WAEC Further Mathematics 2006 past questions and answers

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

Question 1

Sets and Logic

If P = {x : 1 ≤ x ≤ 6} and Q = {x : 2 < x < 9}, where x ∈ R, find P ∩ Q.

  1. A.{x : 2 ≤ x ≤ 6}
  2. B.{x : 2 ≤ x < 6}
  3. C.{x : 2 < x < 6}
  4. D.{x : 2 < x ≤ 6}
Show the answer and explanation

Answer: D ({x : 2 < x ≤ 6})

Explanation

The intersection keeps the numbers in both sets: they must be greater than 2 (from Q) and at most 6 (from P), so P ∩ Q = {x : 2 < x ≤ 6}.

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Question 3

Indices and Logarithms

Simplify √((−1/64)−2/3)

  1. A.−4
  2. B.−¼
  3. C.½
  4. D.4
Show the answer and explanation

Answer: D (4)

Explanation

(−1/64)−2/3 = (−64)2/3 = (−4)² = 16, and √16 = 4.

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Question 4

Binary Operations

A binary operation * is defined on the set R, of real numbers by a * b = ab/4. Find the value of √2 * √6.

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

Answer: C (√3/2)

Explanation

√2 * √6 = (√2 × √6)/4 = √12/4 = 2√3/4 = √3/2.

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Question 5

Polynomials

If (x − 3) is a factor of 2x3 + 3x2 − 17x − 30, find the remaining factors

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

Answer: D ((2x + 5)(x + 2))

Explanation

Dividing by (x − 3) leaves 2x² + 9x + 10, which factorises as (2x + 5)(x + 2).

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Question 6

Functions

Two functions f and g are defined by f : x → 3x − 1 and g : x → 2x3, evaluate fg(−2)

  1. A.−49
  2. B.−47
  3. C.−10
  4. D.−9
Show the answer and explanation

Answer: A (−49)

Explanation

First g(−2) = 2(−8) = −16, then f(−16) = 3(−16) − 1 = −49.

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Question 7

Indices and Logarithms

Given that 1/82−3y = 2y+2, find y.

  1. A.⅕
  2. B.⅞
  3. C.1
  4. D.1⅕
Show the answer and explanation

Answer: C (1)

Explanation

1/82−3y = 2−3(2−3y) = 29y−6. Equate powers: 9y − 6 = y + 2, so 8y = 8 and y = 1.

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Question 8

Surds

Given that (√3 − 5√2)(√3 + √2) = p + q√6, find q.

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

Answer: B (−4)

Explanation

Expand: 3 + √6 − 5√6 − 10 = −7 − 4√6, so p = −7 and q = −4.

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Question 9

Functions

If f(x) = 1/(2 − x), x ≠ 2, find f−1(−½).

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

Answer: A (4)

Explanation

f−1(−½) is the x for which 1/(2 − x) = −½. Then 2 − x = −2, so x = 4.

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Question 10

Binomial Theorem

Find the coefficient of x4 in the binomial expansion of (1 − 2x)6.

  1. A.320
  2. B.240
  3. C.−240
  4. D.−320
Show the answer and explanation

Answer: B (240)

Explanation

The x⁴ term is ⁶C₄(−2x)⁴ = 15 × 16x⁴ = 240x⁴.

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Question 11

Coordinate Geometry

Find the equation of the line passing through (0, −1) and parallel to the y-axis.

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

Answer: C (x = 0)

Explanation

A line parallel to the y-axis has equation x = constant. It passes through a point with x = 0, so the line is x = 0.

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Question 12

Polynomials

The roots of the equation 2x2 + kx + 5 = 0 are α and β, where k is a constant. If α2 + β2 = −1, find the values of k.

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

Answer: C (±4)

Explanation

α + β = −k/2 and αβ = 5/2. α² + β² = (α + β)² − 2αβ = k²/4 − 5 = −1, so k² = 16 and k = ±4.

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Question 13

Sequences and Series

Find the sum of the exponential series 96 + 24 + 6 + ..

  1. A.144
  2. B.128
  3. C.72
  4. D.64
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Answer: B (128)

Explanation

This is a geometric series with a = 96 and r = ¼. Sum to infinity = a/(1 − r) = 96 ÷ ¾ = 128.

