WAEC Mathematics 1999 past questions and answers

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

Question 1

Surds

Evaluate √20 × (√5)3

  1. A.10
  2. B.20
  3. C.25
  4. D.50
Show the answer and explanation

Answer: D (50)

Explanation

√20 = 2√5 and (√5)3 = 5√5, so the product is 2 × 5 × √5 × √5 = 10 × 5 = 50.

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Question 2

Sets

If P = {3, 7, 11, 13} and Q = {2, 4, 8, 16}, which of the following is correct?

  1. A.(P∩Q)' = {2, 3, 4, 13}
  2. B.n(P∪Q) = 4
  3. C.P∪Q = ϕ
  4. D.P∩Q = ϕ
Show the answer and explanation

Answer: D (P∩Q = ϕ)

Explanation

P and Q have no element in common, so P∩Q is the empty set.

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Question 3

Progressions

It is observed that 1 + 3 = 22, 1 + 3 + 5 = 32, 1 + 3 + 5 + 7 = 42. If 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = p2, find P.

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

Answer: C (8)

Explanation

The sum of the first n odd numbers is n2. There are 8 odd numbers here, so p = 8.

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Question 4

Fractions, Decimals and Approximation

Evaluate 3.0 × 101 − 2.8 × 10−1 leaving the answer in standard form.

  1. A.2 × 10−1
  2. B.2 × 102
  3. C.2.972 × 101
  4. D.2.972 × 102
Show the answer and explanation

Answer: C (2.972 × 101)

Explanation

30 − 0.28 = 29.72 = 2.972 × 101.

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Question 5

Commercial Arithmetic

A trader makes a loss of 15% when selling an article. Find the ratio, Selling Price: Cost Price.

  1. A.3 : 20
  2. B.3:17
  3. C.17:20
  4. D.20:23.
Show the answer and explanation

Answer: C (17:20)

Explanation

With a 15% loss the selling price is 85% of the cost price: 85 : 100 = 17 : 20.

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Question 6

Ratio, Percentages and Rates

The side of a square is increased from 20cm to 21cm. Calculate the percentage increase in its area

  1. A.2.5%
  2. B.9.3%
  3. C.10.0%
  4. D.10.25%.
Show the answer and explanation

Answer: D (10.25%.)

Explanation

Area goes from 400 to 441 cm2, an increase of 41. 41/400 × 100% = 10.25%.

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Question 7

Number Bases

Arrange in ascending order of magnitude: 268, 367 and 259.

  1. A.259, 268, 367
  2. B.268, 259, 367
  3. C.367, 268, 259
  4. D.367, 259, 268.
Show the answer and explanation

Answer: B (268, 259, 367)

Explanation

In base ten: 268 = 22, 259 = 23, 367 = 27, so the order is 268, 259, 367.

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Question 8

Fractions, Decimals and Approximation

The side of a rhombus is 10cm long, correct to the nearest whole number. Between what limits should the perimeter P lie?

  1. A.39cm ≤ P ≤ 41cm
  2. B.38cm ≤ P < 42cm
  3. C.38.5 ≤ P < 41.5cm
  4. D.38.5cm ≤ P ≤ 40cm.
Show the answer and explanation

Answer: B (38cm ≤ P < 42cm)

Explanation

The side lies from 9.5 cm up to (but not including) 10.5 cm. Multiplying by 4: 38 ≤ P < 42.

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Question 9

Logarithms

Simplify log78 − log72 + log74

  1. A.0
  2. B.2 log72
  3. C.3 log72
  4. D.4 log72.
Show the answer and explanation

Answer: D (4 log72.)

Explanation

log7(8 ÷ 2 × 4) = log716 = log724 = 4 log72.

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Question 10

Surds

If K√28 + √63 − √7 = 0, find K.

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

Answer: B (−1)

Explanation

√28 = 2√7 and √63 = 3√7, so 2K√7 + 3√7 − √7 = 0, giving 2K = −2 and K = −1.

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Question 11

Probability

From the set A = {3, √2, 2√3, √9, √7}, a number is selected at random. Find the probability that it is a rational number.

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

Answer: B (2/5)

Explanation

Only 3 and √9 = 3 are rational; the other three are irrational. Probability = 2/5.

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Question 13

Data Presentation

The pie chart illustrates the amount of private time a student spends in a week studying various subjects. If he spends 2½ hours a week on Science, find the total number of hours he studies in a week.

