WAEC Mathematics 2008 past questions and answers

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

Question 1

Ratio, Percentages and Rates

If x% of 240 equals 12, find x.

  1. A.x = 1
  2. B.x = 3
  3. C.x = 5
  4. D.x = 7
Show the answer and explanation

Answer: C (x = 5)

Explanation

x/100 × 240 = 12, so 2.4x = 12 and x = 5.

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Question 2

Fractions, Decimals and Approximation

Evaluate ((3.2)2 − (4.8)2)/(3.2 + 4.8)

  1. A.−0.80
  2. B.−1.60
  3. C.−10.24
  4. D.−12.80
Show the answer and explanation

Answer: B (−1.60)

Explanation

Difference of two squares: (3.2 − 4.8)(3.2 + 4.8)/(3.2 + 4.8) = 3.2 − 4.8 = −1.60.

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Question 3

Fractions, Decimals and Approximation

Find the product of 0.0409 and 0.0021 leaving your answer in the standard form.

  1. A.8.6 × 10−6
  2. B.8.6 × 10−5
  3. C.8.6 × 104
  4. D.8.6 × 105
Show the answer and explanation

Answer: B (8.6 × 10−5)

Explanation

0.0409 × 0.0021 = 0.00008589 ≈ 8.6 × 10−5.

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Question 4

Surds

Simplify: √50 + 10/√2

  1. A.10
  2. B.10√2
  3. C.20
  4. D.20√2
Show the answer and explanation

Answer: B (10√2)

Explanation

√50 = 5√2 and 10/√2 = 10√2/2 = 5√2, so the sum is 10√2.

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Question 5

Commercial Arithmetic

P naira invested for 4 years at r % simple interest per annum yields 0.36 P naira interest. Find the value of r.

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

Answer: C (9)

Explanation

I = PRT/100: P × r × 4/100 = 0.36P, so 4r = 36 and r = 9.

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Question 6

Number Bases

Convert 425 to a base three numeral.

  1. A.2013
  2. B.2103
  3. C.2113
  4. D.2223
Show the answer and explanation

Answer: C (2113)

Explanation

425 = 4 × 5 + 2 = 22. 22 = 2 × 9 + 1 × 3 + 1, so it is 2113.

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Question 7

Commercial Arithmetic

A trader bought 100 tubers of yam at 5 for N350.00. She sold them in sets of 4 for N290.00. Find her gain per cent.

  1. A.3.6%
  2. B.3.5%
  3. C.3.4%
  4. D.2.5%
Show the answer and explanation

Answer: A (3.6%)

Explanation

Cost = 20 × N350 = N7,000. Sales = 25 × N290 = N7,250. Gain = N250, and 250/7000 × 100% ≈ 3.6%.

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Question 8

Linear and Simultaneous Equations

If p − 2g + 1 = g + 3p and p − 2 = 0, find g.

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

Answer: B (−1)

Explanation

p = 2, so 2 − 2g + 1 = g + 6. Then −3 = 3g and g = −1.

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Question 10

Indices

Simplify: ∛(27x3y9)

  1. A.9xy3
  2. B.3xy6
  3. C.3xy3
  4. D.9y3
Show the answer and explanation

Answer: C (3xy3)

Explanation

Take the cube root of each factor: ∛27 = 3, ∛x3 = x, ∛y9 = y3.

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Question 11

Algebraic Expressions

Given that x = 2 and y = −1/4, evaluate (x2y − 2xy)/5

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

Answer: A (0)

Explanation

x2y = 4 × (−1/4) = −1 and 2xy = 2 × 2 × (−1/4) = −1. So (−1 − (−1))/5 = 0.

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Question 13

Linear and Simultaneous Equations

If 4y is 9 greater than the sum of y and 3x, by how much is y greater than x?

  1. A.3
  2. B.6
  3. C.9
  4. D.12
Show the answer and explanation

Answer: A (3)

Explanation

4y = y + 3x + 9, so 3y − 3x = 9 and y − x = 3.

