GCE Mathematics 1990 past questions and answers

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

Question 1

Indices

Simplify: 125−1/3 × 49−1/2 × 100

  1. A.350
  2. B.35
  3. C.1/35
  4. D.1/350
  5. E.0
Show the answer and explanation

Answer: C (1/35)

Explanation

125−1/3 = 1/5, 49−1/2 = 1/7 and 100 = 1, so the product is 1/35.

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Question 2

Indices

If 32x = 27, what is x?

  1. A.1
  2. B.1.5
  3. C.4.5
  4. D.18
  5. E.40.5
Show the answer and explanation

Answer: B (1.5)

Explanation

27 = 33, so 2x = 3 and x = 1.5.

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Question 3

Fractions, Decimals and Approximation

Express 0.00562 in standard form

  1. A.5.62 × 10−3
  2. B.5.62 × 10−2
  3. C.0.562 × 10−2
  4. D.5.62 × 102
  5. E.5.62 × 103
Show the answer and explanation

Answer: A (5.62 × 10−3)

Explanation

Move the decimal point 3 places to the right to get 5.62, so the power of 10 is −3.

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Question 4

Logarithms

Given that 1/3 log10P = 1, find the value of P.

  1. A.1/10
  2. B.3
  3. C.10
  4. D.100
  5. E.1000
Show the answer and explanation

Answer: E (1000)

Explanation

Multiply by 3: log10P = 3, so P = 103 = 1000.

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Question 5

Logarithms

Simplify: (log√8)/(log 8)

  1. A.1/3
  2. B.1/2
  3. C.1/3 log√2
  4. D.1/3 log√8
  5. E.1/2 log√2
Show the answer and explanation

Answer: B (1/2)

Explanation

log√8 = log 81/2 = ½ log 8, so the fraction is ½.

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Question 6

Logarithms

Evaluate, using logarithm table, log(0.65)2

  1. A.1.6258
  2. B.0.6272
  3. C.0.6258
  4. D.1̄.6258
  5. E.1.6272
Show the answer and explanation

Answer: D (1̄.6258)

Explanation

log 0.65 = 1̄.8129, and log(0.65)2 = 2 × 1̄.8129 = 1̄.6258.

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Question 7

Logarithms

If log x = 2̄.3675 and log y = 0.9750, what is the value of x + y? Correct to three significant figures.

  1. A.1.18
  2. B.1.31
  3. C.9.03
  4. D.9.44
  5. E.9.46
Show the answer and explanation

Answer: E (9.46)

Explanation

x = antilog 2̄.3675 = 0.02331 and y = antilog 0.9750 = 9.441, so x + y = 9.464 ≈ 9.46.

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Question 8

Ratio, Percentages and Rates

While doing his Physics practicals, Idowu recorded a reading as 1.12cm instead of 1.21cm. Calculate his percentage error.

  1. A.1.17%
  2. B.6.38%
  3. C.7.44%
  4. D.8.05%
  5. E.9.00%
Show the answer and explanation

Answer: C (7.44%)

Explanation

Error = 1.21 − 1.12 = 0.09. Percentage error = 0.09/1.21 × 100 ≈ 7.44%.

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Question 9

Progressions

Find the 4th term of an A.P. whose first term is 2 and the common difference is 0.5.

  1. A.0.5
  2. B.2.5
  3. C.3.5
  4. D.4
  5. E.4.5
Show the answer and explanation

Answer: C (3.5)

Explanation

4th term = a + 3d = 2 + 3(0.5) = 3.5.

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Question 10

Graphs of Functions

From the graph, determine the roots of the equation y = 2x2 + x − 6.

Diagram for this question
  1. A.−3, 4
  2. B.−2, −6
  3. C.−2, 1.5
  4. D.−1, 1
  5. E.2, 1
Show the answer and explanation

Answer: C (−2, 1.5)

Explanation

The roots are where the curve crosses the x-axis: x = −2 and x = 1.5. Check: 2x2 + x − 6 = (2x − 3)(x + 2).