Diagram for this question
  1. A.3⅓ hours
  2. B.5 hours
  3. C.8 hours
  4. D.12 hours.
Show the answer and explanation

Answer: D (12 hours.)

Explanation

Science has 75° of the 360°. Total = 2½ × 360/75 = 12 hours.

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Question 14

Probability

A group of eleven people can speak either English or French or both. Seven can speak English and six can speak French. What is the probability that a person chosen at random can speak both English and French?

  1. A.2/11
  2. B.4/11
  3. C.5/11
  4. D.11/13
Show the answer and explanation

Answer: A (2/11)

Explanation

Number who speak both = 7 + 6 − 11 = 2, so the probability is 2/11.

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Question 15

Angles and Polygons

The interior angle of a regular polygon is twice the exterior angle. How many sides has the polygon?

  1. A.5
  2. B.6
  3. C.8
  4. D.9.
Show the answer and explanation

Answer: B (6)

Explanation

Interior + exterior = 180° and interior = 2 × exterior, so the exterior angle is 60°. Sides = 360 ÷ 60 = 6.

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Question 16

Triangles and Quadrilaterals

Which of the following figures have one line of symmetry only? I. Isosceles Triangle II. Rhombus III. Kite

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

Answer: C (I and III only)

Explanation

An isosceles triangle and a kite each have one line of symmetry; a rhombus has two (its diagonals).

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Question 17

Triangles and Quadrilaterals

In the diagram below, |XR| = |RY| = |YZ| and ∠XRY = ∠YRZ = 62°. Calculate ∠XYZ.

Diagram for this question
  1. A.50°
  2. B.62°
  3. C.112°
  4. D.115°
Show the answer and explanation

Answer: D (115°)

Explanation

Triangle XRY is isosceles with apex 62°, so ∠RYX = 59°. Triangle RYZ has RY = YZ, so ∠YZR = 62° and ∠RYZ = 56°. ∠XYZ = 59 + 56 = 115°.

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Question 18

Circle Theorems

In the diagram, RST is a tangent to circle VSU centre O. ∠SVU = 50° and UV is a diameter. Calculate ∠RSV

Diagram for this question
  1. A.90°
  2. B.50°
  3. C.45°
  4. D.40°.
Show the answer and explanation

Answer: D (40°.)

Explanation

∠VSU = 90° (angle in a semicircle), so ∠SUV = 40°. The angle between tangent and chord, ∠RSV, equals ∠SUV in the alternate segment: 40°.

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Question 19

Triangles and Quadrilaterals

Calculate the value of y from the diagram

Diagram for this question
  1. A.18°
  2. B.42°
  3. C.48°
  4. D.132°.
Show the answer and explanation

Answer: C (48°)

Explanation

Through the vertex of y draw a line parallel to the two given lines. One arm of y makes 180 − 105 = 75° with it and the other makes 180 − 123 = 57° with it on the other side, so y = 180 − 75 − 57 = 48°.

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Question 20

Circle Theorems

In the diagram, PQR is a circle with centre O. ∠QPR = 50°, ∠PQO = 30° and ∠ORP = m. Find m.

Diagram for this question
  1. A.20°
  2. B.25°
  3. C.30°
  4. D.50°.
Show the answer and explanation

Answer: A (20°)

Explanation

OP = OQ, so ∠OPQ = 30° and ∠OPR = 50 − 30 = 20°. OP = OR, so ∠ORP = 20°.

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Question 21

Algebraic Expressions

Given that m = −3 and n = 2, find the value of (3n2 − 2m3)/m

  1. A.−22
  2. B.−14
  3. C.14
  4. D.22.
Show the answer and explanation

Answer: A (−22)

Explanation

3n2 = 12 and 2m3 = −54, so the numerator is 12 + 54 = 66. 66 ÷ (−3) = −22.

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Question 22

Algebraic Expressions

Simplify (2 − 18m2)/(1 + 3m)

  1. A.2(1 + 3m)
  2. B.2(1 + 3m2)
  3. C.2(1 − 3m)
  4. D.2(1 − 3m2).
Show the answer and explanation

Answer: C (2(1 − 3m))

Explanation

2 − 18m2 = 2(1 − 9m2) = 2(1 − 3m)(1 + 3m); cancel (1 + 3m) to get 2(1 − 3m).

