WAEC Mathematics 2007 past questions and answers

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

Question 1

Number Bases

Evaluate (111two)2 and leave your answer in base 2.

  1. A.111001two
  2. B.110001two
  3. C.101001two
  4. D.10010two
Show the answer and explanation

Answer: B (110001two)

Explanation

111two = 7, and 72 = 49 = 32 + 16 + 1 = 110001two.

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Question 2

Fractions, Decimals and Approximation

Find the LCM of 23 × 3 × 52, 2 × 32 × 5 and 22 × 33 × 5.

  1. A.22 × 33 × 52
  2. B.23 × 33 × 5
  3. C.22 × 32 × 52
  4. D.23 × 33 × 52
Show the answer and explanation

Answer: D (23 × 33 × 52)

Explanation

The LCM takes the highest power of each prime: 23, 33 and 52.

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Question 3

Fractions, Decimals and Approximation

Simplify (0.3 × 105) ÷ (0.4 × 107), leaving your answer in the standard form.

  1. A.7.5 × 10−4
  2. B.7.5 × 10−3
  3. C.7.5 × 10−2
  4. D.7.5 × 10−1
Show the answer and explanation

Answer: B (7.5 × 10−3)

Explanation

0.3/0.4 = 0.75 and 105/107 = 10−2, so the result is 0.75 × 10−2 = 7.5 × 10−3.

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Question 4

Variation

If p ∝ 1/q, which of the following is true?

  1. A.q ∝ p2
  2. B.q ∝ 1/p2
  3. C.q ∝ √p
  4. D.q ∝ 1/p
Show the answer and explanation

Answer: D (q ∝ 1/p)

Explanation

p = k/q means pq = k, so q = k/p: q varies inversely as p.

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Question 5

Surds

Simplify 3√45 − 12√5 + 16√20, leaving your answer in surd form.

  1. A.29√5
  2. B.14√15
  3. C.12√15
  4. D.11√5
Show the answer and explanation

Answer: A (29√5)

Explanation

3√45 = 9√5 and 16√20 = 32√5, so 9√5 − 12√5 + 32√5 = 29√5.

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Question 6

Inequalities

Which of the following statements is true for the set P = {−3.2 ≤ x < 5} where x is an integer?

  1. A.Least value of x is −3.2
  2. B.Least value of x is −3
  3. C.Greatest value of x is 4.9
  4. D.Greatest value of x is 5
Show the answer and explanation

Answer: B (Least value of x is −3)

Explanation

The integers with −3.2 ≤ x < 5 are −3, −2, …, 4. The least is −3 and the greatest is 4.

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Question 7

Indices

Simplify: 3x3/(3x)3

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

Answer: C (1/9)

Explanation

(3x)3 = 27x3, so 3x3/27x3 = 1/9.

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Question 8

Ratio, Percentages and Rates

N140,000 is shared between Abu, Kayode and Uche. Abu has twice as much as Kayode, and Kayode has twice as much as Uche. What is Kayode's share?

  1. A.N80,000
  2. B.N40,000
  3. C.N20,000
  4. D.N10,000
Show the answer and explanation

Answer: B (N40,000)

Explanation

If Uche gets u, Kayode gets 2u and Abu 4u: 7u = 140,000, so u = 20,000 and Kayode gets N40,000.

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Question 9

Linear and Simultaneous Equations

I am x years old and my brother is 3 years older. How old was my brother last year?

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

Answer: B ((x + 2) years)

Explanation

The brother is now x + 3, so last year he was x + 3 − 1 = x + 2.

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Question 11

Algebraic Expressions

If x = 3, y = 2 and z = 4, what is the value of 3x2 − 2y + z?

  1. A.17
  2. B.27
  3. C.35
  4. D.71
Show the answer and explanation

Answer: B (27)

Explanation

3(9) − 2(2) + 4 = 27 − 4 + 4 = 27.

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Question 12

Linear and Simultaneous Equations

Solve the simultaneous equations: x + y = 2 and 3x − 2y = 1.