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Question 14

Indices and Logarithms

Evaluate log0.25 8.

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

Answer: D (−3/2)

Explanation

Let log0.25 8 = x, so (¼)x = 8, i.e. 2−2x = 2³. Then −2x = 3 and x = −3/2.

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Question 15

Differentiation

Evaluate limx→1 (1 − x)/(x2 − 3x + 2)

  1. A.1
  2. B.½
  3. C.0
  4. D.−1
Show the answer and explanation

Answer: A (1)

Explanation

x² − 3x + 2 = (x − 1)(x − 2) and 1 − x = −(x − 1). Cancel to get −1/(x − 2); at x = 1 this is −1/(−1) = 1.

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Question 16

Statistics

The mean age of n men in a club is 50 years. Two men, aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of n.

  1. A.30
  2. B.20
  3. C.18
  4. D.14
Show the answer and explanation

Answer: B (20)

Explanation

Total age is 50n. After the two leave: (50n − 118)/(n − 2) = 49, so 50n − 118 = 49n − 98 and n = 20.

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Question 17

Permutations and Combinations

A committee of 4 is to be selected from a group of 5 men and 3 women. In how many ways can this be done if the chairman of the committee must be a man?

  1. A.15
  2. B.40
  3. C.70
  4. D.175
Show the answer and explanation

Answer: D (175)

Explanation

Choose the chairman from the 5 men (5 ways), then the other 3 members from the remaining 7 people (⁷C₃ = 35 ways): 5 × 35 = 175.

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Question 18

Permutations and Combinations

Simplify nP4/nC4

  1. A.24
  2. B.18
  3. C.12
  4. D.6
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Answer: A (24)

Explanation

ⁿP₄ = n!/(n − 4)! and ⁿC₄ = n!/((n − 4)! 4!), so the ratio is 4! = 24.

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Question 19

Matrices and Determinants

Which of the following matrices is a singular matrix?

  1. A.(1 0; 0 1)
  2. B.(2 1; 3 2)
  3. C.(3 8; 6 16)
  4. D.(0 1; 1 0)
Show the answer and explanation

Answer: C ((3 8; 6 16))

Explanation

A singular matrix has determinant 0. For (3 8; 6 16): 3 × 16 − 8 × 6 = 48 − 48 = 0.

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Question 20

Indices and Logarithms

Simplify 8n × 22n ÷ 43n.

  1. A.2−n
  2. B.21−n
  3. C.2n
  4. D.2n+1
Show the answer and explanation

Answer: A (2−n)

Explanation

Write everything in base 2: 23n × 22n ÷ 26n = 23n + 2n − 6n = 2−n.

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Question 21

Trigonometry

The area of a sector of a circle is 3 cm. If the sector subtends an angle of 1.5 radians at the centre, calculate the radius of the circle.

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

Answer: C (2cm)

Explanation

Area of a sector = ½r²θ, so 3 = ½ × r² × 1.5. Then r² = 4 and r = 2 cm.

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Question 22

Dynamics

A particle of mass 2.5 kg is moving at a speed of 12 ms−1. If a force of magnitude 10 N acts against it, find how long it takes to come to rest.

  1. A.1.5s
  2. B.3.0s
  3. C.4.0s
  4. D.6.0s
Show the answer and explanation

Answer: B (3.0s)

Explanation

The retardation is F/m = 10/2.5 = 4 ms⁻². Time to stop = 12 ÷ 4 = 3 s.

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Question 23

Statistics

Calculate the standard deviation of the distribution below.
Age (in years): 1 − 5, 6 − 10, 11 − 15
Frequency: 3, 5, 2

  1. A.1.10
  2. B.2.36
  3. C.3.50
  4. D.7.50
Show the answer and explanation

Answer: C (3.50)

Explanation

Midpoints 3, 8, 13 with frequencies 3, 5, 2 give mean 75/10 = 7.5. Σf(x − x̄)² = 60.75 + 1.25 + 60.5 = 122.5, so SD = √(122.5/10) = √12.25 = 3.5.

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Question 24

Probability

In a firing contest, the probabilities that Kojo and Kwame hit the target are 2/5 and 1/3 respectively. What is the Probability that none of them will hit the target?

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

Answer: B (2/5)

Explanation

Kojo misses with probability 3/5 and Kwame with 2/3. Both miss: 3/5 × 2/3 = 2/5.