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Question 23

Algebraic Expressions

If y = √(ax − b) express x in terms of y, a and b.

  1. A.x = (y2 − b)/a
  2. B.x = (y + b)/a
  3. C.x = (y − b)/a
  4. D.x = (y2 + b)/a
Show the answer and explanation

Answer: D (x = (y2 + b)/a)

Explanation

Square both sides: y2 = ax − b, so ax = y2 + b and x = (y2 + b)/a.

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Question 25

Algebraic Expressions

Simplify 4/(x + 1) − 3/(x − 1)

  1. A.(x + 7)/(x2 − 1)
  2. B.(x − 7)/(x2 + 1)
  3. C.(x − 7)/(x2 − 1)
  4. D.(x − 11)/(x2 − 1)
Show the answer and explanation

Answer: C ((x − 7)/(x2 − 1))

Explanation

Common denominator (x + 1)(x − 1) = x2 − 1. Numerator: 4(x − 1) − 3(x + 1) = x − 7.

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Question 26

Variation

If y varies inversely as x2, how does x vary with y?

  1. A.x varies inversely as y2
  2. B.x varies inversely as √y
  3. C.x varies directly as y2
  4. D.x varies directly as √y
Show the answer and explanation

Answer: B (x varies inversely as √y)

Explanation

y = k/x2 gives x2 = k/y, so x = √k/√y: x varies inversely as √y.

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Question 27

Trigonometry

If tan x = 1/√3, find cos x − sin x such that 0° ≤ x ≤ 90°

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

Answer: C ((√3 − 1)/2)

Explanation

tan x = 1/√3 means x = 30°. cos 30° − sin 30° = √3/2 − 1/2 = (√3 − 1)/2.

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Question 28

Trigonometry

From the top of a cliff 20m high, a boat can be sighted at sea 75m from the foot of the cliff. Calculate the angle of depression of the boat from the top of the cliff.

  1. A.14.9°
  2. B.15.5°
  3. C.74.5°
  4. D.75.1°.
Show the answer and explanation

Answer: A (14.9°)

Explanation

tan θ = 20/75 ≈ 0.2667, so θ ≈ 14.9°.

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Question 29

Circle Theorems

From the diagram below. ABC is a triangle inscribed in a circle centre O. ∠ACB = 40° and |AB| = x cm. Calculate the radius of the circle.

Diagram for this question
  1. A.x/sin 40°
  2. B.x/cos 40°
  3. C.x/(2 sin 40°)
  4. D.x/(2 cos 40°)
Show the answer and explanation

Answer: C (x/(2 sin 40°))

Explanation

By the sine rule, AB/sin C = 2R (the diameter). So 2R = x/sin 40° and R = x/(2 sin 40°).

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Question 30

Bearings

The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040°. What is the bearing of A from C?

Diagram for this question
  1. A.040°
  2. B.070°
  3. C.110°
  4. D.290°
Show the answer and explanation

Answer: D (290°)

Explanation

∠ABC = 40° (from the north–south line at B), so the equal base angles are 70°. A is 70° west of north from C: bearing 360 − 70 = 290°.

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Question 32

Mensuration of Plane Shapes

In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST.

Diagram for this question
  1. A.60cm2
  2. B.50cm2
  3. C.40cm2
  4. D.30cm2
Show the answer and explanation

Answer: D (30cm2)

Explanation

PQ = QR = TS = 4 cm, so PR = 8 cm; the height QT is 5 cm. Area = ½(8 + 4) × 5 = 30 cm2.

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Question 33

Mensuration of Solids

The height of a pyramid on a square base is 15cm. If the volume is 80cm3, find the length of the side of the base

  1. A.3.3cm
  2. B.5.3cm
  3. C.4.0cm
  4. D.8.0cm.
Show the answer and explanation

Answer: C (4.0cm)

Explanation

V = ⅓ × s2 × 15 = 5s2 = 80, so s2 = 16 and s = 4 cm.

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Question 34

Fractions, Decimals and Approximation

Simplify 3.72 × 0.025 and express your answer in the standard form.

  1. A.9.3 × 103
  2. B.9.3 × 102
  3. C.9.3 × 10−2
  4. D.9.3 × 10−3
Show the answer and explanation

Answer: C (9.3 × 10−2)

Explanation

3.72 × 0.025 = 0.093 = 9.3 × 10−2.

