WAEC Mathematics 2018 past questions and answers

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

Question 1

Surds

Simplify: √108 + √125 − √75

  1. A.√3 + 5√5
  2. B.6√3 − 5√5
  3. C.6√3 + √2
  4. D.6√3 − √2
Show the answer and explanation

Answer: A (√3 + 5√5)

Explanation

√108 = 6√3, √125 = 5√5 and √75 = 5√3, so the sum is 6√3 + 5√5 − 5√3 = √3 + 5√5.

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Question 2

Indices

Evaluate: (64½ + 125⅓)2

  1. A.121
  2. B.144
  3. C.169
  4. D.196
Show the answer and explanation

Answer: C (169)

Explanation

64½ = 8 and 125⅓ = 5, so (8 + 5)² = 169.

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Question 3

Variation

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

  1. A.yx2 = 300
  2. B.yx2 = 900
  3. C.y = 100x/9
  4. D.y = 9002
Show the answer and explanation

Answer: B (yx2 = 900)

Explanation

y = k/x², so k = yx² = 100 × 9 = 900 and the equation is yx² = 900.

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Question 4

Number Bases

Find the value of x for which 32four = 22x.

  1. A.three
  2. B.five
  3. C.six
  4. D.seven
Show the answer and explanation

Answer: C (six)

Explanation

32four = 12 + 2 = 14 and 22x = 2x + 2, so 2x + 2 = 14 and x = 6.

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Question 5

Fractions, Decimals and Approximation

Simplify: 2¼ × 3½ ÷ 4⅜.

  1. A.5/9
  2. B.1⅕
  3. C.1¼
  4. D.1⅘
Show the answer and explanation

Answer: D (1⅘)

Explanation

9/4 × 7/2 × 8/35 = 504/280 = 9/5 = 1⅘.

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Question 6

Ratio, Percentages and Rates

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?

  1. A.40.0%
  2. B.42.2%
  3. C.50.0%
  4. D.52.5%
Show the answer and explanation

Answer: D (52.5%)

Explanation

150 boys + 60 girls = 210 of 400 pupils, i.e. 52.5%.

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Question 7

Logarithms

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

  1. A.2
  2. B.3
  3. C.4
  4. D.5
Show the answer and explanation

Answer: C (4)

Explanation

log10((6x − 4)/2) = 1, so (6x − 4)/2 = 10, 6x = 24 and x = 4.

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Question 8

Algebraic Expressions

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

  1. A.30
  2. B.37
  3. C.39
  4. D.41
Show the answer and explanation

Answer: B (37)

Explanation

98.6 − 32 = 66.6 = 9C/5, so C = 66.6 × 5/9 = 37.

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Question 10

Algebraic Expressions

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

  1. A.1½
  2. B.2
  3. C.2½
  4. D.3
Show the answer and explanation

Answer: A (1½)

Explanation

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

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Question 12

Graphs of Functions

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

  1. A.y = x2 − 5x − 6
  2. B.y = x2 + 5x − 6
  3. C.y = x2 + x − 6
  4. D.y = x2 − x − 6
Show the answer and explanation

Answer: D (y = x2 − x − 6)

Explanation

y = (x + 2)(x − 3) = x² − x − 6.

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Question 13

Mensuration of Solids

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

  1. A.14m
  2. B.10m
  3. C.7m
  4. D.5m
Show the answer and explanation

Answer: C (7m)

Explanation

22/7 × r² × 10 = 385 gives r² = 12.25, r = 3.5 m, so the diameter is 7 m.

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Question 14

Mensuration of Solids

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

  1. A.113 cm3
  2. B.131 cm3
  3. C.311 cm3
  4. D.414 cm3
Show the answer and explanation

Answer: A (113 cm3)

Explanation

4 × 22/7 × r² = 792/7 gives r² = 9, r = 3. Volume = 4/3 × 22/7 × 27 ≈ 113 cm³.

