WAEC Mathematics 2016 past questions and answers

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

Question 1

Number Bases

If 23x + 101x = 130x, find the value of x.

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

Answer: D (4)

Explanation

In base ten: (2x + 3) + (x² + 1) = x² + 3x, so 2x + 4 = 3x and x = 4.

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Question 2

Fractions, Decimals and Approximation

Simplify: (¾ − ⅔) × 1⅕.

  1. A.1/60
  2. B.5/72
  3. C.1/10
  4. D.17/10
Show the answer and explanation

Answer: C (1/10)

Explanation

¾ − ⅔ = 1/12, and 1/12 × 6/5 = 6/60 = 1/10.

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Question 3

Surds

Simplify: (10√3/√5 − √15)2

  1. A.75.00
  2. B.15.00
  3. C.8.66
  4. D.3.87
Show the answer and explanation

Answer: B (15.00)

Explanation

10√3/√5 = 10√15/5 = 2√15, so the bracket is 2√15 − √15 = √15 and its square is 15.

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Question 4

Variation

The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45 cm in 3 seconds, how far will it fall in 6 seconds?

  1. A.90 cm
  2. B.135 cm
  3. C.180 cm
  4. D.225 cm
Show the answer and explanation

Answer: C (180 cm)

Explanation

d = kt²: 45 = 9k so k = 5; then d = 5 × 36 = 180 cm.

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Question 5

Logic

Which of the following is a vaild conclusion from the premise: 'Nigerian footballers are good footballers'?

  1. A.Joseph plays football in Nigeria therefore he is a good footballer.
  2. B.Joseph is a good footballer therefore he is a Nigerian footballer.
  3. C.Joseph is a Nigerian footballer therefore he is a good footballer.
  4. D.Joseph plays good football therefore he is a Nigerian footballer.
Show the answer and explanation

Answer: C (Joseph is a Nigerian footballer therefore he is a good footballer.)

Explanation

The set of Nigerian footballers lies inside the set of good footballers, so anyone who is a Nigerian footballer must be a good footballer.

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Question 6

Ratio, Percentages and Rates

On a map, 1 cm represents 5 km. Find the area on the map that respresents 100 km2.

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

Answer: B (4 cm2)

Explanation

1 cm² on the map is 5 × 5 = 25 km², so 100 km² is 100 ÷ 25 = 4 cm².

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Question 7

Indices

Simplify: (3n−1 × 27n+1)/81n

  1. A.32n
  2. B.9
  3. C.3n
  4. D.3n+1
Show the answer and explanation

Answer: B (9)

Explanation

3n − 1 × 33n + 3 ÷ 34n = 32 = 9.

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Question 8

Commercial Arithmetic

What sum of money will amount to D10,400.00 in 5 years at 6% simple interest?

  1. A.D8,000.00
  2. B.D10,000.00
  3. C.D12,000.00
  4. D.D16,000.00
Show the answer and explanation

Answer: A (D8,000.00)

Explanation

Interest for 5 years at 6% is 30% of P, so 1.3P = 10 400 and P = D8,000.

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Question 9

Inequalities

Which of the following number lines illustrates the solution of the inequality 4 ≤ ⅓(2x − 1) < 5?

Diagram for this question
  1. A.Open circle at 6½ with arrow to the left, and open circle at 8 with arrow to the right
  2. B.Closed circles at 6½ and 8
  3. C.Open circles at 6½ and 8
  4. D.Closed circle at 6½ and open circle at 8
Show the answer and explanation

Answer: D (Closed circle at 6½ and open circle at 8)

Explanation

12 ≤ 2x − 1 < 15 gives 6½ ≤ x < 8: a closed (shaded) circle at 6½ and an open circle at 8.

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Question 10

Quadratic Equations

The roots of a quadratic equation are 4/3 and −3/7. Find the equation.