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Question 35

Number Bases

Two numbers 24 and 31 are equal in value when converted to base ten. Find the equation connecting x and y.

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

Answer: A (2x = 3(y − 1))

Explanation

24x = 2x + 4 and 31y = 3y + 1. Setting them equal: 2x = 3y − 3 = 3(y − 1).

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Question 36

Linear and Simultaneous Equations

A car travels at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed.

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

Answer: C ((x + 2y)/3 km h−1)

Explanation

Total distance = x + 2y km in 3 hours, so average speed = (x + 2y)/3 km/h.

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Question 37

Mensuration of Plane Shapes

The height and base of a triangle are in ratio 1 : 3 respectively. If the area of the triangle is 216cm2, find the length of the base.

  1. A.24cm
  2. B.36cm
  3. C.72cm
  4. D.144cm.
Show the answer and explanation

Answer: B (36cm)

Explanation

Let height = h, base = 3h. ½ × 3h × h = 216, so h2 = 144, h = 12 and the base is 36 cm.

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Question 38

Mensuration of Solids

Water flows from a tap into a cylindrical container at the rate of 5π cm3 per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.

  1. A.2cm
  2. B.5cm
  3. C.8cm
  4. D.15cm.
Show the answer and explanation

Answer: B (5cm)

Explanation

Volume after 9 s = 45π cm3 = π × 32 × h, so h = 45/9 = 5 cm.

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Question 39

Inequalities

Amina had m mangoes. She ate 3 and shared the remainder equally with her brother Uche. Each had at least 10. Which of the following inequalities represents the statements above.

  1. A.m/2 − 3 ≤ 10
  2. B.m/2 − 3 ≥ 10
  3. C.(m − 3)/2 ≥ 10
  4. D.(m − 3)/2 ≤ 10.
Show the answer and explanation

Answer: C ((m − 3)/2 ≥ 10)

Explanation

After eating 3 she has m − 3, and each share is (m − 3)/2. 'At least 10' means ≥ 10.

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Question 40

Inequalities

Which of the following pairs of inequalities is represented on the number line?

Diagram for this question
  1. A.x < −2 and x ≥ 1
  2. B.x ≤ −2 and x > 1
  3. C.x ≤ −2 and x ≥ 1
  4. D.x < −2 and x > 1.
Show the answer and explanation

Answer: C (x ≤ −2 and x ≥ 1)

Explanation

Both end points are shaded dots, so −2 and 1 are included: x ≤ −2 and x ≥ 1.

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Question 41

Quadratic Equations

If x2 + 15x + 50 = ax2 + bx + c = 0, which of the following statements is not true?

  1. A.x = −5
  2. B.x = 10
  3. C.x + 10 = 0
  4. D.bc = 750.
Show the answer and explanation

Answer: B (x = 10)

Explanation

x2 + 15x + 50 = (x + 5)(x + 10), so the roots are −5 and −10, not 10. Also b = 15 and c = 50, so bc = 750.

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Question 42

Quadratic Equations

The roots of a quadratic equation in x, are −m and 2n. Find the equation.

  1. A.x2 + x(m − 2n) − 2mn = 0
  2. B.x2 − x(m − 2n) − 2mn = 0
  3. C.x2 − x(2n − m) + 2mn = 0
  4. D.x2 + x(m − 2n) + 2mn = 0.
Show the answer and explanation

Answer: A (x2 + x(m − 2n) − 2mn = 0)

Explanation

(x + m)(x − 2n) = x2 + (m − 2n)x − 2mn = 0.

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Question 43

Probability

A survey shows that 28% of all the men in a village wear size 9 shoes. What is the probability that a man selected at random in the village wears size 9 shoes?

  1. A.7/25
  2. B.41/50
  3. C.1/2
  4. D.1.
Show the answer and explanation

Answer: A (7/25)

Explanation

28% = 28/100 = 7/25.

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Question 44

Inequalities

Solve the inequality 2x + 3 < 5x.

  1. A.x > 1
  2. B.x < 3/7
  3. C.x > 3/7
  4. D.x > −1.
Show the answer and explanation

Answer: A (x > 1)

Explanation

Subtract 2x: 3 < 3x, so x > 1.

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Question 45

Triangles and Quadrilaterals

In the diagram, DE||BC, |AD| = x cm and |DB| = |AE| = y cm. Find |CE| in terms of x and y.

