WAEC Mathematics 1998 past questions and answers

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

Question 1

Sets

If S⊂R, then

  1. A.S∩R = R
  2. B.S∩R = S
  3. C.S∪R = S
  4. D.S∩R = S'
  5. E.S∩R = R'
Show the answer and explanation

Answer: B (S∩R = S)

Explanation

If S is a subset of R, every element of S is already in R, so the common part of S and R is S itself.

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Question 2

Number Bases

Convert 35 to a number in base two.

  1. A.1011two
  2. B.10011two
  3. C.100011two
  4. D.110010two
  5. E.110001two
Show the answer and explanation

Answer: C (100011two)

Explanation

35 = 32 + 2 + 1 = 1×25 + 0 + 0 + 0 + 1×2 + 1, so 35 = 100011two.

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Question 3

Progressions

The nth term of a sequence is represented by 3 × 2(2−n). Write down the first three terms of the sequence

  1. A.3/2, 3, 6
  2. B.6, 3, 3/2
  3. C.3/2, 3, 1/3
  4. D.2/3, 3, 8/3
  5. E.3/2, 0, 6.
Show the answer and explanation

Answer: B (6, 3, 3/2)

Explanation

n = 1: 3 × 21 = 6; n = 2: 3 × 20 = 3; n = 3: 3 × 2−1 = 3/2.

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Question 6

Ratio, Percentages and Rates

A piece of cloth was measured as 6.10m. If the actual length of the cloth is 6.35m, find the percentage error, correct to 2 decimal places.

  1. A.3.05%
  2. B.3.94%
  3. C.15.00%
  4. D.25.00%
  5. E.39.30%.
Show the answer and explanation

Answer: B (3.94%)

Explanation

Error = 6.35 − 6.10 = 0.25 m. Percentage error = 0.25/6.35 × 100% ≈ 3.94%.

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Question 7

Indices

Simplify: (82/3 × 27−1/3)/641/3

  1. A.−3
  2. B.1/9
  3. C.1/3
  4. D.27/8
  5. E.4.
Show the answer and explanation

Answer: C (1/3)

Explanation

82/3 = 4, 27−1/3 = 1/3 and 641/3 = 4, so the value is 4 × 1/3 ÷ 4 = 1/3.

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Question 8

Quadratic Equations

Solve the equation 5x2 − 4x − 1 = 0

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

Answer: D (1, −1/5)

Explanation

5x2 − 4x − 1 = (5x + 1)(x − 1) = 0, so x = 1 or x = −1/5.

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Question 9

Inequalities

p and q are two positive numbers such that p > 2q. Which one of the following statements is not true?

  1. A.−p < −2q
  2. B.−p > −2q
  3. C.−q < 2p
  4. D.q < ½p
  5. E.p2 > 2q2.
Show the answer and explanation

Answer: B (−p > −2q)

Explanation

Multiplying p > 2q by −1 reverses the sign: −p < −2q, so −p > −2q is false.

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Question 10

Algebraic Expressions

For what values of x is the expression (x − 5)/(x2 − 2x − 3) not defined?

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

Answer: C (−1, 3)

Explanation

A fraction is undefined when its denominator is 0: x2 − 2x − 3 = (x − 3)(x + 1) = 0 gives x = 3 or −1.

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Question 11

Algebraic Expressions

Make S the subject of the formula: V = K/√(T − S)

  1. A.S = T − K2/V2
  2. B.S = (K2 − T)/V2
  3. C.S = T − V2/K2
  4. D.S = T(V2 − K2)/V2
  5. E.S = (T + V2)/K2
Show the answer and explanation

Answer: A (S = T − K2/V2)

Explanation

√(T − S) = K/V, so T − S = K2/V2 and S = T − K2/V2.

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Question 13

Linear and Simultaneous Equations

Find (x − y) if 4x − 3y = 7 and 3x − 2y = 5.