  1. A.21x2 − 19x − 12 = 0
  2. B.21x2 + 37x − 12 = 0
  3. C.21x2 − x + 12 = 0
  4. D.21x2 + 7x − 4 = 0
Show the answer and explanation

Answer: A (21x2 − 19x − 12 = 0)

Explanation

(3x − 4)(7x + 3) = 21x² + 9x − 28x − 12 = 21x² − 19x − 12.

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Question 11

Algebraic Expressions

Find the values of y for which the expression (y2 − 9y + 18)/(y2 + 4y − 21) is undefined

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

Answer: C (3, −7)

Explanation

It is undefined when the denominator is zero: y² + 4y − 21 = (y − 3)(y + 7) = 0, so y = 3 or −7.

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Question 13

Algebraic Expressions

Simplify: 2/(1 − x) − 1/x.

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

Answer: B ((3x − 1)/(x(1 − x)))

Explanation

Over x(1 − x): (2x − (1 − x))/(x(1 − x)) = (3x − 1)/(x(1 − x)).

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Question 14

Algebraic Expressions

Make s the subject of the relation: p = s + sm2/nr

  1. A.s = mrp/(nr + m2)
  2. B.s = (nr + m2)/nrp
  3. C.s = nrp/(mr + m2)
  4. D.s = nrp/(nr + m2)
Show the answer and explanation

Answer: D (s = nrp/(nr + m2))

Explanation

p = s(nr + m²)/nr, so s = nrp/(nr + m²).

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Question 15

Algebraic Expressions

Factorize: (2x + 3y)2 − (x − 4y)2.

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

Answer: A ((3x − y)(x + 7y))

Explanation

Difference of two squares: (2x + 3y + x − 4y)(2x + 3y − x + 4y) = (3x − y)(x + 7y).

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Question 16

Mensuration of Solids

The curved surface area of a cylinder 5 cm high, is 110 cm2. Find the radius of its base. [Take π = 22/7]

  1. A.2.6 cm
  2. B.3.5 cm
  3. C.3.6 cm
  4. D.7.0 cm
Show the answer and explanation

Answer: B (3.5 cm)

Explanation

2πrh = 110: 2 × 22/7 × r × 5 = 110, so r = 3.5 cm.

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Question 17

Mensuration of Solids

The volume of a pyramid with height 15 cm is 90 cm3. If its base is a rectangle with dimensions x cm by 6 cm, find the value of x.

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

Answer: A (3)

Explanation

⅓ × 6x × 15 = 30x = 90, so x = 3.

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Question 18

Circle Theorems

In the diagram, YW is a tangent to the circle at X, |UV| = |VX| and ∠VXW = 50°. Find the value of ∠UXY

Diagram for this question
  1. A.70°
  2. B.80°
  3. C.105°
  4. D.110°
Show the answer and explanation

Answer: B (80°)

Explanation

∠VUX = ∠VXW = 50° (alternate segment). Triangle UVX is isosceles, so ∠UXV = 50° and ∠UVX = 80°; by the alternate segment theorem ∠UXY = ∠UVX = 80°.

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Question 19

Angles and Polygons

In the diagram, PF, QT, RG intersect at S and PQ/RG. If ∠SPQ = 113° and ∠RST = 22°, find ∠PSQ.

Diagram for this question
  1. A.22°
  2. B.45°
  3. C.67°
  4. D.89°
Show the answer and explanation

Answer: B (45°)

Explanation

∠QSG = ∠RST = 22° (vertically opposite). PQ//RG makes ∠SPQ + ∠PSG = 180°, so ∠PSG = 67° and ∠PSQ = 67° − 22° = 45°.

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Question 20

Circle Theorems

In the diagram, O, is the centre of circle, ∠XOZ = (10m)° and ∠XWZ = m°. Calculate the value of m.

Diagram for this question
  1. A.30
  2. B.36
  3. C.40
  4. D.72
Show the answer and explanation

Answer: A (30)

Explanation

∠XWZ = m stands on the arc XYZ, so the angle at the centre on that arc is 2m. The two angles at O add to 360°: 10m + 2m = 360, so m = 30.