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

Answer: B (x = 1 and y = 1)

Explanation

Double the first equation (2x + 2y = 4) and add the second: 5x = 5, so x = 1 and y = 1.

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Question 13

Indices

If 8x+1 = 1/4, find x.

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

Answer: A (−5/3)

Explanation

Write both as powers of 2: 23(x+1) = 2−2, so 3x + 3 = −2 and x = −5/3.

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Question 14

Mensuration of Plane Shapes

The area of a square field is 110.25 m2. Find the cost of fencing it round at N75.00 per metre.

  1. A.N1,575.00
  2. B.N3,150.00
  3. C.N4,734.30
  4. D.N8,268.75
Show the answer and explanation

Answer: B (N3,150.00)

Explanation

Side = √110.25 = 10.5 m, perimeter = 4 × 10.5 = 42 m, cost = 42 × 75 = N3,150.00.

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Question 15

Circle Theorems

A chord of length 6 cm is drawn in a circle of radius 5 cm. Find the distance of the chord from the centre of the circle.

  1. A.2.5 cm
  2. B.3.0 cm
  3. C.3.5 cm
  4. D.4.0 cm
Show the answer and explanation

Answer: D (4.0 cm)

Explanation

The perpendicular from the centre bisects the chord: d2 = 52 − 32 = 16, so d = 4 cm.

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Question 16

Mensuration of Solids

A sector of a circle of radius 14 cm containing an angle 60° is folded to form a cone. Calculate the radius of the base of the cone.

  1. A.5½ cm
  2. B.4⅔ cm
  3. C.3½ cm
  4. D.2⅓ cm
Show the answer and explanation

Answer: D (2⅓ cm)

Explanation

Arc length = circumference of base: (60/360) × 2π × 14 = 2πr, so r = 14/6 = 2⅓ cm.

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Question 17

Mensuration of Plane Shapes

In the diagram, |LN| = 4 cm, LN̂M = 90° and tan y = 2/3. What is the area of the ΔLMN?

Diagram for this question
  1. A.24 cm2
  2. B.12 cm2
  3. C.10 cm2
  4. D.6 cm2
Show the answer and explanation

Answer: B (12 cm2)

Explanation

tan y = LN/MN = 4/MN = 2/3, so MN = 6 cm. Area = ½ × 6 × 4 = 12 cm2.

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Question 18

Mensuration of Solids

Find the volume of a solid cylinder with base radius 10 cm and height 14 cm. (Take π = 22/7)

  1. A.220 cm3
  2. B.880 cm3
  3. C.1400 cm3
  4. D.4400 cm3
Show the answer and explanation

Answer: D (4400 cm3)

Explanation

V = πr2h = 22/7 × 100 × 14 = 4400 cm3.

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Question 19

Triangles and Quadrilaterals

From the diagram, find the value of x.

Diagram for this question
  1. A.80°
  2. B.70°
  3. C.55°
  4. D.35°
Show the answer and explanation

Answer: A (80°)

Explanation

The slanting lines are parallel and the horizontal lines are parallel. The 100° angle and the angle between the slanting line and the horizontal going right are on a straight line, so that angle is 80°, and x equals it (corresponding angles).

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Question 20

Triangles and Quadrilaterals

Find the value of x in the diagram.

Diagram for this question
  1. A.281°
  2. B.269°
  3. C.201°
  4. D.179°
Show the answer and explanation

Answer: A (281°)

Explanation

Through the vertex of x draw a line parallel to the two parallel lines. The upper arm makes 45° with it (vertically opposite the 45° and alternate angles) and the lower arm makes 34°, so the angle on the left is 45° + 34° = 79°. The reflex angle x = 360° − 79° = 281°.

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Question 21

Angles and Polygons

Each interior angle of a regular polygon is 162°. How many sides has the polygon?

  1. A.8
  2. B.12
  3. C.16
  4. D.20
Show the answer and explanation

Answer: D (20)

Explanation

Each exterior angle = 180° − 162° = 18°, and n = 360° ÷ 18° = 20.

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Question 22

Triangles and Quadrilaterals

From the diagram, find the value of x.