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Question 11

Graphs of Functions

From the graph of y = 2x2 + x − 6, find the values of x for which y = 5

Diagram for this question
  1. A.−2.5, 2
  2. B.−2.6, 2.1
  3. C.−2.0, 1.5
  4. D.−2.0, 2.0
  5. E.0
Show the answer and explanation

Answer: B (−2.6, 2.1)

Explanation

Draw the line y = 5 and read where it meets the curve: about x = −2.6 and x = 2.1 (solving 2x2 + x − 11 = 0 gives −2.61 and 2.11).

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Question 12

Mensuration of Plane Shapes

An arc of length 22cm subtends an angle of θ at the centre of the circle. What is the value of θ if the radius of the circle is 15cm? (Take π = 22/7)

  1. A.70°
  2. B.84°
  3. C.96°
  4. D.156°
  5. E.168°
Show the answer and explanation

Answer: B (84°)

Explanation

θ/360 × 2 × 22/7 × 15 = 22, so θ = 22 × 360 × 7/(44 × 15) = 84°.

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Question 13

Progressions

Find the sum of the first five terms of the G.P. 2, 6, 18, ...

  1. A.484
  2. B.243
  3. C.242
  4. D.130
  5. E.1
Show the answer and explanation

Answer: C (242)

Explanation

a = 2, r = 3. S5 = 2(35 − 1)/(3 − 1) = 35 − 1 = 242.

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Question 14

Sets

Let J be the set of positive integers. If H = {x : x ∈ J, x2 < 3 and x ≠ 0}, then

  1. A.H = {1}
  2. B.H is an infinite set
  3. C.H = {0, 1, 2}
  4. D.H = { }
  5. E.J ⊂ H
Show the answer and explanation

Answer: A (H = {1})

Explanation

The only positive integer whose square is less than 3 is 1 (22 = 4 is too big), so H = {1}.

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Question 15

Sets

In a class of 80 students, every student had to study Economics or Geography, or both Economics and Geography. If 65 students studied Economics and 50 studied Geography, how many studied both subjects?

  1. A.15
  2. B.30
  3. C.35
  4. D.45
  5. E.50
Show the answer and explanation

Answer: C (35)

Explanation

n(E ∪ G) = n(E) + n(G) − n(E ∩ G), so 80 = 65 + 50 − n(E ∩ G) and n(E ∩ G) = 35.

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Question 16

Inequalities

The area shaded with horizontal lines is the solution set of the inequalities:

Diagram for this question
  1. A.y ≥ x, y + 3 ≥ 2x, x ≤ 3
  2. B.y ≤ x, y + 2x ≥ −3, x ≤ 3
  3. C.y ≤ −x, y + 2x ≤ 3, x ≥ −3
  4. D.y ≥ −x, y + 3 ≤ 2x, x ≥ −3
  5. E.y ≤ x, y ≤ 2x − 3, x ≥ 3
Show the answer and explanation

Answer: B (y ≤ x, y + 2x ≥ −3, x ≤ 3)

Explanation

The horizontal-lined region lies below y = x, above the falling line y + 2x = −3 and to the left of x = 3.

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Question 17

Inequalities

The area shaded with vertical lines is the solution set of the inequalities:

Diagram for this question
  1. A.y ≤ x, y + 2x ≤ −3
  2. B.y ≥ x, y + 2x ≥ −3
  3. C.y ≤ x, y ≤ 2x − 3, x ≤ 3
  4. D.y ≤ −x, y ≤ 2x − 3, x ≤ 3
  5. E.y ≥ x, y + 2x ≥ 3, x ≤ 3
Show the answer and explanation

Answer: A (y ≤ x, y + 2x ≤ −3)

Explanation

The vertical-lined region lies below the line y = x and below the falling line y + 2x = −3, so y ≤ x and y + 2x ≤ −3.

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Question 18

Mensuration of Plane Shapes

In the diagram below, O is the centre of the circle with radius 10cm, and ∠ABC = 30°. Calculate, correct to 1 decimal place, the length of arc AC. (Take π = 3.14)

Diagram for this question
  1. A.5.2cm
  2. B.10.5cm
  3. C.13.2cm
  4. D.20.6cm
  5. E.31.4cm
Show the answer and explanation

Answer: B (10.5cm)

Explanation

The angle at the centre is twice the angle at the circumference: ∠AOC = 60°. Arc AC = 60/360 × 2 × 3.14 × 10 = 10.47 ≈ 10.5cm.