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Question 21

Trigonometry

Kweku walked 8 m up a slope and was 3 m above the ground. If he walks 12 m further up the slope, how far above the ground will he be?

  1. A.4.5 m
  2. B.6.0 m
  3. C.7.5 m
  4. D.9.0 m
Show the answer and explanation

Answer: C (7.5 m)

Explanation

Height is proportional to distance along the slope: 3/8 × 20 = 7.5 m.

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Question 22

Circle Theorems

In the diagram, TS is a tangent to the circle at S, |PR| = |RS| and ∠PQR = 117°. Calculate ∠PST.

Diagram for this question
  1. A.54°
  2. B.44°
  3. C.34°
  4. D.27°
Show the answer and explanation

Answer: A (54°)

Explanation

∠PSR = 180° − 117° = 63° (cyclic quadrilateral). Triangle PRS is isosceles so ∠RPS = 63° and ∠PRS = 54°; ∠PST = ∠PRS = 54° (alternate segment).

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Question 23

Triangles and Quadrilaterals

In the diagram, PR/SV/WY, TX/QY, ∠PQT = 48° and ∠TXW = 60°. Find ∠TQU.

Diagram for this question
  1. A.120°
  2. B.108°
  3. C.72°
  4. D.60°
Show the answer and explanation

Answer: C (72°)

Explanation

Since QY//TX and PR//WY, ∠PQU = 180° − 60° = 120°. Then ∠TQU = 120° − 48° = 72°.

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Question 24

Coordinate Geometry

A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ

  1. A.3/5
  2. B.⅔
  3. C.3/2
  4. D.5/3
Show the answer and explanation

Answer: C (3/2)

Explanation

(8 − 2)/(5 − 1) = 6/4 = 3/2.

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Question 25

Coordinate Geometry

A straight line passes through the points P(1,2) and Q(5,8). Calculate the length PQ

  1. A.4√11
  2. B.4√10
  3. C.2√17
  4. D.2√13
Show the answer and explanation

Answer: D (2√13)

Explanation

PQ = √(4² + 6²) = √52 = 2√13.

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Question 26

Triangles and Quadrilaterals

In the diagram, TX is prependicular to UW, |UX| = 1 cm and |TX| = |WX| = √3 cm. Find ∠UTW.

Diagram for this question
  1. A.135°
  2. B.105°
  3. C.75°
  4. D.60°
Show the answer and explanation

Answer: C (75°)

Explanation

tan∠UTX = 1/√3 so ∠UTX = 30°; triangle TXW is right-angled and isosceles so ∠XTW = 45°. ∠UTW = 75°.

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Question 27

Trigonometry

If cos θ = x and sin 60° = x + 0.5, 0° < θ < 90°, find, correct to the nearest degree, the value of θ.

  1. A.66°
  2. B.67°
  3. C.68°
  4. D.69°
Show the answer and explanation

Answer: D (69°)

Explanation

x = 0.8660 − 0.5 = 0.3660, so θ = cos⁻¹ 0.3660 ≈ 68.5°, which is 69° to the nearest degree.

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Question 28

Measures of Central Tendency

The table shows the ages of students in a club.
Age (Years): 13, 14, 15, 16, 17
Frequency: 10, 24, 8, 5, 3
How many student are in the club?

  1. A.50
  2. B.55
  3. C.60
  4. D.65
Show the answer and explanation

Answer: A (50)

Explanation

10 + 24 + 8 + 5 + 3 = 50.

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Question 29

Measures of Central Tendency

The table shows the ages of students in a club.
Age (Years): 13, 14, 15, 16, 17
Frequency: 10, 24, 8, 5, 3
Find the median age:

  1. A.13
  2. B.14
  3. C.15
  4. D.16
Show the answer and explanation

Answer: B (14)

Explanation

With 50 students the median is the average of the 25th and 26th ages. The first 10 are 13 and the next 24 are 14, so both are 14.

