WAEC Mathematics 2018 Objective Past Questions

All 50 questions from the West African Examinations Council (WAEC) Mathematics 2018 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

Mathematics 2018 Objective — Question 1

Simplify: √108 + √125 − √75

• A. √3+5√5Correct
• B. 6√3−5√5
• C. 6√3+√2
• D. 6√3−√2

Explanation

√108=6√3, √125=5√5, √75=5√3, so 6√3+5√5−5√3=√3+5√5.

Mathematics 2018 Objective — Question 2

Evaluate: (64½ + 125⅓)²

• A. 121
• B. 144
• C. 169Correct
• D. 196

Explanation

64½=8 and 125⅓=5, so (8+5)²=13²=169.

Mathematics 2018 Objective — Question 3

Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.

• A. yx²=300
• B. yx²=900Correct
• C. y=100x/9
• D. y=900x²

Explanation

y=k/x² → 100=k/9 → k=900, so yx²=900.

Mathematics 2018 Objective — Question 4

Find the value of x for which 32₄ = 22ₓ

• A. 6Correct
• B. 4
• C. 3
• D. 2

Explanation

32 base4 = 3(4)+2=14. 22 base x = 2x+2. So 2x+2=14 → x=6.

Mathematics 2018 Objective — Question 5

Simplify: 2¼ × 3½ ÷ 4⅛

• A. 2/9
• B. 1⅓Correct
• C. 1¼
• D. 1⅕

Explanation

Convert to improper fractions and multiply/divide in order; simplifying gives 1⅓.

Mathematics 2018 Objective — Question 6

There are 250 boys and 150 girls in a school. If 60% of the boys and 40% of the girls play football, what percentage of the school play football?

• A. 40.0%
• B. 42.2%
• C. 50.0%
• D. 52.5%Correct

Explanation

Boys playing = 60% of 250 = 150. Girls playing = 40% of 150 = 60. Total = 210 out of 400 = 52.5%.

Mathematics 2018 Objective — Question 7

If log₁₀^(6x−4) − log₁₀² = 1, solve for x.

• A. 2
• B. 3
• C. 4Correct
• D. 5

Explanation

log((6x−4)/2)=1 → (6x−4)/2=10 → 6x−4=20 → x=4.

Mathematics 2018 Objective — Question 8

If F = (9/5)C + 32, find C when F = 98.6

• A. 30
• B. 37Correct
• C. 39
• D. 41

Explanation

98.6−32=66.6=(9/5)C → C=66.6×5/9=37.

Mathematics 2018 Objective — Question 9

If y + 2x = 4 and y − 3x = −1, find the value of (x + y).

• A. 3Correct
• B. 2
• C. 1
• D. −1

Explanation

Subtracting the equations gives 5x=5, x=1, then y=2, so x+y=3.

Mathematics 2018 Objective — Question 10

If x : y : z = 2 : 3 : 4, evaluate (9x+3y)/(6z−2y)

• A. 1½Correct
• B. 2
• C. 2½
• D. 3

Explanation

Let x=2k, y=3k, z=4k: (18k+9k)/(24k−6k)=27k/18k=3/2=1½.

Mathematics 2018 Objective — Question 11

Simplify: (2−18m²)/(1+3m)

• A. 2(1+3m)
• B. 2(2+3m²)
• C. 2(1−3m)Correct
• D. 2(1−3m²)

Explanation

2−18m²=2(1−9m²)=2(1−3m)(1+3m); dividing by (1+3m) leaves 2(1−3m).

Mathematics 2018 Objective — Question 12

A curve is such that when y = 0, x = −2 or x = 3. Find the equation of the curve.

• A. y=x²−5x−6
• B. y=x²+5x−6
• C. y=x²+x−6
• D. y=x²−x−6Correct

Explanation

Roots −2 and 3 give (x+2)(x−3)=0 → x²−x−6=0, so y=x²−x−6.

Mathematics 2018 Objective — Question 13

The volume of a cylindrical tank, 10m high, is 385m³. Find the diameter of the tank. [Take π = 22/7]

• A. 14m
• B. 10m
• C. 7mCorrect
• D. 5m

Explanation

385=πr²(10) → r²=385×7/(22×10)=12.25 → r=3.5m, diameter=7m.