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Question 15

Mensuration of Plane Shapes

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

  1. A.170°
  2. B.177°
  3. C.182°
  4. D.192°
Show the answer and explanation

Answer: B (177°)

Explanation

Arc = 21.4 − 8.4 = 13 cm. θ/360 × 2 × 22/7 × 4.2 = 13 gives θ = 13 × 360/26.4 ≈ 177°.

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Question 16

Triangles and Quadrilaterals

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

Diagram for this question
  1. A.|PN|/|PR|
  2. B.|PM|/|PQ|
  3. C.|PM|/|PR|
  4. D.|PR|/|PQ|
Show the answer and explanation

Answer: C (|PM|/|PR|)

Explanation

The equal-angle marks show triangle PNM is similar to triangle PQR with N matching Q and M matching R, so |PN|/|PQ| = |PM|/|PR|.

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Question 17

Circle Theorems

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

Diagram for this question
  1. A.110°
  2. B.135°
  3. C.165°
  4. D.225°
Show the answer and explanation

Answer: C (165°)

Explanation

In quadrilateral OPSR the radii meet the tangents at 90°, so m = 360° − 180° − 70° = 110°. n is the angle at the circumference on the same arc, so n = 55°. m + n = 165°.

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Question 18

Triangles and Quadrilaterals

Find the value of t in the diagram.

Diagram for this question
  1. A.63°
  2. B.117°
  3. C.126°
  4. D.234°
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Answer: C (126°)

Explanation

WYZ is a straight line, so ∠WYX = 180° − 117° = 63°. The angle at the centre is twice that: t = 126°.

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Question 19

Circle Theorems

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

Diagram for this question
  1. A.40°
  2. B.65°
  3. C.85°
  4. D.95°
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Answer: D (95°)

Explanation

In triangle QRS, ∠QRS = 95°. QT//RS gives ∠TQR = 180° − 95° = 85°, so ∠TQP = 95°. By the alternate segment theorem ∠QST = ∠TQP = 95°.

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Question 21

Coordinate Geometry

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

  1. A.8 units
  2. B.11 units
  3. C.15 units
  4. D.17 units
Show the answer and explanation

Answer: D (17 units)

Explanation

MN = √(8² + 15²) = √289 = 17 units.

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Question 22

Coordinate Geometry

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.

  1. A.1
  2. B.2
  3. C.3
  4. D.9
Show the answer and explanation

Answer: A (1)

Explanation

Gradient = (−2 − 2)/(−8 − 4) = 1/3. Writing y = x/3 + c as 3y = x + 3c gives p = 1.

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Question 23

Trigonometry

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

  1. A.27 m
  2. B.13.5√3 m
  3. C.13.5√2 m
  4. D.9√3 m
Show the answer and explanation

Answer: D (9√3 m)

Explanation

Height = 27 tan 30° = 27/√3 = 9√3 m.

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Question 24

Trigonometry

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

  1. A.1/10
  2. B.⅕
  3. C.5/12
  4. D.1⅖
Show the answer and explanation

Answer: B (⅕)

Explanation

Sides 3, 4, 5: sin x = 4/5, cos x = 3/5, so sin x − cos x = 1/5.

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Question 25

Bearings

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

  1. A.10 m
  2. B.15 m
  3. C.25 m
  4. D.30 m
Show the answer and explanation

Answer: A (10 m)

Explanation

The line XY makes 60° with the north line (360° − 300°), so the north–south distance is 20 cos 60° = 10 m.

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Question 26

Measures of Central Tendency

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

  1. A.2
  2. B.4
  3. C.6
  4. D.8
Show the answer and explanation

Answer: B (4)

Explanation

1 + 3 + 5 + 7 + x = 20, so x = 4.

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Question 28

Probability

The table shows the distribution of goals scored by 25 teams in a football competition.
Number of goals: 1, 2, 3, 4, 5, 6, 7
Number of teams: 3, 1, 6, 6, 4, 2, 3
Calculate the probability that a team selected at random scored at most 3 goals.