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Question 30

Data Presentation

The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?

Diagram for this question
  1. A.Le 2,250,000.00
  2. B.Le 2,700,000.00
  3. C.Le 3,600,000.00
  4. D.Le 4,500,000.00
Show the answer and explanation

Answer: C (Le 3,600,000.00)

Explanation

Food's angle = 360° − (70° + 80° + 90°) = 120°, so food = 120/360 × 10 800 000 = Le 3,600,000.

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Question 31

Probability

A fair die is thrown two times. What is the probability that the sum of the score is at least 10?

  1. A.5/36
  2. B.1/6
  3. C.5/18
  4. D.⅔
Show the answer and explanation

Answer: B (1/6)

Explanation

Sums of 10, 11 or 12 come from (4,6), (5,5), (6,4), (5,6), (6,5), (6,6): 6 of 36 outcomes = 1/6.

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Question 32

Measures of Dispersion

The marks of eight students in a test are: 10, 4, 5, 3, 14, 16 and 7. Find the range.

  1. A.16
  2. B.14
  3. C.13
  4. D.11
Show the answer and explanation

Answer: C (13)

Explanation

Range = highest − lowest = 16 − 3 = 13.

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Question 33

Logarithms

If log2(3x − 1) = 5, find x.

  1. A.2.00
  2. B.3.67
  3. C.8.67
  4. D.11.00
Show the answer and explanation

Answer: D (11.00)

Explanation

3x − 1 = 2⁵ = 32, so 3x = 33 and x = 11.

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Question 34

Mensuration of Solids

A sphere of radius r cm has the same volume as a cylinder of radius 3 cm and height 4 cm. Find the value of r.

  1. A.⅔
  2. B.2
  3. C.3
  4. D.6
Show the answer and explanation

Answer: C (3)

Explanation

⁴⁄₃πr³ = π × 3² × 4 = 36π, so r³ = 27 and r = 3.

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Question 35

Fractions, Decimals and Approximation

Express 1975 correct to 2 significant figures.

  1. A.20
  2. B.1,900
  3. C.1,980
  4. D.2,000
Show the answer and explanation

Answer: D (2,000)

Explanation

The third figure is 7, so 19 rounds up to 20 hundreds: 1975 ≈ 2,000.

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Question 36

Mensuration of Plane Shapes

In the diagram, MOPQ is a trapezium with QP//MO, MQ//NP, NQ//OP, |QP| = 9 cm and the height of ΔQNP = 6 cm, calculate the area of the trapezium.

Diagram for this question
  1. A.96 cm2
  2. B.90 cm2
  3. C.81 cm2
  4. D.27 cm2
Show the answer and explanation

Answer: C (81 cm2)

Explanation

MNPQ and NOPQ are parallelograms, so MN = NO = 9 cm and MO = 18 cm. Area = ½(18 + 9) × 6 = 81 cm².

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Question 37

Mensuration of Plane Shapes

The perimeter of a sector of a circle of radius 21 cm is 64 cm. Find the angle of the sector. [Take π = 22/7]

  1. A.70°
  2. B.60°
  3. C.55°
  4. D.42°
Show the answer and explanation

Answer: B (60°)

Explanation

Arc = 64 − 42 = 22 cm. θ/360 × 2 × 22/7 × 21 = 22 gives θ/360 × 132 = 22, so θ = 60°.

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Question 38

Sets

Determine M′ ∩ N from the Venn diagram.

Diagram for this question
  1. A.{f, g}
  2. B.{e}
  3. C.{c, f, g}
  4. D.{e, f, g}
Show the answer and explanation

Answer: A ({f, g})

Explanation

M′ = {f, g, e} (everything outside M); the part of it inside N is {f, g}.

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Question 39

Modular Arithmetic

If 20 (mod 9) is equivalent to y (mod 6), find y.