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Question 14

Algebraic Expressions

If p = (3/5)√(q/r), express q in terms of p and r.

  1. A.9/(25pr2)
  2. B.(9/25)p2r
  3. C.(25/9)p2r
  4. D.(25/9)pr2
Show the answer and explanation

Answer: C ((25/9)p2r)

Explanation

Square both sides: p2 = 9q/(25r), so q = 25p2r/9.

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Question 15

Algebraic Expressions

Simplify: (2x2 − 5x − 12)/(4x2 − 9)

  1. A.(x + 4)/(2x + 3)
  2. B.(x + 4)/(2x − 3)
  3. C.(x − 4)/(2x + 3)
  4. D.(x − 4)/(2x − 3)
Show the answer and explanation

Answer: D ((x − 4)/(2x − 3))

Explanation

2x2 − 5x − 12 = (2x + 3)(x − 4) and 4x2 − 9 = (2x − 3)(2x + 3). Cancel (2x + 3) to get (x − 4)/(2x − 3).

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Question 16

Triangles and Quadrilaterals

In the diagram, ∠QPR = 90°. If q2 = 25 − r2, find the value of p.

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

Answer: C (5)

Explanation

By Pythagoras, p2 = q2 + r2 = 25, so p = 5.

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Question 17

Mensuration of Plane Shapes

POR is a sector of a circle centre O, radius 4 cm. If PÔR = 30°, find, correct to 3 significant figures, the area of sector POR. [Take π = 22/7]

  1. A.4.19 cm2
  2. B.8.38 cm2
  3. C.10.5 cm2
  4. D.20.9 cm2
Show the answer and explanation

Answer: A (4.19 cm2)

Explanation

Area = 30/360 × 22/7 × 42 = 88/21 ≈ 4.19 cm2.

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Question 18

Mensuration of Solids

If the volume of a cube is 343 cm3, find the length of its side.

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

Answer: C (7 cm)

Explanation

Side = ∛343 = 7 cm.

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Question 19

Mensuration of Plane Shapes

In the diagram, PQRS is a rhombus. |PR| = 10 cm and |QS| = 24 cm. Calculate the perimeter of the rhombus.

Diagram for this question
  1. A.34 cm
  2. B.52 cm
  3. C.56 cm
  4. D.96 cm
Show the answer and explanation

Answer: B (52 cm)

Explanation

The diagonals bisect each other at right angles: side = √(52 + 122) = 13 cm. Perimeter = 4 × 13 = 52 cm.

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Question 20

Triangles and Quadrilaterals

In the diagram, |PQ| = |QR| and |PR| = |RS| = |SP|. Calculate the size of ∠QRS.

Diagram for this question
  1. A.150°
  2. B.120°
  3. C.90°
  4. D.60°
Show the answer and explanation

Answer: C (90°)

Explanation

ΔPRS is equilateral, so ∠PSR = ∠PRS = 60°. PQRS is cyclic, so ∠PQR = 180° − 60° = 120°, and the isosceles ΔPQR has ∠QRP = 30°. ∠QRS = 30° + 60° = 90°.

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Question 21

Angles and Polygons

In the figure shown, PQS is a straight line. What is the value of ∠PRQ?

Diagram for this question
  1. A.128°
  2. B.108°
  3. C.98°
  4. D.78°
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Answer: D (78°)

Explanation

The exterior angle RQS (148°) equals the sum of the two opposite interior angles: 70° + ∠PRQ = 148°, so ∠PRQ = 78°.

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Question 22

Angles and Polygons

In the diagram, PQ // RS, QU // PT and ∠PSR = 42°. Find angle x.

Diagram for this question
  1. A.84°
  2. B.48°
  3. C.42°
  4. D.32°
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Answer: C (42°)

Explanation

Because PQ ∥ RS, the angle PQ makes with PT equals 42° (corresponding angles). Because QU ∥ PT, x is the corresponding angle at Q, so x = 42°.

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Question 23

Triangles and Quadrilaterals

The angles of a quadrilateral are (x + 10)°, 2y°, 90° and (100 − y)°. Find y in terms of x.