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

Answer: C (2)

Explanation

Solving gives x = 1 and y = −1 (check: 4 + 3 = 7, 3 + 2 = 5), so x − y = 2.

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Question 15

Algebraic Expressions

If h(m + n) = m(h + r) find h in terms of m, n and r

  1. A.h = mr/(2m + n)
  2. B.h = mr/(n + m)
  3. C.h = (m + r)/n
  4. D.h = (m + n)/m
  5. E.h = mr/n
Show the answer and explanation

Answer: E (h = mr/n)

Explanation

Expand: hm + hn = hm + mr. The hm terms cancel, leaving hn = mr, so h = mr/n.

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Question 16

Linear and Simultaneous Equations

Given that (6x − y)/(x + 2y) = 2, find the value of x/y

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

Answer: D (5/4)

Explanation

6x − y = 2x + 4y, so 4x = 5y and x/y = 5/4.

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Question 17

Latitude and Longitude

The radius of a geographical globe is 60cm. Find the length of the parallel of Latitude 60°N.

  1. A.66πcm
  2. B.60πcm
  3. C.30πcm
  4. D.15πcm
  5. E.6πcm.
Show the answer and explanation

Answer: B (60πcm)

Explanation

Radius of the parallel = 60 cos 60° = 30 cm, so its length = 2π × 30 = 60π cm.

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Question 18

Mensuration of Plane Shapes

In the diagram below, O is the centre of the two concentric circles of radii 13cm and 10cm respectively. Find the area of the shaded portion in the sector with an angle of 120° at the centre.

Diagram for this question
  1. A.1/3πcm2
  2. B.πcm2
  3. C.3πcm2
  4. D.23πcm2
  5. E.46πcm2.
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Answer: D (23πcm2)

Explanation

Shaded area = 120/360 × π(132 − 102) = 1/3 × 69π = 23π cm2.

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Question 19

Algebraic Expressions

Express as a single fraction: x/(x − 2) − (x + 2)/(x + 3)

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

Answer: D ((3x + 4)/((x − 2)(x + 3)))

Explanation

Numerator = x(x + 3) − (x + 2)(x − 2) = x2 + 3x − x2 + 4 = 3x + 4.

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Question 20

Data Presentation

The following numbers represent a set of scores for a class of 32 students, where the maximum score possible was 12, 6, 5, 9, 4, 4, 8, 7, 5, 6, 3, 2, 5, 4, 6, 9, 10, 4, 3, 2, 3, 4, 6, 8, 7, 4, 2, 1, 8, 7, 7, 6, 11. What percentage of the class, correct to the nearest whole number, scored above 6?

  1. A.34%
  2. B.35%
  3. C.37%
  4. D.38%
  5. E.50%.
Show the answer and explanation

Answer: A (34%)

Explanation

The first 12 is the maximum possible score; the 32 scores after it include 11 above 6 (9, 8, 7, 9, 10, 8, 7, 8, 7, 7, 11). 11/32 × 100% ≈ 34%.

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Question 21

Probability

A die with faces numbered 1 to 6 is rolled once. What is the probability of obtaining 3 or 4?

  1. A.1/36
  2. B.1/6
  3. C.1/3
  4. D.1/2
  5. E.2/3.
Show the answer and explanation

Answer: C (1/3)

Explanation

Two of the six faces are wanted: 2/6 = 1/3.

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Question 23

Graphs of Functions

Use the graph to answer this question. The graph is that of the quadratic expression

Diagram for this question
  1. A.y = x2 + x − 2
  2. B.y = x2 − x + 2
  3. C.y = x2 − x − 2
  4. D.y = x + 2x + 2
  5. E.y = x2 + 2x − 2
Show the answer and explanation

Answer: C (y = x2 − x − 2)

Explanation

The curve crosses the x-axis at −1 and 2, so y = (x + 1)(x − 2) = x2 − x − 2 (it also passes through (0, −2)).