  1. A.3/25
  2. B.⅕
  3. C.6/25
  4. D.⅖
Show the answer and explanation

Answer: D (⅖)

Explanation

3 + 1 + 6 = 10 teams scored 3 or fewer: 10/25 = 2/5.

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Question 29

Probability

The table shows the distribution of goals scored by 25 teams in a football competition.
Number of goals: 1, 2, 3, 4, 5, 6, 7
Number of teams: 3, 1, 6, 6, 4, 2, 3
Find the probability that a team selected random scored either 4 or 7 goals.

  1. A.9/25
  2. B.11/25
  3. C.⅗
  4. D.18/25
Show the answer and explanation

Answer: A (9/25)

Explanation

6 teams scored 4 and 3 scored 7: (6 + 3)/25 = 9/25.

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Question 30

Triangles and Quadrilaterals

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

Diagram for this question
  1. A.Trapezium
  2. B.Prism
  3. C.Rectangle
  4. D.Rhombus
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Answer: D (Rhombus)

Explanation

Joining the midpoints of a rectangle gives four equal sides (each 5 cm) but no right angles: a rhombus.

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Question 31

Mensuration of Plane Shapes

In the diagram, WXYZ is a rectangle with dimension 8 cm by 6 cm. P, Q, R and S are the midpoints of the sides of the rectangle as shown. Calculate the area of the part of the rectangle that is not shaded.

Diagram for this question
  1. A.25 cm2
  2. B.24 cm2
  3. C.16 cm2
  4. D.12 cm2
Show the answer and explanation

Answer: B (24 cm2)

Explanation

The four corner triangles each have area ½ × 4 × 3 = 6 cm², so the unshaded part is 24 cm² (half of 48).

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Question 32

Mensuration of Solids

The total surface area of a hemisphere is 75π cm2. Find is radius.

  1. A.5.0 cm
  2. B.7.0 cm
  3. C.8.5 cm
  4. D.12.0 cm
Show the answer and explanation

Answer: A (5.0 cm)

Explanation

3πr² = 75π, so r² = 25 and r = 5 cm.

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Question 33

Algebraic Expressions

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

  1. A.0 or 5
  2. B.−5 or 5
  3. C.−1 or 5
  4. D.0 or 1
Show the answer and explanation

Answer: D (0 or 1)

Explanation

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

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Question 34

Quadratic Equations

Solve the equation 2x2 − x − 6 = 0.

  1. A.x = −3/2 or 2
  2. B.x = −2 or 3/2
  3. C.x = −3 or 2
  4. D.x = 3 or 2
Show the answer and explanation

Answer: A (x = −3/2 or 2)

Explanation

2x² − x − 6 = (2x + 3)(x − 2) = 0, so x = −3/2 or 2.

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Question 36

Progressions

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

  1. A.2n × 3(n − 1)
  2. B.2n × 3n
  3. C.2n × 3n
  4. D.2n × 3n−1
Show the answer and explanation

Answer: B (2n × 3n)

Explanation

The first factors are 2, 4, 8, 16 = 2n and the second are 3, 6, 9, 12 = 3n.

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Question 37

Modular Arithmetic

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

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

Answer: C (3)

Explanation

Try values: 3 × 3 = 9 ≡ 4 (mod 5), so x = 3.

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Question 38

Inequalities

The solution of x − 2 ≥ 2x − 1 illustrated on the number line as

Diagram for this question
  1. A.Number line −3 to 2: dot at 1, arrow to the left
  2. B.Number line −3 to 2: dot at 1, arrow to the right
  3. C.Number line −3 to 2: dot at −1, arrow to the left
  4. D.Number line −3 to 2: dot at −1, arrow to the right
Show the answer and explanation

Answer: C (Number line −3 to 2: dot at −1, arrow to the left)

Explanation

x − 2 ≥ 2x − 1 gives −1 ≥ x, i.e. x ≤ −1: a solid dot at −1 with the arrow to the left.