  1. A.y = 160 + x
  2. B.y = 100 + x
  3. C.y = 160 − x
  4. D.y = x − 100
Show the answer and explanation

Answer: C (y = 160 − x)

Explanation

The angles add to 360°: x + 10 + 2y + 90 + 100 − y = 360, so x + y = 160 and y = 160 − x.

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Question 24

Mensuration of Plane Shapes

In the figure |PX| = |XQ|, PQ // YZ and XY // QR. What is the ratio of the area of XYZQ to the area of ΔYZR?

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

Explanation

X is the midpoint of PQ, so Y and Z are midpoints of PR and QR (midpoint theorem). The triangle splits into four equal triangles: XYZQ is two of them and YZR is one, so the ratio is 2 : 1.

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Question 25

Circle Theorems

In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.

Diagram for this question
  1. A.25°
  2. B.65°
  3. C.115°
  4. D.130°
Show the answer and explanation

Answer: B (65°)

Explanation

The 65° is an exterior angle at Q, so the interior ∠PQR = 180° − 65° = 115°. Opposite angles of a cyclic quadrilateral add to 180°, so x = 180° − 115° = 65°.

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Question 26

Angles and Polygons

If x° is obtuse, which of the following is true?

  1. A.x > 90
  2. B.180 < x < 270
  3. C.x < 90
  4. D.90 < x < 180
Show the answer and explanation

Answer: D (90 < x < 180)

Explanation

An obtuse angle is bigger than 90° but less than 180°.

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Question 27

Trigonometry

If tan x = 1, evaluate sin x + cos x, leaving your answer in the surd form.

  1. A.2√2
  2. B.½√2
  3. C.√2
  4. D.2
Show the answer and explanation

Answer: C (√2)

Explanation

tan x = 1 gives x = 45°, so sin x + cos x = 1/√2 + 1/√2 = 2/√2 = √2.

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Question 28

Trigonometry

If cos (x + 25)° = sin 45°, find the value of x.

  1. A.20
  2. B.30
  3. C.45
  4. D.60
Show the answer and explanation

Answer: A (20)

Explanation

sin 45° = cos 45°, so x + 25 = 45 and x = 20.

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Question 30

Indices

If 2n = 128, find the value of (2n−1)(5n−2).

  1. A.5(106)
  2. B.2(106)
  3. C.5(105)
  4. D.2(105)
Show the answer and explanation

Answer: D (2(105))

Explanation

27 = 128, so n = 7. Then 26 × 55 = 2 × (2 × 5)5 = 2(105).

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Question 31

Variation

If x ∝ (45 + ½y), which of the following is true?

  1. A.x varies directly as y
  2. B.x varies inversely as y
  3. C.x is partly constant and partly varies as y
  4. D.x varies jointly as 45 and directly as y
Show the answer and explanation

Answer: C (x is partly constant and partly varies as y)

Explanation

x = k(45 + ½y) = 45k + (k/2)y: a constant part plus a part that varies with y, i.e. partial variation.

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Question 32

Logarithms

Simplify: log√8/(log 4 − log 2)

  1. A.2/3
  2. B.½ log 2
  3. C.3/2
  4. D.log 2
Show the answer and explanation

Answer: C (3/2)

Explanation

log√8 = log 23/2 = (3/2) log 2 and log 4 − log 2 = log 2, so the value is 3/2.

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Question 33

Sets

Every staff in an office owns either a Mercedes and/or a Toyota car. 20 own Mercedes, 15 own Toyota and 5 own both. How many staff are there in the office?

  1. A.25
  2. B.30
  3. C.35
  4. D.45
Show the answer and explanation

Answer: B (30)

Explanation

n(M ∪ T) = 20 + 15 − 5 = 30.

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Question 34

Ratio, Percentages and Rates

A train travels 60 km in M minutes. If its average speed is 400 km per hour, find the value of M.

  1. A.15
  2. B.12
  3. C.10
  4. D.9
Show the answer and explanation

Answer: D (9)

Explanation

Time = 60/400 h = 0.15 h = 0.15 × 60 = 9 minutes.