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Question 24

Graphs of Functions

Use the graph to answer this question. The values of x when y = −1 are approximately

Diagram for this question
  1. A.−1.0 and 2.0
  2. B.−1.3 and 2.3
  3. C.−0.6 and 2.6
  4. D.−0.6 and 1.6
  5. E.1.6 and 1.6.
Show the answer and explanation

Answer: D (−0.6 and 1.6)

Explanation

Draw the line y = −1 across the graph; it meets the curve at about x = −0.6 and x = 1.6 (x2 − x − 1 = 0).

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Question 26

Trigonometry

In ΔPQR, ∠PQR is a right angle. |QR| = 2cm and ∠PRQ = 60°. Find |PR|.

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

Answer: B (4cm)

Explanation

QR is adjacent to the 60° angle: cos 60° = QR/PR, so PR = 2 ÷ ½ = 4 cm.

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Question 28

Trigonometry

In sin (X + 30)° = Cos 40°, find X.

  1. A.10°
  2. B.20°
  3. C.50°
  4. D.60°
  5. E.90°.
Show the answer and explanation

Answer: B (20°)

Explanation

sin A = cos B when A + B = 90°, so X + 30 + 40 = 90 and X = 20°.

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Question 30

Trigonometry

A ladder 5cm long rests against a wall such that its foot makes an angle 30° with the horizontal. How far is the foot of the ladder from the wall?

  1. A.5√2/3 m
  2. B.2½m
  3. C.5√3/2 m
  4. D.10√3/3 m
  5. E.10√3m
Show the answer and explanation

Answer: C (5√3/2 m)

Explanation

The ladder is the hypotenuse. Distance from wall = 5 × cos 30° = 5 × √3/2 = 5√3/2 m.

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Question 31

Circle Theorems

A chord of a circle of radius 26cm is 10cm from the centre of the circle. Calculate the length of the chord

  1. A.16cm
  2. B.27.86cm
  3. C.32cm
  4. D.48cm
  5. E.55.72cm.
Show the answer and explanation

Answer: D (48cm)

Explanation

Half the chord = √(262 − 102) = √576 = 24 cm, so the chord is 48 cm.

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Question 32

Circle Theorems

Two chords PQ and RS of a circle intersect at right angles at a point inside the circle. If ∠QPR = 35°, find ∠PQS

  1. A.35°
  2. B.45°
  3. C.55°
  4. D.70°
  5. E.90°.
Show the answer and explanation

Answer: C (55°)

Explanation

In the triangle formed by P, R and the crossing point, the angles are 35° and 90°, so ∠PRS = 55°. ∠PQS = ∠PRS (angles on the same arc PS) = 55°.

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Question 33

Circle Theorems

In the diagram, O is the centre of the circle and the reflex angle ROS is 264°. Find ∠RTS.

Diagram for this question
  1. A.48°
  2. B.53°
  3. C.60°
  4. D.132°
  5. E.192°.
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Answer: A (48°)

Explanation

The non-reflex angle ROS = 360 − 264 = 96°. The angle at the circumference is half the angle at the centre: 48°.

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Question 34

Angles and Polygons

If the exterior angles of a quadrilateral are y°, (y + 5)°, (y + 10)° and (y + 25)°, find y.

  1. A.10°
  2. B.20°
  3. C.40°
  4. D.60°
  5. E.80°.
Show the answer and explanation

Answer: E (80°.)

Explanation

Exterior angles of any polygon add to 360°: 4y + 40 = 360, so y = 80°.

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Question 35

Circle Theorems

In the diagram PQ is a diameter of circle PMQN centre O. If ∠PQM = 63°, find ∠MNQ.

Diagram for this question
  1. A.58.5°
  2. B.44°
  3. C.35°
  4. D.31.5°
  5. E.27°.
Show the answer and explanation

Answer: E (27°.)

Explanation

∠PMQ = 90° (angle in a semicircle), so ∠MPQ = 27°. ∠MNQ = ∠MPQ (same arc MQ) = 27°.

