WAEC Further Mathematics 2015 past questions and answers

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

Question 1

Surds

Simplify: (1 − 2√5)/(2 + 3√2)

  1. A.14(2√2 + 6√5 − 4√10)
  2. B.(1/14)(2 − 3√2 − 4√5 − 6√10)
  3. C.(1/14)(3√2 + 4√5 − 6√10 − 2)
  4. D.14(2 + 3√2 − 6√5 + 4√10)
Show the answer and explanation

Answer: C ((1/14)(3√2 + 4√5 − 6√10 − 2))

Explanation

Multiply top and bottom by (2 − 3√2): numerator 2 − 3√2 − 4√5 + 6√10, denominator 4 − 18 = −14. Dividing by −14 gives (1/14)(3√2 + 4√5 − 6√10 − 2).

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Question 2

Trigonometry

Solve: 2cos x − 1 = 0, for 0 ≤ x ≤ 2π

  1. A.(2π/3, 4π/3)
  2. B.(π/6, 5π/6)
  3. C.(π/5, 2π/5)
  4. D.(π/3, 5π/3)
Show the answer and explanation

Answer: D ((π/3, 5π/3))

Explanation

cos x = ½, which happens at x = π/3 and x = 2π − π/3 = 5π/3.

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Question 3

Indices and Logarithms

Solve 4(2x²) = 8x.

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

Answer: A (1 and 2)

Explanation

2^(2 + x²) = 2^(3x), so x² − 3x + 2 = 0, (x − 1)(x − 2) = 0: x = 1 or 2.

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Question 4

Indices and Logarithms

If log3 x = log9 3, find the value of x.

  1. A.32
  2. B.3½
  3. C.31/3
  4. D.21/3
Show the answer and explanation

Answer: B (3½)

Explanation

log₉ 3 = ½ (since 9^½ = 3). So log₃ x = ½ and x = 3^½.

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Question 5

Binomial Theorem

Find the third term of (x/2 − 1)8 in descending order of x.

  1. A.x7/8
  2. B.7x6/16
  3. C.7x5/4
  4. D.(35/8)x4
Show the answer and explanation

Answer: B (7x6/16)

Explanation

Third term = ⁸C₂(x/2)⁶(−1)² = 28x⁶/64 = 7x⁶/16.

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Question 6

Functions

Given that f: x → x2 and g: x → x + 3 where x ∈ R find fog(2).

  1. A.25
  2. B.9
  3. C.7
  4. D.5
Show the answer and explanation

Answer: A (25)

Explanation

g(2) = 5, then f(5) = 25.

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Question 8

Matrices and Determinants

Given that P = (−2 1 / 3 4) and Q = (5 −3 / 2 −1), find PQ − QP. (Matrices written row by row; / separates the rows.)

  1. A.(0 0 / 0 0)
  2. B.(27 12 / 16 −15)
  3. C.(−20 −6 / 12 −8)
  4. D.(11 12 / 30 −11)
Show the answer and explanation

Answer: D ((11 12 / 30 −11))

Explanation

PQ = (−8 5 / 23 −13) and QP = (−19 −7 / −7 −2). PQ − QP = (11 12 / 30 −11).

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Question 9

Polynomials

Which of the following is a factor of the polynomial 6x4 + 2x3 + 15x + 5?

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

Answer: A (3x + 1)

Explanation

Group: 2x³(3x + 1) + 5(3x + 1) = (3x + 1)(2x³ + 5).

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Question 10

Functions

Given that f: x → (2x − 1)/(x + 2), x ≠ −2, find f−1, the inverse of f

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

Answer: A (f−1: x → (1 + 2x)/(2 − x), x ≠ 2)

Explanation

y(x + 2) = 2x − 1 gives x(y − 2) = −1 − 2y, so x = (1 + 2y)/(2 − y). f⁻¹(x) = (1 + 2x)/(2 − x), x ≠ 2.

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Question 11

Sequences and Series

If 36, p, 9/4 and q are consecutive terms of an exponential sequence (G. P), find the sum of p and q.

  1. A.9/16
  2. B.81/16
  3. C.9
  4. D.9 9/16
Show the answer and explanation

Answer: D (9 9/16)

Explanation

r² = (9/4)/36 = 1/16, so r = ¼ (taking the positive ratio): p = 9 and q = 9/16. p + q = 9 9/16.