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Question 35

Mensuration of Plane Shapes

An arc of a circle, radius 14 cm, is 18.33 cm long. Calculate, correct to the nearest degree, the angle which the arc subtends at the centre of the circle. [Take π = 22/7]

  1. A.11°
  2. B.20°
  3. C.22°
  4. D.75°
Show the answer and explanation

Answer: D (75°)

Explanation

θ/360 × 2 × 22/7 × 14 = 18.33, so θ = 18.33 × 360/88 ≈ 75°.

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Question 36

Mensuration of Solids

What is the length of an edge of a cube whose total surface area is X cm2 and whose volume is X/2 cm3?

  1. A.3
  2. B.6
  3. C.9
  4. D.12
Show the answer and explanation

Answer: A (3)

Explanation

With edge a: 6a2 = X and a3 = X/2 = 3a2, so a = 3.

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Question 37

Mensuration of Plane Shapes

XY is a chord of circle centre O and radius 7 cm. The chord XY which is 8 cm long subtends an angle of 120° at the centre of the circle. Calculate the perimeter of the minor segment. [Take π = 22/7]

  1. A.14.67 cm
  2. B.22.67 cm
  3. C.29.33 cm
  4. D.37.33 cm
Show the answer and explanation

Answer: B (22.67 cm)

Explanation

Minor arc = 120/360 × 2 × 22/7 × 7 = 14.67 cm. Perimeter of the segment = arc + chord = 14.67 + 8 = 22.67 cm.

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Question 38

Mensuration of Plane Shapes

If the perimeter of ΔPQR in the diagram is 24 cm, what is the area of ΔPRS?

Diagram for this question
  1. A.19.5 cm2
  2. B.15.0 cm2
  3. C.13.0 cm2
  4. D.9.3 cm2
Show the answer and explanation

Answer: B (15.0 cm2)

Explanation

|PR| = 24 − 6 − 8 = 10 cm. The height from S to PR is 3 cm, so area of ΔPRS = ½ × 10 × 3 = 15 cm2.

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Question 39

Linear and Simultaneous Equations

If p = ½ and 1/(p − 1) = 2/(p + x), find the value of x.

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

Answer: B (−1½)

Explanation

1/(½ − 1) = −2, so 2/(½ + x) = −2, giving ½ + x = −1 and x = −1½.

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Question 40

Quadratic Equations

Find the quadratic equation whose roots are c and −c.

  1. A.x2 + c2 = 0
  2. B.x2 − c2 = 0
  3. C.x2 + 2cx + c2 = 0
  4. D.x2 − 2cx + c2 = 0
Show the answer and explanation

Answer: B (x2 − c2 = 0)

Explanation

(x − c)(x + c) = 0 gives x2 − c2 = 0.

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Question 43

Inequalities

Which of the number lines below best represents the inequality −3 ≤ x < 4?

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

Answer: C (Diagram C)

Explanation

−3 is included (filled dot at −3) and 4 is not (open circle at 4), with the line joining them.

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Question 44

Circle Theorems

In the diagram, PX is a tangent to the circle and RST is an equilateral triangle. Calculate ∠PTS.

Diagram for this question
  1. A.60°
  2. B.90°
  3. C.120°
  4. D.150°
Show the answer and explanation

Answer: C (120°)

Explanation

By the alternate segment theorem, ∠STX = ∠TRS = 60°. P, T, X lie on a straight line, so ∠PTS = 180° − 60° = 120°.

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Question 45

Angles and Polygons

In the diagram, ∠QPR = 60°, ∠PQR = 50°, ∠QRS = 2x°, ∠SRP = 3x°, ∠UQP = y° and RS // TU. Calculate y.

Diagram for this question
  1. A.102°
  2. B.78°
  3. C.70°
  4. D.60°
Show the answer and explanation

Answer: A (102°)

Explanation

In ΔPQR, 5x = 180 − 60 − 50 = 70, so x = 14 and ∠QRS = 28°. RS ∥ TU gives ∠TQR = 28° (alternate). On the straight line UQT: y + 50 + 28 = 180, so y = 102°.