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Question 36

Mensuration of Plane Shapes

In the diagram, O is the centre of circle PRQ. The radius is 3.5cm and ∠POQ = 50°. (Take π = 3.142) Calculate, correct to one decimal place, the length of arc PQ.

Diagram for this question
  1. A.175.0 cm
  2. B.157.1cm
  3. C.37.7 cm
  4. D.11.0cm
  5. E.3.1cm.
Show the answer and explanation

Answer: E (3.1cm.)

Explanation

Arc PQ = 50/360 × 2 × 3.142 × 3.5 ≈ 3.05, which is 3.1 cm to one decimal place.

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Question 37

Mensuration of Plane Shapes

In the diagram, O is the centre of circle PRQ. The radius is 3.5cm and ∠POQ = 50°. (Take π = 3.142) Calculate, correct to three significant figures, the area of sector OPQ.

Diagram for this question
  1. A.3.403cm2
  2. B.5.350cm2
  3. C.10.99cm2
  4. D.157.1cm2
  5. E.175.0cm2.
Show the answer and explanation

Answer: B (5.350cm2)

Explanation

Area = 50/360 × 3.142 × 3.52 ≈ 5.35 cm2.

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Question 41

Data Presentation

The pie chart below shows the distribution of how students travelled to a certain school on a particular day. What percentage, to the nearest whole number, of the students travelled to school on foot?

Diagram for this question
  1. A.24%
  2. B.23%
  3. C.20%
  4. D.15%
  5. E.14%.
Show the answer and explanation

Answer: A (24%)

Explanation

The foot sector is 85°: 85/360 × 100% ≈ 23.6%, which is 24% to the nearest whole number.

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Question 42

Mensuration of Plane Shapes

The diagram below is a rectangle. If the perimeter is 36m, find the area of the rectangle.

Diagram for this question
  1. A.64m2
  2. B.65m2
  3. C.84m2
  4. D.124m2
  5. E.275m2
Show the answer and explanation

Answer: B (65m2)

Explanation

2[(2x + 5) + (x + 1)] = 36 gives 3x + 6 = 18, x = 4. The sides are 13 m and 5 m, so the area is 65 m2.

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Question 43

Triangles and Quadrilaterals

In Δ PQR, T is a point on QR such that ∠QPT = 39° and ∠PTR = 83°. Calculate ∠PQT.

  1. A.34°
  2. B.44°
  3. C.45°
  4. D.97°
  5. E.107°.
Show the answer and explanation

Answer: B (44°)

Explanation

∠PTR is an exterior angle of triangle PQT, so it equals ∠QPT + ∠PQT: ∠PQT = 83 − 39 = 44°.

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Question 47

Mensuration of Solids

Find the curved surface area of a cone of radius 3cm and slant height 7 cm. (Take π = 22/7)

  1. A.22cm2
  2. B.44cm2
  3. C.66cm2
  4. D.132cm2
  5. E.198cm2
Show the answer and explanation

Answer: C (66cm2)

Explanation

Curved surface = πrl = 22/7 × 3 × 7 = 66 cm2.

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Question 48

Triangles and Quadrilaterals

In the diagram below, |PQ| = |PR| = |RS| and ∠RSP = 35°. Find ∠QPR.

Diagram for this question
  1. A.70°
  2. B.50°
  3. C.45°
  4. D.40°
  5. E.35°.
Show the answer and explanation

Answer: D (40°)

Explanation

PR = RS, so ∠RPS = ∠RSP = 35° and the exterior angle ∠PRQ = 70°. PQ = PR, so ∠PQR = 70° too, and ∠QPR = 180 − 140 = 40°.

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Question 49

Variation

P varies inversely as Q. The table below shows the value of Q for some selected values of P.
P: 6, 8, 9, 12
Q: 24, 18, ?, 12
What is the missing value of Q in the table?