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

Answer: B (2)

Explanation

20 = 2 × 9 + 2, so 20 ≡ 2 (mod 9); y = 2.

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Question 40

Algebraic Expressions

Simplify: ((p − r)2 − r2)/(2p2 − 4pr)

  1. A.½
  2. B.p − 2r
  3. C.1/(p − 2r)
  4. D.2p/(p − 2r)
Show the answer and explanation

Answer: A (½)

Explanation

Numerator = (p − r − r)(p − r + r) = p(p − 2r); denominator = 2p(p − 2r). The fraction is ½.

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Question 41

Circle Theorems

In the diagram, O is the centre of the circle, ∠QPS = 100°, ∠PSQ = 60° and ∠QSR = 80°. Calculate ∠SQR.

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

Answer: A (20°)

Explanation

∠QRS = 180° − 100° = 80° (opposite angles of cyclic quadrilateral). In triangle QRS, ∠SQR = 180° − 80° − 80° = 20°.

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Question 42

Probability

A bag contains 5 red and 4 blue identical balls. If two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second blue.

  1. A.2/9
  2. B.5/18
  3. C.20/81
  4. D.5/9
Show the answer and explanation

Answer: C (20/81)

Explanation

With replacement: 5/9 × 4/9 = 20/81.

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Question 44

Progressions

Find the next three terms of the sequence: 0, 1, 1, 2, 3, 5, 8,…

  1. A.13, 19, 23
  2. B.9, 11, 13
  3. C.11, 15, 19
  4. D.13, 21, 34
Show the answer and explanation

Answer: D (13, 21, 34)

Explanation

Each term is the sum of the two before it: 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34.

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Question 45

Measures of Dispersion

Find the lower quartile of the distribution illustrated by the cumulative frequency curve.

Diagram for this question
  1. A.17.5
  2. B.19.0
  3. C.27.5
  4. D.28.0
Show the answer and explanation

Answer: B (19.0)

Explanation

Total frequency is 600, so the lower quartile is at cumulative frequency 150; reading across to the curve and down gives about 19.

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Question 46

Angles and Polygons

The ratio of the exterior angle to the interior angle of a regular polygon is 1:11. How many sides has the polygon?

  1. A.30
  2. B.24
  3. C.18
  4. D.12
Show the answer and explanation

Answer: B (24)

Explanation

Exterior = 1/12 × 180° = 15°, so the number of sides = 360 ÷ 15 = 24.

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Question 47

Linear and Simultaneous Equations

Halima is n years old. Her brother's age is 5 years more than half of her age. How old is her brother?

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

Answer: D (n/2 + 5)

Explanation

Half her age is n/2; 5 more is n/2 + 5.

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Question 48

Circle Theorems

In the diagram, MN is a chord of a circle KMN centre O and radius 10 cm. If ∠MON = 140°, find correct to the nearest cm, the length of the chord MN.

Diagram for this question
  1. A.19 cm
  2. B.18cm
  3. C.17 cm
  4. D.12 cm
Show the answer and explanation

Answer: A (19 cm)

Explanation

MN = 2 × 10 × sin 70° ≈ 18.8 ≈ 19 cm.

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Question 50

Triangles and Quadrilaterals

From the diagram, which of the following is true?

Diagram for this question
  1. A.m + n + p = 180°
  2. B.m + n = 180°
  3. C.m − p + n
  4. D.n = m + p
Show the answer and explanation

Answer: D (n = m + p)

Explanation

n is an exterior angle of the triangle, and p equals the triangle's top angle (vertically opposite), so n = m + p.

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Theory question 101

Fractions, Decimals and Approximation

(a) Without using Mathematical tables or calculators, evaluate: (0.09 × 1.21)/(3.3 × 0.00025), leaving the answer in standard form (Scientific Notation)
(b) A principal of GH¢5,600 was deposited for 3 years at compound interest. If the interest earned was GH¢1,200.00, find, correct to 3 significant figures, the interest rate per annum.