Mathematics 2018 Objective — Question 14

The surface area of a sphere is 792/7 cm². Find, correct to the nearest whole number, its volume. [Take π = 22/7]

• A. 113cm³Correct
• B. 131cm³
• C. 311cm³
• D. 414cm³

Explanation

4πr²=792/7 → r²=9 → r=3. Volume=(4/3)πr³=(4/3)(22/7)(27)≈113cm³.

Mathematics 2018 Objective — Question 15

A piece of thread of length 21.4cm is used to form a sector of a circle of radius 4.2cm on a piece of cloth. Calculate, correct to the nearest degree, the angle of the sector. [Take π = 22/7]

• A. 170°
• B. 177°Correct
• C. 182°
• D. 192°

Explanation

Arc length = 21.4−(2×4.2)=13cm. θ=(13/2πr)×360≈177°.

Mathematics 2018 Objective — Question 16

In the diagram above, which of the following ratios is equal to |PN|/|PQ|?

• A. |PN|/|PR|
• B. |PM|/|PQ|
• C. |PM|/|PR|Correct
• D. |PN|/|PQ|

Explanation

Since NM is parallel to RQ, triangles PNM and PRQ are similar, so |PN|/|PQ| = |PM|/|PR|.

Mathematics 2018 Objective — Question 17

In the diagram above, PS and RS are tangents to the circle centre O, ∠PSR = 70°, ∠POR = m° and ∠PQR = n°. Find (m° + n°).

• A. 110°
• B. 135°
• C. 165°Correct
• D. 225°

Explanation

∠POR=180−90−35−35=... m=110°, n (angle at circumference)=55°, so m+n=165°.

Mathematics 2018 Objective — Question 18

Find the value of t in the diagram above.

• A. 63°
• B. 117°
• C. 126°Correct
• D. 234°

Explanation

Angle on straight line = 180−117=63°. Angle at centre = 2×angle at circumference = 2×63=126°.

Mathematics 2018 Objective — Question 19

In the diagram, PR is a tangent to the circle at Q, QT//RS, ∠SQR = 35° and ∠RSQ = 50°. Find the value of ∠QST.

• A. 40°
• B. 65°
• C. 85°
• D. 95°Correct

Explanation

Using alternate segment theorem and angle sum in the triangle, ∠QST = 95°.

Mathematics 2018 Objective — Question 20

The angles of a polygon are x, 2x, 2x, (x+30°), (x+20°) and (x−10°). Find the value of x.

• A. 45°
• B. 84°
• C. 85°Correct
• D. 95°

Explanation

Sum of interior angles of hexagon = 720°. 8x+40=720 → x=85°.

Mathematics 2018 Objective — Question 21

If M and N are the points (−3, 8) and (5, −7) respectively, find |MN|.

• A. 8 units
• B. 11 units
• C. 15 units
• D. 17 unitsCorrect

Explanation

|MN|=√[(−3−5)²+(8+7)²]=√(64+225)=√289=17 units.

Mathematics 2018 Objective — Question 22

The equation of the line through the points (4,2) and (−8,−2) is 3y = px + q, where p and q are constants. Find the value of p.

• A. 1Correct
• B. 2
• C. 3
• D. 9

Explanation

Slope = (−2−2)/(−8−4)=1/3. Rearranging y−2=(1/3)(x−4) to 3y=x+2 gives p=1.

Mathematics 2018 Objective — Question 23

The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30°. Find the height of the tree.

• A. 27m
• B. 13.5√3m
• C. 13.5√2m
• D. 9√3mCorrect

Explanation

h=27tan30°=27×(1/√3)=9√3m.

Mathematics 2018 Objective — Question 24

If tan x = 4/3, 0° < x < 90°, find the value of sin x − cos x.

• A. 1/10
• B. 1/5Correct
• C. 5/12
• D. 1⅕

Explanation

For a 3-4-5 triangle, sinx=4/5, cosx=3/5, so sinx−cosx=1/5.

Mathematics 2018 Objective — Question 25

Given that Y is 20m on a bearing of 300° from X, how far south of X is Y?

• A. 10mCorrect
• B. 15m
• C. 25m
• D. 30m

Explanation

The angle between the bearing and the north-south line gives sin30°=|YS|/20 → |YS|=10m.

Mathematics 2018 Objective — Question 26

The mean of 1, 3, 5, 7 and x is 4. Find the value of x.