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Question 39

Logic

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

  1. A.If p is true and q is true
  2. B.If p is true and q is false
  3. C.If p is false and q is false
  4. D.If p is false and q is true
Show the answer and explanation

Answer: B (If p is true and q is false)

Explanation

An implication 'if p then q' is false only when p is true and q is false.

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Question 40

Circle Theorems

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

Diagram for this question
  1. A.90°
  2. B.60°
  3. C.45°
  4. D.30°
Show the answer and explanation

Answer: B (60°)

Explanation

Equal chords cut equal angles at O, so each is 180° ÷ 3 = 60°. Triangle OYZ is isosceles (radii) with apex angle 60°, so it is equilateral and m = 60°.

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Question 41

Measures of Dispersion

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

  1. A.6
  2. B.7
  3. C.8
  4. D.9
Show the answer and explanation

Answer: C (8)

Explanation

With 11 values Q₁ is the 3rd value (4) and Q₃ is the 9th value (12); 12 − 4 = 8.

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Question 42

Measures of Central Tendency

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.
No. of envelopes: 4, 7, 20, 9, 4, 5, 3, 1, 2
Amount (N): 5000, 2000, 1000, 700, 500, 100, 50, 2, 10
How much was the total donation?

  1. A.N26,792.00
  2. B.N26,972.00
  3. C.N62,792.00
  4. D.N62,972.00
Show the answer and explanation

Answer: D (N62,972.00)

Explanation

20 000 + 14 000 + 20 000 + 6 300 + 2 000 + 500 + 150 + 2 + 20 = N62,972.

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Question 43

Angles and Polygons

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

Diagram for this question
  1. A.2 : 3 : 4
  2. B.3 : 4 : 5
  3. C.4 : 5 : 6
  4. D.5 : 6 : 7
Show the answer and explanation

Answer: D (5 : 6 : 7)

Explanation

m + n + r = 180° (angles on a line). So r = 70°, m = 50° and n = 60°, giving 50 : 60 : 70 = 5 : 6 : 7.

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Question 44

Ratio, Percentages and Rates

If x : y = ¼ : ⅜ and y : z = ⅓ : 4/9, find x : z.

  1. A.2 : 3
  2. B.3 : 4
  3. C.3 : 8
  4. D.1 : 2
Show the answer and explanation

Answer: D (1 : 2)

Explanation

x : y = 2 : 3 and y : z = 3 : 4, so x : z = 2 : 4 = 1 : 2.

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Question 45

Measures of Dispersion

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

  1. A.8
  2. B.9
  3. C.10
  4. D.12
Show the answer and explanation

Answer: B (9)

Explanation

Mean = 350/10 = 35. The distances from 35 add to 15 + 5 + 10 + 5 + 0 + 15 + 10 + 5 + 15 + 10 = 90, so the mean deviation is 9.

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Question 46

Sets

M and N are two subsets of the universal set (U). If n(U) = 48, n(M) = 20, n(N) = 30 and n(M ∪ N) = 40, find n(M ∩ N)′.

  1. A.18
  2. B.20
  3. C.30
  4. D.38
Show the answer and explanation

Answer: D (38)

Explanation

n(M ∩ N) = 20 + 30 − 40 = 10, so n(M ∩ N)′ = 48 − 10 = 38.

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Question 47

Fractions, Decimals and Approximation

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

  1. A.153/250
  2. B.68/111
  3. C.61/100
  4. D.21/33
Show the answer and explanation

Answer: A (153/250)

Explanation

0.612 = 612/1000 = 153/250.

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Question 48

Angles and Polygons

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

Diagram for this question
  1. A.x = 240° − y − z
  2. B.x = 180° − y − z
  3. C.x = 360° + y − z
  4. D.x = 360° − y − z
Show the answer and explanation

Answer: D (x = 360° − y − z)

Explanation

Use co-interior angles between the parallel lines on each crossing line, then the angle sum of the triangle containing x: this gives x = 360° − y − z.

