WAEC Mathematics 1995 past questions and answers

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

Question 1

Number Bases

Find (1012)2, expressing the answer in base 2.

  1. A.10101
  2. B.11001
  3. C.10010
  4. D.11101
  5. E.10110.
Show the answer and explanation

Answer: B (11001)

Explanation

1012 = 5, and 5² = 25. In base 2, 25 = 16 + 8 + 1 = 110012.

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Question 2

Ratio, Percentages and Rates

If 3 children share N10.50 among themselves in ratio 6:7:8, how much is the largest share?

  1. A.N3.00
  2. B.N3.50
  3. C.N4.00
  4. D.N4.50
  5. E.N5.00
Show the answer and explanation

Answer: C (N4.00)

Explanation

Total parts = 6 + 7 + 8 = 21. Largest share = 8/21 × N10.50 = N4.00.

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Question 3

Fractions, Decimals and Approximation

Express 0.000834 in standard form.

  1. A.8.34 × 10-4
  2. B.8.34 × 10-3
  3. C.8.34 × 103
  4. D.8.34 × 104
  5. E.8.34 × 106
Show the answer and explanation

Answer: A (8.34 × 10-4)

Explanation

Move the decimal point 4 places to the right to get 8.34, so 0.000834 = 8.34 × 10-4.

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Question 4

Indices

Simplify: 0.027-1/3

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

Answer: A (3 1/3)

Explanation

0.0271/3 = 0.3, so 0.027-1/3 = 1/0.3 = 10/3 = 3 1/3.

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Question 5

Logarithms

Given that log2a = log84, find a.

  1. A.21/2
  2. B.41/3
  3. C.42/3
  4. D.22/3
  5. E.23
Show the answer and explanation

Answer: D (22/3)

Explanation

log84 = 2/3 because 82/3 = 4. So log2a = 2/3 and a = 22/3.

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Question 6

Commercial Arithmetic

By selling some crates of soft drinks for N600.00, a dealer makes a profit of 50%. How much did the dealer pay for the drinks?

  1. A.N1,200.00
  2. B.N900.00
  3. C.N450.00
  4. D.N400.00
  5. E.N200.00
Show the answer and explanation

Answer: D (N400.00)

Explanation

Selling price = 150% of cost price, so cost = 600 × 100/150 = N400.00.

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Question 7

Progressions

Find the nth term Un of the A.P., 11, 4, −3,........

  1. A.Un = 19 + 7n
  2. B.Un = 19 − 7n
  3. C.Un = 18 − 7n
  4. D.Un = 18 + 7n
  5. E.Un = 17 − 7n.
Show the answer and explanation

Answer: C (Un = 18 − 7n)

Explanation

a = 11 and d = −7, so Un = 11 + (n − 1)(−7) = 18 − 7n.

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Question 8

Progressions

If 16/9, x, 1, y are in Geometric Progression (G.P.), find the product of x and y.

  1. A.9/16
  2. B.3/4
  3. C.1
  4. D.4/3
  5. E.16/6
Show the answer and explanation

Answer: C (1)

Explanation

1 is the third term: 16/9 × r² = 1 gives r = 3/4. Then x = 16/9 × 3/4 = 4/3 and y = 1 × 3/4 = 3/4, so xy = 1.

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Question 9

Sets

If R = [2, 4, 6, 7] and S = [1, 2, 4, 8], then R∪S equal

  1. A.[1,2,4,6,7,8]
  2. B.[1,2,4,7,8]
  3. C.[1,4,7,8]
  4. D.[2,6,7]
  5. E.[2,4].
Show the answer and explanation

Answer: A ([1,2,4,6,7,8])

Explanation

The union contains every element in R or S: {1, 2, 4, 6, 7, 8}.

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Question 10

Sets

In the diagram below, the shaded portion is

Diagram for this question
  1. A.P'∩R
  2. B.Q∩R
  3. C.P'∩Q∩R
  4. D.(P∪Q)'∩R.
  5. E.(P∩Q) U (P'∩R).
Show the answer and explanation

Answer: A (P'∩R)

Explanation

The shaded part is all of R except the part inside P (it includes the part of R inside Q), which is P'∩R.

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Question 11

Algebraic Expressions

Find the values of x for which the expression is underlined: (6x − 1)/(x2 + 4x − 5)

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

Answer: B (−5 or +1)

Explanation

The expression is undefined when the denominator is zero: x² + 4x − 5 = (x + 5)(x − 1) = 0, so x = −5 or x = 1.

