) time; (ii) distance travelled; when the particle is momentarily at rest.

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Model answer

(a) r = 2m + n = 4i + j; |r| = √17 ≈ 4.12.
(b)(i) v = 3 − t² = 0 at t = √3 ≈ 1.73 s.
(ii) S(√3) = 3√3 − √3 + 9 = 9 + 2√3 ≈ 12.46 m; since S = 9 at t = 0, the particle has travelled 2√3 ≈ 3.46 m by then.

Explanation

Momentarily at rest means v = dS/dt = 0.

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