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Question 12

Differentiation

Find the minimum value of y = x2 + 6x − 12.

  1. A.−21
  2. B.−12
  3. C.−6
  4. D.−3
Show the answer and explanation

Answer: A (−21)

Explanation

dy/dx = 2x + 6 = 0 at x = −3; y = 9 − 18 − 12 = −21.

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Question 13

Coordinate Geometry

A line passes through the origin and the point (1¼, 2½). What is the gradient of the line?

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

Answer: B (2)

Explanation

Gradient = 2½ ÷ 1¼ = 2.

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Question 14

Coordinate Geometry

A line passes through the origin and the point (1¼, 2½). Find the y coordinate of the line when x = 4.

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

Answer: D (8)

Explanation

The line is y = 2x, so y = 8 when x = 4.

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Question 15

Permutations and Combinations

In how many ways can a committee of five be selected from eight students if two particular students are to be included?

  1. A.20
  2. B.28
  3. C.54
  4. D.58
Show the answer and explanation

Answer: A (20)

Explanation

The two are already in; choose the other 3 from the remaining 6: ⁶C₃ = 20.

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Question 16

Vectors

If x = i − 3j and y = 6i + j, calculate the angle between x and y.

  1. A.60°
  2. B.75°
  3. C.81°
  4. D.85°
Show the answer and explanation

Answer: C (81°)

Explanation

x·y = 6 − 3 = 3; |x| = √10, |y| = √37. cos θ = 3/√370 ≈ 0.156, θ ≈ 81°.

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Question 17

Integration

The gradient of a curve at the point (−2, 0) is 3x2 − 4x. Find the equation of the curve.

  1. A.y = 6x − 4
  2. B.y = 6x2 − 4 + 12
  3. C.y = x3 − 2x2
  4. D.y = x3 − 2x2 + 16
Show the answer and explanation

Answer: D (y = x3 − 2x2 + 16)

Explanation

y = x³ − 2x² + k; through (−2, 0): −8 − 8 + k = 0, k = 16.

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Question 18

Polynomials

If α and β are the roots of x2 + x − 2 = 0, find the value of (1/α2 + 1/β2)

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

Answer: A (5/4)

Explanation

α + β = −1, αβ = −2. (α² + β²)/(αβ)² = (1 + 4)/4 = 5/4.

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Question 19

Polynomials

Given that x2 + 4x + k ≡ (x + r)2 + 1, find the values of k and r

  1. A.k = 5, r = −1
  2. B.k = 5, r = 2
  3. C.k = 2, r = 5
  4. D.k = 1, r = 5
Show the answer and explanation

Answer: B (k = 5, r = 2)

Explanation

(x + r)² + 1 = x² + 2rx + r² + 1, so 2r = 4 (r = 2) and k = 4 + 1 = 5.

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Question 20

Sets and Logic

Given the statements: p : the subject is difficult. q: I will do my best. Which of the following is equivalent to "Although the subject is difficult, I will do my best".

  1. A.p ∨ q
  2. B.~p ∨ q
  3. C.p ∧ ~q
  4. D.p ∧ q
Show the answer and explanation

Answer: D (p ∧ q)

Explanation

'Although … ,' joins two statements that are both true, so it means 'and': p ∧ q.

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Question 21

Statics

Forces of magnitude 8 N and 5 N act on a body as shown in the diagram. Calculate, correct to 2 decimal places, the resultant force acting at O.

Diagram for this question
  1. A.14.06 N
  2. B.13.00 N
  3. C.9.83 N
  4. D.8.26 N
Show the answer and explanation

Answer: D (8.26 N)

Explanation

Horizontal: 8 cos 45° − 5 cos 30° = 5.657 − 4.330 = 1.327; vertical: 8 sin 45° + 5 sin 30° = 5.657 + 2.5 = 8.157. Resultant = √(1.327² + 8.157²) ≈ 8.26 N.

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Question 22

Statics

Forces of magnitude 8 N and 5 N act on a body as shown in the diagram. Calculate, correct to 2 decimal places, the angle that the resultant makes with the horizontal.

Diagram for this question
  1. A.80.76°
  2. B.75.00°
  3. C.71.99°
  4. D.15.00°
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Answer: A (80.76°)

Explanation

tan θ = 8.157/1.327 ≈ 6.15, so θ ≈ 80.76°.