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Question 13

Inequalities

Solve the inequality: 1/3(2x − 1) < 5.

  1. A.x < −5
  2. B.x < −6
  3. C.x < 7
  4. D.x < 8
  5. E.x < 16
Show the answer and explanation

Answer: D (x < 8)

Explanation

Multiply by 3: 2x − 1 < 15, so 2x < 16 and x < 8.

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Question 14

Quadratic Equations

What is the smaller value of x for which x2 − 3x + 2 = 0?

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

Answer: A (1)

Explanation

x² − 3x + 2 = (x − 1)(x − 2) = 0, so x = 1 or 2. The smaller is 1.

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Question 15

Algebraic Expressions

Factorize the expression x(a − c) + y(c − a)

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

Answer: B ((a − c)(x − y))

Explanation

y(c − a) = −y(a − c), so the expression is (a − c)x − (a − c)y = (a − c)(x − y).

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Question 16

Variation

If y ∝ 1/x2 and x = 3 when y = 4, find y when x = 2

  1. A.1
  2. B.3
  3. C.9
  4. D.18
  5. E.21
Show the answer and explanation

Answer: C (9)

Explanation

y = k/x²; 4 = k/9 gives k = 36. When x = 2, y = 36/4 = 9.

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Question 17

Quadratic Equations

Solve the equation 3x2 + 25x − 18 = 0

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

Answer: D (−9,2/3)

Explanation

3x² + 25x − 18 = (3x − 2)(x + 9) = 0, so x = 2/3 or x = −9.

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Question 18

Quadratic Equations

Solve the equation (x + 2)(x − 7) = 0

  1. A.x = 1 or 8
  2. B.x = −2 or 7
  3. C.x = −4 or 5
  4. D.x = −3 or 6
  5. E.x = −5 or −m2.
Show the answer and explanation

Answer: B (x = −2 or 7)

Explanation

Either x + 2 = 0 or x − 7 = 0, so x = −2 or x = 7.

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Question 19

Linear and Simultaneous Equations

Solve the following simultaneous equations: x + y = 3/2; x − y = 5/2 and use your result to find the value of 2y + x

  1. A.−2
  2. B.−1
  3. C.½
  4. D.1
  5. E.3½
Show the answer and explanation

Answer: D (1)

Explanation

Adding the equations gives 2x = 4, so x = 2 and y = −½. Then 2y + x = −1 + 2 = 1.

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Question 20

Graphs of Functions

Use the graph of y = 3x2 + x − 7 to answer this question. What is the minimum value of y?

Diagram for this question
  1. A.−10
  2. B.−7
  3. C.−4
  4. D.−1
  5. E.2.
Show the answer and explanation

Answer: B (−7)

Explanation

The lowest point of the curve is just below the y-axis at about y = −7 (exactly −7 1/12 at x = −1/6).

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Question 21

Graphs of Functions

Use the graph of y = 3x2 + x − 7 to answer this question. Find the roots of the equation y = 3x2 + x − 7

Diagram for this question
  1. A.1.0 and −1.2
  2. B.1.1 and −1.3
  3. C.1.4 and −1.7
  4. D.2.0 and −1.9
  5. E.2.4 and −2.0.
Show the answer and explanation

Answer: C (1.4 and −1.7)

Explanation

The roots are where the curve cuts the x-axis: about x = 1.4 and x = −1.7 (the formula gives (−1 ± √85)/6).

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Question 22

Triangles and Quadrilaterals

In the diagram below. |AB| = 12cm, |AE| = 8cm, |DC| = 9cm and AB||DC. Calculate |EC|.

Diagram for this question
  1. A.10cm
  2. B.9cm
  3. C.8cm
  4. D.7.5
  5. E.6cm.
Show the answer and explanation

Answer: E (6cm.)

Explanation

Triangles ABE and CDE are similar (AB ∥ DC), so EC/AE = DC/AB: EC = 8 × 9/12 = 6 cm.

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Question 23

Mensuration of Plane Shapes

The diagonals AC and BD of a rhombus ABCD are 16cm and 12cm long respectively. Calculate the area of the rhombus.