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Question 46

Triangles and Quadrilaterals

In the diagram, find the size of the angle marked a°.

Diagram for this question
  1. A.60°
  2. B.80°
  3. C.120°
  4. D.160°
Show the answer and explanation

Answer: B (80°)

Explanation

The reflex angle at O is 280°, so ∠ROT inside the shape is 80° and the angle at S on the circle is ½ × 280° = 140°. The four angles of RO TS add to 360°: 60 + 80 + 140 + a = 360, so a = 80°.

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Question 47

Probability

Given the sets A = {2, 4, 6, 8} and B = {2, 3, 5, 9}. If a number is selected at random from set B, what is the probability that the number is prime?

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

Answer: B (3/4)

Explanation

2, 3 and 5 are prime; 9 is not. Probability = 3/4.

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Question 48

Probability

Given the sets A = {2, 4, 6, 8} and B = {2, 3, 5, 9}. If a number is picked at random from each of the two sets, what is the probability that their product is odd?

  1. A.1
  2. B.3/4
  3. C.1/4
  4. D.0
Show the answer and explanation

Answer: D (0)

Explanation

Every number in A is even, and an even number times anything is even, so an odd product is impossible: probability 0.

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Question 49

Probability

Given the sets A = {2, 4, 6, 8} and B = {2, 3, 5, 9}. If a number is picked at random from each of the two sets, what is the probability that their difference is 6 or 7?

  1. A.1/256
  2. B.1/16
  3. C.1/8
  4. D.1/2
Show the answer and explanation

Answer: C (1/8)

Explanation

There are 4 × 4 = 16 pairs. Difference 6: (8, 2); difference 7: (2, 9). So 2/16 = 1/8.

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Question 50

Measures of Dispersion

Find the mean deviation of these numbers: 10, 12, 14, 15, 17, 19.

  1. A.2.5
  2. B.2.6
  3. C.2.7
  4. D.2.8
Show the answer and explanation

Answer: A (2.5)

Explanation

Mean = 87/6 = 14.5. Absolute deviations: 4.5, 2.5, 0.5, 0.5, 2.5, 4.5, total 15. Mean deviation = 15/6 = 2.5.

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Theory question 101

Logarithms

(a) Without using calculator or tables, find the value of log 3.6 given that log 2 = 0.3010, log 3 = 0.4771 and log 5 = 0.6990.
(b) If all numbers in the equation y/(y + 101) = 11/10010 are in base two, solve for y.

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Model answer

(a) 3.6 = 18/5 = (2 × 32)/5.
log 3.6 = log 2 + 2 log 3 − log 5 = 0.3010 + 0.9542 − 0.6990 = 0.5562.
(b) In base ten: 101two = 5, 11two = 3, 10010two = 18.
y/(y + 5) = 3/18 → 18y = 3y + 15 → 15y = 15 → y = 1.

Explanation

Write 3.6 as a product and quotient of 2, 3 and 5.

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Theory question 102

Inequalities

(a) Solve the inequality (2/5)(x − 2) − (1/6)(x + 5) ≤ 0
(b) Given that P = (x2 − y2)/(x2 + xy), (i) express P in its simplest form; (ii) find the value of P if x = −4 and y = −6.

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Model answer

(a) Multiply by 30: 12(x − 2) − 5(x + 5) ≤ 0 → 7x − 49 ≤ 0 → x ≤ 7.
(b)(i) P = (x − y)(x + y)/[x(x + y)] = (x − y)/x.
(ii) P = (−4 − (−6))/(−4) = 2/(−4) = −½.

Explanation

Clear the fractions with the LCM of 5 and 6.

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Theory question 103

Data Presentation

The table below shows how a man spends his income in a month.
Food: ₦4500; House rent: ₦3000; Provisions: ₦2500; Electricity: ₦2000; Transportation: ₦5000; Others: ₦3000
(a) Represent the information on a pie chart.
(b) What percentage of his income is spent on transportation?

