WAEC Mathematics 2000 past questions and answers

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

Question 1

Ratio, Percentages and Rates

Express 7/19 as a percentage, correct to 1 decimal place.

  1. A.2.7%
  2. B.3.7%
  3. C.27.1%
  4. D.36.8%.
Show the answer and explanation

Answer: D (36.8%.)

Explanation

7/19 × 100% = 36.84…% ≈ 36.8%.

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Question 2

Fractions, Decimals and Approximation

Express 398753 correct to three significant figures.

  1. A.398000
  2. B.398700
  3. C.398800
  4. D.399000
Show the answer and explanation

Answer: D (399000)

Explanation

Keep 3, 9, 8; the next digit is 7, so round up: 399000.

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Question 3

Surds

Simplify 10/√32

  1. A.(5/4)√2
  2. B.(4/5)√2
  3. C.(5/16)√2
  4. D.(16/5)√2
Show the answer and explanation

Answer: A ((5/4)√2)

Explanation

√32 = 4√2, so 10/(4√2) = 10√2/8 = (5/4)√2.

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Question 4

Number Bases

Find the missing number in the addition of the following numbers, in base seven.
4321 + 1234 + **** = 12341

  1. A.3453
  2. B.5556
  3. C.6016
  4. D.13453
Show the answer and explanation

Answer: A (3453)

Explanation

In base ten: 123417 = 3263, 43217 = 1534, 12347 = 466. The missing number is 3263 − 2000 = 1263 = 34537.

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Question 5

Fractions, Decimals and Approximation

What fraction must be subtracted from the sum of 2 1/6 and 2 7/12 to give 3 1/4?

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

Answer: D (1 1/2)

Explanation

2 1/6 + 2 7/12 = 4 9/12 = 4 3/4. Then 4 3/4 − 3 1/4 = 1 1/2.

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Question 6

Indices

Simplify (16/81)−3/4 × √(100/81)

  1. A.80/243
  2. B.1/64
  3. C.25/6
  4. D.15/4
Show the answer and explanation

Answer: D (15/4)

Explanation

(16/81)−3/4 = (81/16)3/4 = (3/2)3 = 27/8, and √(100/81) = 10/9. 27/8 × 10/9 = 15/4.

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Question 7

Fractions, Decimals and Approximation

Which of the following numbers is a perfect cube?

  1. A.350
  2. B.504
  3. C.950
  4. D.1728.
Show the answer and explanation

Answer: D (1728.)

Explanation

1728 = 12 × 12 × 12 = 123.

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Question 8

Number Bases

If 104x = 68, find the value of x

  1. A.5
  2. B.7
  3. C.8
  4. D.9.
Show the answer and explanation

Answer: C (8)

Explanation

104x = x2 + 4 = 68, so x2 = 64 and x = 8.

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Question 9

Ratio, Percentages and Rates

The ages of three men are in the ratio 3 : 4 : 5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men.

  1. A.45 years
  2. B.72 years
  3. C.108 years
  4. D.216 years.
Show the answer and explanation

Answer: C (108 years)

Explanation

The difference is 5 − 3 = 2 parts = 18 years, so 1 part = 9. Sum = 12 parts = 108 years.

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Question 10

Logarithms

Given that log4x = −3, find x.

  1. A.1/81
  2. B.1/64
  3. C.64
  4. D.81
Show the answer and explanation

Answer: B (1/64)

Explanation

x = 4−3 = 1/64.

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Question 11

Logarithms

Given that the logarithm of a number is 1̄.8732, find, correct to 2 significant figures, the square root of the number.

  1. A.0.29
  2. B.0.75
  3. C.0.86
  4. D.0.93
Show the answer and explanation

Answer: C (0.86)

Explanation

The number is antilog 1̄.8732 = 0.747. Its square root is about 0.864 ≈ 0.86 (or halve the log: 1̄.9366 gives 0.864).

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Question 12

Ratio, Percentages and Rates

A car moves at an average speed of 30kmh−1, how long does it take to cover 200 metres?

  1. A.2.4 sec
  2. B.24 sec
  3. C.144 sec
  4. D.240 sec.
Show the answer and explanation

Answer: B (24 sec)

Explanation

30 km/h = 30000 m ÷ 3600 s = 25/3 m/s. Time = 200 ÷ 25/3 = 24 seconds.