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Question 23

Vectors

Given that r = 2i − j, s = 3i + 5j and t = 6i − 2j, find the magnitude of (2r + s − t)

  1. A.√15
  2. B.4
  3. C.√24
  4. D.√26
Show the answer and explanation

Answer: D (√26)

Explanation

2r + s − t = (4 + 3 − 6)i + (−2 + 5 + 2)j = i + 5j; magnitude √26.

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Question 24

Statistics

The table shows the distribution of marks scored by students in a test.
Marks: 0, 1, 2, 3, 4, 5
No. of candidates: 6, 4, 8, 10, 9, 3
How many candidates scored above the median mark?

  1. A.3
  2. B.12
  3. C.22
  4. D.30
Show the answer and explanation

Answer: B (12)

Explanation

40 candidates; the 20th and 21st marks are both 3 (cumulative 6, 10, 18, 28), so the median is 3. Above it: 9 + 3 = 12.

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Question 25

Statistics

The table shows the distribution of marks scored by students in a test.
Marks: 0, 1, 2, 3, 4, 5
No. of candidates: 6, 4, 8, 10, 9, 3
Find the inter-quartile range of the distribution.

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

Answer: B (3)

Explanation

The lower quartile (10th mark) is 1 and the upper quartile (30th mark) is 4, so the inter-quartile range is 4 − 1 = 3.

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Question 26

Dynamics

A mass of 75 kg is placed on a lift. Find the force exerted by the floor of the lift on the mass when the lift is moving up with a constant velocity. [Take g = 9.8 ms−2]

  1. A.750 N
  2. B.745 N
  3. C.735 N
  4. D.98 N
Show the answer and explanation

Answer: C (735 N)

Explanation

Constant velocity means no acceleration, so R = mg = 75 × 9.8 = 735 N.

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Question 27

Probability

Each of the 90 students in a class speak at least Igbo and Hausa. If 56 students speak Igbo and 50 speak Hausa, find the probability that a student selected at random from the class speaks Igbo only.

  1. A.28/45
  2. B.4/9
  3. C.8/45
  4. D.1/9
Show the answer and explanation

Answer: B (4/9)

Explanation

Both languages: 56 + 50 − 90 = 16, so Igbo only = 40. P = 40/90 = 4/9.

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Question 28

Matrices and Determinants

If |1 + 2x, −1 / 6, 3 − x| = −3, find the values of x. (Determinant written row by row; / separates the rows.)

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

Answer: D (x = 4, −3/2)

Explanation

(1 + 2x)(3 − x) + 6 = −3 gives 2x² − 5x − 12 = 0, (2x + 3)(x − 4) = 0: x = 4 or −3/2.

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Question 29

Integration

Find ∫ (x3 + 5x + 1)/x3 dx.

  1. A.x2 + 10x + c
  2. B.x + (5/3)x3 + x4 + c
  3. C.x − 5x2 − 2x2 + c
  4. D.x − 5/x − 1/(2x2) + c
Show the answer and explanation

Answer: D (x − 5/x − 1/(2x2) + c)

Explanation

∫(1 + 5x⁻² + x⁻³) dx = x − 5/x − 1/(2x²) + c.

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Question 31

Dynamics

A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by a = (3t − 2) ms−2. Find the other time when the velocity would be zero.

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

Answer: C (4/3 seconds)

Explanation

v = 3t²/2 − 2t (v = 0 at t = 0). v = 0 again when 3t/2 = 2, t = 4/3 s.

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Question 32

Dynamics

A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by a = (3t − 2) ms−2. Find the distance covered after 3 seconds

  1. A.10 m
  2. B.9 m
  3. C.13/3 m
  4. D.9/2 m
Show the answer and explanation

Answer: D (9/2 m)

Explanation

s = ∫v dt = t³/2 − t² (s = 0 at t = 0). At t = 3: 27/2 − 9 = 9/2 m from the start.

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Question 33

Differentiation

Given that y = 4 − 9x and Δx = 0.1, calculate Δy.

  1. A.9.0
  2. B.0.9
  3. C.−0.3
  4. D.−0.9
Show the answer and explanation

Answer: D (−0.9)

Explanation

Δy = −9Δx = −0.9.

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Question 34

Probability

Four fair coins are tossed once. Calculate the probability of obtaining equal number of heads and tails.