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Question 19

Algebraic Expressions

Factorize x2 + 4x − 192

  1. A.(x − 4)(x + 48)
  2. B.(x − 48)(x + 4)
  3. C.(x − 12)(x + 16)
  4. D.(x − 12)(x − 16)
  5. E.(x + 12)(x + 16)
Show the answer and explanation

Answer: C ((x − 12)(x + 16))

Explanation

Find two numbers with product −192 and sum 4: 16 and −12. So x2 + 4x − 192 = (x − 12)(x + 16).

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Question 20

Algebraic Expressions

Factorize 2e2 − 3e + 1

  1. A.(2e − 1)(e − 1)
  2. B.(e + 1)(2e + 1)
  3. C.(2e + 3)(e − 2)
  4. D.(2e − 3)(e − 1)
  5. E.(e2 − 3)(2e − 1)
Show the answer and explanation

Answer: A ((2e − 1)(e − 1))

Explanation

(2e − 1)(e − 1) = 2e2 − 2e − e + 1 = 2e2 − 3e + 1.

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Question 21

Quadratic Equations

Solve the equation 7y2 = 3y.

  1. A.y = 3 or 7
  2. B.y = 0 or 7
  3. C.y = 0 or 3/7
  4. D.y = 0 or 9
  5. E.y = 0 or 10
Show the answer and explanation

Answer: C (y = 0 or 3/7)

Explanation

7y2 − 3y = 0 gives y(7y − 3) = 0, so y = 0 or y = 3/7.

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Question 22

Graphs of Functions

Use the graph to answer this question. Locate the points where y = x − 3 intersects y = x2 − 2x − 3

Diagram for this question
  1. A.(0,1), (0,2)
  2. B.(0,2), (0,3)
  3. C.(0,3), (0,4)
  4. D.(3,0), (0,−3)
  5. E.(−3,0), (0,3)
Show the answer and explanation

Answer: D ((3,0), (0,−3))

Explanation

Set x2 − 2x − 3 = x − 3: x2 − 3x = 0, so x = 0 or 3, giving the points (0, −3) and (3, 0).

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Question 23

Quadratic Equations

Find the value of m which makes x2 + 8x + m a perfect square.

  1. A.2
  2. B.4
  3. C.8
  4. D.12
  5. E.16
Show the answer and explanation

Answer: E (16)

Explanation

m must be (half of 8)2 = 16, giving (x + 4)2.

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Question 24

Quadratic Equations

Solve the equation 2a2 − 3a − 27 = 0.

  1. A.3/2, 9
  2. B.−2/3, 9
  3. C.3, 9/2
  4. D.−3, −9/2
  5. E.−3, 9/2
Show the answer and explanation

Answer: E (−3, 9/2)

Explanation

2a2 − 3a − 27 = (2a − 9)(a + 3) = 0, so a = 9/2 or a = −3.

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Question 25

Mensuration of Plane Shapes

A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector, correct to the nearest degree. (Take π = 22/7)

  1. A.6°
  2. B.26°
  3. C.52°
  4. D.103°
  5. E.206°
Show the answer and explanation

Answer: D (103°)

Explanation

θ/360 × 22/7 × 72 = 44 gives θ/360 × 154 = 44, so θ = 102.9° ≈ 103°.

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Question 26

Mensuration of Solids

Calculate the surface area of a sphere of radius 7cm. (Take π = 22/7)

  1. A.86cm2
  2. B.154cm2
  3. C.616cm2
  4. D.1434cm2
  5. E.4312cm2
Show the answer and explanation

Answer: C (616cm2)

Explanation

Surface area = 4πr2 = 4 × 22/7 × 49 = 616cm2.

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Question 27

Latitude and Longitude

If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ?

  1. A.75°
  2. B.60°
  3. C.45°
  4. D.30°
  5. E.0°
Show the answer and explanation

Answer: D (30°)

Explanation

The radius of a parallel of latitude is R cos(latitude). R cos 30° = R cos θ, so θ = 30°.

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Question 28

Mensuration of Solids

A cylindrical container closed at both ends, has a radius of 7cm and height 5cm. (Take π = 22/7)
Find the total surface area of the container.