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Model answer

Total = ₦20,000. Angle = amount/20,000 × 360°.
(a) Food 81°, House rent 54°, Provisions 45°, Electricity 36°, Transportation 90°, Others 54° (total 360°). Draw and label the sectors.
(b) 5000/20,000 × 100% = 25%.

Explanation

Each ₦1,000 is 18° of the pie.

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Theory question 104

Angles and Polygons

(a) A pentagon is such that one of its exterior angles is 60°. Two others are (90 − m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.
(b) In the diagram, PQR is a straight line, QR = √3 cm and SQ = 2 cm. Calculate, correct to one decimal place, ∠PQS.

Diagram for this question
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Model answer

(a) Exterior angles add to 360°: 60 + 2(90 − m) + 2(30 + 2m) = 360 → 300 + 2m = 360 → m = 30.
(b) In right-angled ΔSQR (right angle at R): cos ∠SQR = √3/2, so ∠SQR = 30°.
∠PQS = 180° − 30° = 150.0°.

Explanation

The exterior angles of any polygon add up to 360°.

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Theory question 105

Mensuration of Plane Shapes

(a) A rectangular field is l metres long and b metres wide. Its perimeter is 280 metres. If the length is two and a half times its breadth, find the values of l and b.
(b) The base of a pyramid is a 4.5 m by 2.5 m rectangle. The height of the pyramid is 4 m. Calculate its volume.

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Model answer

(a) 2(l + b) = 280 → l + b = 140, and l = 2.5b. So 3.5b = 140 → b = 40 m and l = 100 m.
(b) V = ⅓ × base area × height = ⅓ × 4.5 × 2.5 × 4 = 15 m3.

Explanation

Substitute l = 2.5b into the perimeter equation.

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Theory question 106

Indices

(a) If 2x+y = 16 and 4x−y = 1/32, find the values of x and y.
(b) P, Q and R are related in such a way that P ∝ Q2/R. When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.

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Model answer

(a) 2x+y = 24 → x + y = 4. 22(x−y) = 2−5 → x − y = −5/2.
Adding: 2x = 3/2 → x = ¾; y = 4 − ¾ = 3¼.
(b) P = kQ2/R: 36 = 9k/4 → k = 16.
200 = 16Q2/2 → Q2 = 25 → Q = 5 (or −5).

Explanation

Write every number as a power of 2 and equate the indices.

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Theory question 107

Quadratic Equations

(a) Solve, correct to two decimal places, the equation 4x2 = 11x + 21.
(b) A man invests £1500 for two years at compound interest. After one year, his money amounts to £1560. Find the: (i) rate of interest; (ii) interest for the second year.
(c) A car costs ₦300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.

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Model answer

(a) 4x2 − 11x − 21 = 0. x = [11 ± √(121 + 336)]/8 = (11 ± √457)/8 = (11 ± 21.378)/8.
x ≈ 4.05 or x ≈ −1.30.
(b)(i) Interest in year 1 = £60; rate = 60/1500 × 100% = 4%.
(ii) Compound interest: second-year interest = 4% of £1560 = £62.40.
(c) After year 1: 300,000 × 0.75 = ₦225,000. After year 2: 225,000 × 0.80 = ₦180,000.00.

Explanation

With compound interest the second year's interest is on the new amount, £1560.

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Theory question 108

Data Presentation

The ages, in years, of 50 teachers in a school are given below:
21 37 49 27 49 42 26 33 46 40
50 29 23 24 29 31 36 22 27 38
30 28 42 39 34 23 21 32 41 46
46 31 33 29 28 43 47 40 34 44
26 38 34 49 45 27 25 33 39 40
(a) Form a frequency distribution of the data using the intervals: 21 − 25, 26 − 30, 31 − 35, etc.
(b) Draw the histogram of the distribution.
(c) Use your histogram to estimate the mode.
(d) Calculate the mean age.