  1. A.4
  2. B.9
  3. C.15
  4. D.16
  5. E.144.
Show the answer and explanation

Answer: D (16)

Explanation

For inverse variation PQ is constant: 6 × 24 = 144. So Q = 144 ÷ 9 = 16.

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Question 50

Angles and Polygons

Three of the angles of a hexagon are each x°. The others are each 3x°. Find x.

  1. A.30°
  2. B.40°
  3. C.60°
  4. D.80°
  5. E.120°.
Show the answer and explanation

Answer: C (60°)

Explanation

Hexagon angles add to 720°: 3x + 9x = 720, so x = 60°.

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Theory question 101

Sets

A, B and C are subsets of the universal set U such that:
U = {0, 1, 2, 3, ……, 12}
A = {x: 0 ≤ x ≤ 7}
B = {4, 6, 8, 10, 12}
C = {1 < y < 8}, where y is a prime number
(a) Draw a Venn diagram to illustrate the information given above.
(b) Find: (i) (B∪C)' (ii) A'∩B∩C

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Model answer

A = {0, 1, 2, 3, 4, 5, 6, 7}, B = {4, 6, 8, 10, 12}, C = {2, 3, 5, 7}.
(a) C lies inside A; A and B overlap in {4, 6}; B and C do not overlap. Place 0, 1 in A only; 2, 3, 5, 7 in C (inside A); 4, 6 in A∩B; 8, 10, 12 in B only; 9, 11 outside all three.
(b)(i) B∪C = {2, 3, 4, 5, 6, 7, 8, 10, 12}, so (B∪C)' = {0, 1, 9, 11}.
(ii) B∩C = ∅ (no even number in C is prime except 2, which is not in B), so A'∩B∩C = ∅ = { }.

Explanation

List each set in full before drawing the Venn diagram.

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Theory question 102

Logarithms

(a) Given that log102 = 0.3010, log107 = 0.8451 and log105 = 0.6990, evaluate without using logarithm tables. (i) log1035; (ii) log102.8
(b) Given that N0.8942 = 2.8, use your result in (a)(ii) to find the value of N.

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Model answer

(a)(i) log 35 = log 7 + log 5 = 0.8451 + 0.6990 = 1.5441.
(ii) log 2.8 = log (7 × 4/10) = log 7 + 2 log 2 − log 10 = 0.8451 + 0.6020 − 1 = 0.4471.
(b) 0.8942 log N = log 2.8 = 0.4471, so log N = 0.5.
N = 100.5 = √10 ≈ 3.162.

Explanation

Break each number into factors whose logs you are given.

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Theory question 103

Quadratic Equations

(a) The value of the expression 2Ax − Kx2 is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.
(b) Solve the equation x2 − 3x − 1 = 0, giving your answers correct to 1 decimal place.

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Model answer

(a) x = 1: 2A − K = 7. x = 2: 4A − 4K = 4, i.e. A − K = 1.
Subtract: A = 6, then K = 5.
(b) x = [3 ± √(9 + 4)]/2 = (3 ± √13)/2 = (3 ± 3.606)/2.
x ≈ 3.3 or x ≈ −0.3.

Explanation

Substitute each given x value to get two simultaneous equations.

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Theory question 104

Quadratic Equations

The area of a rectangular floor is 13.5m2. One side is 1.5m longer than the other.
(a) Calculate the dimensions of the floor
(b) If it costs N250.00 per square metre to carpet the floor and only N2,200.00 is available, what area of the floor can be covered with carpet?

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Model answer

(a) Let the shorter side be x m: x(x + 1.5) = 13.5, so 2x2 + 3x − 27 = 0, (2x + 9)(x − 3) = 0.
x = 3 (a length cannot be negative). The floor is 3 m by 4.5 m.
(b) Area covered = 2200 ÷ 250 = 8.8 m2.

Explanation

Form a quadratic from length × width = area and reject the negative root.