  1. A.24cm2
  2. B.36cm2
  3. C.48cm2
  4. D.60cm2
  5. E.96cm2
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Answer: E (96cm2)

Explanation

Area of a rhombus = ½ × product of diagonals = ½ × 16 × 12 = 96 cm².

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Question 24

Mensuration of Solids

A water tank of height ½m has a square base of side 1½m. If it is filled with water from a water tanker holding 1500 litres, how many litres of water are left in the water tanker? [1000 litres = 1m3]

  1. A.37.5litres
  2. B.375 litres
  3. C.3750 litres
  4. D.37500 litres
  5. E.375000 litres.
Show the answer and explanation

Answer: B (375 litres)

Explanation

Tank volume = 1.5 × 1.5 × 0.5 = 1.125 m³ = 1125 litres. Water left = 1500 − 1125 = 375 litres.

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Question 27

Mensuration of Solids

Find the total surface area of solid circular cone with base radius 3cm and slant height 4cm. [Take π = 22/7]

  1. A.37 5/7cm2
  2. B.66cm2
  3. C.75 3/7cm2
  4. D.78 2/7cm2
  5. E.88cm2
Show the answer and explanation

Answer: B (66cm2)

Explanation

Total surface area = πr² + πrl = π(9 + 12) = 22/7 × 21 = 66 cm².

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Question 28

Mensuration of Solids

A hollow sphere has a volume of kcm3 and a surface area of kcm2. Calculate the diameter of the sphere.

  1. A.3cm
  2. B.6cm
  3. C.9cm
  4. D.12cm
  5. E.More information is needed.
Show the answer and explanation

Answer: B (6cm)

Explanation

Set 4/3 πr³ = 4πr²; dividing by 4πr² gives r/3 = 1, so r = 3 cm and the diameter is 6 cm.

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Question 29

Mensuration of Solids

The diagram above shows a cone with the dimensions of its frustrum indicated. Calculate the height of the cone.

Diagram for this question
  1. A.12cm
  2. B.15cm
  3. C.18cm
  4. D.24cm
  5. E.30cm.
Show the answer and explanation

Answer: D (24cm)

Explanation

The small cone cut off is similar to the whole cone. With height H: (H − 12)/H = 6/12, so 2H − 24 = H and H = 24 cm.

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Question 30

Latitude and Longitude

The positions of two countries P and Q are (15°N, 12°E) and (65°N, 12°E) respectively. What is the difference in latitude?

  1. A.104°
  2. B.100°
  3. C.80°
  4. D.50°
  5. E.24°
Show the answer and explanation

Answer: D (50°)

Explanation

Both are in the northern hemisphere, so subtract: 65° − 15° = 50°.

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Question 31

Mensuration of Solids

A 120° sector of a circle of radius 21cm is bent to form a cone. What is the base radius of the cone?

  1. A.3½cm
  2. B.7cm
  3. C.10½cm
  4. D.14cm
  5. E.21cm
Show the answer and explanation

Answer: B (7cm)

Explanation

The arc becomes the base circumference: 120/360 × 2π × 21 = 2πr, so r = 21/3 = 7 cm.

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Question 32

Mensuration of Plane Shapes

The angle of a sector of a circle is 108°. If the radius of the circle is 3½cm, find the perimeter of the sector

  1. A.6 3/5cm
  2. B.6 4/5cm
  3. C.7 1/10cm
  4. D.10 2/5cm
  5. E.13 3/5cm.
Show the answer and explanation

Answer: E (13 3/5cm.)

Explanation

Arc = 108/360 × 2 × 22/7 × 3.5 = 6.6 cm. Perimeter = 6.6 + 3.5 + 3.5 = 13.6 cm = 13 3/5 cm.

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Question 33

Circle Theorems

In the diagram below; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35°, find ∠BAO.

Diagram for this question
  1. A.20°
  2. B.25°
  3. C.30°
  4. D.35°
  5. E.40°
Show the answer and explanation

Answer: A (20°)

Explanation

∠DBC = 35° (isosceles triangle), so the exterior angle ∠BDA = 35° + 35° = 70°. AD is a diameter, so ∠ABD = 90°. Then ∠BAO = 180° − 90° − 70° = 20°.

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Question 34

Triangles and Quadrilaterals

In the diagram above, AO is perpendicular to OB. Find x.

