WAEC Mathematics 2014 past questions and answers

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

Question 1

Fractions, Decimals and Approximation

Simplify: 10⅖ − 6⅔ + 3.

  1. A.6 4/15
  2. B.6 11/15
  3. C.7 4/15
  4. D.7 11/15
Show the answer and explanation

Answer: B (6 11/15)

Explanation

52/5 − 20/3 + 3 = (156 − 100 + 45)/15 = 101/15 = 6 11/15.

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Question 2

Number Bases

If 23x = 325, find the value of x.

  1. A.7
  2. B.6
  3. C.5
  4. D.4
Show the answer and explanation

Answer: A (7)

Explanation

2x + 3 = 3 × 5 + 2 = 17, so 2x = 14 and x = 7.

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Question 4

Data Presentation

The bar chart shows the scores of some students in a test. How many students took the test?

Diagram for this question
  1. A.18
  2. B.19
  3. C.20
  4. D.22
Show the answer and explanation

Answer: C (20)

Explanation

Add the bar heights for scores 0 to 4: 4 + 2 + 5 + 6 + 3 = 20.

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Question 5

Probability

The bar chart shows the scores of some students in a test. If one student is selected at random, find the probability that he/she scored at most 2 marks.

Diagram for this question
  1. A.11/18
  2. B.11/20
  3. C.7/22
  4. D.5/19
Show the answer and explanation

Answer: B (11/20)

Explanation

Scores 0, 1 or 2: 4 + 2 + 5 = 11 of the 20 students, so the probability is 11/20.

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Question 6

Surds

Simplify: √12(√48 − √3).

  1. A.18
  2. B.16
  3. C.14
  4. D.12
Show the answer and explanation

Answer: A (18)

Explanation

√12 = 2√3 and √48 − √3 = 4√3 − √3 = 3√3. 2√3 × 3√3 = 18.

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Question 8

Triangles and Quadrilaterals

In the diagram, the value of x + y = 220°. Find the value of n.

Diagram for this question
  1. A.20°
  2. B.40°
  3. C.60°
  4. D.80°
Show the answer and explanation

Answer: B (40°)

Explanation

x and y are exterior angles at the base. The interior base angles are 180° − x and 180° − y, which add to 360° − 220° = 140°. So n = 180° − 140° = 40°.

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Question 9

Inequalities

Given that x > y and 3 < y, which of the following is/are true?
I. y > 3
II. x < 3
III. x > y > 3

  1. A.I only
  2. B.I and II 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

3 < y means y > 3 (I), and x > y > 3 (III). Then x > 3, so II is false.

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Question 10

Linear and Simultaneous Equations

Three quarters of a number added to two and a half of that number gives 13. Find the missing number.

  1. A.4
  2. B.5
  3. C.6
  4. D.7
Show the answer and explanation

Answer: A (4)

Explanation

¾n + 2½n = 13 gives (13/4)n = 13, so n = 4.

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Question 11

Sets

If X = {0, 2, 4, 6}, Y = {1, 2, 3, 4} and Z = {1, 3} are subsets of U = {x: 0 ≤ x ≤ 6}, find X ∩ (Y′ ∪ Z).

  1. A.{0, 2, 6}
  2. B.{1, 3}
  3. C.{0, 6}
  4. D.{ }
Show the answer and explanation

Answer: C ({0, 6})

Explanation

Y′ = {0, 5, 6}, so Y′ ∪ Z = {0, 1, 3, 5, 6}. The elements of X in it are 0 and 6.

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Question 12

Quadratic Equations

Find the truth set of the equation x2 = 3(2x + 9).

  1. A.{x : x = 3, x = 9}
  2. B.{x : x = −3, x = −9}
  3. C.{x : x = 3, x = −9}
  4. D.{x : x = −3, x = 9}
Show the answer and explanation

Answer: D ({x : x = −3, x = 9})

Explanation

x2 − 6x − 27 = 0 factorises as (x − 9)(x + 3) = 0, so x = 9 or x = −3.

