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?
A.15
B.40
C.70
D.175
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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.
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?
A.1/5
B.2/5
C.3/5
D.4/5
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Answer: B (2/5)
Explanation
Kojo misses with probability 3/5 and Kwame with 2/3. Both miss: 3/5 × 2/3 = 2/5.
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
A.18.91
B.19.74
C.25.63
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.
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
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)
A.32
B.36
C.48
D.64
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Answer: A (32)
Explanation
For upward acceleration, R − pg = pa, so R = p(g + a). 480 = 15p, giving p = 32.
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]
A.1s
B.3s
C.7s
D.9s
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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.
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.
A.14.11
B.13.81
C.12.06
D.11.05
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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.
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.
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.
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).
(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.
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
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.
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.
(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.
(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).
(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.
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.
(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?
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.
(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)?
(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.
(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.
(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.