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Theory question 105

Probability

(a) A number is selected at random from each of the sets {2,3,4} and {1,3,5}. What is the probability that the sum of the two numbers will be less than 7 but greater than 3?
(b) In the diagram above, ABCD is a circle. DAE, CBE, ABF and DCF are straight lines. If y + m = 90°, find the value of x.

Diagram for this question
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Model answer

(a) The 9 equally likely sums are 3, 5, 7, 4, 6, 8, 5, 7, 9. Those greater than 3 and less than 7 are 5, 4, 6, 5: probability = 4/9.
(b) Let m = ∠AEB and x = ∠ABE = ∠CBF (vertically opposite).
Exterior angle of triangle ABE: ∠DAB = x + m. Exterior angle of triangle BCF: ∠DCB = x + y.
ABCD is cyclic, so (x + m) + (x + y) = 180°: 2x + 90° = 180°, x = 45°.

Explanation

Use exterior angles of the two small triangles, then opposite angles of the cyclic quadrilateral.

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Theory question 106

Progressions

The third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the first four terms.

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Model answer

ar2 = 360 and ar5 = 1215. Divide: r3 = 1215/360 = 27/8, so r = 3/2.
a = 360 ÷ (3/2)2 = 360 × 4/9 = 160.
First four terms: 160, 240, 360, 540.

Explanation

Dividing the two term equations removes a and leaves r³.

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Theory question 107

Commercial Arithmetic

(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sale by 10%. Find (i) the final sale price (ii) the percentage profit.
(b) Solve the equation 2(2x+1) − 9(2x) + 4 = 0

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Model answer

(a)(i) Marked price = 130% of N6,000 = N7,800. Sale price = 90% of N7,800 = N7,020.
(ii) Profit = 7,020 − 6,000 = N1,020; percentage profit = 1020/6000 × 100% = 17%.
(b) 22x+1 = 2(2x)2. Let y = 2x: 2y2 − 9y + 4 = 0, (2y − 1)(y − 4) = 0.
y = ½ or 4, so 2x = 2−1 or 22: x = −1 or x = 2.

Explanation

Percentage profit is always worked out on the cost price.

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Theory question 108

Graphs of Functions

(a) Copy and complete the table of values for the relation y = 5 − 7x − 6x2 for −3 ≤ x ≤ 2.
x: −3, −2, −1, −0.5, 0, 1, 2
y: −28, , 6, , 5, ,
(b) Using a scale of 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis, draw the: (i) graph of y = 5 − 7x − 6x2 (ii) line y = 3 on the same axes.
(c) Use your graph to find the; (i) Roots of the equation 2 − 7x − 6x2 = 0 (ii) Maximum value of y = 5 − 7x − 6x2

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Model answer

(a) x = −2: y = −5; x = −0.5: y = 7; x = 1: y = −8; x = 2: y = −33 (with −28, 6 and 5 given).
(b) Plot the points, join with a smooth upside-down parabola, and draw the horizontal line y = 3.
(c)(i) 2 − 7x − 6x2 = 0 is the same as 5 − 7x − 6x2 = 3, so read x where the curve meets y = 3: x ≈ −1.4 and x ≈ 0.2 (exact: −1.40 and 0.24).
(ii) Maximum value ≈ 7.0 (at x ≈ −0.6).

Explanation

Rewrite the new equation so one side matches the drawn curve; the other side is the line to draw.

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Theory question 109

Constructions

Using a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 5cm and |QT| = 4.5cm (b) a quadrilateral QRSP which has a common base QR with ΔQRT such that QTP is a straight line, PQ||SR, |QP| = 9cm and |RS| = 4.5cm (i) Measure |PS| (ii) Find the perpendicular distance between RS and PQ. (iii) What is QRSP?