Diagram for this question
  1. A.x
  2. B.x2/y
  3. C.y/x2
  4. D.y2/x
Show the answer and explanation

Answer: D (y2/x)

Explanation

A line parallel to BC divides the sides in the same ratio: AD/DB = AE/EC, so x/y = y/CE and CE = y2/x.

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Question 46

Loci

Describe the locus L shown in the diagram below.

Diagram for this question
  1. A.Locus of points equidistant from X and Y.
  2. B.Locus of points equidistant from X and Z
  3. C.Locus of points equidistant from ZX and ZY
  4. D.Locus of points equidistant from XY and ZY
Show the answer and explanation

Answer: C (Locus of points equidistant from ZX and ZY)

Explanation

L is the bisector of angle XZY, and every point on an angle bisector is the same distance from the two arms ZX and ZY.

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Question 47

Triangles and Quadrilaterals

Which of the following statements is true from the diagram below?

Diagram for this question
  1. A.m + n = 90°
  2. B.m = n
  3. C.n > 45°
  4. D.m < 45°.
Show the answer and explanation

Answer: A (m + n = 90°)

Explanation

The co-interior angles between the parallel lines add to 180°: 2m + 2n = 180°, so m + n = 90°.

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Question 48

Circle Theorems

In the diagram, O is the centre of the circle. ∠PQR = 75°, ∠OPS = y° and OR is parallel to PS. Find y.

Diagram for this question
  1. A.15
  2. B.30
  3. C.45
  4. D.75.
Show the answer and explanation

Answer: B (30)

Explanation

∠POR = 2 × 75° = 150° (angle at centre). OR ∥ PS, so ∠OPS and ∠POR are co-interior: y = 180 − 150 = 30.

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Question 49

Triangles and Quadrilaterals

The sides of a right-angle triangle, in ascending order of magnitude are 8cm, (x − 2) cm and x cm. Find x.

  1. A.16
  2. B.17
  3. C.34
  4. D.90.
Show the answer and explanation

Answer: B (17)

Explanation

By Pythagoras, 82 + (x − 2)2 = x2: 64 + x2 − 4x + 4 = x2, so 4x = 68 and x = 17.

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Question 50

Circle Theorems

In the diagram, CDE is a triangle, ABEF is cyclic quadrilateral, AB || CD, ∠BAF = 65° and ∠AFE = 85°. Calculate ∠CDE

Diagram for this question
  1. A.20°
  2. B.30°
  3. C.50°
  4. D.65°
Show the answer and explanation

Answer: B (30°)

Explanation

Opposite angles of ABEF: ∠ABE = 95° and ∠BEF = 115°, so ∠CED = 65° and ∠ABC = 85°. AB ∥ CD gives ∠BCD = 85°, so ∠CDE = 180 − 85 − 65 = 30°.

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Theory question 101

Logarithms

(b) Evaluate log5(3/5) + 3log5(15/2) − log5(81/8)

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Model answer

Combine: log5[3/5 × (15/2)3 ÷ 81/8] = log5[3/5 × 3375/8 × 8/81].
3 × 3375/(5 × 81) = 10125/405 = 25.
log525 = log552 = 2.

Explanation

Turn the coefficient 3 into a power first, then combine into a single log.

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Theory question 102

Sets

(a) If a is the set {1, 2, 3, ……, 19, 20} and A, B and C are subsets of a such that A = {multiples of five}, B = {multiples of four}, C = {multiples of three}, list the elements of (i) A (ii) B (iii) C
(b) Find; (i) A∩B (ii) A∩C (iii) B∪C
(c) Using your results in (b) show that (A∩B) ∪ (A∩C) = A∩(B∪C).

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Model answer

(a)(i) A = {5, 10, 15, 20} (ii) B = {4, 8, 12, 16, 20} (iii) C = {3, 6, 9, 12, 15, 18}.
(b)(i) A∩B = {20} (ii) A∩C = {15} (iii) B∪C = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}.
(c) (A∩B) ∪ (A∩C) = {20} ∪ {15} = {15, 20}.
A∩(B∪C): the elements of A that are in B∪C are 15 and 20, so A∩(B∪C) = {15, 20}. The two sets are equal.

Explanation

Work out both sides separately and show they contain the same elements.