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Question 13

Commercial Arithmetic

A man bought a television set on hire purchase for N25,000, out of which he paid N10,000. If he is allowed to pay the balance in eight equal instalments, find the value of each instalment.

  1. A.N1250
  2. B.N1578
  3. C.N1875
  4. D.3125.
Show the answer and explanation

Answer: C (N1875)

Explanation

Balance = 25000 − 10000 = N15000; 15000 ÷ 8 = N1875.

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Question 14

Bearings

A tree is 8km due south of a building. Kofi is standing 8km west of the tree. How far is Kofi from the building?

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

Answer: C (8√2km)

Explanation

The two 8 km distances are at right angles, so the distance is √(82 + 82) = 8√2 km.

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Question 15

Bearings

A tree is 8km due south of a building. Kofi is standing 8km west of the tree. Find the bearing of Kofi from the building.

  1. A.315°
  2. B.270°
  3. C.225°
  4. D.135°
Show the answer and explanation

Answer: C (225°)

Explanation

Kofi is equally far south and west of the building, so he is exactly south-west: bearing 225°.

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Question 16

Bearings

Which of the following bearings is equivalent to S50°W?

  1. A.040°
  2. B.130°
  3. C.220°
  4. D.230°.
Show the answer and explanation

Answer: D (230°.)

Explanation

Start at south (180°) and turn 50° towards west: 180 + 50 = 230°.

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Question 18

Data Presentation

The bar chart shows the distribution of marks scored by a group of students in a test. How many students scored 4 marks and above?

Diagram for this question
  1. A.15
  2. B.11
  3. C.10
  4. D.7.
Show the answer and explanation

Answer: B (11)

Explanation

Reading the bars for marks 4 to 10: 4 + 2 + 2 + 0 + 1 + 1 + 1 = 11 students.

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Question 19

Data Presentation

The bar chart shows the distribution of marks scored by a group of students in a test. How many students took the test?

Diagram for this question
  1. A.38
  2. B.22
  3. C.15
  4. D.11.
Show the answer and explanation

Answer: B (22)

Explanation

Add all the bars (marks 0 to 10): 1 + 3 + 2 + 5 + 4 + 2 + 2 + 0 + 1 + 1 + 1 = 22.

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Question 20

Measures of Dispersion

Calculate the standard deviation of the following marks: 2, 3, 6, 2, 5, 0, 4, 2.

  1. A.1.5
  2. B.1.7
  3. C.1.8
  4. D.1.9.
Show the answer and explanation

Answer: C (1.8)

Explanation

Mean = 24/8 = 3. Squared deviations add to 1 + 0 + 9 + 1 + 4 + 9 + 1 + 1 = 26. Variance = 26/8 = 3.25, so SD ≈ 1.8.

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Question 21

Probability

The probabilities that Kodjo and Adoga pass an examination are 3/4 and 3/5 respectively. Find the probability of both boys failing the examination.

  1. A.1/10
  2. B.3/10
  3. C.9/20
  4. D.2/3
Show the answer and explanation

Answer: A (1/10)

Explanation

P(both fail) = (1 − 3/4)(1 − 3/5) = 1/4 × 2/5 = 1/10.

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Question 22

Triangles and Quadrilaterals

Which of the following triangles is congruent?

Diagram for this question
  1. A.I and III only
  2. B.II and IV only
  3. C.I and II only
  4. D.II and III only.
Show the answer and explanation

Answer: D (II and III only.)

Explanation

In II and III the 50° angle lies between the 2 cm and 3 cm sides, so they are congruent by SAS. In I and IV the 50° angle is not between those two sides.

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Question 23

Angles and Polygons

Which of the following statements is/are not true about a rectangle? I. Each diagonal cuts the rectangle into two congruent triangles. II. A rectangle has four lines of symmetry. III. The diagonals intersect at right angles.

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

Answer: D (II and III only.)

Explanation

A rectangle has only two lines of symmetry and its diagonals are not perpendicular (unless it is a square). Statement I is true.

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Question 24

Circle Theorems

In the diagram, PQRS is a circle centre O. POR is a diameter and ∠PRQ = 40°. Calculate ∠QSR.

Diagram for this question
  1. A.30°
  2. B.40°
  3. C.45°
  4. D.50°
Show the answer and explanation

Answer: D (50°)

Explanation

∠PQR = 90° (angle in a semicircle), so ∠QPR = 50°. ∠QSR = ∠QPR (same arc QR) = 50°.