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Model answer

(a) Draw QR = 8 cm. With centre Q radius 4.5 cm and centre R radius 5 cm draw arcs meeting at T; join QT and RT.
(b) Produce QT to P with QP = 9 cm. At R construct a line parallel to QP (copy ∠RQP at R on the same side) and mark S on it with RS = 4.5 cm. Join PS.
(i) |PS| ≈ 5.0 cm (TPSR is a parallelogram, so PS = TR = 5 cm).
(ii) Drop a perpendicular from R to QP: about 4.5 cm (calculated 4.55 cm).
(iii) QRSP is a trapezium (one pair of parallel sides, PQ ∥ SR).

Explanation

Notice TP = 9 − 4.5 = 4.5 cm = RS, which makes TPSR a parallelogram — a useful check on the measurement.

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Theory question 110

Bearings

A surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point, W, a distance of 180m in the south-east direction, he observes the bearing of Z and Y to be 337° and 050° respectively. (a) Calculate, correct to the nearest metre (i) |XY| (ii) |ZW|. (b) If N is on XY such that XZ = ZN, find the bearing of Z from N.

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Model answer

(a)(i) In triangle XWY: ∠YXW = 45° (east to south-east), ∠XWY = (360 − 315) + 50 = 95° (X is on bearing 315° from W), so ∠XYW = 180 − 45 − 95 = 40°.
Sine rule: XY/sin 95° = 180/sin 40°, XY = 180 × 0.9962/0.6428 ≈ 279 m.
(ii) In triangle XWZ: ∠ZXW = 135 − 46 = 89°, ∠XWZ = 337 − 315 = 22°, so ∠XZW = 69°.
ZW/sin 89° = 180/sin 69°, ZW = 180 × 0.9998/0.9336 ≈ 193 m.
(b) ∠ZXN = 90 − 46 = 44°. XZ = ZN, so ∠ZNX = 44° too. From N, X is due west (270°), so the bearing of Z from N is 270° + 44° = 314°.

Explanation

Draw a clear sketch with north lines at X and W before finding any angles.

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Theory question 111

Mensuration of Plane Shapes

In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.
(a) Calculate (i) the obtuse angle SOT, (ii) arc SRT (iii) |BT|.
(b) Find, correct to the nearest cm, the length of the string. [Take π = 22/7].

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Model answer

(a)(i) ASOM is a square (radii meet tangents at 90°), so ∠SOM = 90°. In OMBT, ∠MOT = 360 − 90 − 90 − 60 = 120°. Obtuse ∠SOT = 360 − 90 − 120 = 150°.
(ii) Arc SRT = 150/360 × 2 × 22/7 × 10 ≈ 26.19 cm.
(iii) OB bisects ∠TBM, so ∠OBT = 30° and BT = 10/tan 30° = 10√3 ≈ 17.32 cm.
(b) String = AS + arc SRT + TB = 10 + 26.19 + 17.32 = 53.51 ≈ 54 cm.

Explanation

Tangents from a point are equal, and a radius meets a tangent at 90°.

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Theory question 112

Data Presentation

The table below shows the number of eggs laid by chicken in a man's farm in a year.
No. of Eggs Per Year: 45–49, 50–54, 55–59, 60–64, 65–69, 70–74
No. of Chicken: 10, 36, 64, 52, 28, 10
(a) Draw a cumulative frequency curve of the distribution. (b) Use your graph to find the interquartile range. (c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?

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Model answer

(a) Upper class boundaries 49.5, 54.5, 59.5, 64.5, 69.5, 74.5 with cumulative frequencies 10, 46, 110, 162, 190, 200; plot (44.5, 0) and these points and draw a smooth curve.
(b) Q1 at 50th chicken ≈ 54.5 + 4/64 × 5 ≈ 54.8; Q3 at 150th ≈ 59.5 + 40/52 × 5 ≈ 63.3. Interquartile range ≈ 63.3 − 54.8 ≈ 8.5 (accept values read from the graph).
(c) Chickens laying 60 or more eggs = 52 + 28 + 10 = 90, so P = 90/200 = 9/20 = 0.45.

Explanation

For quartiles of 200 items, read across at 50 and 150 on the cumulative frequency axis.

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