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Model answer

(a) = (9 × 10⁻² × 1.21)/(3.3 × 2.5 × 10⁻⁴) = (10.89/8.25) × 10² = 1.32 × 10²
(b) Amount = 5600 + 1200 = GH¢6,800
5600(1 + r/100)³ = 6800 → (1 + r/100)³ = 1.2143
1 + r/100 = 1.0669 → r ≈ 6.69% per annum

Explanation

In (b) use the compound interest formula, not simple interest.

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Theory question 102

Inequalities

(a) Solve: 7(x + 4) − ⅔(x − 6) ≤ 2[x − 3(x + 5)].
(b) A transport company has a total 20 vehicles made up of tricycles and taxicabs. Each tricycle carries 2 passengers while each taxicab carries 4 passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?

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Model answer

(a) 7x + 28 − ⅔x + 4 ≤ 2(−2x − 15) = −4x − 30
7x − ⅔x + 4x ≤ −62 → (31/3)x ≤ −62 → x ≤ −6
(b) Let t = tricycles, c = taxicabs: t + c = 20 and 2t + 4c = 66
2t + 4(20 − t) = 66 → 80 − 2t = 66 → t = 7
The company has 7 tricycles (and 13 taxicabs).

Explanation

Remove the inner bracket first in (a).

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Theory question 103

Circle Theorems

(a) In the diagram, ∠RTS = 28°, ∠VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find ∠VUS.
(b) A cylindrical tin, 7 cm high, is closed at one end. If its total surface area is 462 cm2, calculate its radius. [Take π = 22/7]

Diagram for this question
Show the model answer
Model answer

(a) ∠VUR = ∠VRM = 46° (angle in the alternate segment)
∠RUS = ∠RTS = 28° (angles in the same segment)
∠VUS = 46° + 28° = 74°
(b) Total surface area = 2πrh + πr² = 462
22/7 × r(14 + r) = 462 → r² + 14r − 147 = 0 → (r − 7)(r + 21) = 0
r = 7 cm (r cannot be negative)

Explanation

Only one circular end is included because the tin is open at the other end.

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Theory question 104

Probability

The table shows the outcome when a die is thrown a number of times.
Scores: 1, 2, 3, 4, 5, 6
Frequency: 25, 30, x, 28, 40, 32
If the probability of obtaining a 3 is 0.225: (a) how many times was the die thrown? (b) calculate the probability that a trial chosen at random gives a score of an even number or a prime number.

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Model answer

(a) Total = 155 + x; x/(155 + x) = 0.225 → x = 0.225 × 155/0.775 = 45
The die was thrown 155 + 45 = 200 times.
(b) Even or prime scores are 2, 3, 4, 5, 6 (every score except 1)
Frequency = 200 − 25 = 175 → probability = 175/200 = 7/8

Explanation

Count each score once, even though 2 is both even and prime.

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Theory question 105

Mensuration of Plane Shapes

(a) In the diagram, PQST is a parallelogram, PQR is a straight line, |TS| = 8 cm, |SM| = 6 cm and area of triangle PSR = 36 cm2. Find the value of |QR|
(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angles of 45° and 60° respectively. If the tree is 10.65 m high, calculate, correct to 3 significant figures, the height of the flagpole.

Diagram for this question
Show the model answer
Model answer

(a) Area of triangle PSR = ½ × |PR| × 6 = 36 → |PR| = 12 cm
|PQ| = |TS| = 8 cm (opposite sides of parallelogram), so |QR| = 12 − 8 = 4 cm
(b) Horizontal distance d = 10.65/tan 60° ≈ 6.149 m
Drop from the bird to the top of the pole = d tan 45° ≈ 6.149 m
Height of flagpole = 10.65 − 6.149 ≈ 4.50 m

Explanation

Use the 60° angle (to the bottom) to find the horizontal distance first.