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Question 13

Coordinate Geometry

The coordinates of points P and Q are (4,3) and (2, −1) respectively. Find the shortest distance between P and Q.

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

Answer: D (2√5)

Explanation

|PQ| = √((4 − 2)2 + (3 + 1)2) = √(4 + 16) = √20 = 2√5.

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Question 14

Algebraic Expressions

Make u the subject of the formula, E = (m/2g)(v2 − u2).

  1. A.u = √(v2 − 2Eg/m)
  2. B.u = √(v2/m − 2Eg/4)
  3. C.u = √(v − 2Eg/m)
  4. D.u = √(2v2Eg/m)
Show the answer and explanation

Answer: A (u = √(v2 − 2Eg/m))

Explanation

2gE = m(v2 − u2), so v2 − u2 = 2Eg/m and u = √(v2 − 2Eg/m).

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Question 15

Triangles and Quadrilaterals

In the diagram, ∠QPT = ∠PTS = 90°, ∠PQR = 110° and ∠TSR = 20°. Find the size of the obtuse angle QRS.

Diagram for this question
  1. A.140°
  2. B.130°
  3. C.120°
  4. D.110°
Show the answer and explanation

Answer: B (130°)

Explanation

PQRST is a pentagon, so its angles add to 540°: the reflex angle at R = 540° − 90° − 90° − 110° − 20° = 230°. The obtuse angle QRS = 360° − 230° = 130°.

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Question 16

Variation

If x varies inversely as y and y varies directly as z, what is the relationship between x and z?

  1. A.x ∝ z
  2. B.x ∝ 1/z
  3. C.x ∝ z2
  4. D.x ∝ 1/z2
Show the answer and explanation

Answer: B (x ∝ 1/z)

Explanation

x = k/y and y = cz, so x = k/(cz): x varies inversely as z.

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Question 17

Coordinate Geometry

Find the gradient of the line joining the points (2, −3) and (2, 5).

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

Answer: D (undefined.)

Explanation

Gradient = (5 + 3)/(2 − 2) = 8/0, which is undefined (the line is vertical).

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Question 18

Polynomials

If (x − a) is a factor of bx − ax + x2 − ab, find the other factor.

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

Answer: A ((x + b))

Explanation

x2 + bx − ax − ab = x(x + b) − a(x + b) = (x − a)(x + b).

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Question 19

Measures of Central Tendency

The table shows the distribution of the height of plants in a nursery.
Height: 2, 3, 4, 5, 6
Frequency: 2, 4, 5, 3, 1
Calculate the mean height of the plants.

  1. A.3.8
  2. B.3.0
  3. C.2.8
  4. D.2.3
Show the answer and explanation

Answer: A (3.8)

Explanation

Σfx = 4 + 12 + 20 + 15 + 6 = 57 and Σf = 15. Mean = 57/15 = 3.8.

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Question 20

Angles and Polygons

In the diagram, PQR is a straight line, (m + n) = 120° and (n + r) = 100°. Find (m + r).

Diagram for this question
  1. A.110°
  2. B.120°
  3. C.140°
  4. D.160°
Show the answer and explanation

Answer: C (140°)

Explanation

m + n + r = 180°. Adding the two given sums: m + 2n + r = 220°, so n = 40° and m + r = 140°.

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Question 21

Angles and Polygons

In the diagram, SR is parallel to UW, ∠WVT = x°, ∠VUT = y°, ∠RSV = 45° and ∠VTU = 20°. Find the value of x.

Diagram for this question
  1. A.20
  2. B.45
  3. C.65
  4. D.135
Show the answer and explanation

Answer: B (45)

Explanation

SVT is a straight line crossing the parallel lines SR and UW, so x equals the corresponding angle ∠RSV = 45°.

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Question 22

Angles and Polygons

In the diagram, SR is parallel to UW, ∠WVT = x°, ∠VUT = y°, ∠RSV = 45° and ∠VTU = 20°. Calculate the value of y.