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Question 25

Angles and Polygons

Each side of a regular convex polygon subtends an angle of 30° at its centre. Calculate each interior angle

  1. A.75°
  2. B.150°
  3. C.160°
  4. D.168°
Show the answer and explanation

Answer: B (150°)

Explanation

The central angle equals the exterior angle, 30°, so the interior angle is 180 − 30 = 150°.

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Question 26

Angles and Polygons

If the interior angles of a hexagon are 107°, 2x°, 150°, 95°, (2x − 15)° and 123°, find x.

  1. A.57½°
  2. B.65°
  3. C.106°
  4. D.120°
Show the answer and explanation

Answer: B (65°)

Explanation

Hexagon angles add to 720°: 475 + 4x − 15 = 720, so 4x = 260 and x = 65°.

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Question 27

Angles and Polygons

In the diagram, POS and ROT are straight lines. OPQR is a parallelogram. |OS| = |OT| and ∠OST = 50°. Calculate ∠OPQ.

Diagram for this question
  1. A.160°
  2. B.140°
  3. C.120°
  4. D.100°
Show the answer and explanation

Answer: D (100°)

Explanation

Triangle OST is isosceles, so ∠SOT = 180 − 100 = 80°, and ∠POR = 80° (vertically opposite). Adjacent angles of a parallelogram add to 180°, so ∠OPQ = 100°.

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Question 28

Algebraic Expressions

Given that x = 1/2 and y = 4, evaluate 3x2y + xy2.

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

Answer: D (11.)

Explanation

3x2y = 3 × 1/4 × 4 = 3 and xy2 = 1/2 × 16 = 8, so the total is 11.

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Question 29

Indices

Given that 27(1 + x) = 9, find x.

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

Answer: B (−1/3)

Explanation

33(1 + x) = 32, so 3 + 3x = 2 and x = −1/3.

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Question 30

Variation

If x varies inversely as y and x = 2/3 when y = 9, find the value of y when x = 3/4

  1. A.1/18
  2. B.8/81
  3. C.9/2
  4. D.8
Show the answer and explanation

Answer: D (8)

Explanation

xy is constant: 2/3 × 9 = 6. So y = 6 ÷ 3/4 = 8.

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Question 31

Polynomials

Given that (2x + 7) is a factor of 2x2 + 3x − 14, find the other factor.

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

Answer: C (x − 2)

Explanation

(2x + 7)(x − 2) = 2x2 − 4x + 7x − 14 = 2x2 + 3x − 14.

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Question 32

Algebraic Expressions

Simplify 1/(x − 3) − 3(x − 1)/(x2 − 9)

  1. A.(x − 1)/(x − 3)
  2. B.−2/(x + 3)
  3. C.(x − 1)/(x + 3)
  4. D.4x/(x2 − 9)
Show the answer and explanation

Answer: B (−2/(x + 3))

Explanation

Over (x − 3)(x + 3): (x + 3) − 3(x − 1) = −2x + 6 = −2(x − 3). Cancel (x − 3) to get −2/(x + 3).

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Question 33

Inequalities

Which of the following number lines represents the inequality 2 ≤ x < 9?

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

Answer: A (Number line A)

Explanation

2 is included (shaded dot) and 9 is not (open circle). Only line A has a shaded dot at 2 and an open circle at 9.

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Question 34

Inequalities

For an inequality for a distance d metres which is more than 18m, but not more than 23m.

  1. A.18 ≤ d ≤ 23
  2. B.18 < d ≤ 23
  3. C.18 ≤ d < 23
  4. D.d < 18 or d > 23.
Show the answer and explanation

Answer: B (18 < d ≤ 23)

Explanation

'More than 18' means d > 18 and 'not more than 23' means d ≤ 23.

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Question 35

Quadratic Equations

Find the equation whose roots are −8 and 5.

  1. A.x2 + 13x + 40 = 0
  2. B.x2 − 13x − 40 = 0
  3. C.x2 − 3x + 40 = 0
  4. D.x2 + 3x − 40 = 0.
Show the answer and explanation

Answer: D (x2 + 3x − 40 = 0.)

Explanation

(x + 8)(x − 5) = x2 + 3x − 40 = 0.

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Question 36

Algebraic Expressions

Make t the subject of the formula k = m√((t − p)/r)

  1. A.(rk2 + p)/m2
  2. B.(rk2 + pm2)/m2
  3. C.(rk2 − p)/m2
  4. D.(rk2 − p2)/m2
Show the answer and explanation

Answer: B ((rk2 + pm2)/m2)

Explanation

k2/m2 = (t − p)/r, so t = rk2/m2 + p = (rk2 + pm2)/m2.