  1. A.1/4
  2. B.3/8
  3. C.1/2
  4. D.15/16
Show the answer and explanation

Answer: B (3/8)

Explanation

Two heads and two tails: ⁴C₂ = 6 of the 16 outcomes, 6/16 = 3/8.

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Question 35

Statistics

In calculating the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers. If he obtained 20 as the mean, find the correct mean.

  1. A.24
  2. B.23
  3. C.21
  4. D.19
Show the answer and explanation

Answer: C (21)

Explanation

Wrong total 160; correct total 160 − 17 + 25 = 168, mean 168/8 = 21.

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Question 36

Permutations and Combinations

Simplify: nCr ÷ nCr−1

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

Answer: D ((n − r + 1)/r)

Explanation

ⁿCᵣ/ⁿCᵣ₋₁ = [(r − 1)!(n − r + 1)!]/[r!(n − r)!] = (n − r + 1)/r.

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Question 37

Trigonometry

If 2 sin2 θ = 1 + Cos θ, 0° ≤ θ ≤ 90°, find the value of θ.

  1. A.90°
  2. B.60°
  3. C.45°
  4. D.30°
Show the answer and explanation

Answer: B (60°)

Explanation

2(1 − cos²θ) = 1 + cos θ gives 2cos²θ + cos θ − 1 = 0, (2cos θ − 1)(cos θ + 1) = 0. cos θ = ½, θ = 60°.

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Question 38

Dynamics

A 24 N force acts on a body such that it changes its velocity from 5 ms−1 to 9 ms−1 in 2 seconds. If the body is travelling in a straight line, calculate the distance covered during the period.

  1. A.22 m
  2. B.18 m
  3. C.14 m
  4. D.10 m
Show the answer and explanation

Answer: C (14 m)

Explanation

s = ½(u + v)t = ½(5 + 9)(2) = 14 m.

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Question 39

Sequences and Series

The sum, Sn, of a sequence is given by Sn = 2n2 − 5. Find the 6th term.

  1. A.112
  2. B.67
  3. C.45
  4. D.22
Show the answer and explanation

Answer: D (22)

Explanation

T₆ = S₆ − S₅ = (72 − 5) − (50 − 5) = 67 − 45 = 22.

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Question 40

Statics

Forces F1 (8 N, 030°) and F2 (10 N, 150°) act on a particle. Find the horizontal component of the resultant force.

  1. A.1.7 N
  2. B.4.5 N
  3. C.9.0 N
  4. D.13.0 N
Show the answer and explanation

Answer: C (9.0 N)

Explanation

The angles are bearings (from north, clockwise). Horizontal (east) components: 8 sin 30° + 10 sin 150° = 4 + 5 = 9.0 N.

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Theory question 101

Binary Operations

A binary operation Δ is defined on the set of real numbers, R, by a Δ b = a3 − b3. Without using calculator, find the value of (√3 + √2) Δ (√3 − √2), leaving the answer in surd form.

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Model answer

(√3 + √2)³ = 3√3 + 9√2 + 6√3 + 2√2 = 9√3 + 11√2.
(√3 − √2)³ = 9√3 − 11√2.
Difference = 22√2.

Explanation

Expand each cube with the binomial pattern.

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Theory question 102

Coordinate Geometry

Points (2, 1) and (6, 7) are opposite verticals of a square which is inscribed in a circle. Find the: (a) centre of the circle; (b) equation of the circle.

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Model answer

The diagonal of the square is a diameter.
(a) Centre = midpoint = (4, 4).
(b) r² = (2 − 4)² + (1 − 4)² = 13. (x − 4)² + (y − 4)² = 13, i.e. x² + y² − 8x − 8y + 19 = 0.

Explanation

A square's diagonal is a diameter of its circumscribed circle.

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Theory question 103

Integration

If f″(x) = 2, f′(1) = 0 and f(0) = −8, find f(x).

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Model answer

f′(x) = 2x + k; f′(1) = 0 gives k = −2.
f(x) = x² − 2x + c; f(0) = −8 gives c = −8.
f(x) = x² − 2x − 8.

Explanation

Integrate twice, using one condition after each step.

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Theory question 104

Trigonometry

Solve: tan(2x − 15)° − 1 = 0, for values of x such that 0° ≤ x ≤ 360°.

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Model answer

tan(2x − 15)° = 1. For 0 ≤ x ≤ 360, 2x − 15 runs from −15 to 705.
2x − 15 = 45, 225, 405, 585.
x = 30°, 120°, 210°, 300°.