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Question 49

Mensuration of Plane Shapes

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

  1. A.42 cm
  2. B.48 cm
  3. C.52 cm
  4. D.60 cm
Show the answer and explanation

Answer: C (52 cm)

Explanation

The diagonals bisect at right angles, so a side is √(5² + 12²) = 13 cm and the perimeter is 52 cm.

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Question 50

Graphs of Functions

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

  1. A.(0, 0), (1, 1)
  2. B.(0, 0), (0, 1)
  3. C.(1, 0), (0, 0)
  4. D.(0, 0), (0, 0)
Show the answer and explanation

Answer: A ((0, 0), (1, 1))

Explanation

x² = x gives x = 0 or 1, so the points are (0, 0) and (1, 1).

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Theory question 101

Commercial Arithmetic

A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of that year. If the car was bought on 1st March, 2011, calculate, correct to the nearest hundred naira, the value of the car on 28th February, 2015.

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Model answer

1 March 2011 to 28 February 2015 is 4 years.
After year 1: 900 000 × 0.70 = N630,000
After 3 more years: 630 000 × 0.78³ = 630 000 × 0.474552 = N298,967.76
Value ≈ N299,000 (nearest hundred naira)

Explanation

Count the years carefully: the fourth year ends on 28th February 2015.

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Theory question 102

Graphs of Functions

(a) The graph of y = 2px2 − p2x − 14 passes through the point (3, 10). Find the values of p.
(b) Two lines, 3y − 2x = 21 and 4y + 5x = 5, intersect at the point Q. Find the coordinates of Q.

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Model answer

(a) 10 = 18p − 3p² − 14 → 3p² − 18p + 24 = 0 → p² − 6p + 8 = 0 → (p − 2)(p − 4) = 0
p = 2 or p = 4
(b) ×4 and ×3: 12y − 8x = 84 and 12y + 15x = 15 → 23x = −69 → x = −3
4y − 15 = 5 → y = 5; Q = (−3, 5)

Explanation

Substitute x = 3 and y = 10 to get a quadratic in p.

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Theory question 103

Mensuration of Plane Shapes

(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.
(b) Given that sin x = ⅗, 0° < x < 90°, find the value of 5 cos x − 4 tan x.

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Model answer

(a) Diagonals bisect at right angles: half-diagonals 5.1 cm and 4.65 cm
Side = √(5.1² + 4.65²) = √47.6325 ≈ 6.902 cm
Perimeter = 4 × 6.902 ≈ 27.6 cm
(b) Sides 3, 4, 5: cos x = 4/5, tan x = 3/4
5 cos x − 4 tan x = 4 − 3 = 1

Explanation

A rhombus's diagonals cut each other in half at right angles.

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Theory question 104

Circle Theorems

(a) In the diagram, QOS is a diameter, ∠RQS = x° and ∠QST = (3x + 15)°. Find: (i) the value of x. (ii) ∠RSQ.
(b) If 2N4seven = 15Nnine, find the value of N.

Diagram for this question
Show the model answer
Model answer

(a)(i) ∠QRS = 90° (angle in a semicircle)
∠QST is an exterior angle of triangle QRS: 3x + 15 = 90 + x → x = 37.5°
(ii) ∠RSQ = 180° − (3 × 37.5 + 15)° = 52.5° (or 90° − 37.5°)
(b) 2 × 49 + 7N + 4 = 81 + 45 + N → 102 + 7N = 126 + N → N = 4

Explanation

Use the angle in a semicircle and the exterior angle of a triangle.

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Theory question 105

Measures of Dispersion

(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m − 4), (n − 3), (s + 6), (p − 2) and (q − 8).
(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?

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Model answer

(a) m + n + s + p + q = 60
New total = 60 − 4 − 3 + 6 − 2 − 8 = 49; new mean = 49/5 = 9.8
(b) 75% − 25% = 50% of the people: 0.5 × 500 = 250 people

Explanation

Find the total first; adding or subtracting from each value changes the total by the sum of the changes.