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Question 37

Quadratic Equations

Solve the equation 3y2 = 27y.

  1. A.y = 0 or 3
  2. B.y = 0 or 9
  3. C.y = −3 or 3
  4. D.y = 3 or 9
Show the answer and explanation

Answer: B (y = 0 or 9)

Explanation

3y2 − 27y = 3y(y − 9) = 0, so y = 0 or y = 9.

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Question 38

Linear and Simultaneous Equations

Find the value of x such that the expression 1/x + 4/(3x) − 5/(6x) + 1 equals zero

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

Answer: C (−3/2)

Explanation

Multiply by 6x: 6 + 8 − 5 + 6x = 0, so 6x = −9 and x = −3/2.

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Question 39

Variation

Given that p varies directly as q while q varies inversely as r, which of the following statements is true?

  1. A.r varies directly as p
  2. B.p varies inversely as r
  3. C.p varies directly as r
  4. D.q varies inversely as p.
Show the answer and explanation

Answer: B (p varies inversely as r)

Explanation

p = kq and q = c/r, so p = kc/r: p varies inversely as r.

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Question 40

Circle Theorems

In the diagram, PQS is a circle with centre O. RST is a tangent at S and ∠SOP = 96°. Find ∠PST

Diagram for this question
  1. A.42°
  2. B.48°
  3. C.60°
  4. D.66°
Show the answer and explanation

Answer: B (48°)

Explanation

The angle between a tangent and a chord is half the angle the chord subtends at the centre: ½ × 96° = 48°.

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Question 41

Mensuration of Plane Shapes

A bicycle wheel of radius 42cm is rolled over a distance 66 metres. How many revolutions does it make? [Take π = 22/7]

  1. A.2.5
  2. B.5
  3. C.25
  4. D.50
Show the answer and explanation

Answer: C (25)

Explanation

Circumference = 2 × 22/7 × 42 = 264 cm. 6600 ÷ 264 = 25 revolutions.

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Question 42

Mensuration of Solids

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

  1. A.8cm2
  2. B.9.6cm2
  3. C.16cm2
  4. D.25cm2
Show the answer and explanation

Answer: C (16cm2)

Explanation

V = ⅓ × base area × 15 = 5 × base area = 80, so the base area is 16 cm2.

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Question 43

Ratio, Percentages and Rates

A tap leaks at the rate of 2cm3 per second. How long will it take the tap to fill a container of 45 litres capacity? (1 litre = 1000cm3)

  1. A.8 hours
  2. B.6hr 15min.
  3. C.4hr 25 min.
  4. D.3hr.
Show the answer and explanation

Answer: B (6hr 15min.)

Explanation

45 litres = 45000 cm3; at 2 cm3/s that is 22500 s = 6.25 hours = 6 hr 15 min.

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Question 44

Mensuration of Plane Shapes

The lengths of the parallel sides of a trapezium are 5cm and 7cm. If its area is 120cm2, find the perpendicular distance between the parallel sides.

  1. A.5.0cm
  2. B.6.9cm
  3. C.10.0cm
  4. D.20.0cm.
Show the answer and explanation

Answer: D (20.0cm.)

Explanation

½ × (5 + 7) × h = 120, so 6h = 120 and h = 20 cm.

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Question 45

Mensuration of Plane Shapes

The arc of a circle 50cm long, subtends an angle of 75° at the centre of the circle. Find, correct to 3 significant figures, the radius of the circle. [Take π = 22/7]

  1. A.8.74cm
  2. B.38.2 cm
  3. C.61.2 cm
  4. D.76.4cm.
Show the answer and explanation

Answer: B (38.2 cm)

Explanation

50 = 75/360 × 2 × 22/7 × r, so r = 50 × 360 × 7/(75 × 44) ≈ 38.2 cm.

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Question 46

Triangles and Quadrilaterals

In the diagram, |PQ| = |PS|. Which of the following statements is true?

Diagram for this question
  1. A.∠QPS = ∠QRS
  2. B.|PO| = |RO|
  3. C.QR||PS
  4. D.∠PQR = ∠PSR
Show the answer and explanation

Answer: D (∠PQR = ∠PSR)

Explanation

PQRS is a kite (PQ = PS and RQ = RS), symmetric about PR, so the angles at Q and S are equal: ∠PQR = ∠PSR.