Diagram for this question
  1. A.7.5°
  2. B.15°
  3. C.22.5°
  4. D.30°
  5. E.38.6°
Show the answer and explanation

Answer: D (30°)

Explanation

Angles at a point add up to 360°: 4x + 3x + 2x + 90° = 360°, so 9x = 270° and x = 30°.

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Question 35

Angles and Polygons

In the diagram above, PQ is parallel to TU, ∠PQR = 50°, ∠QRS = 86° and ∠STU = 64°. Calculate the value of x.

Diagram for this question
  1. A.136°
  2. B.120°
  3. C.108°
  4. D.100°
  5. E.96°
Show the answer and explanation

Answer: D (100°)

Explanation

Draw lines through R and S parallel to PQ. Using alternate angles, ∠QRS splits into 50° + 36°, and ∠RST splits into 36° + 64°. So x = 36° + 64° = 100°.

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Question 36

Logarithms

If log10x = 2̄.3675 and log10y = 2̄.9738, what is the value of x + y, correct to three significant figures?

  1. A.0.117
  2. B.0.118
  3. C.0.903
  4. D.0.944
  5. E.0.946.
Show the answer and explanation

Answer: A (0.117)

Explanation

x = antilog 2̄.3675 = 0.02331 and y = antilog 2̄.9738 = 0.09415. x + y = 0.11746 ≈ 0.117.

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Question 37

Angles and Polygons

In the diagram below, AB//CD, the bisector of ∠BAC and ∠ACD meet at E. Find the value of ∠AEC.

Diagram for this question
  1. A.30°
  2. B.45°
  3. C.60°
  4. D.75°
  5. E.90°
Show the answer and explanation

Answer: E (90°)

Explanation

∠BAC + ∠ACD = 180° (co-interior angles), so their halves add up to 90°. In triangle AEC, ∠AEC = 180° − 90° = 90°.

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Question 38

Angles and Polygons

In the diagram above, AB//CD. What is the size of the angle marked x?

Diagram for this question
  1. A.103°
  2. B.93°
  3. C.77°
  4. D.62°
  5. E.52°
Show the answer and explanation

Answer: B (93°)

Explanation

The 52° angle at A is repeated where the upper line crosses CD (corresponding angles). In the triangle formed with the 35° angle, x = 180° − 52° − 35° = 93°.

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Question 39

Loci

The locus of a point which is equidistant from two given fixed points is the

  1. A.perpendicular bisector of the straight line joining them
  2. B.angle bisector of the straight lines joining the points to the origin
  3. C.perpendicular to the straight line joining them
  4. D.parallel line to the straight line joining them
  5. E.a line making an acute angle with the line joining the two points.
Show the answer and explanation

Answer: A (perpendicular bisector of the straight line joining them)

Explanation

Every point equally far from two fixed points lies on the perpendicular bisector of the line joining them.

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Question 41

Angles and Polygons

Which of the following angles is an exterior angle of a regular polygon?

  1. A.95°
  2. B.85°
  3. C.78°
  4. D.75°
  5. E.72°
Show the answer and explanation

Answer: E (72°)

Explanation

An exterior angle of a regular polygon must divide 360° exactly. 360° ÷ 72° = 5 (a regular pentagon); none of the others divide 360° exactly.

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Question 42

Triangles and Quadrilaterals

In ΔABC above, BC is produced to D, |AB| = |AC| and ∠BAC = 50°: Find ∠ACD

Diagram for this question
  1. A.50°
  2. B.60°
  3. C.65°
  4. D.100°
  5. E.115°
Show the answer and explanation

Answer: E (115°)

Explanation

The base angles are equal: ∠ACB = (180° − 50°)/2 = 65°. Then ∠ACD = 180° − 65° = 115° (angles on a straight line).

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Question 43

Trigonometry

Without using tables, find the value of sin 20°/cos 70° + cos 25°/sin 65°

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

Answer: A (2)

Explanation

sin 20° = cos 70° and cos 25° = sin 65° (complementary angles), so each fraction equals 1 and the sum is 2.

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Question 44

Trigonometry

In the diagram below, ∠PRQ = 90°, ∠QPR = 30° and |PQ| = 12cm. Find y.

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

Answer: A (6cm)

Explanation

y is opposite the 30° angle and PQ is the hypotenuse, so y = 12 sin 30° = 12 × ½ = 6 cm.

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Question 45

Bearings

The bearing of two points Q and R from a point P are 030° and 120° respectively. If |PQ| = 12m and |PR| = 5m, find the distance QR.