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Question 25

Statistics

The equation of the line of best fit for variables x and y is y = 19.33 + 0.42x, where x is the independent variable. Estimate the value of y when x = 15

  1. A.18.91
  2. B.19.74
  3. C.25.63
  4. D.38.23
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Answer: C (25.63)

Explanation

Substitute x = 15: y = 19.33 + 0.42 × 15 = 19.33 + 6.3 = 25.63.

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Question 26

Differentiation

Find the coordinates of the point on the curve y = x2 + 4x − 2, where the gradient is zero.

  1. A.(−2, 10)
  2. B.(−2, 2)
  3. C.(−2, −2)
  4. D.(−2, −6)
Show the answer and explanation

Answer: D ((−2, −6))

Explanation

dy/dx = 2x + 4 = 0 gives x = −2. Then y = 4 − 8 − 2 = −6, so the point is (−2, −6).

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Question 27

Differentiation

Find the least value of the function f(x) = 3x2 + 18x + 32.

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

Answer: A (5)

Explanation

f'(x) = 6x + 18 = 0 at x = −3. f(−3) = 27 − 54 + 32 = 5, and since the x² coefficient is positive this is the least value.

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Question 28

Dynamics

A force of 32 Newtons is applied to an object of mass m kg which is at rest on a smooth horizontal surface. If the acceleration produced is 8ms−2, find the value of m

  1. A.16
  2. B.12
  3. C.6
  4. D.4
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Answer: D (4)

Explanation

F = ma, so 32 = 8m and m = 4 kg.

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Question 29

Matrices and Determinants

Given that (2 −3; 1 4)(−6; p) = (3; −26). Find the value of p.

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

Answer: B (−5)

Explanation

The first row gives 2(−6) − 3p = 3, so −3p = 15 and p = −5. The second row checks: −6 + 4(−5) = −26.

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Question 30

Coordinate Geometry

Find the coordinates of the centre of the circle 4x2 + 4y2 − 5x + 3y − 2 = 0.

  1. A.(−5/4, 3/4)
  2. B.(3/8, −5/8)
  3. C.(5/8, −3/8)
  4. D.(5/4, −3/4)
Show the answer and explanation

Answer: C ((5/8, −3/8))

Explanation

Divide by 4: x² + y² − (5/4)x + (3/4)y − ½ = 0. The centre is (−g, −f) = (5/8, −3/8).

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Question 32

Coordinate Geometry

The parallelogram PQRS has vertices P(−2,3), Q (1,4), R (2,6) and S (−1,5). Find the coordinates of the point of intersection of the diagonals.

  1. A.(−1,5)
  2. B.(−½, 3½)
  3. C.(0, 4½)
  4. D.(1½, 5)
Show the answer and explanation

Answer: C ((0, 4½))

Explanation

The diagonals of a parallelogram bisect each other, so the point is the midpoint of PR: ((−2 + 2)/2, (3 + 6)/2) = (0, 4½).

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Question 34

Statistics

The mean and median of integers x, y, z and t are 5 and z respectively. If x < y < z < t and y = 4, find (x + t)

  1. A.12
  2. B.11
  3. C.10
  4. D.8
Show the answer and explanation

Answer: A (12)

Explanation

The sum is 4 × 5 = 20. The median (y + z)/2 equals z, so z = y = 4. Then x + t = 20 − 4 − 4 = 12.

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Question 35

Vectors

If a = (3; 2) and b = (−3; 5), find a vector c such that 4a + 3c = b.

  1. A.(3; −1)
  2. B.(−5; −1)
  3. C.(−5; 1)
  4. D.(−5; −9)
Show the answer and explanation

Answer: B ((−5; −1))

Explanation

3c = b − 4a = (−3 − 12; 5 − 8) = (−15; −3), so c = (−5; −1).

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Question 36

Dynamics

A lift moving upwards with a uniform acceleration of 5ms−2, carries a body of mass p kg. If the reaction on the floor is 480 N, find the value of p. (take g = 10ms−2)

  1. A.32
  2. B.36
  3. C.48
  4. D.64
Show the answer and explanation

Answer: A (32)

Explanation

For upward acceleration, R − pg = pa, so R = p(g + a). 480 = 15p, giving p = 32.