  1. A.35cm3
  2. B.154cm3
  3. C.220cm3
  4. D.528cm3
  5. E.770cm3
Show the answer and explanation

Answer: D (528cm3)

Explanation

Total surface area = 2πr2 + 2πrh = 2(154) + 2 × 22/7 × 7 × 5 = 308 + 220 = 528cm2 (the options print the unit as cm3).

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Question 29

Mensuration of Solids

A cylindrical container closed at both ends, has a radius of 7cm and height 5cm. (Take π = 22/7)
What is the volume of the container?

  1. A.35cm3
  2. B.154cm3
  3. C.220cm3
  4. D.528cm3
  5. E.770cm3
Show the answer and explanation

Answer: E (770cm3)

Explanation

Volume = πr2h = 22/7 × 49 × 5 = 770cm3.

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Question 30

Triangles and Quadrilaterals

In the diagram below, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR.

Diagram for this question
  1. A.72°
  2. B.64°
  3. C.52°
  4. D.32°
  5. E.26°
Show the answer and explanation

Answer: C (52°)

Explanation

Triangle PRS is isosceles, so ∠RSP = 32° and the exterior angle ∠PRQ = 32° + 32° = 64°. Triangle PQR is isosceles, so ∠PQR = 64° and ∠QPR = 180° − 128° = 52°.

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Question 31

Triangles and Quadrilaterals

In the diagram below, WXYZ is a rhombus and ∠WYX = 20°. What is the value of ∠XZY?

Diagram for this question
  1. A.20°
  2. B.30°
  3. C.45°
  4. D.60°
  5. E.70°
Show the answer and explanation

Answer: E (70°)

Explanation

A diagonal of a rhombus bisects its angles, so ∠XYZ = 40°. Triangle XYZ is isosceles (YX = YZ), so ∠XZY = (180° − 40°)/2 = 70°.

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Question 32

Mensuration of Plane Shapes

In the diagram below, PS||RQ, |RQ| = 6.4cm and perpendicular PH = 3.2cm. Find the area of SQR.

Diagram for this question
  1. A.5.12cm2
  2. B.9.60cm2
  3. C.10.24cm2
  4. D.20.48cm2
  5. E.40.96cm2
Show the answer and explanation

Answer: C (10.24cm2)

Explanation

S lies on the line PS, which is parallel to RQ, so the height of triangle SQR is also 3.2cm. Area = ½ × 6.4 × 3.2 = 10.24cm2.

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Question 33

Triangles and Quadrilaterals

Use the diagram below. Which angle is equal to ∠VWZ?

Diagram for this question
  1. A.∠VXZ
  2. B.∠VYZ
  3. C.∠XZW
  4. D.∠YXZ
  5. E.∠VYW
Show the answer and explanation

Answer: A (∠VXZ)

Explanation

W, X, Z and V lie on the circle, so ∠VWZ and ∠VXZ stand on the same chord VZ in the same segment and are equal. Y is not on the circle.

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Question 34

Triangles and Quadrilaterals

Use the diagram below. Which triangle is equal in area to ΔVWZ?

Diagram for this question
  1. A.ΔVXZ
  2. B.ΔVYZ
  3. C.ΔXYV
  4. D.ΔWYV
  5. E.ΔWYZ
Show the answer and explanation

Answer: B (ΔVYZ)

Explanation

WY is parallel to VZ, so triangles VWZ and VYZ have the same base VZ and lie between the same parallels; their areas are equal.

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Question 35

Circle Theorems

In the diagram below, PQRS is a cyclic quadrilateral, ∠PSR = 86° and ∠QPR = 38°. Calculate ∠PRQ.

Diagram for this question
  1. A.58°
  2. B.53°
  3. C.48°
  4. D.43°
  5. E.38°
Show the answer and explanation

Answer: C (48°)

Explanation

Opposite angles of a cyclic quadrilateral add to 180°, so ∠PQR = 94°. In triangle PQR, ∠PRQ = 180° − 94° − 38° = 48°.

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Question 36

Circle Theorems

In the diagram below, O is the centre of the circle. If ∠QOR = 290°, find the size of ∠QPR.