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Question 47

Mensuration of Plane Shapes

The area of a circle is 38.5cm2. Find its diameter. [Take π = 22/7]

  1. A.22cm
  2. B.14cm
  3. C.7cm
  4. D.6cm.
Show the answer and explanation

Answer: C (7cm)

Explanation

r2 = 38.5 × 7/22 = 12.25, so r = 3.5 cm and the diameter is 7 cm.

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Question 49

Quadratic Equations

Solve the equation 3 + 5x − 2x2 = 0

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

Answer: D (−1/2, 3)

Explanation

2x2 − 5x − 3 = (2x + 1)(x − 3) = 0, so x = −1/2 or 3.

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Question 50

Commercial Arithmetic

If the simple interest on N2,000 after 9 months is N60, at what rate per annum is the interest charged?

  1. A.2¼%
  2. B.4%
  3. C.5%
  4. D.6%.
Show the answer and explanation

Answer: B (4%)

Explanation

I = PRT/100 with T = 9/12 year: 60 = 2000 × R × 0.75/100, so R = 4%.

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Theory question 101

Fractions, Decimals and Approximation

(a) Evaluate and express your answer in standard form; (4.56 × 3.6)/0.12
(b) Without using mathematical tables or calculator, evaluate (73.8)2 − (26.2)2
(c) Simplify √(1 19/81), expressing your answer in the form a/b where a and b are positive integers.

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Model answer

(a) (4.56 × 3.6)/0.12 = 4.56 × 30 = 136.8 = 1.368 × 102.
(b) Difference of two squares: (73.8 − 26.2)(73.8 + 26.2) = 47.6 × 100 = 4760.
(c) 1 19/81 = 100/81, so √(100/81) = 10/9.

Explanation

Spot the difference of two squares: a² − b² = (a − b)(a + b).

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Theory question 102

Trigonometry

(a) Given that cos x = 0.7431, 0 < x < 90°, use tables to find the values of (i) 2 sin x; (ii) tan (x/2)
(b) The interior angles of a pentagon are in ratio 2 : 3 : 4 : 4 : 5. Find the value of the largest angle.

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Model answer

(a) cos x = 0.7431 gives x = 42°.
(i) 2 sin 42° = 2 × 0.6691 = 1.3382.
(ii) tan 21° = 0.3839.
(b) Angles of a pentagon add to 540°. Ratio total = 18, so the largest angle = 5/18 × 540° = 150°.

Explanation

Find x from the cosine table first, then use it for the other ratios.

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Theory question 103

Quadratic Equations

(a) Given the expression y = ax2 − bx − 12, find the values of x when a = 1, b = 2 and y = 3
(b) If √(x2 + 1) = 5/4 find the positive value of x.

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Model answer

(a) x2 − 2x − 12 = 3, so x2 − 2x − 15 = 0, (x − 5)(x + 3) = 0: x = 5 or x = −3.
(b) x2 + 1 = 25/16, so x2 = 9/16 and the positive value is x = 3/4.

Explanation

Substitute the given values, then rearrange to a quadratic equal to zero.

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Theory question 104

Constructions

Using ruler and a pair of compasses only, (a) construct ΔPQR such that |PQ| = 7cm, |PR| = 6cm and ∠PQR = 60° (b) Locate point M, the mid-point of PQ. (c) Measure ∠RMQ.

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Model answer

(a) Draw PQ = 7 cm. At Q construct 60° (equilateral-triangle arc). With centre P, radius 6 cm, draw an arc to meet the 60° arm at R.
Note: the distance from P to that arm is 7 sin 60° ≈ 6.06 cm, so a 6 cm arc only just touches it; R is at (or very near) the foot of the perpendicular from P, with QR ≈ 3.5 cm.
(b) Bisect PQ (perpendicular bisector) to get M, so QM = 3.5 cm.
(c) Triangle MQR has QM ≈ QR ≈ 3.5 cm and ∠Q = 60°, so it is (nearly) equilateral: ∠RMQ ≈ 60°.

Explanation

Construct 60° with compasses by drawing an equilateral triangle on the line.

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Theory question 105

Data Presentation

The pie chart shows the distribution of marks scored by 200 pupils in a test. (a) How many pupils scored (i) between 41 and 50 marks? (ii) above 60 marks? (b) What fraction of the pupils scored at most 50 marks? (c) What is the modal class?