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Question 38

Dynamics

A particle is projected vertically upwards from a height 45 metres above the ground with a velocity of 40ms−1. How long does it take to hit the ground? [Take g = 10ms−2]

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

Answer: D (9s)

Explanation

Taking up as positive, the ground is at −45 m: −45 = 40t − 5t². So t² − 8t − 9 = 0, (t − 9)(t + 1) = 0, and t = 9 s.

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Question 39

Statics

Two forces, each of magnitude 16 N, are inclined to each other at an angle of 60°. Calculate the magnitude of their resultant.

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

Answer: B (16√3)

Explanation

R² = 16² + 16² + 2(16)(16)cos 60° = 256 + 256 + 256 = 768, so R = √768 = 16√3 N.

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Question 40

Statics

ABCD is a square. Forces of magnitude 14N, 4N, 2N and 2√2N act along the sides AB, BC, CD and DA respectively. Find in Newtons the magnitude of resultant of the forces.

  1. A.14.11
  2. B.13.81
  3. C.12.06
  4. D.11.05
Show the answer and explanation

Answer: C (12.06)

Explanation

Take AB along the x-axis. Along x: 14 − 2 = 12; along y: 4 − 2√2 ≈ 1.17. R = √(12² + 1.17²) ≈ 12.06 N.

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Theory question 101

Sequences and Series

The sum of the 2nd and 5th terms of an arithmetic progression (A.P) is 42. If the difference between the 6th and the 3rd terms is 12, find the (a) common difference; (b) first term; (c) 20th term.

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Model answer

T₂ + T₅ = (a + d) + (a + 4d) = 2a + 5d = 42
T₆ − T₃ = 3d = 12
(a) d = 4
(b) 2a + 20 = 42, so a = 11
(c) T₂₀ = a + 19d = 11 + 76 = 87

Explanation

Write each term as a + (n − 1)d and solve the two equations.

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Theory question 102

Indices and Logarithms

If 22x−3y = 32 and logyx = 2, find the values of x and y.

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Model answer

22x−3y = 2⁵, so 2x − 3y = 5
logyx = 2 means x = y²
2y² − 3y − 5 = 0, (2y − 5)(y + 1) = 0, y = 5/2 or y = −1
A log base must be positive, so reject y = −1
y = 5/2, x = 25/4

Explanation

Turn the log statement into index form, then substitute.

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Theory question 103

Polynomials

If (x + 2) and (x − 1) are factors of f(x) = 6x4 + mx3 − 13x2 + nx + 14, find the; (a) values of m and n; (b) remainder when f(x) is divided by (x + 1).

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Model answer

(a) f(−2) = 0: 96 − 8m − 52 − 2n + 14 = 0, so 4m + n = 29
f(1) = 0: 6 + m − 13 + n + 14 = 0, so m + n = −7
Subtract: 3m = 36, m = 12, n = −19
(b) f(x) = 6x⁴ + 12x³ − 13x² − 19x + 14
Remainder = f(−1) = 6 − 12 − 13 + 19 + 14 = 14

Explanation

Use the factor theorem: f(a) = 0 when (x − a) is a factor; the remainder on dividing by (x + 1) is f(−1).

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Theory question 104

Differentiation

Find the equation of the tangent to the curve y = (x − 1)/(2x + 1), x ≠ −½, at the point (1, 0).

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Model answer

Quotient rule: dy/dx = [(2x + 1) − 2(x − 1)]/(2x + 1)² = 3/(2x + 1)²
At x = 1: gradient = 3/9 = 1/3
Tangent: y − 0 = ⅓(x − 1)
3y = x − 1, i.e. x − 3y − 1 = 0

Explanation

Find the gradient with the quotient rule, then use y − y₁ = m(x − x₁).

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Theory question 105

Permutations and Combinations

(a) There are six points in a plane. How many triangles can be formed with the points?
(b) A family of six is to be seated in a row. In how many ways can this be done, if the father and mother are not to sit together?

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Model answer

(a) Any 3 of the 6 points (no three in a line) make a triangle: ⁶C₃ = 20 triangles
(b) All arrangements: 6! = 720
Father and mother together (treat as one unit, 2 orders): 2 × 5! = 240
Not together: 720 − 240 = 480 ways

Explanation

For 'not together', subtract the 'together' arrangements from the total.