Diagram for this question
  1. A.40
  2. B.30
  3. C.25
  4. D.20
Show the answer and explanation

Answer: D (20)

Explanation

The pentagon's angles add to 540°. Its interior angles are (180 − 5x)°, 120°, 130°, 90° and (180 − 3x)°: 700 − 8x = 540, so x = 20.

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Question 23

Triangles and Quadrilaterals

In the diagram, |SP| = |SR| and ∠PRS = 50°. Calculate ∠PQR.

Diagram for this question
  1. A.120°
  2. B.110°
  3. C.100°
  4. D.80°
Show the answer and explanation

Answer: C (100°)

Explanation

Triangle PSR is isosceles, so ∠SPR = 50° and ∠PSR = 180° − 100° = 80°. PQRS is cyclic, so ∠PQR = 180° − 80° = 100°.

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Question 24

Circle Theorems

In the diagram, O is the centre of the circle and ∠PQR = 106°. Find the value of y.

Diagram for this question
  1. A.16
  2. B.37
  3. C.74
  4. D.127
Show the answer and explanation

Answer: C (74)

Explanation

The reflex angle POR at the centre is twice ∠PQR = 212°. So 2y = 360° − 212° = 148° and y = 74.

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Question 25

Triangles and Quadrilaterals

In a ΔPQR, ∠PQR = ∠PRQ = 45°. Which of the following statements is/are correct?
I. ΔPQR is an equilateral triangle
II. ΔPQR is an isosceles triangle
III. ΔPQR is a right-angled triangle

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

Answer: C (II and III only)

Explanation

Two equal angles make it isosceles, and the third angle is 180° − 90° = 90°, so it is also right-angled. It is not equilateral.

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Question 26

Measures of Central Tendency

The table gives the distribution of the marks of a number of students in a test.
Mark: 1, 2, 3, 4, 5, 6
Frequency: 5, 3, 6, 4, 2, 5
Find the mode of the distribution.

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

Answer: B (3)

Explanation

The mode is the mark with the highest frequency: mark 3 occurs 6 times.

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Question 27

Probability

If the probability of an event occurring is x, what is the probability of the event not occurring?

  1. A.1 − x
  2. B.x − 1
  3. C.0
  4. D.1/x
Show the answer and explanation

Answer: A (1 − x)

Explanation

P(not E) = 1 − P(E) = 1 − x.

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Question 28

Measures of Central Tendency

The table shows the ages (in years) of twenty children chosen at random from a community.
Age (years): 1, 2, 3, 4, 5
Number of children: 2, 3, 5, 6, 4
What is the mean age?

  1. A.4.46 years
  2. B.3.35 years
  3. C.3.30 years
  4. D.3.00 years
Show the answer and explanation

Answer: B (3.35 years)

Explanation

Σfx = 2 + 6 + 15 + 24 + 20 = 67 and Σf = 20, so the mean = 67/20 = 3.35 years.

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Question 29

Measures of Central Tendency

The table shows the ages (in years) of twenty children chosen at random from a community.
Age (years): 1, 2, 3, 4, 5
Number of children: 2, 3, 5, 6, 4
What is the median of the distribution?

  1. A.3.5 years
  2. B.3.0 years
  3. C.2.5 years
  4. D.2.0 years
Show the answer and explanation

Answer: A (3.5 years)

Explanation

With 20 children the median is the average of the 10th and 11th ages. Cumulative frequencies 2, 5, 10, 16: the 10th is 3 and the 11th is 4, so the median is 3.5 years.

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Question 30

Measures of Central Tendency

Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3 by their mean.

  1. A.9
  2. B.8
  3. C.7
  4. D.6
Show the answer and explanation

Answer: A (9)

Explanation

Mean = sum ÷ 9, so sum ÷ mean = 9, the number of values (sum 39, mean 39/9).

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Question 31

Bearings

Q is 32 km away from P on a bearing 042° and R is 25 km from P on a bearing of 132°. Calculate the bearing of R from Q.