Diagram for this question
  1. A.20
  2. B.25
  3. C.45
  4. D.65
Show the answer and explanation

Answer: B (25)

Explanation

x = 45° is an exterior angle of triangle VUT, so x = y + 20°. Hence y = 25.

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Question 23

Mensuration of Solids

The area of a sector of a circle with diameter 12 cm is 66 cm2. If the sector is folded to form a cone, calculate the radius of the base of the cone. [Take π = 22/7]

  1. A.3.0 cm
  2. B.3.5 cm
  3. C.7.0 cm
  4. D.7.5 cm
Show the answer and explanation

Answer: B (3.5 cm)

Explanation

The sector's radius 6 cm becomes the slant height. The cone's curved area πrl = 66, so 22/7 × r × 6 = 66 and r = 3.5 cm.

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Question 24

Circle Theorems

A chord, 7 cm long, is drawn in a circle with radius 3.7 cm. Calculate the distance of the chord from the centre of the circle.

  1. A.0.7 cm
  2. B.1.2 cm
  3. C.2.0 cm
  4. D.2.5 cm
Show the answer and explanation

Answer: B (1.2 cm)

Explanation

Half the chord is 3.5 cm. Distance = √(3.72 − 3.52) = √1.44 = 1.2 cm.

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Question 25

Measures of Dispersion

Which of the following is a measure of dispersion?

  1. A.Range
  2. B.Percentile
  3. C.Median
  4. D.Quartile
Show the answer and explanation

Answer: A (Range)

Explanation

The range measures how spread out the data are; the others describe positions in the data.

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Question 26

Linear and Simultaneous Equations

A box contains 13 currency notes, all of which are either ₦50 or ₦20 notes. The total value of the currency notes is ₦530. How many ₦50 notes are in the box?

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

Answer: D (9)

Explanation

50x + 20(13 − x) = 530 gives 30x + 260 = 530, so x = 9.

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Question 27

Graphs of Functions

The graph is for the relation y = 2x2 + x − 1. What are the coordinates of the point S?

Diagram for this question
  1. A.(1, 0.2)
  2. B.(1, 0.4)
  3. C.(1, 2.0)
  4. D.(1, 4.0)
Show the answer and explanation

Answer: C ((1, 2.0))

Explanation

All options have x = 1. On the curve, y = 2(1)2 + 1 − 1 = 2, so S is (1, 2.0).

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Question 29

Bearings

A ship sails x km due east to a point E and continues x km due north to F. Find the bearing of F from the starting point.

  1. A.045°
  2. B.090°
  3. C.135°
  4. D.225°
Show the answer and explanation

Answer: A (045°)

Explanation

Equal distances east and north make a 45° angle with north, so the bearing is 045°.

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Question 30

Ratio, Percentages and Rates

If x : y = 3 : 2 and y : z = 5 : 4, find the value of x in the ratio x : y : z.

  1. A.8
  2. B.10
  3. C.15
  4. D.20
Show the answer and explanation

Answer: C (15)

Explanation

Make y equal: x : y = 15 : 10 and y : z = 10 : 8, so x : y : z = 15 : 10 : 8 and x = 15.

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Question 31

Commercial Arithmetic

A trader bought sachet water for GH¢ 55.00 per dozen and sold them at 10 for GH¢ 50.00. Calculate, correct to 2 decimal places, his percentage gain.

  1. A.8.00%
  2. B.8.30%
  3. C.9.09%
  4. D.10.00%
Show the answer and explanation

Answer: C (9.09%)

Explanation

Cost per sachet = 55/12 ≈ GH¢4.583; selling price = GH¢5.00. Gain = 0.417/4.583 × 100% ≈ 9.09%.

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Question 32

Circle Theorems

In the figure, PQ is a tangent to the circle at R and UT is parallel to PQ. If ∠TRQ = x°, find ∠URT in terms of x.