  1. A.13m
  2. B.11m
  3. C.9m
  4. D.7m
  5. E.5m.
Show the answer and explanation

Answer: A (13m)

Explanation

The angle QPR = 120° − 30° = 90°, so by Pythagoras QR = √(12² + 5²) = √169 = 13 m.

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Question 46

Measures of Central Tendency

What is the mode of the numbers 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8 and 14?

  1. A.8
  2. B.9
  3. C.10
  4. D.11
  5. E.14.
Show the answer and explanation

Answer: A (8)

Explanation

8 appears four times, more than any other number, so the mode is 8.

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Question 47

Measures of Central Tendency

The mean of 20 observations in an experiment is 4. If the observed largest value is 23, find the mean of the remaining observations.

  1. A.4
  2. B.3
  3. C.2.85
  4. D.2.60
  5. E.2.56.
Show the answer and explanation

Answer: B (3)

Explanation

Total = 20 × 4 = 80. Removing 23 leaves 57 for 19 observations, so the mean is 57 ÷ 19 = 3.

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Question 48

Measures of Central Tendency

Find the median of the following numbers: 2.64, 2.50, 2.72, 2.91 and 2.35.

  1. A.2.91
  2. B.2.72
  3. C.2.64
  4. D.2.50
  5. E.2.35.
Show the answer and explanation

Answer: C (2.64)

Explanation

In order: 2.35, 2.50, 2.64, 2.72, 2.91. The middle (3rd) value is 2.64.

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Question 49

Probability

Two fair dice are tossed together once. Find the probability that the sum of the outcome is at least 10.

  1. A.1/12
  2. B.5/36
  3. C.1/6
  4. D.1/4
  5. E.5/18
Show the answer and explanation

Answer: C (1/6)

Explanation

Sums of 10 or more: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), which is 6 of the 36 outcomes, so 6/36 = 1/6.

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Question 50

Probability

From a box containing 2 red, 6 white and 5 black balls, a ball is randomly selected. What is the probability that the selected ball is black?

  1. A.2/13
  2. B.5/13
  3. C.5/11
  4. D.5/6
  5. E.11/13
Show the answer and explanation

Answer: B (5/13)

Explanation

There are 2 + 6 + 5 = 13 balls and 5 are black, so the probability is 5/13.

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Theory question 101

Algebraic Expressions

(a) Factorize: px − 2qx − 4qy + 2py.
(b) Given that the universal set U = [1,2,3,4,5,6,7,8,9,10], P = [1,2,4,6,10] and Q = [2,3,6,9], show that [P∪Q]' = P'∩Q'.

Show the model answer
Model answer

(a) Group: x(p − 2q) + 2y(p − 2q) = (p − 2q)(x + 2y)
(b) P∪Q = {1, 2, 3, 4, 6, 9, 10}, so (P∪Q)' = {5, 7, 8}
P' = {3, 5, 7, 8, 9}, Q' = {1, 4, 5, 7, 8, 10}, so P'∩Q' = {5, 7, 8}
Both sets are {5, 7, 8}, so (P∪Q)' = P'∩Q'

Explanation

Group the terms in pairs that share a common factor.

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Theory question 102

Variation

The quantity y is partly constant and partly varies inversely as the square of x. (a) Write down the relationship between x and y. (b) When x = 1, y = 11 and when x = 2, y = 5, find the value of y when x = 4.

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Model answer

(a) y = k + a/x2, where k and a are constants
(b) x = 1, y = 11: k + a = 11
x = 2, y = 5: k + a/4 = 5, i.e. 4k + a = 20
Subtract: 3k = 9, k = 3, a = 8, so y = 3 + 8/x2
When x = 4: y = 3 + 8/16 = 3½

Explanation

'Partly constant and partly varies' means a sum of a constant and a varying part.

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Theory question 103

Triangles and Quadrilaterals

(a) In the diagram below, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If |QY| = 2cm and RS = 3cm, find |PZ|.
(b) P and Q are two towns on the earth's surface on latitude 56°N. Their longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and π = 22/7].

Diagram for this question
Show the model answer
Model answer

(a) Opposite sides of a parallelogram are equal: |PQ| = |RS| = 3 cm and |YZ| = |RS| = 3 cm
|PZ| = 3 + 2 + 3 = 8 cm
(b) Difference in longitude = 95° − 25° = 70°
PQ = 70/360 × 2 × 22/7 × 6400 × cos 56°
= 7822.2 × 0.5592 ≈ 4374 km

Explanation

Along a parallel of latitude, use radius R cos(latitude).