Explanation

tan repeats every 180°, and 2x − 15 covers almost two full turns.

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Theory question 105

Dynamics

A car moving with an initial velocity, u, travels in a straight line with a constant acceleration of 3 ms−2 until it attains a velocity of 33 ms−1 after 6 seconds. Calculate the distance travelled by the car.

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Model answer

u = v − at = 33 − 18 = 15 ms⁻¹.
s = ½(u + v)t = ½(15 + 33)(6) = 144 m.

Explanation

Find u first, then the distance.

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Theory question 106

Statistics

Five finalists in a beauty pageant were ranked by two judges X and Y as shown in the table.
Finalist: Anne, Linda, Susan, Rose, Erica
X: 1, 4, 3, 5, 2
Y: 3, 2, 4, 5, 1
Calculate the Spearman's rank correlation co-efficient.

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Model answer

d = −2, 2, −1, 0, 1; Σd² = 10.
ρ = 1 − 6(10)/(5 × 24) = 1 − 0.5 = 0.5.

Explanation

The scores are already ranks, so subtract directly.

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Theory question 107

Probability

There are 8 boys and 6 girls in a class. If two students are selected at random from the class, find the probability that they are of: (a) The same sex; (b) Different sexes.

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Model answer

(a) (8/14)(7/13) + (6/14)(5/13) = (56 + 30)/182 = 43/91.
(b) 1 − 43/91 = 48/91.

Explanation

Selecting two without replacement: the second draw is out of 13.

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Theory question 108

Vectors

Forces of magnitude 3N, 4 N and 2 N act along the vectors j, −i + j and i + j respectively. Calculate, correct to one decimal place, the magnitude of the resultant of the forces.

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Model answer

Unit vectors: j, (−i + j)/√2, (i + j)/√2.
Forces: 3j; 4(−i + j)/√2 = −2.828i + 2.828j; 2(i + j)/√2 = 1.414i + 1.414j.
Resultant = −1.414i + 7.243j; magnitude = √(2 + 52.46) ≈ 7.4 N.

Explanation

Scale each direction to a unit vector before multiplying by the force.

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Theory question 109

Functions

(a) The functions f: x → x2 + 1 and g: x → 5 − 3x are defined on the set of real numbers, R. (i) State the domain of f−1, the inverse of f. (ii) Find g−1(2).
(b) Evaluate ∫02 (x + 3)/√(x2 + 6x + 9) dx

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Model answer

(a)(i) f⁻¹(x) = √(x − 1) (taking x ≥ 0 for f), so the domain of f⁻¹ is {x : x ≥ 1} (the range of f).
(ii) g⁻¹(x) = (5 − x)/3, so g⁻¹(2) = 1.
(b) √(x² + 6x + 9) = x + 3 for 0 ≤ x ≤ 2, so the integrand is 1 and the integral is 2.

Explanation

Notice x² + 6x + 9 = (x + 3)².

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Theory question 110

Surds

(a) If (√5 + 4)/(3 − 2√5) − (2 + √5)/(4 − 2√5) = a + b√5, find the values of a and b.
(b)(i) Evaluate: |2 −1 2 / 1 3 4 / 1 2 1| (rows separated by /). (ii) Using the result in 10(b)(i), find, correct to two decimal places the value of x in the system of equations: 2x − y + 2z + 5 = 0, x + 3y + 4z − 1 = 0, x + 2y + z + 2 = 0.

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Model answer

(a) (√5 + 4)/(3 − 2√5) = (√5 + 4)(3 + 2√5)/(−11) = −2 − √5.
(2 + √5)/(4 − 2√5) = (2 + √5)(4 + 2√5)/(−4) = −9/2 − 2√5.
Difference = 5/2 + √5: a = 5/2, b = 1.
(b)(i) 2(3 − 8) + 1(1 − 4) + 2(2 − 3) = −15.
(ii) Right-hand sides −5, 1, −2. Δx = |−5 −1 2 / 1 3 4 / −2 2 1| = 25 + 9 + 16 = 50.
x = 50/(−15) ≈ −3.33.

Explanation

Cramer's rule: x = Δx/Δ.