Diagram for this question
  1. A.2x°
  2. B.(90 − x)°
  3. C.(90 + x)°
  4. D.(180 − 2x)°
Show the answer and explanation

Answer: D ((180 − 2x)°)

Explanation

UT ∥ PQ gives ∠UTR = ∠TRQ = x (alternate). The alternate segment theorem gives ∠TUR = ∠TRQ = x. So ∠URT = 180° − 2x.

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Question 33

Trigonometry

Given that cos x = 12/13, evaluate (1 − tan x)/tan x.

  1. A.5/13
  2. B.5/7
  3. C.7/5
  4. D.13/5
Show the answer and explanation

Answer: C (7/5)

Explanation

tan x = 5/12, so (1 − 5/12)/(5/12) = (7/12) × (12/5) = 7/5.

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Question 34

Fractions, Decimals and Approximation

Approximate 0.0033780 to 3 significant figures.

  1. A.338
  2. B.0.338
  3. C.0.00338
  4. D.0.003
Show the answer and explanation

Answer: C (0.00338)

Explanation

The significant figures start at 3: 3, 3, 7, and the next digit 8 rounds the 7 up, giving 0.00338.

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Question 35

Indices

Simplify: √((82 × 4n+1)/(22n × 16))

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

Answer: C (4)

Explanation

In powers of 2: (26 × 22n+2)/(22n × 24) = 24 = 16, and √16 = 4.

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Question 36

Algebraic Expressions

If 2/(x − 3) − 3/(x − 2) is equal to P/((x − 3)(x − 2)), find P.

  1. A.−x − 5
  2. B.−(x + 3)
  3. C.5x − 13
  4. D.5 − x
Show the answer and explanation

Answer: D (5 − x)

Explanation

2(x − 2) − 3(x − 3) = 2x − 4 − 3x + 9 = 5 − x.

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Question 37

Algebraic Expressions

Subtract ½(a − b − c) from the sum of ½(a − b + c) and ½(a + b − c).

  1. A.½(a + b + c)
  2. B.½(a − b − c)
  3. C.½(a − b + c)
  4. D.½(a + b − c)
Show the answer and explanation

Answer: A (½(a + b + c))

Explanation

The sum is ½(2a) = a. Then a − ½(a − b − c) = ½(a + b + c).

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Question 38

Trigonometry

A man's eye level is 1.7 m above the horizontal ground and 13 m from a vertical pole. If the pole is 8.3 m high, calculate, correct to the nearest degree, the angle of elevation of the top of the pole from his eyes.

  1. A.33°
  2. B.32°
  3. C.27°
  4. D.26°
Show the answer and explanation

Answer: C (27°)

Explanation

The top is 8.3 − 1.7 = 6.6 m above eye level. tan θ = 6.6/13 ≈ 0.5077, so θ ≈ 27°.

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Question 39

Mensuration of Plane Shapes

A chord subtends an angle of 120° at the centre of a circle of radius 3.5 cm. Find the perimeter of the minor sector containing the chord. [Take π = 22/7].

  1. A.14⅓ cm
  2. B.12⅚ cm
  3. C.8⅐ cm
  4. D.7⅓ cm
Show the answer and explanation

Answer: A (14⅓ cm)

Explanation

Arc = 120/360 × 2 × 22/7 × 3.5 = 22/3 cm. Perimeter = 22/3 + 3.5 + 3.5 = 14⅓ cm.

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Question 40

Mensuration of Plane Shapes

In parallelogram, PQRS, QR is produced to M such that |QR| = |RM|. What fraction of the area of PQMS is the area of PRMS?

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

Answer: C (2/3)

Explanation

With base QR = b and height h: PQMS is a trapezium of area ½(b + 2b)h = 1.5bh. PRMS is a parallelogram (RM = PS and RM ∥ PS) of area bh. bh ÷ 1.5bh = 2/3.

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Question 41

Triangles and Quadrilaterals

Determine the value of m in the diagram.