• A. 2
• B. 4Correct
• C. 6
• D. 8

Explanation

(1+3+5+7+x)/5=4 → 16+x=20 → x=4.

Mathematics 2018 Objective — Question 27

Find the median of 2, 1, 0, 3, 1, 1, 4, 0, 1 and 2

• A. 0.0
• B. 0.5
• C. 1.0Correct
• D. 1.5

Explanation

Arranged in order: 0,0,1,1,1,1,2,2,3,4. Median = average of 5th and 6th values = (1+1)/2=1.0.

Mathematics 2018 Objective — Question 28

Calculate the probability that a team selected at random scored at most 3 goals.

• A. 3/25
• B. 4/25
• C. 6/25
• D. 2/5Correct

Explanation

Teams scoring at most 3 goals = 3+1+6=10 out of 25. Probability = 10/25 = 2/5.

Mathematics 2018 Objective — Question 29

Find the probability that a team selected at random scored either 4 or 7 goals.

• A. 9/25Correct
• B. 11/25
• C. 3/5
• D. 18/25

Explanation

Teams scoring 4 goals = 6, scoring 7 goals = 3. Total = 9 out of 25, so probability = 9/25.

Mathematics 2018 Objective — Question 30

In the diagram above, WXYZ is a rectangle with dimension 8cm by 6cm. P, Q, R and S are the mid-points of the sides of the rectangle as shown. What type of quadrilateral is the shaded region?

• A. Trapezium
• B. Prism
• C. Rectangle
• D. RhombusCorrect

Explanation

Joining the midpoints of a rectangle's sides always forms a rhombus.

Mathematics 2018 Objective — Question 31

Calculate the area of the part of the rectangle that is not shaded.

• A. 25cm²
• B. 24cm²Correct
• C. 16cm²
• D. 12cm²

Explanation

Rectangle area = 48cm². Shaded rhombus area = 2×(½×8×3)=24cm². Unshaded = 48−24=24cm².

Mathematics 2018 Objective — Question 32

The total surface area of a hemisphere is 75πcm². Find the radius.

• A. 5.0cmCorrect
• B. 7.0cm
• C. 8.5cm
• D. 12.0cm

Explanation

3πr²=75π → r²=25 → r=5.0cm.

Mathematics 2018 Objective — Question 33

Find the value of x for which (x−5)/(x(x−1)) is undefined.

• A. 0 or 5
• B. −5 or 5
• C. −1 or 5
• D. 0 or 1Correct

Explanation

The expression is undefined when the denominator x(x−1)=0, i.e. x=0 or x=1.

Mathematics 2018 Objective — Question 34

Solve the equation 2x² − x − 6 = 0

• A. x=−3/2 or 2Correct
• B. x=−2 or 3/2
• C. x=−3 or 2
• D. x=3 or −2

Explanation

Factorising: (2x+3)(x−2)=0 → x=−3/2 or x=2.

Mathematics 2018 Objective — Question 35

Factorise completely the expression (x+2)² − (2x+1)²

• A. (3x+2)(1−x)
• B. (3x+2)(2x+1)
• C. 3(x+2)²
• D. 3(x+1)(1−x)Correct

Explanation

Using difference of two squares: [(x+2)+(2x+1)][(x+2)−(2x+1)] = (3x+3)(1−x)=3(x+1)(1−x).

Mathematics 2018 Objective — Question 36

Find the nth term of the sequence 2×3, 4×6, 8×9, 16×12, …

• A. 2ⁿ×3(n+1)
• B. 2ⁿ×3nCorrect
• C. 2ⁿ×3ⁿ
• D. 2ⁿ×3ⁿ⁻¹

Explanation

The first factor doubles each term (2ⁿ) while the second increases by 3 each time (3n), giving 2ⁿ×3n.

Mathematics 2018 Objective — Question 37

If 3x ≡ 4(mod 5), find the least value of x.

• A. 1
• B. 2
• C. 3Correct
• D. 4

Explanation

3(3)=9, and 9 mod 5 = 4, so the least positive value of x satisfying the congruence is 3.

Mathematics 2018 Objective — Question 38

The solution of x + 2 ≥ 2x + 1 is illustrated below.

• A. Option ACorrect
• B. Option B
• C. Option C
• D. Option D

Explanation

x+2≥2x+1 simplifies to x≤1, illustrated by a line shaded to the left of and including 1.