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Theory question 104

Probability

(a) A pack of 52 playing cards in shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine. [Hint: There are 4 fives and 2 red nines in a pack of 52 cards].
(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If |PQ| = 5cm and |QR| = 3m, find the bearing of R from P, correct to the nearest degree.

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Model answer

(a) P(five or red nine) = 4/52 + 2/52 = 6/52 = 3/26
(b) Sketch: the bearing of P from Q is 330°, so ∠PQR = 360° − 330° + 60° = 90°
tan ∠QPR = 3/5 = 0.6, so ∠QPR ≈ 31°
R lies to the east side of PQ, so the bearing of R from P = 150° − 31° = 119°

Explanation

Find the angle at Q from the back bearing first; it is a right angle here.

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Theory question 105

Data Presentation

(a) The frequency table shows the marks scored by 32 students in a test.
Marks scored: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
No. of Students: 2, 3, 4, 4, 4, 4, 5, 3, 2, 1
Find the: (a) (i) Mean (ii) Median (iii) Mode
(b)(i) Find the number of students that scored at least 8 marks. (b) Caculate the percentage that scored at least 8 marks.

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Model answer

(a)(i) Σfx = 2 + 6 + 12 + 16 + 20 + 24 + 35 + 24 + 18 + 10 = 167; mean = 167/32 ≈ 5.2
(ii) Median is the average of the 16th and 17th marks; both are 5, so median = 5
(iii) Mode = 7 (frequency 5)
(b) At least 8 marks: 3 + 2 + 1 = 6 students
Percentage = 6/32 × 100% = 18.75%

Explanation

With 32 values the median is the mean of the 16th and 17th values in order.

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Theory question 106

Logarithms

(a) Using mathematical tables, find; (i) 2 sin 63.35° (ii) log cos 44.74°
(b) Find the value of K given that log K − log (K − 2) = log 5.
(c) Use logarithm tables to evaluate (3.68)2 × 6.705/√0.3581

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Model answer

(a)(i) sin 63.35° = 0.8938, so 2 sin 63.35° = 1.7876
(ii) cos 44.74° = 0.7103, log 0.7103 = 1̄.8514
(b) log(K/(K − 2)) = log 5, so K = 5(K − 2), 4K = 10, K = 2.5
(c) log 3.68 = 0.5658; × 2 = 1.1316; + log 6.705 (0.8264) = 1.9580
log 0.3581 = 1̄.5540; ÷ 2 for the root = 1̄.7770
1.9580 − 1̄.7770 = 2.1810; antilog ≈ 151.7

Explanation

To take a square root with logs, halve the log (handle the negative characteristic carefully).

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Theory question 107

Algebraic Expressions

(a) Given that p = x + ym3, find m in terms of p, x and y.
(b) Using the method of completing the square, find the roots of the equations x2 − 6x + 7 = 0, correct to 1 decimal place.
(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.

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(a) ym3 = p − x, m3 = (p − x)/y, m = ∛((p − x)/y)
(b) x2 − 6x = −7; add 9: (x − 3)2 = 2
x − 3 = ±√2 = ±1.414, x = 4.4 or 1.6
(c) Let the numbers be n and n + 2: n(n + 2) = 195
n2 + 2n − 195 = 0, (n − 13)(n + 15) = 0
n = 13 (n = −15 rejected, not positive); the numbers are 13 and 15

Explanation

To complete the square, add (half the coefficient of x)² to both sides.

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Theory question 108

Trigonometry

(a) Copy and complete the table for the relation y = 2 cos 2x − 1
x: 0°, 30°, 60°, 90°, 120°, 150°, 180°
y = 2cos 2x − 1: 10, 0, __, __, __, __, 1.0
(b) Using a scale of 2 cm = 30° on the x-axis and 2cm = 1 unit on the y-axis, draw the graph of y = 2 cos 2x − 1 for 0° ≤ x ≤ 180°
(c) On the same axis draw the graph of y = 1/180 (x − 360).
(d) Use your graphs to find the: (i) values of x for which 2 cos 2x + ½ = 0; (ii) roots of the equation 2 cos 2x − x/180 + 1 = 0.