  1. A.122°
  2. B.184°
  3. C.190°
  4. D.226°
Show the answer and explanation

Answer: B (184°)

Explanation

∠QPR = 132° − 42° = 90°, and tan ∠PQR = 25/32 gives ∠PQR ≈ 38°. The bearing of P from Q is 222°, so the bearing of R from Q is 222° − 38° = 184°.

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Question 32

Triangles and Quadrilaterals

In the diagram, |PQ| = 4 cm, |QR| = 6 cm, |RS| = 12 cm and ∠QRS = 90°. Find the value of x.

Diagram for this question
  1. A.27
  2. B.26
  3. C.18
  4. D.13
Show the answer and explanation

Answer: D (13)

Explanation

tan ∠PSR = 10/12 gives 39.8°, and tan ∠QSR = 6/12 gives 26.6°. x = 39.8° − 26.6° ≈ 13°.

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Question 33

Bearings

The bearing of a point P from another point Q is 310°. If |PQ| = 200 m, how far west of Q is P?

  1. A.128.6 m
  2. B.153.2 m
  3. C.167.8 m
  4. D.187.9 m
Show the answer and explanation

Answer: B (153.2 m)

Explanation

A bearing of 310° is 50° west of north. Distance west = 200 sin 50° = 200 × 0.766 ≈ 153.2 m.

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Question 34

Logarithms

Find the value of a if log10a + log10a2 = 0.9030.

  1. A.4.0
  2. B.2.0
  3. C.1.6
  4. D.0.0
Show the answer and explanation

Answer: B (2.0)

Explanation

log a + 2 log a = 3 log a = 0.9030, so log a = 0.3010 and a = 2.0.

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Question 35

Linear and Simultaneous Equations

A messenger was paid N2.50 an hour during the normal working hours and N4.00 an hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime?

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

Answer: C (4)

Explanation

Let t be overtime hours: 2.5(10 − t) + 4t = 31, so 25 + 1.5t = 31 and t = 4.

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Question 36

Commercial Arithmetic

At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years?

  1. A.4%
  2. B.3½%
  3. C.3%
  4. D.2½%
Show the answer and explanation

Answer: D (2½%)

Explanation

R = 100I/(PT) = (100 × 39)/(520 × 3) = 3900/1560 = 2.5%.

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Question 37

Ratio, Percentages and Rates

If the ratio x : y = 3 : 5 and y : z = 4 : 7, find the ratio x : y : z.

  1. A.15 : 28 : 84
  2. B.12 : 20 : 35
  3. C.3 : 5 : 4
  4. D.5 : 4 : 7
Show the answer and explanation

Answer: B (12 : 20 : 35)

Explanation

Make y the same in both: x : y = 12 : 20 and y : z = 20 : 35, so x : y : z = 12 : 20 : 35.

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Question 38

Sets

The Venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x.

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

Answer: A (5)

Explanation

Add all the regions: 5 + 6 + 8 + 4 + 2 + 5 + x = 30 + x = 35, so x = 5.

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Question 39

Sets

The Venn diagram shows the choice of food of a number of visitors to a canteen. How many people took at least two kinds of food?

Diagram for this question
  1. A.10
  2. B.12
  3. C.15
  4. D.17
Show the answer and explanation

Answer: D (17)

Explanation

Add the overlap regions: 6 + 4 + 2 + x = 12 + 5 = 17 (x = 5 from the total of 35).

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Question 40

Triangles and Quadrilaterals

PQR is an equilateral triangle with side 2√3 cm. Calculate its height.

  1. A.1.7 cm
  2. B.3.0 cm
  3. C.3.2 cm
  4. D.3.9 cm
Show the answer and explanation

Answer: B (3.0 cm)

Explanation

The height bisects the base: h2 = (2√3)2 − (√3)2 = 12 − 3 = 9, so h = 3 cm.

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Question 41

Polynomials

Which of the following is a factor of 2 − x − x2?

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

Answer: A (1 − x)

Explanation

2 − x − x2 = (1 − x)(2 + x), so 1 − x is a factor.