Diagram for this question
  1. A.110°
  2. B.70°
  3. C.55°
  4. D.35°
  5. E.20°
Show the answer and explanation

Answer: D (35°)

Explanation

The 290° is the reflex angle on P's side, so the angle at O facing arc QR is 360° − 290° = 70°. The angle at the circumference is half of it: 35°.

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Question 37

Circle Theorems

In the diagram, PQ is a tangent at T to the circle ABT. ABC is a straight line and TC bisects ∠BTQ. Find x.

Diagram for this question
  1. A.20°
  2. B.30°
  3. C.35°
  4. D.40°
  5. E.50°
Show the answer and explanation

Answer: E (50°)

Explanation

By the alternate segment theorem, ∠BTQ = ∠BAT = 40° and ∠ABT = ∠ATP = 70°. TC bisects ∠BTQ, so ∠BTC = 20°. In triangle BTC the exterior angle ∠ABT = ∠BTC + x, so x = 70° − 20° = 50°.

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Question 40

Trigonometry

Use Mathematical tables to evaluate (cos 40° − sin 30°).

  1. A.−0.2660
  2. B.−0.0266
  3. C.0.0266
  4. D.0.2660
  5. E.1.5320
Show the answer and explanation

Answer: D (0.2660)

Explanation

cos 40° = 0.7660 and sin 30° = 0.5000, so the difference is 0.2660.

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Question 41

Trigonometry

If the shadow of a pole 7m high is ½ its length, what is the angle of elevation of the sun, correct to the nearest degree?

  1. A.90°
  2. B.63°
  3. C.60°
  4. D.26°
  5. E.0°
Show the answer and explanation

Answer: B (63°)

Explanation

The shadow is 3.5m, so tan θ = 7/3.5 = 2 and θ = 63.4° ≈ 63°.

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Question 42

Trigonometry

From the top of a building 10m high, the angle of depression of a stone lying on the horizontal ground is 69°. Calculate, correct to 1 decimal place, the distance of the stone from the foot of the building.

  1. A.3.6m
  2. B.3.8m
  3. C.6.0m
  4. D.9.3m
  5. E.26.1m
Show the answer and explanation

Answer: B (3.8m)

Explanation

The angle of elevation of the top from the stone is 69°, so distance = 10/tan 69° = 10/2.605 = 3.84 ≈ 3.8m.

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Question 43

Bearings

The bearing of a point X from a point Y is 074°. What is the bearing of Y from X?

  1. A.106°
  2. B.148°
  3. C.164°
  4. D.254°
  5. E.286°
Show the answer and explanation

Answer: D (254°)

Explanation

The back bearing differs by 180°: 074° + 180° = 254°.

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Question 44

Trigonometry

Using the graph above, find the value of sin 270° + 2 cos 270°.

Diagram for this question
  1. A.−2
  2. B.−1
  3. C.0
  4. D.1
  5. E.2
Show the answer and explanation

Answer: B (−1)

Explanation

From the graphs, sin 270° = −1 and cos 270° = 0, so the value is −1 + 0 = −1.

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Question 45

Data Presentation

The annual salary of Mr. Johnson Mohammed for 1989 was N12,000.00. He spent this on agricultural projects, education of his children, food items, savings, maintenance and miscellaneous items as shown in the pie chart.
How much did he spend on food items?

Diagram for this question
  1. A.N9,700.00
  2. B.N6,700.00
  3. C.N4,000.00
  4. D.3,700.00
  5. E.2,300.00
Show the answer and explanation

Answer: E (2,300.00)

Explanation

Food angle = 360° − (120° + 80° + 18° + 30° + 43°) = 69°. Amount = 69/360 × N12,000 = N2,300.

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Question 46

Data Presentation

The annual salary of Mr. Johnson Mohammed for 1989 was N12,000.00. He spent this on agricultural projects, education of his children, food items, savings, maintenance and miscellaneous items as shown in the pie chart.
How much money did he invest in agriculture.

Diagram for this question
  1. A.N1,200.00
  2. B.N1,440.00
  3. C.N2,000.00
  4. D.N2,910.00
  5. E.N4,000.00
Show the answer and explanation

Answer: E (N4,000.00)

Explanation

Agric. project has 120°: 120/360 × N12,000 = N4,000.