Diagram for this question
  1. A.80°
  2. B.90°
  3. C.110°
  4. D.150°
Show the answer and explanation

Answer: B (90°)

Explanation

The marked sides make the cyclic quadrilateral a kite symmetric about the diagonal through the 80° angle, so the angle opposite m equals m. Opposite angles of a cyclic quadrilateral add to 180°, so 2m = 180° and m = 90°.

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Question 42

Measures of Dispersion

In a cumulative frequency graph, the lower quartile is 18 years while the 60th percentile is 48 years. What percentage of the distribution is at most 18 years or greater than 48 years.

  1. A.15%
  2. B.35%
  3. C.65%
  4. D.85%
Show the answer and explanation

Answer: C (65%)

Explanation

25% are at most 18 (lower quartile) and 100% − 60% = 40% are above 48. Total 65%.

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Question 43

Probability

If a number is selected at random from each of the sets P = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime.

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

Answer: B (4/9)

Explanation

The 9 sums are 3, 4, 6, 4, 5, 7, 5, 6, 8. The primes are 3, 5, 7 and 5, so the probability is 4/9.

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Question 44

Circle Theorems

In the diagram, O is the centre of the circle, PR is a tangent to the circle at Q and ∠SOQ = 86°. Calculate the value of ∠SQR.

Diagram for this question
  1. A.43°
  2. B.47°
  3. C.54°
  4. D.86°
Show the answer and explanation

Answer: A (43°)

Explanation

OS = OQ, so ∠OQS = (180° − 86°)/2 = 47°. The radius OQ is perpendicular to the tangent, so ∠SQR = 90° − 47° = 43°.

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Question 45

Logarithms

If log 5.957 = 0.7750, find log ∛0.0005957.

  1. A.4̄.1986
  2. B.2̄.9250
  3. C.1̄.5917
  4. D.1̄.2853
Show the answer and explanation

Answer: B (2̄.9250)

Explanation

log 0.0005957 = 4̄.7750 = −3.225. Divide by 3: −1.075 = 2̄.9250.

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Question 46

Probability

The probability of an event P happening is 1/5 and that of event Q is 1/4. If the events are independent, what is the probability that neither of them happens?

  1. A.4/5
  2. B.3/4
  3. C.3/5
  4. D.1/20
Show the answer and explanation

Answer: C (3/5)

Explanation

P(neither) = (1 − 1/5)(1 − 1/4) = 4/5 × 3/4 = 3/5.

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Question 47

Angles and Polygons

Each exterior angle of a polygon is 30°. Calculate the sum of the interior angles.

  1. A.540°
  2. B.720°
  3. C.1080°
  4. D.1800°
Show the answer and explanation

Answer: D (1800°)

Explanation

n = 360 ÷ 30 = 12 sides. Sum = (12 − 2) × 180° = 1800°.

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Question 48

Progressions

Find the number of terms in the Arithmetic Progression (A.P.) 2, −9, −20, …, −141.

  1. A.11
  2. B.12
  3. C.13
  4. D.14
Show the answer and explanation

Answer: D (14)

Explanation

a = 2, d = −11: 2 + (n − 1)(−11) = −141, so n − 1 = 13 and n = 14.

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Question 49

Modular Arithmetic

In what modulus is it true that 9 + 8 = 5?

  1. A.mod 10
  2. B.mod 11
  3. C.mod 12
  4. D.mod 13
Show the answer and explanation

Answer: C (mod 12)

Explanation

9 + 8 = 17 and 17 − 12 = 5, so 17 ≡ 5 (mod 12).

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Question 50

Mensuration of Solids

The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in P is 10 cm high, what would be the height of the same quantity of water in Q?

  1. A.2.5 cm
  2. B.5.0 cm
  3. C.7.5 cm
  4. D.20.0cm
Show the answer and explanation

Answer: A (2.5 cm)

Explanation

πr2 × 10 = π(2r)2 × h = 4πr2h, so h = 10/4 = 2.5 cm.