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Model answer

(a) x = 0°: 1.0; 30°: 0; 60°: −2; 90°: −3; 120°: −2; 150°: 0; 180°: 1.0
(b) Plot the points and join with a smooth curve.
(c) Straight line through (0°, −2) and (180°, −1).
(d)(i) 2 cos 2x + ½ = 0 means 2 cos 2x − 1 = −1½: draw y = −1½ and read x ≈ 52° and x ≈ 128°
(ii) The equation rearranges to 2 cos 2x − 1 = (x − 360)/180, so read where the curve meets the line: x ≈ 55° and x ≈ 131°

Explanation

Rearrange each equation so one side is the plotted curve; the other side gives the line to draw.

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Theory question 109

Constructions

(a) Using a ruler and a pair of compasses only construct triangle ABC with |AB| = 7.5cm, |BC| = 8.1cm and ∠ABC = 105°.
(b) Locate a point D on BC such that |BD| : |DC| is 3:2
(c) Through D, construct a line l perpendicular to BC.
(d) If the line l meets AC at P, measure |BP|

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Model answer

(a) Draw BC = 8.1 cm; construct 105° at B (90° + 15°, i.e. bisect between 90° and 120°); mark A with BA = 7.5 cm; join AC.
(b) Divide BC in the ratio 3 : 2: draw a line from B, step off 5 equal units, join the 5th to C and draw a parallel through the 3rd mark; BD ≈ 4.86 cm.
(c) Construct the perpendicular to BC at D.
(d) |BP| ≈ 5.4 cm

Explanation

Show all construction arcs; dividing a line in a ratio uses equal steps and parallel lines.

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Theory question 110

Bearings

(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of: (i) Y from Z (ii) Z from X.
(b) An aircraft flies due south from an airfield on latitude 36°N longitude 138°E to an airfield on latitude 36°S longitude 138°E. (i) Calculate the distance travelled, correct to three significant figures (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour. [Take π = 22/7 R = 6400km].

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Model answer

(a) ∠YXZ = 90° − 60° = 30°; ∠XYZ = 240° − 195° = 45°; ∠XZY = 105°
(i) YZ = 20 sin 30°/sin 105° = 10/0.9659 ≈ 10.4 km
(ii) XZ = 20 sin 45°/sin 105° = 14.14/0.9659 ≈ 14.6 km
(b)(i) Angle between the latitudes = 36° + 36° = 72° along a meridian (a great circle)
Distance = 72/360 × 2 × 22/7 × 6400 ≈ 8046 km ≈ 8050 km
(ii) Time = 8046 ÷ 800 ≈ 10 hours

Explanation

Draw north lines at X and Y to find the angles of the triangle, then use the sine rule.

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Theory question 111

Data Presentation

The table below shows the scores of 2000 candidates in an entrance examination into a private secondary school.
% Marks: 11-20, 21-30, 31-40, 41-50, 51-60, 61-70, 71-80, 81-90
No. of Pupils: 68, 184, 294, 402, 480, 310, 164, 98
(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the: (i) cut-off mark, if 300 candidates are to be offered admission. (ii) probability that a candidate picked at random scored at least 45%.

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Model answer

(a) Upper class boundaries 20.5, 30.5, 40.5, 50.5, 60.5, 70.5, 80.5, 90.5 with cumulative frequencies 68, 252, 546, 948, 1428, 1738, 1902, 2000; plot and draw a smooth ogive
(b)(i) 300 admitted means 2000 − 300 = 1700 below the cut-off; reading at 1700 gives about 69%
(ii) About 710 candidates scored below 45%, so about 1290 scored at least 45%
Probability ≈ 1290/2000 ≈ 0.65

Explanation

Plot cumulative frequency against upper class boundaries; the top 300 lie above the 1700th candidate.

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Theory question 112

Probability

A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining; (a) two red balls: (b) two blue balls or two black balls; (c) one black ball and one red ball in any order.

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Model answer

Total = 10 balls; draws are independent (with replacement)
(a) P(two red) = 2/10 × 2/10 = 1/25
(b) P(two blue) + P(two black) = (5/10)2 + (3/10)2 = 25/100 + 9/100 = 17/50
(c) P(black, red) + P(red, black) = 2 × 3/10 × 2/10 = 12/100 = 3/25

Explanation

'In any order' means add both orders.

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