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Question 42

Algebraic Expressions

Make w the subject of the relation (a + bc)/(wd + f) = g

  1. A.(a + bc − fg)/dg
  2. B.(a − bc + fg)/fg
  3. C.(a + bc − f)/dg
  4. D.(a + bc − dg)/dg
Show the answer and explanation

Answer: A ((a + bc − fg)/dg)

Explanation

a + bc = g(wd + f) = gwd + fg, so gwd = a + bc − fg and w = (a + bc − fg)/dg.

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Question 43

Algebraic Expressions

Expand the expression (3a − xy)(3a + xy).

  1. A.9a2 − x2y2
  2. B.9a2 + x2y2
  3. C.9a2 − xy
  4. D.9a2 + x2y
Show the answer and explanation

Answer: A (9a2 − x2y2)

Explanation

Difference of two squares: (3a)2 − (xy)2 = 9a2 − x2y2.

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Question 44

Inequalities

For what range of values of x is 4x − 3(2x − 1) > 1?

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

Answer: C (x < 1)

Explanation

4x − 6x + 3 > 1 gives −2x > −2. Dividing by −2 reverses the sign: x < 1.

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Question 45

Fractions, Decimals and Approximation

Find the smallest value of k such that 22 × 33 × 5 × k is a perfect square.

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

Answer: C (15)

Explanation

Every prime needs an even power. 33 needs one more 3 and 5 needs one more 5, so k = 3 × 5 = 15.

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Question 47

Mensuration of Plane Shapes

Two circles have radii 16 cm and 23 cm. What is the difference between their circumferences? (Take π = 22/7)

  1. A.244.92 cm
  2. B.149.92 cm
  3. C.44.00 cm
  4. D.43.96 cm
Show the answer and explanation

Answer: C (44.00 cm)

Explanation

Difference = 2π(23 − 16) = 2 × 22/7 × 7 = 44.00 cm.

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Question 48

Mensuration of Solids

The volume of a cylinder is 1200 cm3 and the area of its base is 150 cm2. Find the height of the cylinder.

  1. A.80.00 cm
  2. B.8.00 cm
  3. C.0.80 cm
  4. D.0.08 cm
Show the answer and explanation

Answer: B (8.00 cm)

Explanation

Volume = base area × height, so height = 1200 ÷ 150 = 8 cm.

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Question 49

Triangles and Quadrilaterals

In the diagram, |XR| = 4 cm, |RZ| = 12 cm, |SR| = n, |YZ| = m and SR || YZ. Find m in terms of n.

Diagram for this question
  1. A.m = 2n
  2. B.m = 3n
  3. C.m = 4n
  4. D.m = 5n
Show the answer and explanation

Answer: C (m = 4n)

Explanation

SR ∥ YZ, so triangles XSR and XYZ are similar: SR/YZ = XR/XZ = 4/16. So n/m = 1/4 and m = 4n.

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Question 50

Mensuration of Plane Shapes

In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4 cm. Find the area of the shaded region.

Diagram for this question
  1. A.43.36 cm2
  2. B.32.07 cm2
  3. C.18.21 cm2
  4. D.6.93 cm2
Show the answer and explanation

Answer: C (18.21 cm2)

Explanation

Radius = 4 cm. Shaded = semicircle − triangle = ½ × 22/7 × 16 − ½ × 4 × 4 × sin 60° = 25.14 − 6.93 = 18.21 cm2.

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Theory question 101

Fractions, Decimals and Approximation

(a) Evaluate, without using mathematical tables or calculator, (3.69 × 105) ÷ (1.64 × 10−3), leaving your answer in the standard form.
(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of a year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.

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Model answer

(a) 3.69 ÷ 1.64 = 2.25 and 105 ÷ 10−3 = 108, so the answer is 2.25 × 108.
(b) Interest from A = 20,000 × y/100 = 200y; from B = 25,000 × 1.5y/100 = 375y.
200y + 375y = 4600, so 575y = 4600 and y = 8.

Explanation

Divide the numbers and subtract the powers of 10 separately.

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Theory question 102

Progressions

(a) Simplify: (x2 − 8x + 16)/(x2 − 7x + 12)
(b) If 1/2, 1/x, 1/3 are successive terms of an arithmetic progression (A.P.), show that (2 − x)/(x − 3) = 2/3.