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Theory question 101

Fractions, Decimals and Approximation

(a) Without using tables or calculator, simplify: (0.6 × 32 × 0.004)/(1.2 × 0.008 × 0.16) leaving the answer in standard form (scientific notation).
(b) In the diagram, EF is parallel to GH. If ∠AEF = 3x°, ∠ABC = 120° and ∠CHG = 7x°, find the value of ∠GHB.

Diagram for this question
Show the model answer
Model answer

(a) Numerator = 0.0768; denominator = 0.001536. 0.0768 ÷ 0.001536 = 50 = 5.0 × 101.
(b) Through B draw a line parallel to EF and GH. BA makes 3x with it (corresponding to ∠AEF), so BC makes 120° − 3x with it.
∠GHB = 120° − 3x (alternate angles).
B, H, C are on a straight line: (120 − 3x) + 7x = 180 → x = 15.
∠GHB = 120° − 45° = 75°.

Explanation

Draw a line through the bend parallel to the given lines.

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Theory question 102

Surds

(a) Simplify 3√75 − √12 + √108, leaving the answer in surd form (radicals).
(b) If 124n = 232five, find n.

Show the model answer
Model answer

(a) 3√75 = 15√3, √12 = 2√3, √108 = 6√3. 15√3 − 2√3 + 6√3 = 19√3.
(b) 232five = 50 + 15 + 2 = 67. n2 + 2n + 4 = 67 → n2 + 2n − 63 = 0 → (n + 9)(n − 7) = 0. A base must be positive, so n = 7.

Explanation

Write each surd as a multiple of √3.

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Theory question 103

Linear and Simultaneous Equations

(a) Solve the simultaneous equations: 1/x + 1/y = 5; 1/y − 1/x = 1.
(b) A man drives from Ibadan to Oyo, a distance of 48 km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometres of good surface.

Show the model answer
Model answer

(a) Adding: 2/y = 6, so y = ⅓. Then 1/x = 5 − 3 = 2, so x = ½.
(b) Let x km be good surface: x/72 + (48 − x)/48 = ¾.
Multiply by 144: 2x + 3(48 − x) = 108 → 144 − x = 108 → x = 36 km.

Explanation

Time = distance ÷ speed for each part; the times add to 45 minutes = ¾ hour.

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Theory question 104

Mensuration of Plane Shapes

(a) In the diagram, O is the centre of the circle radius r cm and ∠XOY = 90°. If the area of the shaded part is 504 cm2, calculate the value of r. [Take π = 22/7]
(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If ∠PQR = 66° and ∠PSQ = 109°, calculate the value of ∠RQS.

Diagram for this question
Show the model answer
Model answer

(a) Segment = sector − triangle: (90/360)(22/7)r2 − ½r2 = (11/14 − 7/14)r2 = (2/7)r2 = 504.
r2 = 1764, so r = 42 cm.
(b) In isosceles ΔPQS (PS = QS): ∠SQP = (180° − 109°)/2 = 35.5°.
∠RQS = ∠PQR + ∠PQS = 66° + 35.5° = 101.5°.

Explanation

The shaded part is a minor segment: sector minus the right-angled triangle.

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Theory question 105

Probability

A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3. What is the probability that he will win: (a) both contracts; (b) exactly one of the contracts; (c) neither of the contracts?

Show the model answer
Model answer

P(X) = 0.5, P(not X) = 0.5, P(Y) = 0.7, P(not Y) = 0.3.
(a) 0.5 × 0.7 = 0.35.
(b) 0.5 × 0.3 + 0.5 × 0.7 = 0.15 + 0.35 = 0.50.
(c) 0.5 × 0.3 = 0.15.

Explanation

For independent events, multiply the probabilities.

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Theory question 106

Linear and Simultaneous Equations

(a) If 3/(2p − ½) = (⅓)/(¼p + 1), find p.
(b) A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.