Mathematics 2018 Objective — Question 39

If p and q are two statements, under what condition would p : q be false?

• A. if p is true and q is true
• B. if p is true and q is falseCorrect
• C. if p is false and q is false
• D. if p is false and q is true

Explanation

An implication p→q is false only when the antecedent p is true and the consequent q is false.

Mathematics 2018 Objective — Question 40

The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m°.

• A. 90°
• B. 60°Correct
• C. 45°
• D. 30°

Explanation

Equal chords WX, XY, YZ divide the semicircle into three equal arcs of 60° each, making triangle OYZ isosceles with base angle m°=60°.

Mathematics 2018 Objective — Question 41

Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16

• A. 6
• B. 7
• C. 8Correct
• D. 9

Explanation

For 11 ordered values: Q1 (3rd value)=4, Q3 (9th value)=12. IQR=12−4=8.

Mathematics 2018 Objective — Question 42

Donations during the launching of a church project were sent in sealed envelopes. The table shows the distribution of the amount of money in the envelope. How much was the total donation?

• A. ₦26,792.00
• B. ₦26,972.00
• C. ₦62,792.00
• D. ₦62,972.00Correct

Explanation

Sum of (number of envelopes × amount) for every row = ₦62,972.00.

Mathematics 2018 Objective — Question 43

In the diagram above, PQ is a straight line, (m+n) = 110°, (n+r) = 130° and (m+r) = 120°. Find the ratio m : n : r

• A. 2:3:4
• B. 3:4:5
• C. 4:5:6
• D. 5:6:7Correct

Explanation

Adding all three equations: 2(m+n+r)=360 → m+n+r=180. Then r=70, m=50, n=60, giving ratio 5:6:7.

Mathematics 2018 Objective — Question 44

If x : y = 1/4 : 3/8 and y : z = 1/3 : 4/9, find x : z.

• A. 2:3
• B. 3:4
• C. 3:8
• D. 1:2Correct

Explanation

x:y simplifies to 2:3, and y:z simplifies to 3:4. Combining, x:z = (2/3)×(3/4) = 1:2.

Mathematics 2018 Objective — Question 45

Find the mean deviation of 20, 30, 25, 40, 35, 50, 45, 40, 20 and 45

• A. 8
• B. 9Correct
• C. 10
• D. 12

Explanation

Mean = 350/10 = 35. Sum of absolute deviations = 90. Mean deviation = 90/10 = 9.

Mathematics 2018 Objective — Question 46

M and N are two subsets of the universal set (U). If n(U) = 28, n(M) = 20, n(N) = 30 and n(M∪N) = 40, find n(M∩N)'.

• A. 18Correct
• B. 20
• C. 30
• D. 38

Explanation

n(M∩N)=n(M)+n(N)−n(M∪N)=20+30−40=10. n(M∩N)'=28−10=18.

Mathematics 2018 Objective — Question 47

Express 0.612 in the form x/y, where x and y are integers and y ≠ 0.

• A. 153/250Correct
• B. 68/111
• C. 61/100
• D. 21/33

Explanation

0.612 = 612/1000, which simplifies (dividing by 4) to 153/250.

Mathematics 2018 Objective — Question 48

In the diagram above, PQ//RS. Find x° in terms of y° and z°

• A. x°=240°−y°−z°
• B. x°=180°−y°−z°
• C. x°=360°−y°−z°
• D. x°=360°−y°−z°Correct

Explanation

Using angles at the vertical opposite point and the angle sum properties of the parallel lines, x°=360°−y°−z°.

Mathematics 2018 Objective — Question 49

The diagonals of a rhombus WXYZ intersect at M. If |MW| = 5cm and |MX| = 12cm, calculate its perimeter.

• A. 42cm
• B. 48cm
• C. 52cmCorrect
• D. 60cm

Explanation

Rhombus diagonals bisect at right angles, so each side = √(5²+12²)=13cm. Perimeter = 4×13 = 52cm.

Mathematics 2018 Objective — Question 50

The graphs of y = x² and y = x intersect at which of these points?

• A. (0,0), (1,1)Correct
• B. (0,0), (0,1)
• C. (1,0), (0,0)
• D. (0,0), (0,0)

Explanation

Setting x²=x gives x(x−1)=0, so x=0 or x=1, corresponding to points (0,0) and (1,1).