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Model answer

(a) 21–25: 7; 26–30: 11; 31–35: 9; 36–40: 9; 41–45: 6; 46–50: 8 (total 50).
(b) Bars on class boundaries 20.5, 25.5, …, 50.5 with heights equal to the frequencies.
(c) Modal class 26–30. From the histogram (joining the corners of the tallest bar to the neighbouring bars): mode ≈ 25.5 + 5 × 4/(4 + 2) ≈ 28.8 years.
(d) Class marks 23, 28, 33, 38, 43, 48: Σfx = 161 + 308 + 297 + 342 + 258 + 384 = 1750. Mean = 1750/50 = 35 years.

Explanation

Tally each age into its class first and check the frequencies add to 50.

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Theory question 109

Trigonometry

(a) The triangle ABC has sides AB = 17 m, BC = 12 m and AC = 10 m. Calculate the: (i) largest angle of the triangle; (ii) area of the triangle.
(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38 m high, is 63°. Calculate, correct to the nearest metre, the distance between T and S.

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(a)(i) The largest angle is C, opposite AB. cos C = (122 + 102 − 172)/(2 × 12 × 10) = −45/240 = −0.1875, so C ≈ 100.8°.
(ii) Area = ½ × 12 × 10 × sin 100.8° ≈ 58.9 m2.
(b) tan 63° = 38/|TS|, so |TS| = 38/1.9626 ≈ 19.4 ≈ 19 m.

Explanation

The largest angle is opposite the longest side; a negative cosine means it is obtuse.

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Theory question 110

Constructions

(a) Using ruler and a pair of compasses only, construct: (i) a quadrilateral PQRS such that |PQ| = 7 cm, ∠QPS = 60°, |PS| = 6.5 cm, ∠PQR = 135° and |QS| = |QR|; (ii) locus l1 of points equidistant from P and Q; (iii) locus l2 of points equidistant from P and S.
(b)(i) Label the point T where l1 and l2 intersect. (ii) With centre T and radius |TP|, construct a circle l3.

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Model answer

(a)(i) Draw PQ = 7 cm; construct 60° at P and mark PS = 6.5 cm. Construct 135° at Q (90° + 45°). With centre Q and radius QS, cut the 135° arm at R. Join SR.
(ii) l1: perpendicular bisector of PQ. (iii) l2: perpendicular bisector of PS.
(b)(i) T is where l1 and l2 meet.
(ii) The circle centre T, radius TP, passes through P, Q and S (radius ≈ 3.9 cm).

Explanation

T is equidistant from P, Q and S, so the circle passes through all three.

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Theory question 111

Angles and Polygons

(a) In the diagram, AB//CD and BC//FE. CD̂E = 75° and DÊF = 26°. Find the angles marked x and y.
(b) The diagram shows a circle ABCD with centre O and radius 7 cm. The reflex angle AOC = 190° and angle DAO = 35°. Find: (i) ∠ABC; (ii) ∠ADC.
(c) Using the diagram in 11(b) above, calculate, correct to 3 significant figures, the length of: (i) arc ABC; (ii) the chord AD. [Take π = 3.142]

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(a) Through D draw a line parallel to BC and FE. Its angle with DE = 26° (alternate), so its angle with DC = 75° − 26° = 49°. Hence ∠BCD = 49° (alternate) and x = ∠ABC = ∠BCD = 49° (AB ∥ CD).
y = 360° − 49° = 311°.
(b) ∠AOC (non-reflex) = 360° − 190° = 170°.
(i) ∠ABC = ½ × 190° = 95°. (ii) ∠ADC = ½ × 170° = 85°.
(c)(i) Arc ABC = 170/360 × 2 × 3.142 × 7 ≈ 20.8 cm.
(ii) ΔAOD is isosceles with ∠OAD = ∠ODA = 35°, so ∠AOD = 110°. AD = 2 × 7 × sin 55° ≈ 11.5 cm.

Explanation

An angle at the circumference is half the angle at the centre on the same arc.