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(a) Cross-multiply: 3(¼p + 1) = ⅓(2p − ½) → ¾p + 3 = ⅔p − ⅙ → (1/12)p = −19/6 → p = −38.
(b) Cost price = 760 × 100/120 ≈ GH¢633.33. Selling price = 95% of 760 = GH¢722.
Profit = 722 − 633.33 = GH¢88.67. Percentage profit = 88.67/633.33 × 100% ≈ 14%.

Explanation

Find the cost price from the marked price first.

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Theory question 107

Trigonometry

(a) Copy and complete the table of values for the relation y = 2 sin x + 1.
x: 0°, 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°
y: 1.0, __, __, __, 2.7, __, __, 0.0, −0.7, __
(b) Using scales of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of y = 2 sin x + 1 for 0° ≤ x ≤ 270°.
(c) Use the graph to find the values of x for which sin x = ¼.

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Model answer

(a) 30°: 2.0; 60°: 2.7; 90°: 3.0; 150°: 2.0; 180°: 1.0; 270°: −1.0.
(b) Plot the points and draw a smooth curve.
(c) sin x = ¼ means 2 sin x + 1 = 1.5. Draw y = 1.5: x ≈ 15° and x ≈ 165°.

Explanation

Change the equation so it matches the graph: 2(¼) + 1 = 1.5.

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Theory question 108

Modular Arithmetic

(a) Copy and complete the following table for multiplication modulo 11.
⊗: 1, 5, 9, 10
1: 1, 5, 9, 10
5: 5, __, __, __
9: 9, __, __, __
10: 10, __, __, __
Use the table to: (i) evaluate (9 ⊗ 5) ⊗ (10 ⊗ 10); (ii) find the truth set of I. 10 ⊗ m = 2; II. n ⊗ n = 4.
(b) When a fraction is reduced to its lowest term, it is equal to ¾. The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.

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(a) Row 5: 5, 3, 1, 6. Row 9: 9, 1, 4, 2. Row 10: 10, 6, 2, 1.
(i) 9 ⊗ 5 = 1 and 10 ⊗ 10 = 1, so the answer is 1 ⊗ 1 = 1.
(ii) I. 10 ⊗ m = 2 → m = 9, truth set {9}. II. n ⊗ n = 4 → n = 9, truth set {9}.
(b) Let the fraction be 3k/4k. 2(3k) = 4k + 34 → k = 17. The fraction is 51/68.

Explanation

Each entry is the remainder when the ordinary product is divided by 11.

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Theory question 109

Sets

(a) In the Venn diagram, P, Q and R are subsets of the universal set U. If n(U) = 125, find: (i) the value of x; (ii) n(P ∪ Q ∩ R′).
(b) In the diagram, O is the centre of the circle. If WX is parallel to YZ and ∠WXY = 50°, find the value of: (i) ∠WYZ; (ii) ∠YEZ.

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(a)(i) (16 − 2x) + 5x + (6 + x) + 4x + 8x + 7x + (19 − 3x) + 4 = 125 → 45 + 20x = 125 → x = 4.
(ii) (P ∪ Q) ∩ R′ = the parts of P and Q outside R: (16 − 2x) + 5x + (6 + x) = 22 + 4x = 38.
(b) XW is a diameter, so ∠XYW = 90° and ∠XWY = 180° − 90° − 50° = 40°.
(i) WX ∥ YZ, so ∠WYZ = ∠XWY = 40° (alternate).
(ii) ∠WOZ = 2 × ∠WYZ = 80°. In ΔOEW: ∠OEW = 180° − 80° − 40° = 60°. ∠YEZ = ∠OEW = 60° (vertically opposite).

Explanation

The angle in a semicircle is 90°; the angle at the centre is twice the angle at the circumference.

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Theory question 110

Mensuration of Plane Shapes

(a) Solve: (x − 2)(x − 3) = 12.
(b) In the diagram, M and N are the centres of two circles of equal radii 7 cm. The circles intercept at P and Q. If ∠PMQ = ∠PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take π = 22/7].