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Theory question 106

Statistics

The table shows the distribution of the ages of a group of people in a village.
Ages (in years): 15 − 18, 19 − 22, 23 − 26, 27 − 30, 31 − 34, 35 − 38
Freq.: 40, 33, 25, 10, 8, 4
Using an assumed mean of 24.5, calculate the mean of the distribution.

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Model answer

Class midpoints: 16.5, 20.5, 24.5, 28.5, 32.5, 36.5
d = x − 24.5: −8, −4, 0, 4, 8, 12
fd: −320, −132, 0, 40, 64, 48
Σf = 120, Σfd = −300
Mean = 24.5 + (−300/120) = 24.5 − 2.5 = 22 years

Explanation

Mean = assumed mean + Σfd/Σf.

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Theory question 107

Dynamics

A body of mass 5kg resting on a smooth horizontal plane, is acted upon by force 6i + 2j, 5i + 4j and 4 − j. Calculate the;
(a) velocity of the body;
(b) magnitude of its velocity, after 4 seconds.

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Model answer

Resultant force = (6 + 5 + 4)i + (2 + 4 − 1)j = 15i + 5j N
a = F/m = (15i + 5j)/5 = 3i + j ms⁻²
(a) Starting from rest, v = at = 4(3i + j) = 12i + 4j ms⁻¹
(b) |v| = √(12² + 4²) = √160 = 4√10 ≈ 12.65 ms⁻¹

Explanation

Add the forces, use F = ma, then v = u + at with u = 0.

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Theory question 108

Dynamics

An object is projected vertically upwards with a velocity of 80 ms−1. Find the;
(a) maximum height reached; (b) time taken to return to the point of projection. [Take g = 10 ms−2]

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Model answer

(a) At the top v = 0: 0 = 80² − 2(10)s, so s = 6400/20 = 320 m
(b) Time to the top: 0 = 80 − 10t, t = 8 s
Time to return = 2 × 8 = 16 s

Explanation

Use v² = u² − 2gs for the height and v = u − gt for the time up; the time down is the same.

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Theory question 109

Surds

(a) simplify (√75 − 3)/(√3 + 1), leaving your answers in the form a + b√c, where a, b and c are rational numbers.
(b) The points (7,3), (2,8) and (−3, 3) lie on a circle. Find the: (i) equation; (ii) radius, of the circle.

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Model answer

(a) √75 = 5√3; multiply top and bottom by (√3 − 1):
(5√3 − 3)(√3 − 1)/(3 − 1) = (15 − 5√3 − 3√3 + 3)/2 = (18 − 8√3)/2 = 9 − 4√3
(b) x² + y² + 2gx + 2fy + c = 0 through the points:
14g + 6f + c = −58; 4g + 16f + c = −68; −6g + 6f + c = −18
Solving: g = −2, f = −3, c = −12
(i) x² + y² − 4x − 6y − 12 = 0
(ii) centre (2, 3), r = √(4 + 9 + 12) = 5 units

Explanation

Rationalise with the conjugate; for the circle, substitute each point into the general equation.

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Theory question 110

Differentiation

(a) The gradient of the tangent to the curve y = 4x3 at points P and Q is 108. Find the coordinates of P and Q.
(b) Given that  = 45°, B̂ = 30°, sin(A + B) = sin A cos B + sin B cos A and cos(A + B) = cos A cos B − sin A sin B
(i) show that: sin 15° = (√6 − √2)/4 and cos 15° = (√6 + √2)/4
(ii) hence find tan 15°

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Model answer

(a) dy/dx = 12x² = 108, x² = 9, x = ±3
x = 3: y = 108; x = −3: y = −108
P(3, 108) and Q(−3, −108) (either order)
(b)(i) sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° = (√2/2)(√3/2) − (√2/2)(½) = (√6 − √2)/4
cos 15° = cos 45° cos 30° + sin 45° sin 30° = (√6 + √2)/4
(ii) tan 15° = (√6 − √2)/(√6 + √2) = (√6 − √2)²/(6 − 2) = (8 − 4√3)/4 = 2 − √3

Explanation

Set the derivative equal to 108; for 15°, use 45° − 30° (replace B by −B in the given formulas).