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Model answer

(a) (x − 4)(x − 4)/[(x − 4)(x − 3)] = (x − 4)/(x − 3).
(b) Equal common difference: 1/x − 1/2 = 1/3 − 1/x.
So (2 − x)/2x = (x − 3)/3x. Multiply by 6x: 3(2 − x) = 2(x − 3).
Hence (2 − x)/(x − 3) = 2/3.

Explanation

In an AP the difference between consecutive terms is constant.

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Theory question 103

Mensuration of Plane Shapes

The diagram shows the cross-section of a railway tunnel. If |AB| = 100 m and the radius of the arc is 56 m, calculate, correct to the nearest metre, the perimeter of the cross-section.

Diagram for this question
Show the model answer
Model answer

Let O be the centre. Half of AB = 50 m, so sin ½∠AOB = 50/56 and ½∠AOB ≈ 63.23°, giving ∠AOB ≈ 126.5°.
The arc of the tunnel is the major arc: 360° − 126.5° = 233.5°.
Arc length = 233.5/360 × 2 × 22/7 × 56 ≈ 228.4 m.
Perimeter = 228.4 + 100 ≈ 328 m.

Explanation

The tunnel's curved part is the major arc, so use the reflex angle at the centre.

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Theory question 104

Bearings

Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the: (a) distance of Z from Y; (b) bearing of Z from Y.

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Model answer

(a) ∠YXZ = 225° − 135° = 90°, so |ZY| = √(602 + 802) = 100 km.
(b) tan ∠XYZ = 80/60, so ∠XYZ ≈ 53.1°. The bearing of X from Y is 315°.
Bearing of Z from Y = 315° − 53.1° ≈ 262°.

Explanation

Draw a sketch with north lines at X and Y; the angle at X is a right angle.

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Theory question 105

Probability

Out of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that: (a) the first two are good and the third is bad; (b) all the three are bad; (c) all the three are good.

Show the model answer
Model answer

P(bad) = 6/24 = 1/4, P(good) = 3/4; with replacement the draws are independent.
(a) 3/4 × 3/4 × 1/4 = 9/64.
(b) (1/4)3 = 1/64.
(c) (3/4)3 = 27/64.

Explanation

With replacement the probabilities stay the same for each draw, so multiply.

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Theory question 106

Fractions, Decimals and Approximation

(a) Simplify: (3 1/12 + 7/8)/(2 1/4 − 1/6)
(b) If p = m/2 − n2/5m, (i) make n the subject of the relation; (ii) find, correct to three significant figures, the value of n when p = 14 and m = −8.

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(a) Numerator: 37/12 + 7/8 = 74/24 + 21/24 = 95/24. Denominator: 9/4 − 1/6 = 27/12 − 2/12 = 25/12.
95/24 ÷ 25/12 = 95/24 × 12/25 = 19/10 = 1 9/10.
(b)(i) n2/5m = m/2 − p, so n2 = 5m(m − 2p)/2 and n = ±√[5m(m − 2p)/2].
(ii) n2 = 5(−8)(−8 − 28)/2 = (−40)(−36)/2 = 720, so n = ±26.8.

Explanation

Isolate the n2 term first, then take the square root.

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Theory question 107

Logarithms

(a) With the aid of four-figure logarithm tables, evaluate (0.004592)1/3.
(b) If log10y + 3log10x = 2, express y in terms of x.
(c) Solve the equations: 3x − 2y = 21; 4x + 5y = 5.

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(a) log 0.004592 = 3̄.6620. Divide by 3: (−3 + 0.6620)/3 = 1̄.2207. Antilog = 0.1663.
(b) log(yx3) = 2, so yx3 = 100 and y = 100/x3.
(c) 15x − 10y = 105 and 8x + 10y = 10. Adding: 23x = 115, so x = 5; then 3(5) − 2y = 21 gives y = −3.

Explanation

When dividing a log with a negative characteristic, make the characteristic a multiple of 3 first.