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Theory question 103

Trigonometry

ABC is a triangle right-angled at C. P is the mid-point of AC, ∠PBC = 37° and |BC| = 5cm. Calculate: (a) |AC| correct to 3 significant figures (b) ∠PBA.

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Model answer

(a) In triangle PBC: PC = 5 tan 37° = 5 × 0.7536 = 3.768 cm.
AC = 2 × PC = 7.536 ≈ 7.54 cm.
(b) tan ∠ABC = AC/BC = 7.536/5 = 1.507, so ∠ABC ≈ 56.4°.
∠PBA = ∠ABC − ∠PBC = 56.4 − 37 = 19.4°.

Explanation

∠PBA is the part of angle B between BA and BP, so subtract 37° from the whole angle at B.

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Theory question 104

Mensuration of Plane Shapes

In the diagram, ABCD is a trapezium in which AD//BC and ∠ABC is a right angle. If |AD| = 15cm, |BD| = 17cm and |BC| = 9cm, calculate (a) |AB|. (b) the area of the triangle BCD. (c) |CD|. (d) the perimeter of the trapezium.

Diagram for this question
Show the model answer
Model answer

(a) ∠BAD = 90° (AD ∥ BC), so AB = √(172 − 152) = √64 = 8 cm.
(b) Area BCD = ½ × BC × AB = ½ × 9 × 8 = 36 cm2.
(c) Drop a perpendicular from C to AD: horizontal gap = 15 − 9 = 6 cm, height 8 cm, so CD = √(62 + 82) = 10 cm.
(d) Perimeter = 15 + 8 + 9 + 10 = 42 cm.

Explanation

The height of triangle BCD on base BC is the trapezium's height AB.

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Theory question 105

Linear and Simultaneous Equations

(a) Solve the simultaneous equations 3y − 2x = 21; 4y + 5x = 5.
(b) Six identical cards numbered 1–6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability that he: (i) loses (ii) loses $20.00.

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(a) Multiply the first by 5 and the second by 2: 15y − 10x = 105 and 8y + 10x = 10. Add: 23y = 115, y = 5.
Then 2x = 3(5) − 21 = −6, x = −3.
(b)(i) He loses with any card from 1 to 5: P = 5/6.
(ii) He loses $20 only with card 2: P = 1/6.

Explanation

Make the x coefficients equal and opposite, then add.

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Theory question 106

Measures of Central Tendency

The table gives the frequency distribution of marks obtained by a group of students in a test.
Marks: 3, 4, 5, 6, 7, 8
Frequency: 5, x−1, x, 9, 4, 1
If the mean mark is 5, (a) Calculate the value of x. (b) find the: (i) mode: (ii) median of the distribution (c) If one of the students is selected at random, find the probability that he scored at least 7 marks

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(a) Σf = 2x + 18; Σfx = 15 + 4(x − 1) + 5x + 54 + 28 + 8 = 9x + 101.
(9x + 101)/(2x + 18) = 5 gives 9x + 101 = 10x + 90, so x = 11.
(b) Frequencies are 5, 10, 11, 9, 4, 1 (total 40).
(i) Mode = 5 (highest frequency, 11).
(ii) The 20th and 21st marks are both 5 (cumulative 5, 15, 26…), so median = 5.
(c) At least 7 marks: 4 + 1 = 5 students. P = 5/40 = 1/8.

Explanation

Set up mean = Σfx ÷ Σf with x still unknown, then solve.

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Theory question 107

Mensuration of Solids

(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5cm and external radius 5cm, calculate: (i) the volume, in cm3, of metal used in making the pipe; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take π = 22/7].
(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.

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(a) Length = 28 m = 2800 cm.
(i) Metal = π(R2 − r2)h = 22/7 × (25 − 12.25) × 2800 = 112,200 cm3.
(ii) Water = πr2h = 22/7 × 12.25 × 2800 = 107,800 cm3 = 107.8 litres.
(b) ∠PMN = ∠MLN (angle between tangent and chord equals the angle in the alternate segment).
LN ∥ MP, so ∠PMN = ∠MNL (alternate angles).
Hence ∠MLN = ∠MNL, so ML = MN and triangle LMN is isosceles.

Explanation

Convert the length to centimetres before multiplying; 1000 cm³ = 1 litre.