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Theory question 111

Binomial Theorem

(a)(i) Write down the Binomial expansion of (1 + x)4. (ii) Use the result in 11(a)(i) to evaluate, correct to three decimal places (5/4)4.
(b) The first, second and fifth terms of a linear sequence (A. P.) are three consecutive terms of an exponential sequence (G. P.). If the first term of the linear sequence is 7, find its common difference.

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Model answer

(a)(i) 1 + 4x + 6x² + 4x³ + x⁴.
(ii) x = ¼: 1 + 1 + 0.375 + 0.0625 + 0.0039 ≈ 2.441.
(b) (7 + d)² = 7(7 + 4d) gives d² = 14d, so d = 14 (d ≠ 0).

Explanation

For G.P. terms, (middle)² = first × third.

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Theory question 112

Statistics

The table shows the distribution of heights of a group of people.
Height (m): 0.4 − 0.5, 0.6 − 0.9, 1.0 − 1.2, 1.3 − 1.4, 1.5 − 1.7
No. of People: 2, 8, 12, 6, 6
(a) Draw a histogram to illustrate the distribution. (b) Using an assumed mean of 1.1 m, find, correct to one decimal place, the mean height of the group.

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Model answer

(a) Boundaries (widths): 0.35–0.55 (0.2), 0.55–0.95 (0.4), 0.95–1.25 (0.3), 1.25–1.45 (0.2), 1.45–1.75 (0.3). Unequal widths: plot frequency per 0.1 m – 1, 2, 4, 3, 2 – as touching bars.
(b) Class marks 0.45, 0.75, 1.1, 1.35, 1.6; d = −0.65, −0.35, 0, 0.25, 0.5.
Σfd = −1.3 − 2.8 + 0 + 1.5 + 3.0 = 0.4; Σf = 34.
Mean = 1.1 + 0.4/34 ≈ 1.1 m.

Explanation

With unequal class widths, plot frequency densities.

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Theory question 113

Probability

(a) Edem and his wife were invited to a dinner by a family of 5. They all sat in such a way that Edem sat next to his wife. Find the number of ways of seating them in a row.
(b) A bag contains 4 red and 5 black identical balls. If 5 balls are selected at random from the bag, one after the other with replacement, find the probability that: (i) a red ball was picked 3 times; (ii) a black ball was picked at most 2 times.

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Model answer

(a) 7 people; treat Edem and his wife as one unit: 6! × 2! = 1440.
(b) P(red) = 4/9, P(black) = 5/9, n = 5.
(i) ⁵C₃(4/9)³(5/9)² = 16 000/59 049 ≈ 0.271.
(ii) Black 0, 1 or 2 times: [(4/9)⁵ + 5(5/9)(4/9)⁴ + 10(5/9)²(4/9)³] = (1024 + 6400 + 16 000)/59 049 = 23 424/59 049 ≈ 0.397.

Explanation

Glue the couple together, then let them swap places.

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Theory question 114

Vectors

(a) Given that AB = (4, 5) and BC = (−3, 5) (column vectors), find the: (i) angle between the vectors AB and BC;
(b) P, Q, R and M are points in the OXY plane. If PQ = 2i + 8j, PR = 11i − 12j and M divides QR internally in the ratio 3 : 7, find PM.

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Model answer

(a) AB·BC = −12 + 25 = 13; |AB| = √41, |BC| = √34.
cos θ = 13/√1394 ≈ 0.348, θ ≈ 69.6°.
(b) QR = PR − PQ = 9i − 20j. QM = (3/10)QR = 2.7i − 6j.
PM = PQ + QM = 4.7i + 2j.

Explanation

M is 3/10 of the way from Q to R.

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Theory question 115

Dynamics

(a) A bucket full of water with a mass of 8 kg is pulled out of a well with a light inextensible rope. Find its acceleration when the tension in the rope is 150 N. [Take g = 10 ms−2]
(b) A mass of 12 kg is acted upon by a force F, changing its speed from 15 ms−1 to 25 ms−1 after covering a distance of 50 m. Find the: (i) value of F; (ii) distance covered when its speed is 35 ms−1

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(a) T − mg = ma: 150 − 80 = 8a, a = 8.75 ms⁻² upwards.
(b)(i) 25² = 15² + 2a(50), so a = 4 ms⁻² and F = 12 × 4 = 48 N.
(ii) 35² = 15² + 2(4)s, so s = 1000/8 = 125 m (from where its speed was 15 ms⁻¹).

Explanation

Use v² = u² + 2as when time is not involved.

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