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Question 47

Measures of Central Tendency

The table shows the amount of money (in naira) collected through voluntary donations in a secondary school.
Amount (in naira): 3, 6, 9, 12, 15, 18
No. of Students: 3, 9, 6, 15, 3, 12
What is the mode?

  1. A.N3
  2. B.N6
  3. C.N9
  4. D.N12
  5. E.15
Show the answer and explanation

Answer: D (N12)

Explanation

The amount given by the most students (15 students) is N12.

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Question 48

Measures of Central Tendency

The table shows the amount of money (in naira) collected through voluntary donations in a secondary school.
Amount (in naira): 3, 6, 9, 12, 15, 18
No. of Students: 3, 9, 6, 15, 3, 12
Find the median of distribution

  1. A.N3.00
  2. B.N9.00
  3. C.N12.00
  4. D.N15.00
  5. E.N18.00
Show the answer and explanation

Answer: C (N12.00)

Explanation

There are 48 students, so the median is between the 24th and 25th. Cumulative frequencies are 3, 12, 18, 33, … so both lie in the N12 group.

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Question 49

Probability

Two groups of male students cast their vote on a particular proposal. The results are as follows:
Group A: In favour 128, Against 32
Group B: In favour 96, Against 48
If a student in favour of the proposal is selected for a post, what is the probability that he is from group A?

  1. A.8/19
  2. B.16/35
  3. C.10/19
  4. D.4/7
  5. E.4/5
Show the answer and explanation

Answer: D (4/7)

Explanation

224 students are in favour and 128 of them are in group A: 128/224 = 4/7.

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Question 50

Probability

Two groups of male students cast their vote on a particular proposal. The results are as follows:
Group A: In favour 128, Against 32
Group B: In favour 96, Against 48
If a student is chosen at random, what is probability that he is against the proposal?

  1. A.3/19
  2. B.4/19
  3. C.5/19
  4. D.9/19
  5. E.10/19
Show the answer and explanation

Answer: C (5/19)

Explanation

80 of the 304 students are against: 80/304 = 5/19.

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Theory question 101

Fractions, Decimals and Approximation

(a) Simplify (0.016 × 0.084)/0.48 leaving your answer in standard form.
(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.

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Model answer

(a) 0.016 × 0.084 = 0.001344; 0.001344 ÷ 0.48 = 0.0028 = 2.8 × 10−3
(b) a + d = 2 and a + 5d = 5, so 4d = 3, d = 0.75m and a = 1.25m
Lengths: 1.25m, 2m, 2.75m, 3.5m, 4.25m, 5m, 5.75m, 6.5m

Explanation

In (b), subtract the two term equations to find d first.

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Theory question 102

Inequalities

Show on a graph, the area which gives the solution set of the inequalities:
y − 2x ≤ 4, 3y + x ≥ 6, y ≥ 7x − 9.

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Model answer

Draw the boundary lines y = 2x + 4 (through (0, 4) and (−2, 0)), x + 3y = 6 (through (0, 2) and (6, 0)) and y = 7x − 9 (through (1, −2) and (2, 5)).
Shade the unwanted side of each line: above y = 2x + 4, below x + 3y = 6, below y = 7x − 9 (or shade the wanted region, stating which).
The solution set is the triangle with vertices about (−0.86, 2.29), (1.5, 1.5) and (2.6, 9.2).
Check with a point inside, e.g. (1, 4): 4 − 2 ≤ 4, 12 + 1 ≥ 6, 4 ≥ −2, all true.

Explanation

Test a point such as (1, 4) in each inequality to know which side to keep.

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Theory question 103

Triangles and Quadrilaterals

(a) Prove that the sum of the angles of a triangle is two right angles
(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that ∠CAE + ∠CBD = 2∠CEB.

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(a) Given triangle ABC. Draw a line through A parallel to BC, making angles p (beside B) and q (beside C) at A.
p = ∠B and q = ∠C (alternate angles). p + ∠A + q = 180° (angles on a straight line). So ∠A + ∠B + ∠C = 180° = two right angles.
(b) Let ∠ACE = ∠ECB = c (CE bisects ∠ACB) and ∠CAE = a.
∠CBD is an exterior angle of triangle ABC: ∠CBD = a + 2c.
∠CEB is an exterior angle of triangle AEC: ∠CEB = a + c.
So ∠CAE + ∠CBD = a + a + 2c = 2(a + c) = 2∠CEB.