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Theory question 111

Integration

(a) Using the substitution U = 5 − x2, evaluate ∫12 x/√(5 − x2) dx
(b) If y = px2 + qx, dy/dx = 7 and d2y/dx2 = 6, find the values of p and q.

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Model answer

(a) U = 5 − x², dU = −2x dx; x = 1 → U = 4, x = 2 → U = 1
Integral = ∫41 −½U−½ dU = ½∫14 U−½ dU = [√U] from 1 to 4 = 2 − 1 = 1
(b) dy/dx = 2px + q, d²y/dx² = 2p = 6, so p = 3
Then 6x + q = 7, so q = 7 − 6x; if the condition is at x = 1 (as intended), q = 1

Explanation

Change the limits when you substitute; d²y/dx² gives p straight away.

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Theory question 112

Matrices and Determinants

Question 12 is for candidates in Sierra Leone and Gambia
(a) Given that A = (3 4; 2 5) and X = (3 4), evaluate XA
(b) A linear transformation T maps A = (3 4) → A1(−6, 8) and B(5, 2) → B(−10, 4). Find the: (i) Matrix of T, (ii) Image of the point k(1, −1) under T.

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Model answer

(a) XA = (3×3 + 4×2 3×4 + 4×5) = (17 32)
(b)(i) Let T = (a b; c d)
3a + 4b = −6 and 5a + 2b = −10 give a = −2, b = 0
3c + 4d = 8 and 5c + 2d = 4 give c = 0, d = 2
T = (−2 0; 0 2)
(ii) T(1; −1) = (−2; −2), so the image is (−2, −2)

Explanation

Write T with unknown entries and form simultaneous equations from the two given images.

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Theory question 113

Probability

(a) A man P has 5 red, 3 blue and 2 white buses. Another man Q has 3 red, 2 blue and 4 white buses. A bus owned by P is involved in an accident with a bus belonging to Q. Calculate the probability that the two buses are not of the same colour.
(b) A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?

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Model answer

(a) P(same colour) = (5/10)(3/9) + (3/10)(2/9) + (2/10)(4/9) = (15 + 6 + 8)/90 = 29/90
P(not same colour) = 1 − 29/90 = 61/90
(b) Going: 6 airlines × 4 shipping lines; returning: 3 shipping lines × 5 airlines
Number of ways = 6 × 4 × 3 × 5 = 360

Explanation

Find P(same colour) first and subtract from 1; on the return trip one airline and one shipping line are already used.

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Theory question 114

Statistics

The table below shows the distribution of hours spent at work by the employees of a factory in a week.
Time (hours): 20 − 29, 30 − 39, 40 − 49, 50 − 59, 60 − 69, 70 − 79
No. of Persons: 8, 11, 23, 25, 8, 5
(a) Draw an ogive for the distribution.
(b) Using your graph, estimate the (i) median; (ii) lower quartile; (iii) 40th percentile; (iv) number of employees that spent at least 50 hours 30 minutes.

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Model answer

Cumulative frequencies at upper class boundaries 29.5, 39.5, 49.5, 59.5, 69.5, 79.5: 8, 19, 42, 67, 75, 80
(a) Plot these points (plus (19.5, 0)) and join with a smooth S-shaped curve
(b)(i) Median at 40th: 39.5 + (21/23) × 10 ≈ 48.6 hours
(ii) Lower quartile at 20th: 39.5 + (1/23) × 10 ≈ 39.9 hours
(iii) 40th percentile at 32nd: 39.5 + (13/23) × 10 ≈ 45.2 hours
(iv) Cumulative frequency at 50.5 ≈ 42 + 2.5 = 44.5, so about 80 − 44.5 ≈ 35 employees
(Readings from a graph are accepted within a small tolerance.)

Explanation

Plot cumulative frequency against upper class boundaries; read positions N/2, N/4 and 0.4N.

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Theory question 115

Probability

(a) Two pupils are chosen at random from a group of 4 boys and 5 girls. Find the probability that the two pupils chosen would be boys.
(b) Twenty percent of the total production of transistors produced by a machine are below standard. If a random sample of 6 transistors produced by the machine is taken, what is the probability of getting; (i) exactly 2; (ii) exactly 1; (iii) at least 2; (iv) at most 2, standard transistor(s)?