Diagram for this question
Show the model answer
Model answer

Angles from the chart: 31–40: 54°, 41–50: 90°, 51–60: 108°, 61–70: 54°, 71–80: 36°, 81–90: 18°.
(a)(i) 90/360 × 200 = 50 pupils.
(ii) Above 60: 54 + 36 + 18 = 108°, so 108/360 × 200 = 60 pupils.
(b) At most 50: 54 + 90 = 144°, so 144/360 = 2/5 of the pupils (80 pupils).
(c) The modal class is 51–60 (largest sector, 108°).

Explanation

Number in a class = angle/360 × total.

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Theory question 106

Probability

(a) The table shows the number of limes and apples of the same size in a bag.
Good: Limes 10, Apples 8
Bad: Limes 6, Apples 6
If two of the fruits are picked at random, one at a time, without replacement, find the probability that: (i) both are good limes; (ii) both are bad fruits; (iii) one is a good apple and the other a bad lime
(b) Solve the equation log3(4x + 1) − log3(3x − 5) = 2

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Model answer

Total fruits = 30.
(a)(i) P = 10/30 × 9/29 = 3/29.
(ii) There are 12 bad fruits: P = 12/30 × 11/29 = 22/145.
(iii) Good apple then bad lime or bad lime then good apple: 2 × (8/30 × 6/29) = 16/145.
(b) log3[(4x + 1)/(3x − 5)] = 2, so (4x + 1)/(3x − 5) = 9.
4x + 1 = 27x − 45, 23x = 46, x = 2 (check: 3x − 5 = 1 > 0).

Explanation

Without replacement, the second fraction has one fewer in the denominator (29).

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Theory question 107

Commercial Arithmetic

(a) A man earns N150,000 per annum. He is allowed a tax free pay of N40,000. If he pays 25 kobo in the naira as tax on his taxable income, how much has he left?
(b) A bookshop had 650 copies of a book for sale. The books were marked at N75 per copy in order to make a profit of 30%. A bookseller bought 300 copies at 5% discount. If the remaining copies are sold at N75 each, calculate the percentage profit the bookshop would make on the whole?

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(a) Taxable income = 150,000 − 40,000 = N110,000. Tax = 0.25 × 110,000 = N27,500. Left = 150,000 − 27,500 = N122,500.
(b) Cost price per copy = 75 ÷ 1.3; total cost = 650 × 75/1.3 = N37,500.
Sales: 300 × 75 × 0.95 = N21,375 and 350 × 75 = N26,250; total N47,625.
Profit = 47,625 − 37,500 = N10,125; percentage profit = 10,125/37,500 × 100% = 27%.

Explanation

The marked price is 130% of cost, so divide by 1.3 to get the cost.

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Theory question 108

Graphs of Functions

(a) Copy and complete the following table of values for the relation y = x2 − 2x − 5.
x: −3, −2, −1, 0, 1, 2, 3, 4, 5
y: , , −2, , −6, , −2, 3, 10
(b) Draw the graph of the relation y = x2 − 2x − 5, using a scale of 2cm to 1 unit on the x-axis, and 2cm to 2 units on the y-axis.
(c) Using the same axes, draw the graph of y = 2x − 3.
(d) Obtain in the form ax2 + bx + c = 0 where a, b and c are integers, the equation which is satisfied by the x-coordinate of the points of intersection of the two graphs.
(e) From your graphs, determine the roots of the equation obtained in (d) above.

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(a) x = −3: 10; x = −2: 3; x = 0: −5; x = 2: −5.
(b) Plot and join with a smooth U-shaped curve (lowest point (1, −6)).
(c) Straight line through (0, −3) and (3, 3).
(d) x2 − 2x − 5 = 2x − 3 gives x2 − 4x − 2 = 0.
(e) Read the x-values where the line meets the curve: x ≈ −0.4 and x ≈ 4.4 (exact 2 ± √6 = −0.45, 4.45).

Explanation

At an intersection both equations give the same y, so set them equal.

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Theory question 109

Measures of Central Tendency

(a) The mean of 1, 2, x, 11, y, 14 is 8 and the median is 9. Find the values of x and y.
(b) In the diagram, MN||PQ, |LM| = 3cm and |LP| = 4cm. If the area of ΔLMN is 18cm2, find the area of quadrilateral MPQN.