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Theory question 108

Mensuration of Solids

(a) A cylinder with radius 3.5 cm has its two ends closed. If the total surface area is 209 cm2, calculate the height of the cylinder. (Take π = 22/7)
(b) In the diagram, O is the centre of the circle and ABC is a tangent at B. If BD̂F = 66° and DB̂C = 57°, calculate, (i) EB̂F (ii) BĜF.

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(a) Total surface area = 2πr2 + 2πrh = 2 × 22/7 × 12.25 + 2 × 22/7 × 3.5 × h = 77 + 22h = 209, so h = 6 cm.
(b)(i) EB is a diameter, so ∠EDB = 90° (angle in a semicircle). ∠EDF = 90° − 66° = 24°, and ∠EBF = ∠EDF = 24° (same segment).
(ii) The radius OB is perpendicular to the tangent, so ∠EBD = 90° − 57° = 33°. In ΔGBD: ∠BGD = 180° − 33° − 66° = 81°, so ∠BGF = 180° − 81° = 99°.

Explanation

Use the angle in a semicircle and the right angle between a tangent and a radius.

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Theory question 109

Constructions

(a) Using a ruler and a pair of compasses only, construct: (i) a triangle PQR such that |PQ| = 10 cm, |QR| = 7 cm and PQ̂R = 90°; (ii) the locus l1 of points equidistant from Q and R; (iii) the locus l2 of points equidistant from P and Q.
(b) Locate the point O equidistant from P, Q and R.
(c) With O as centre, draw the circumcircle of the triangle PQR.
(d) Measure the radius of the circumcircle.

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(a)(i) Draw PQ = 10 cm, construct 90° at Q and mark QR = 7 cm; join PR.
(ii) l1 is the perpendicular bisector of QR. (iii) l2 is the perpendicular bisector of PQ.
(b) O is where l1 and l2 meet (it is the midpoint of PR).
(c) Circle with centre O through P, Q and R.
(d) Radius ≈ 6.1 cm (half of PR = ½√149 ≈ 6.10 cm).

Explanation

Points equidistant from two points lie on their perpendicular bisector.

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Theory question 110

Algebraic Expressions

(a) Simplify (x2 − y2)/(3x + 3y)
(b) In the diagram, PQRS is a rectangle. |PK| = 15 cm, |SK| = |KR| and PK̂S = 37°. Calculate, correct to three significant figures: (i) |PS|; (ii) |SK|; (iii) the area of the shaded portion.

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(a) (x − y)(x + y)/3(x + y) = (x − y)/3.
(b)(i) |PS| = 15 sin 37° ≈ 9.03 cm.
(ii) |SK| = 15 cos 37° ≈ 12.0 cm (11.98 cm).
(iii) |SR| = 2 × 11.98 = 23.96 cm. Rectangle = 23.96 × 9.027 ≈ 216.3 cm2; ΔPSK = ½ × 11.98 × 9.027 ≈ 54.1 cm2.
Shaded area ≈ 216.3 − 54.1 ≈ 162 cm2.

Explanation

The shaded part is the rectangle minus the right triangle PSK.

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Theory question 111

Angles and Polygons

(a) In the diagram, AOB is a straight line, ∠AOC = 3(x + y)°, ∠COB = 45°, ∠AOD = (5x + y)° and ∠DOB = y°. Find the values of x and y.
(b) From two points on opposite sides of a pole 33 m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base of the pole are on the same horizontal level, calculate, correct to three significant figures, the distance between the two points.

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(a) Angles on a straight line: 3(x + y) + 45 = 180, so x + y = 45.
(5x + y) + y = 180, so 5x + 2y = 180.
Substituting x = 45 − y: 225 − 3y = 180, so y = 15 and x = 30.
(b) Distances from the pole: 33/tan 53° ≈ 24.87 m and 33/tan 67° ≈ 14.01 m.
Distance between the points ≈ 24.87 + 14.01 ≈ 38.9 m.

Explanation

The points are on opposite sides, so add the two horizontal distances.