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Model answer

(a) ⁴C₂/⁹C₂ = 6/36 = 1/6
(b) Let X = number of standard transistors; p(standard) = 0.8, q = 0.2, n = 6
(i) P(X = 2) = ⁶C₂(0.8)²(0.2)⁴ = 15 × 0.64 × 0.0016 = 0.01536
(ii) P(X = 1) = 6(0.8)(0.2)⁵ = 0.001536
(iii) P(X ≥ 2) = 1 − P(0) − P(1) = 1 − 0.000064 − 0.001536 = 0.9984
(iv) P(X ≤ 2) = 0.000064 + 0.001536 + 0.01536 = 0.01696

Explanation

Use the binomial distribution with p = 0.8 for a standard transistor.

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Theory question 116

Statics

(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings XP and YP. The strings are inclined at 60° and 40° respectively to the downward vertical. Find, correct to two decimal places, the tensions in the strings. [Take g = 10 ms−2]
(b) The height h metres, of a ball thrown into the air is 2 + 20t + kt2, after t seconds. If it takes 2 seconds for the ball to reach its highest point, find: (i) the value of k; (ii) its highest point from the point of throw.

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Model answer

(a) Weight = 150 N. Angle between the strings = 100°; angle between XP and the weight = 120°, between YP and the weight = 140°
Lami: TXP/sin 140° = TYP/sin 120° = 150/sin 100°
TXP = 150 × 0.6428/0.9848 ≈ 97.91 N
TYP = 150 × 0.8660/0.9848 ≈ 131.91 N
(b)(i) dh/dt = 20 + 2kt = 0 at t = 2, so k = −5
(ii) h(2) = 2 + 40 − 20 = 22 m above the reference level, i.e. 20 m above the point of throw (where h = 2)

Explanation

Use Lami's theorem (or resolve horizontally and vertically); the highest point is where dh/dt = 0.

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Theory question 117

Dynamics

(a) A ball P moving with velocity 2u ms−1, collides with a similar ball Q, of different mass, which is at rest. After collision, Q moves with velocity u ms−1 and P with velocity ½u ms−1, in the opposite direction. Find the ratio of the masses of P and Q.
(b) Two forces of magnitudes 3N and 7N have a resultant of magnitude 5N. Calculate, correct to one decimal place, the angle between the two forces.
(c) AB = (−4; 6) and CB = (2; −3) are two vectors in the XY plane. If V is the mid-point of AB, find CV.

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Model answer

(a) Momentum: mP(2u) + 0 = mP(−½u) + mQ(u)
2.5mP = mQ, so mP : mQ = 2 : 5
(b) 5² = 3² + 7² + 2(3)(7)cos θ, so 42 cos θ = −33, cos θ = −0.7857
θ ≈ 141.8°
(c) VB = ½AB = (−2; 3)
CV = CB + BV = CB − VB = (2; −3) − (−2; 3) = (4; −6)

Explanation

Conserve momentum with P's final velocity negative; for (b) use the resultant formula R² = P² + Q² + 2PQ cos θ.

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Theory question 118

Statics

(a) A body of mass 5kg is placed on a smooth plane inclined at an angle 30° to the horizontal. Find the magnitude of the force; (i) acting parallel to the plane; (ii) required to keep the body in equilibrium. [Take g = 10 ms−2]
(b) A uniform plank PQ of length 10m and mass m kg rests on two supports A and B, where |PA| = |BQ| = 1m. A load of mass 8kg is placed on the plank at point C such that |AC| = 3.5 m. If the reaction at B is 100 N, calculate the (i) value of m; (ii) reaction at A. [Take g = 10ms−2]

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Model answer

(a)(i) Component of the weight down the plane = mg sin 30° = 5 × 10 × ½ = 25 N
(ii) A force of 25 N up the plane (parallel to it) keeps the body in equilibrium (normal reaction = 50 cos 30° ≈ 43.3 N)
(b) AB = 8 m; the plank's weight 10m N acts at its centre, 4 m from A; the load 80 N acts 3.5 m from A
(i) Moments about A: 100 × 8 = 80 × 3.5 + 10m × 4, so 800 = 280 + 40m, m = 13 kg
(ii) RA + 100 = 80 + 130, so RA = 110 N

Explanation

On a smooth plane only the weight component along the plane needs balancing; for the plank take moments about one support.

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