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Theory question 106

Progressions

(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5, … 101.
(b) Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all three means of transport. The number, x, of travellers who travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train: (i) represent this information in a Venn diagram; (ii) calculate the I. value of x; II. number who travelled by at least two means.

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(a) a = 1, d = 2: 101 = 1 + 2(n − 1) → n = 51
S = n/2 (a + l) = 51/2 × 102 = 2601
(b)(i) Three overlapping circles B, T, C in a universal set of 95: B only = x, B∩C only = x, B∩T only = 7, T∩C only = 3, all three = 8.
(ii) I. Bus: x + x + 7 + 8 = 47 → x = 16
II. At least two means = 7 + 3 + 8 + 16 = 34

Explanation

Fill the Venn diagram from the centre (all three) outwards.

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Theory question 107

Quadratic Equations

(a) Using completing the squares method, solve, correct to 2 decimal places, (x − 2)/4 = (x + 2)/2x.
(b) In the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, |QR| = |RS| and ∠QTS = 52°, find: (i) ∠SQT; (ii) ∠PQT.

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(a) 2x(x − 2) = 4(x + 2) → 2x² − 8x − 8 = 0 → x² − 4x = 4
x² − 4x + 4 = 8 → (x − 2)² = 8 → x − 2 = ±2.828
x ≈ 4.83 or x ≈ −0.83
(b)(i) ∠QRS = 180° − 52° = 128° (opposite angles of cyclic quadrilateral QRST)
Triangle QRS is isosceles: ∠RSQ = (180° − 128°)/2 = 26°
∠SQT = ∠RSQ = 26° (alternate angles, RS//QT)
(ii) ∠PQS = 90° (angle in a semicircle), so ∠PQT = 90° − 26° = 64°

Explanation

Add the square of half the coefficient of x to both sides.

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Theory question 108

Triangles and Quadrilaterals

(a) In the diagram, ∠KLM = x, ∠LMK = y, ∠KJH = r and ∠KGF = 110°. If 2x = r = y, find the value of x.
(b) Ten boys and twelve girls collected donations for a project. The total amount colected by the boys was N600.00 greater than that collected by the girls. If the average collection of the boys was N100.00 greater than the average collection of the girls, how much was collected by the two groups?

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(a) ∠KGJ = 180° − 110° = 70°
∠JKM = x + y (exterior angle of triangle KLM)
In triangle JKG: r + (x + y) + 70° = 180°
With r = y = 2x: 2x + x + 2x = 110° → x = 22°
(b) Let the girls collect N g; boys collect N(g + 600)
(g + 600)/10 = g/12 + 100 → 6(g + 600) = 5g + 6000 → g = 2400
Girls N2,400, boys N3,000; together N5,400

Explanation

Use the exterior angle of a triangle twice in (a).

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Theory question 109

Measures of Dispersion

The weight (in kg) of 50 contestants at a competition is as follows:
65 66 67 66 64 66 65 63 65 68
64 62 66 64 67 65 64 66 65 67
65 67 66 64 65 64 66 65 64 65
66 65 64 65 63 63 67 65 63 64
66 64 68 65 63 65 64 67 66 64
(a) Construct a frequency table for the discrete data.
(b) Calculate, correct to 2 decimal places, the: (i) mean; (ii) standard deviation of the data.

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Model answer

(a) Weight: 62, 63, 64, 65, 66, 67, 68
Frequency: 1, 5, 12, 14, 10, 6, 2 (total 50)
(b)(i) Σfx = 3253, mean = 3253/50 = 65.06 kg
(ii) Σf(x − x̄)² = 92.82
SD = √(92.82/50) = √1.8564 ≈ 1.36 kg

Explanation

Multiply each squared deviation by its frequency before adding.

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Theory question 110

Constructions

Using ruler and a pair of compasses only. (a) construct: (i) ΔXYZ such that |XY| = 10 cm, ∠XYZ = 30° and ∠YXZ = 45°; (ii) locus, l1, of points equidistant from Y and Z; (iii) locus, l2, of points parallel to XY through Z. (b) Locate point M, the point of intersection of l1 and l2. (c) Measure ∠ZMY.