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Theory question 106

Sets

In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only − 28: Clutch and Steering − 14; Clutch, Steering and Brakes − 8: Clutch and Brakes − 20: Brakes and Steering only − 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only. (a) Draw a Venn diagram to illustrate this information. (b) How many cars had: (i) faulty Brakes? (ii) only one fault?

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Model answer

Failed = 40% of 240 = 96
(a) Three circles C, B, S: C only 28; C∩S only 14 − 8 = 6; C∩B only 20 − 8 = 12; all three 8; B∩S only 6; B only x; S only 2x.
28 + 6 + 12 + 8 + 6 + x + 2x = 96 → 3x = 36 → x = 12
(b)(i) Faulty brakes = 12 + 12 + 8 + 6 = 38
(ii) Only one fault = 28 + 12 + 24 = 64

Explanation

'Clutch and Steering − 14' includes the 8 cars with all three faults.

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Theory question 107

Coordinate Geometry

(a) Find the equation of the line passing through the points (2,5) and (−4, −7).
(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P from R is 300°. If |PQ| = 5 km and |PR| = 8 km, (i) illustrate the information in a diagram. (ii) calculate, correct to three significant figures, the: (I) distance between Q and R; (II) bearing of R from Q.

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Model answer

(a) Gradient = (−7 − 5)/(−4 − 2) = 2; y − 5 = 2(x − 2) → y = 2x + 1
(b)(i) R is on a bearing of 120° from P, so ∠QPR = 120° − 30° = 90°.
(ii)(I) QR = √(5² + 8²) = √89 ≈ 9.43 km
(II) tan∠PQR = 8/5 → ∠PQR ≈ 58.0°
Bearing of P from Q = 210°, so bearing of R from Q = 210° − 58.0° ≈ 152°

Explanation

Turn the back bearing (300° − 180° = 120°) into the bearing of R from P.

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Theory question 108

Commercial Arithmetic

(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book to James at a profit of x%. If James paid N(6x + ¾) more for the book than what Lamin paid, find the value of x.
(b) Find the range of values of x which satisfies the inequality 3x − 2 < 10 + x < 2 + 5x.

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(a) James paid 300(1 + x/100)²
300(1 + x/100)² − 300 = 6x + ¾ → 6x + 0.03x² = 6x + ¾
0.03x² = 0.75 → x² = 25 → x = 5
(b) 3x − 2 < 10 + x → x < 6; 10 + x < 2 + 5x → x > 2
So 2 < x < 6

Explanation

Split the double inequality into two and solve each.

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Theory question 109

Mensuration of Plane Shapes

In the diagram, |PT| = 4 cm, |TS| = 6 cm |PQ| = 6 cm and ∠SPR = 30°. Calculate, correct to the nearest whole number: (a) |SR|; (b) area of TQRS.

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TQ//SR, so triangles PTQ and PSR are similar: |PS| = 10 cm and |PR| = 6 × 10/4 = 15 cm
(a) |SR|² = 10² + 15² − 2 × 10 × 15 cos 30° ≈ 65.19 → |SR| ≈ 8.07 ≈ 8 cm
(b) Area PSR = ½ × 10 × 15 × sin 30° = 37.5 cm²
Area PTQ = ½ × 4 × 6 × sin 30° = 6 cm²
Area TQRS = 37.5 − 6 = 31.5 ≈ 32 cm²

Explanation

Find |PR| from similar triangles, then use the cosine rule.

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Theory question 110

Trigonometry

(a) In ΔPQS, PQ = 12 cm, PS = 5 cm, ∠SPQ = ∠PRQ = 90°. Find, correct to three significant figures, PR.
(b) The lengths of two ladders, L and M are 10 m and 12 m respectively. They are placed against a wall such that each ladder makes the same angle with the horizontal ground. If the foot of L is 8 m from the foot of the wall, (i) draw a diagram to illustrate this information; (ii) calculate the height at which M touches the wall.