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(a) Sum = 6 × 8 = 48, so x + y = 20.
With x < 11 < y the order is 1, 2, x, 11, y, 14 and the median = (x + 11)/2 = 9, so x = 7 and y = 13.
(b) Triangles LMN and LPQ are similar with ratio 3 : 4, so areas are in ratio 9 : 16.
Area LPQ = 18 × 16/9 = 32 cm2; area MPQN = 32 − 18 = 14 cm2.

Explanation

Areas of similar figures are in the ratio of the squares of their sides.

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Theory question 110

Trigonometry

(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises
(b) In the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5cm and |BC| = 8cm. Calculate, correct to the nearest degree, ∠BAC.
(c) Two boats, 70 metres apart and on opposite sides of a light house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take tan 71.6° = 3]

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(a) Height = 100 sin 24° = 100 × 0.4067 ≈ 41 m.
(b) The perpendicular from A bisects BC: sin(½∠BAC) = 4/5 = 0.8, so ½∠BAC = 53.13° and ∠BAC ≈ 106°.
(c) Let the height be h. Distances from the foot: h/tan 71.6° = h/3 and h/tan 45° = h.
h/3 + h = 70, so 4h/3 = 70 and h = 52.5 m.

Explanation

In (c) the two horizontal distances add up to the 70 m between the boats.

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Theory question 111

Mensuration of Solids

(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) Calculate, in m3, the volume of soil dug out (ii) If the soil is used to raise the level of the rectangular floor of a room 4m by 12m, calculate, correct to the nearest cm, the thickness of the new layer of soil. [Take π = 22/7]
(b) The diagram shows a quadrilateral ABCD in which ∠DAB is a right angle. |AB| = 3.3cm, |BC| = 3.9cm, |CD| = 5.2cm and |DA| = 5.6cm (i) find the length of BD (ii) show that ∠BCD = 90°

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(a)(i) V = πr2h = 22/7 × 1 × 8 ≈ 25.14 m3.
(ii) Thickness = 25.14 ÷ (4 × 12) = 0.524 m ≈ 52 cm.
(b)(i) BD = √(3.32 + 5.62) = √(10.89 + 31.36) = √42.25 = 6.5 cm.
(ii) BC2 + CD2 = 15.21 + 27.04 = 42.25 = BD2, so by the converse of Pythagoras' theorem ∠BCD = 90°.

Explanation

The soil's volume does not change: volume = floor area × thickness.

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Theory question 112

Progressions

(a) The first term of an Arithmetic progression (A.P) is −8. The ratio of the 7th term to the 9th term is 5:8. Calculate the common difference of the progression.
(b) A sphere of radius 2cm is of mass 11.2g. Find: (i) the volume of the sphere, (ii) the density of the sphere, (iii) the mass of a sphere of the same material but with radius 3cm, [Take π = 22/7]

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(a) (−8 + 6d)/(−8 + 8d) = 5/8, so 8(−8 + 6d) = 5(−8 + 8d): −64 + 48d = −40 + 40d, 8d = 24, d = 3.
(b)(i) V = 4/3 × 22/7 × 23 = 704/21 ≈ 33.5 cm3.
(ii) Density = 11.2 ÷ 33.52 ≈ 0.334 g/cm3.
(iii) Mass = density × volume = 0.334 × (4/3 × 22/7 × 27) ≈ 37.8 g (or 11.2 × 27/8 = 37.8 g).

Explanation

Same material means same density; the mass scales with the cube of the radius.

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Theory question 113

Latitude and Longitude

(a) Two places X and Y on the equator are on longitude 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far is X from the North pole? [Take π = 22/7] and radius of the Earth as 6400 km).
(b) In the diagram, PQR is a circle centre O. N is the midpoint of chord PQ. |PQ| = 8 cm, |ON| = 3cm and ∠ONR = 20°. Calculate the size of ∠ORN to the nearest degree.

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(a)(i) Difference in longitude = 123 − 67 = 56°. Distance = 56/360 × 2 × 22/7 × 6400 ≈ 6258 km.
(ii) From the equator to the North Pole is 90° of a great circle: 90/360 × 2 × 22/7 × 6400 ≈ 10057 km.
(b) ON ⟂ PQ, so the radius OP = √(42 + 32) = 5 cm = OR.
In triangle ONR, sine rule: sin ∠ORN/3 = sin 20°/5, so sin ∠ORN = 0.2052 and ∠ORN ≈ 12°.

Explanation

Both places are on the equator, so the equator itself is the circle to use.

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