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Theory question 112

Progressions

(a) The 3rd and 8th terms of an arithmetic progression (A.P.) are −9 and 26 respectively. Find the: (i) common difference; (ii) first term.
(b) In the diagram, PQ||YZ, |XP| = 2 cm, |PY| = 3 cm, |PQ| = 6 cm and the area of ΔXPQ = 24 cm2. Calculate the area of the trapezium PQZY.

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(a) a + 2d = −9 and a + 7d = 26. Subtracting: 5d = 35.
(i) d = 7. (ii) a = −9 − 14 = −23.
(b) ΔXPQ ~ ΔXYZ with ratio XP : XY = 2 : 5, so the areas are in the ratio 4 : 25.
Area of ΔXYZ = 24 × 25/4 = 150 cm2.
Area of PQZY = 150 − 24 = 126 cm2.

Explanation

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

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Theory question 113

Measures of Dispersion

In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table.
Number of absentees: 0–4, 5–9, 10–14, 15–19, 20–24
Number of days: 1, 5, 10, 9, 5
Calculate the: (a) mean; (b) standard deviation, correct to two decimal places.

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Class marks x: 2, 7, 12, 17, 22; Σf = 30.
(a) Σfx = 2 + 35 + 120 + 153 + 110 = 420. Mean = 420/30 = 14.00.
(b) Deviations d = x − 14: −12, −7, −2, 3, 8. Σfd2 = 144 + 245 + 40 + 81 + 320 = 830.
SD = √(830/30) = √27.67 ≈ 5.26.

Explanation

Use the class mid-points as x.

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Theory question 114

Vectors and Transformations

(Questions 14 and 15 are for candidates in Ghana only.)
(a) Using a scale of 2 cm to 4 units on both axes, draw on a graph sheet, two perpendicular axes Ox and Oy for the intervals −16 ≤ x ≤ 16, −16 ≤ y ≤ 16.
(b) Draw the triangle ABC with the vertices A(2, 2), B(8, 2) and C(6, 6).
(c) Draw the image ΔA1B1C1 of ΔABC through a rotation of 180° about the origin where A → A1, B → B1 and C → C1.
(d) Draw the image ΔA2B2C2 of ΔABC under the translation by the vector (−8, 8) where A → A2, B → B2 and C → C2.
(e) Draw the image ΔA3B3C3 of the ΔA1B1C1 under the reflection in the line x − 2 = 0, where A1 → A3, B1 → B3 and C1 → C3.
(f) Describe the transformation that will map ΔABC onto ΔA3B3C3.

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(a), (b) Axes to scale; A(2, 2), B(8, 2), C(6, 6).
(c) (x, y) → (−x, −y): A1(−2, −2), B1(−8, −2), C1(−6, −6).
(d) Add (−8, 8): A2(−6, 10), B2(0, 10), C2(−2, 14).
(e) Reflection in x = 2: (x, y) → (4 − x, y): A3(6, −2), B3(12, −2), C3(10, −6).
(f) ABC → A3B3C3 is (x, y) → (x + 4, −y): a reflection in the x-axis followed by a translation of 4 units in the positive x-direction (vector (4, 0)).

Explanation

Write each transformation as a rule on (x, y) and combine them.

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Theory question 115

Inequalities

(a) Given that f(x) = 2x − 1 and g(x) = x2 + 1: (i) find f(1 + x); (ii) find the range of values of x for which f(x) < −3; (iii) simplify f(x) − g(x).
(b) Given that A(1, 3), B(−2, −1) and C(2, 3m) where m is a constant; find: (i) |BA|; (ii) the value of m if BC = (4, −5).

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(a)(i) f(1 + x) = 2(1 + x) − 1 = 2x + 1.
(ii) 2x − 1 < −3, so 2x < −2 and x < −1.
(iii) f(x) − g(x) = 2x − 1 − x2 − 1 = −x2 + 2x − 2.
(b)(i) |BA| = √[(1 + 2)2 + (3 + 1)2] = √25 = 5 units.
(ii) BC = (2 − (−2), 3m − (−1)) = (4, 3m + 1) = (4, −5), so 3m + 1 = −5 and m = −2.

Explanation

A vector from B to C is (position of C) − (position of B).

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