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Theory question 112

Trigonometry

(a) Copy and complete the table of values for y = 3 sin x + 2 cos x for 0° ≤ x ≤ 360°.
x: 0°, 60°, 120°, 180°, 240°, 300°, 360°
y: 2.00, –, –, –, –, –, 2.00
(b) Using a scale of 2 cm to 60° on x-axis and 2 cm to 1 unit on y-axis, draw the graph of y = 3 sin x + 2 cos x for 0° ≤ x ≤ 360°.
(c) Use your graph to solve the equation: 3 sin x + 2 cos x = 1.5.
(d) Find the range of values of x for which 3 sin x + 2 cos x < −1.

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(a) 0°: 2.00; 60°: 3.60; 120°: 1.60; 180°: −2.00; 240°: −3.60; 300°: −1.60; 360°: 2.00.
(b) Plot the points and join with a smooth curve.
(c) Where y = 1.5 meets the curve: x ≈ 122° and x ≈ 351°.
(d) The curve is below y = −1 for about 162° < x < 310°.
(Accept graph readings within a few degrees.)

Explanation

For example at 60°: 3(0.866) + 2(0.5) = 3.60.

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Theory question 113

Progressions

(a) If 3, x, y, 18 are in Arithmetic Progression (A.P), find the values of x and y.
(b)(i) The sum of the second and third terms of a geometric progression is six times the fourth term. Find the two possible values of the common ratio. (ii) If the second term is 8 and the common ratio is positive, find the first six terms.

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(a) 18 = 3 + 3d → d = 5, so x = 8 and y = 13.
(b)(i) ar + ar2 = 6ar3 → 6r2 − r − 1 = 0 → (3r + 1)(2r − 1) = 0 → r = ½ or r = −⅓.
(ii) r = ½ and ar = 8, so a = 16. First six terms: 16, 8, 4, 2, 1, ½.

Explanation

Divide the GP equation by ar (not zero) to get a quadratic in r.

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Theory question 114

Graphs of Functions

(Questions 14 and 15 are for candidates in Ghana, Sierra Leone and The Gambia only.)
(a) The function f is defined as f: x → 3x2 − 5x. (i) Evaluate f(−2). (ii) Find the values of x for which f(x) = −7/4.
(b) A rhombus ABCD has A at (0, 2) and B at (5, 3). If the diagonals intersect at (0, 3), find the coordinates of C and D.

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(a)(i) f(−2) = 3(4) + 10 = 22.
(ii) 3x2 − 5x = −7/4 → 12x2 − 20x + 7 = 0 → (6x − 7)(2x − 1) = 0 → x = 7/6 or x = ½.
(b) The diagonals bisect each other at M(0, 3). C = 2M − A = (0, 4); D = 2M − B = (−5, 3).

Explanation

The point where the diagonals meet is the midpoint of AC and of BD.

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Theory question 115

Vectors and Transformations

(a) Using a scale of 2 cm to 2 units on both axes, draw two perpendicular axes OX and OY for the interval −8 ≤ x ≤ 8 and −10 ≤ y ≤ 10.
(b) Draw, clearly labelling all the vertices and coordinates, (i) triangle PQR with P(−3, 6), Q(−1, 1) and R(4, 2); (ii) ΔP1Q1R1, with P1(5, 6), Q1(3, 1) and R1(−2, 2); (iii) ΔP2Q2R2, the image of ΔPQR under a rotation through 180° about the origin where P → P2, Q → Q2 and R → R2; (iv) the point Q3, the image of Q under a translation by the vector (−4, 3).
(c) What single transformation maps ΔPQR onto ΔP1Q1R1?

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Model answer

(a), (b)(i), (ii) Axes to scale; plot and label PQR and P1Q1R1.
(iii) (x, y) → (−x, −y): P2(3, −6), Q2(1, −1), R2(−4, −2).
(iv) Q3 = (−1 − 4, 1 + 3) = (−5, 4).
(c) Each point keeps its y-coordinate and the midpoints of PP1, QQ1, RR1 all have x = 1: a reflection in the line x = 1.

Explanation

Check the midpoint of each point and its image to find the mirror line.

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