Explanation

An exterior angle of a triangle equals the sum of the two interior opposite angles.

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Theory question 104

Fractions, Decimals and Approximation

A carpenter was told to make a rectangular desk with top, of dimension 50cm by 40cm. The carpenter actually made the desk 60cm by 35cm.
(a) Calculate the percentage error in the: (i) length and the breadth; (ii) area of the table top.
(b) Find the product of the two errors in (a)(i).

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(a)(i) Length: error = 10cm, 10/50 × 100 = 20%. Breadth: error = 5cm, 5/40 × 100 = 12.5%
(ii) Intended area = 2000cm2, actual = 2100cm2; error = 100cm2; 100/2000 × 100 = 5%
(b) 20 × 12.5 = 250 (as fractions 0.2 × 0.125 = 0.025)

Explanation

Percentage error = error ÷ the correct (intended) value × 100.

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Theory question 105

Trigonometry

The feet of two vertical poles of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30°. Calculate the: (i) distance RT; (ii) distance of the foot of the taller pole from P, correct to three significant figures; (iii) angle of elevation of T from P, correct to one decimal place.

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Height of T above R = 7 − 3 = 4m
(i) sin 30° = 4/RT, so RT = 4/0.5 = 8m
(ii) Distance between the poles = 4/tan 30° = 6.928m; distance from P = 20 + 6.928 = 26.9m
(iii) tan θ = 7/26.93 = 0.2600, θ = 14.6°

Explanation

Draw the horizontal through R; the right-angled triangle above it has height 4m.

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Theory question 106

Mensuration of Solids

(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. (Take π = 3.142)
(b) A hemispherical tank of diameter 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.

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(a) V = ⅓πr2h = ⅓ × 3.142 × 16 × 6 = 100.5cm3 (≈ 101cm3)
(b) Tank volume = ⅔πr3 = ⅔ × 3.142 × 53 = 261.8m3
Water per second = π(0.2)2 × 2 = 0.2514m3
Time = 261.8 ÷ 0.2514 = 1041.7s = 17.4 minutes

Explanation

Change 20cm to 0.2m before working out the flow.

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Theory question 107

Measures of Dispersion

The weights to the nearest kilogram of a group of 50 students in a College of Technology are given below:
65, 70, 60, 46, 51, 55, 59, 63, 68, 53
47, 53, 72, 53, 67, 62, 64, 70, 57, 56
73, 56, 48, 51, 58, 63, 65, 62, 49, 64
53, 59, 63, 50, 48, 72, 67, 56, 61, 64
66, 52, 49, 62, 71, 58, 53, 69, 63, 59
(a) Prepare a grouped frequency table with class intervals 45 − 49, 50 − 54, 55 − 59, etc.
(b) Using an assumed mean of 62 or otherwise, calculate the mean and the standard deviation of the grouped data, correct to one decimal place.

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(a) 45−49: 6; 50−54: 9; 55−59: 10; 60−64: 12; 65−69: 7; 70−74: 6 (total 50)
(b) Class marks 47, 52, 57, 62, 67, 72; d = x − 62: −15, −10, −5, 0, 5, 10
Σfd = −90 − 90 − 50 + 0 + 35 + 60 = −135; mean = 62 − 135/50 = 59.3kg
Σfd2 = 1350 + 900 + 250 + 0 + 175 + 600 = 3275
SD = √(3275/50 − (−2.7)2) = √(65.5 − 7.29) = √58.21 = 7.6kg

Explanation

Use tally marks so that no weight is missed or counted twice.

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Theory question 108

Constructions

ABCD is a trapezium in which AB||DC, |AB| = 8cm, ∠ABC = 60°, |BC| = 5.5cm and |BD| = 8.3cm.
(a) Using a ruler and a pair of compasses only, construct (i) the trapezium ABCD; (ii) a rectangle PQCD, where P, Q are two points on AB.
(b) Measure |AC| and |QB|

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(a)(i) Draw AB = 8cm; construct 60° at B and mark C with BC = 5.5cm; construct a line through C parallel to AB; with centre B and radius 8.3cm cut this line at D; join AD. Construction arcs must be shown.
(ii) Drop perpendiculars from D and C to AB, meeting AB at P and Q; PQCD is the rectangle.
(b) |AC| ≈ 7.1cm; |QB| ≈ 2.75cm (accept 2.7−2.8cm)

Explanation

To draw the parallel through C, copy the angle at B or construct two perpendiculars.