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(a) x2 − 5x + 6 = 12 → x2 − 5x − 6 = 0 → (x − 6)(x + 1) = 0 → x = 6 or x = −1.
(b) One segment = sector − triangle = 60/360 × 22/7 × 72 − ½ × 72 × sin 60° = 25.667 − 21.218 = 4.449 cm2.
The shaded part is two such segments: 2 × 4.449 ≈ 8.9 ≈ 9 cm2.

Explanation

The shaded lens is made of two equal segments, one from each circle.

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Theory question 111

Probability

The table shows the distribution of outcomes when a die is thrown 50 times.
Scores: 1, 2, 3, 4, 5, 6
Frequency: 2, 5, 13, 11, 9, 10
Calculate the: (a) mean deviation of the distribution; (b) probability that a score selected at random is at least a 4.

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Model answer

(a) Σfx = 2 + 10 + 39 + 44 + 45 + 60 = 200, so the mean = 200/50 = 4.
|x − 4|: 3, 2, 1, 0, 1, 2. Σf|x − 4| = 6 + 10 + 13 + 0 + 9 + 20 = 58.
Mean deviation = 58/50 = 1.16.
(b) Scores 4, 5, 6: 11 + 9 + 10 = 30. Probability = 30/50 = 3/5 (0.6).

Explanation

Mean deviation is the average distance of the scores from the mean.

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Theory question 112

Trigonometry

(a) Given that 5 cos (x + 8.5)° − 1 = 0, 0° ≤ x ≤ 90°, calculate, correct to the nearest degree, the value of x.
(b) The bearing of Q from P is 150° and the bearing of P from R is 015°. If Q and R are 24 km and 32 km respectively from P: (i) represent this information in a diagram; (ii) calculate the distance between Q and R, correct to two decimal places; (iii) find the bearing of R from Q, correct to the nearest degree.

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Model answer

(a) cos(x + 8.5)° = 0.2 → x + 8.5 = 78.46 → x ≈ 70°.
(b)(i) Sketch with north lines at P, Q and R: PQ = 24 km on bearing 150° from P; R is 32 km from P on bearing 195° (back bearing of 015°).
(ii) ∠QPR = 195° − 150° = 45°. QR2 = 322 + 242 − 2(32)(24) cos 45° = 1600 − 1086.1 = 513.9, so QR ≈ 22.67 km.
(iii) sin ∠PQR = 32 sin 45°/22.67 ≈ 0.998, so ∠PQR ≈ 86.5° (the angles 45° + 86.5° + 48.5° add to 180°).
The bearing of P from Q is 330°, so the bearing of R from Q = 330° − 86.5° ≈ 243.5°, i.e. about 243° (244° accepted).

Explanation

Find the angle at P from the two bearings, then use the cosine rule.

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Theory question 113

Algebraic Expressions

(a) Two functions, f and g, are defined by f : x → 2x2 − 1 and g : x → 3x + 2 where x is a real number. (i) If f(x − 1) − 7 = 0, find the values of x. (ii) Evaluate: f(−½) . g(3)/(f(4) − g(5)).
(b) An operation (∗) is defined on the set R, of real numbers by m ∗ n = −n/(m2 + 1), where m, n ∈ R. If −3, −10 ∈ R, show whether or not (∗) is commutative.

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(a)(i) 2(x − 1)2 − 1 − 7 = 0 → (x − 1)2 = 4 → x − 1 = ±2 → x = 3 or x = −1.
(ii) f(−½) = −½, g(3) = 11, f(4) = 31, g(5) = 17. (−½ × 11)/(31 − 17) = −11/28.
(b) −3 ∗ −10 = 10/(9 + 1) = 1. −10 ∗ −3 = 3/(100 + 1) = 3/101.
Since 1 ≠ 3/101, the operation is not commutative.

Explanation

An operation is commutative only if swapping the two numbers never changes the result.

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