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(a)(i) Draw XY = 10 cm; construct 30° at Y and 45° at X; the arms meet at Z.
(ii) l₁ is the perpendicular bisector of YZ.
(iii) l₂ is the line through Z parallel to XY (e.g. copy the angle at Y onto Z).
(b) M is where l₁ meets l₂.
(c) ∠ZMY ≈ 120° (MZ = MY and ∠MZY = 30°, so ∠ZMY = 180° − 60°); accept 120° ± 1°.

Explanation

Leave all construction arcs showing.

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Theory question 111

Ratio, Percentages and Rates

(a) If (3p + 4q)/(3p − 4q) = 2, find p : q
(b) The diagram shows the cross section of a bridge with a semi circular hollow in the middle. If the perimeter of the cross section is 34 m, calculate the: (i) length PQ; (ii) area of the cross section [Take π = 22/7]

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(a) 3p + 4q = 6p − 8q → 3p = 12q → p : q = 4 : 1
(b)(i) Let |PQ| = L; diameter of semicircle TS = L − 4, radius (L − 4)/2
Perimeter = L + 4 + 2 + 2 + 4 + π(L − 4)/2 = 34
L + 11(L − 4)/7 = 22 → 18L = 198 → |PQ| = 11 m
(ii) Radius = 3.5 m; area = 11 × 4 − ½ × 22/7 × 3.5² = 44 − 19.25 = 24.75 m²

Explanation

The perimeter includes the semicircular arc, not the straight gap TS.

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Theory question 112

Trigonometry

x: 0°, 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°, 330°, 360°
y: 2.0, __, __, 3.0, 1.6, __, −2.0, __, −3.6, −3.0, __, __, 2.0
(a) Copy and complete the table of values, correct to one decimal place, for the relation y = 3sin x + 2cos x for 0° ≤ x ≤ 360°.
(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 the relation y = 3sin x + 2cos x for 0° ≤ x ≤ 360°.
(c) Use the graph to solve: (i) 3sin x + 2cos x = 0; (ii) 2 + 2cos x + 3sin x = 0.

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(a) x = 30°: 3.2; 60°: 3.6; 150°: −0.2; 210°: −3.2; 300°: −1.6; 330°: 0.2
(b) Plot the points and join them with a smooth wave (maximum about 3.6 near 56°, minimum about −3.6 near 236°).
(c)(i) Where the curve cuts y = 0: x ≈ 146° and x ≈ 326°
(ii) 3sin x + 2cos x = −2: where the curve meets y = −2: x ≈ 180° and x ≈ 293°

Explanation

For (c)(ii) rearrange to 3sin x + 2cos x = −2 and draw the line y = −2.

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Theory question 113

Coordinate Geometry

(a) Find the equation of a straight line which passes through the point (2, −3) and is parallel to the line 2x + y = 6.
(b) The operation Δ is defined on the set T = {2, 3, 5, 7} by x Δ y = (x + y + xy) mod 8. (i) Construct modulo 8 table for the operation Δ on the set T. (ii) Use the table to find: I. 2 Δ (5 Δ 7); II. 2 Δ n = 5 Δ 7.

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(a) Gradient of 2x + y = 6 is −2; y + 3 = −2(x − 2) → y = −2x + 1 (or 2x + y = 1)
(b)(i) Rows x = 2, 3, 5, 7; columns y = 2, 3, 5, 7:
2: 0, 3, 1, 7
3: 3, 7, 7, 7
5: 1, 7, 3, 7
7: 7, 7, 7, 7
(ii) I. 5 Δ 7 = 7, so 2 Δ 7 = 7
II. 5 Δ 7 = 7; from row 2, 2 Δ n = 7 when n = 7

Explanation

Parallel lines have equal gradients.

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