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Theory question 109

Probability

(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.
(b) A bag contains 10 balls that differ only in colour; 4 are blue and the others are red. Two balls are picked, one after the other with replacement. What is the probability that (i) both are red? (ii) both are the same colour?

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(a) 21 integers. Multiples of 5: 5, 10, 15, 20, 25. Multiples of 3: 6, 9, 12, 15, 18, 21, 24. 15 is in both, so 11 numbers. P = 11/21
(b) P(red) = 6/10, P(blue) = 4/10
(i) 6/10 × 6/10 = 9/25
(ii) 9/25 + (4/10 × 4/10) = 9/25 + 4/25 = 13/25

Explanation

Count a number like 15 that is a multiple of both only once.

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Theory question 110

Graphs of Functions

The following is an incomplete table for the relation y = 2x2 − 5x + 1
x: −3, −2, −1, 0, 1, 2, 3, 4, 5
y: _, _, 8, 1, −2, _, _, _, 26
(a) Copy and complete the table.
(b) Using a scale of 2cm to 1 unit on the x axis and 2cm to 10 units on the y axis, draw a graph of the relation y = 2x2 − 5x + 1 for −3 ≤ x ≤ 5.
(c) Using the same scale and axes, draw the graph of y = x + 6.
(d) Estimate from your graphs, correct to one decimal place: (i) the least value of y and the value of x for which it occurs; (ii) the solution of the equation 2x2 − 5x + 1 = x + 6.

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(a) y = 34, 19, 8, 1, −2, −1, 4, 13, 26
(b) Smooth U-shaped curve through the plotted points with the given scales
(c) Straight line through e.g. (−3, 3), (0, 6), (5, 11)
(d)(i) Least value of y ≈ −2.1 at x ≈ 1.3 (exact: −2.125 at x = 1.25)
(ii) Read x at the intersections: x ≈ −0.7 or x ≈ 3.7 (since 2x2 − 6x − 5 = 0)

Explanation

The solutions of the equation are the x-values where the curve and the line cross.

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Theory question 111

Sets

In a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speaks at least one of these three languages. If 8 speak Hausa and Igbo, 11 speak Hausa and Yoruba and 6 speak Igbo and Yoruba
(a) Draw a Venn diagram to illustrate this information, using x to represent the number of students that speak all three;
(b) calculate the value of x.

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(a) Three overlapping circles H, I, Y in a rectangle of 40. Regions: H∩I only = 8 − x; H∩Y only = 11 − x; I∩Y only = 6 − x; centre = x; H only = 25 − (8 − x) − (11 − x) − x = 6 + x; I only = 2 + x; Y only = 4 + x
(b) Sum of regions = 40: (6 + x) + (2 + x) + (4 + x) + (8 − x) + (11 − x) + (6 − x) + x = 37 + x = 40
x = 3

Explanation

Fill the centre first, then the pairs, then the 'only' regions.

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Theory question 112

Latitude and Longitude

An aeroplane flies from a town P (lat. 40°N, 38°E) to another town Q (lat. 40°N, 22°W). It later flies to a third town T (28°N, 22°W). Calculate the: (a) distance between P and Q along their parallel of latitude; (b) distance between Q and T along the line of longitudes. (c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to three significant figures. (Take the radius of the earth to be 6400km and π = 3.142)

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(a) Difference in longitude = 38° + 22° = 60°
PQ = 60/360 × 2 × 3.142 × 6400 × cos 40° = 5135km ≈ 5130km
(b) Difference in latitude = 40° − 28° = 12°
QT = 12/360 × 2 × 3.142 × 6400 = 1341km ≈ 1340km
(c) Total distance ≈ 6475km; speed = 6475 ÷ 12 ≈ 540km/h

Explanation

Along a parallel of latitude use the radius R cos(latitude); along a meridian use R.

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