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(a) SQ = √(5² + 12²) = 13 cm
Area: ½ × 5 × 12 = ½ × 13 × PR → PR = 60/13 ≈ 4.62 cm
(b)(i) Two right-angled triangles on the same wall and ground, ladders 10 m and 12 m at equal angles θ to the ground.
(ii) cos θ = 8/10, so sin θ = 0.6
Height of M = 12 × 0.6 = 7.2 m

Explanation

PR is the height of the right-angled triangle PQS onto its hypotenuse.

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Theory question 111

Graphs of Functions

(a) Copy and complete the table of values for y = 2x2 + x − 10 for −5 ≤ x ≤ 4.
x: −5, −4, −3, −2, −1, 0, 1, 2, 3, 4
y: __, __, 5, __, −9, −10, __, 0, __, __
(b) Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x2 + x − 10 for −5 ≤ x ≤ 4.
(c) Use the graph to find the solution of: (i) 2x2 + x = 10; (ii) 2x2 + x − 10 = 2x.

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Model answer

(a) x = −5: 35; −4: 18; −2: −4; 1: −7; 3: 11; 4: 26
(b) Plot the points and join with a smooth U-shaped curve.
(c)(i) 2x² + x − 10 = 0: where the curve cuts the x-axis, x = −2.5 or x = 2
(ii) Draw the line y = 2x; it meets the curve at x = −2 and x = 2.5

Explanation

For (c)(ii) draw the straight line y = 2x on the same axes.

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Theory question 112

Vectors and Transformations

(a) If x = (2 / 3), y = (5 / −2) and z = (−4 / 13), find scalars p and q such that px + qy = z.
(b)(i) Using a scale of 2 cm to 2 units on both axes, draw on a graph paper two perpendicular axes Ox and Oy for −5 ≤ x ≤ 5, −5 ≤ y ≤ 5 respectively. (ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates, (I) the quadrilateral WXYZ with W(2, 3), X(4, −1), Y(−3, −4) and Z(−3, 2); (II) the image W1X1Y1Z1 of the quadrilateral WXYZ under an anticlockwise rotation of 90° about the origin where W→W1, X→X1, Y→Y1 and Z→Z1.

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(a) 2p + 5q = −4 and 3p − 2q = 13
Solving: 19p = 57 → p = 3, q = −2
(b)(i) Draw and label the axes with the given scale.
(ii)(I) Plot W(2, 3), X(4, −1), Y(−3, −4), Z(−3, 2) and join in order.
(II) Rotation of 90° anticlockwise: (x, y) → (−y, x)
W₁(−3, 2), X₁(1, 4), Y₁(4, −3), Z₁(−2, −3); plot and join.

Explanation

Column vectors are written (top / bottom).

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Theory question 113

Data Presentation

Marks: 10, 20, 30, 40, 50, 60, 70, 80, 90
Frequency: 1, 1, x, 5, y, 1, 4, 3, 1
The frequency table shows the marks distribution of a class of 30 students in an examination. The mean mark of the distribution is 52. (a) Find the values of x and y. (b) Construct a group frequency distribution table starting with a lower class limit of 1 and a class interval of 10. (c) Draw a histogram for the distribution. (d) Use the histogram to estimate the mode.

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Model answer

(a) 16 + x + y = 30 → x + y = 14
Σfx = 900 + 30x + 50y = 52 × 30 = 1560 → 3x + 5y = 66
Solving: y = 12, x = 2
(b) Classes 1–10, 11–20, …, 81–90 with frequencies 1, 1, 2, 5, 12, 1, 4, 3, 1 (class boundaries 0.5–10.5, …, 80.5–90.5)
(c) Bars on the class boundaries with heights equal to the frequencies.
(d) Modal class 40.5–50.5: draw the crossing lines in the tallest bar; mode ≈ 40.5 + 7/(7 + 11) × 10 ≈ 44.4

Explanation

In (d) join the top corners of the tallest bar to the neighbouring bars and read where the lines cross.

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