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Theory question 108

Trigonometry

(a) Given that sin (A + B) = sin A cos B + cos A sin B, without using mathematical tables or calculator, evaluate sin 105°, leaving your answer in the surd form. (You may use 105° = 60° + 45°).
(b) The houses on the side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?
(c) The 1st and 3rd term of a Geometric Progression (G.P.) are 2 and 2/9 respectively. Find: (i) the common difference, (ii) the 5th term

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(a) sin 105° = sin 60° cos 45° + cos 60° sin 45° = (√3/2)(√2/2) + (1/2)(√2/2) = (√6 + √2)/4.
(b) An A.P. with a = 11, d = 2: Sn = n/2[22 + 2(n − 1)] = n(n + 10) = 551.
n2 + 10n − 551 = 0, (n + 29)(n − 19) = 0, so n = 19 houses.
(c)(i) ar2 = 2/9 with a = 2, so r2 = 1/9 and the common ratio r = ±1/3.
(ii) T5 = ar4 = 2 × 1/81 = 2/81.

Explanation

In a G.P. you look for a common ratio, not a common difference.

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Theory question 109

Trigonometry

In the diagram, a ladder TF, 10m long is placed against a wall at an angle of 70° to the horizontal. (a) How high up the wall, correct to the nearest metre, does the ladder reach? (b) If the foot (F) of the ladder is pulled from the wall to F1 by 1 metre. (i) how far, correct to 2 significant figures, does the top T, slide down the wall to T1 (ii) Calculate, correct to the nearest degree, ∠QF1T1.

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(a) QT = 10 sin 70° = 9.397 ≈ 9 m.
(b) QF = 10 cos 70° = 3.420 m, so QF1 = 4.420 m.
(i) QT1 = √(102 − 4.4202) = √80.46 = 8.970 m. Slide = 9.397 − 8.970 = 0.427 ≈ 0.43 m.
(ii) cos ∠QF1T1 = 4.420/10 = 0.442, so the angle ≈ 64°.

Explanation

The ladder's length stays 10 m, so use Pythagoras for the new height.

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Theory question 110

Variation

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00. (a) Find the cost when there are 44 pupils. (b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can be run without loss?

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Let C = a + kn.
a + 50k = 15705 and a + 40k = 13305. Subtract: 10k = 2400, k = 240; a = 15705 − 12000 = 3705.
(a) C = 3705 + 240 × 44 = $14,265.00.
(b) Need 360n ≥ 3705 + 240n, so 120n ≥ 3705 and n ≥ 30.9. Least number = 31 pupils.

Explanation

'Partly constant and partly varies' means C = a + kn.

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Theory question 111

Loci

On a graph sheet, using a scale of 2cm to 2 units on both axes. (a) draw the straight line joining points P (−5,3) and Q (2,3); (b) construct the locus L of points equidistant from P and Q; (c) by construction, locate points R and S on L, such that PRQS form a rhombus of sides 5cm; (d) find; (i) coordinates of R and S; (ii) area of the rhombus in cm2.

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(a) PQ is horizontal, length 7 units.
(b) L is the perpendicular bisector of PQ: the line x = −1.5.
(c) With centres P and Q and radius 5 units, draw arcs cutting L at R (above) and S (below).
(d)(i) Vertical distance = √(52 − 3.52) = √12.75 ≈ 3.57, so R ≈ (−1.5, 6.6) and S ≈ (−1.5, −0.6).
(ii) Area = ½ × PQ × RS = ½ × 7 × 7.14 ≈ 25.0 cm2 (1 unit = 1 cm on this scale).

Explanation

Every point of a rhombus diagonal's perpendicular bisector is equidistant from its ends, so R and S lie on L.

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Theory question 112

Commercial Arithmetic

(a) A shop-owner marked a shirt at a price to enable him make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer for N864.00. What was the original cost to the shop-owner?
(b) A rectangular lawn of length (x + 5) metres is (x − 2) metres wide. If the diagonal is (x + 6) metres, find; (i) the value of x (ii) the area of the lawn.

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(a) Marked price = 1.2C; sale price = 0.9 × 1.2C = 1.08C = 864, so C = N800.00.
(b)(i) (x + 5)2 + (x − 2)2 = (x + 6)2 gives x2 − 6x − 7 = 0, (x − 7)(x + 1) = 0, so x = 7 (x = −1 makes the width negative).
(ii) Length 12 m, width 5 m: area = 60 m2.

Explanation

Work back from the sale price: each percentage change multiplies the price.

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