Given that P = {x : x is a multiple of 5}, Q = {x : x is a multiple of 3} and R = {x : x is an odd number} are subsets of μ = {x : 20 ≤ x ≤ 35}, find (P ∪ Q) ∩ R.
A.{20, 21, 25, 30, 33}
B.{21, 25, 27, 33, 35}
C.{20, 21, 25, 27, 33, 35}
D.{21, 25, 27, 30, 33, 35}
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Answer: B ({21, 25, 27, 33, 35})
Explanation
P = {20, 25, 30, 35}, Q = {21, 24, 27, 30, 33}. Their union's odd members are 21, 25, 27, 33, 35.
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting. Distance (km): 3, 4, 5, 6, 7, 8 Frequency: 5, 4, x, 9, 2x, 1 If a hunter is selected at random, find the probability that the hunter covered at least 6 km.
A.3/5
B.2/5
C.3/8
D.9/40
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Answer: A (3/5)
Explanation
19 + 3x = 40 gives x = 7. Hunters covering 6 km or more: 9 + 14 + 1 = 24. P = 24/40 = 3/5.
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting. Distance (km): 3, 4, 5, 6, 7, 8 Frequency: 5, 4, x, 9, 2x, 1 Find the mode of the distribution
A.5
B.6
C.7
D.8
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Answer: C (7)
Explanation
x = 7, so 7 km has frequency 2x = 14, the highest. Mode = 7.
A body of mass 18 kg moving with velocity 4 ms−1 collides with another body of mass 6 kg moving in the opposite direction with velocity 10 ms−1. If they stick together after collision, find their common velocity.
A.−½ ms−1
B.⅓ ms−1
C.2 ms−1
D.3 ms−1
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Answer: A (−½ ms−1)
Explanation
Taking the 6 kg body's direction as positive: 18(−4) + 6(10) = 24v, so v = −12/24 = −½ ms⁻¹ (0.5 ms⁻¹ in the direction the 18 kg body was moving).
The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67 m, 1.50 m and 1.40 m respectively. Find the mean height of all the students.
The probability that a student will graduate from a college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?
A.0.06
B.0.22
C.0.78
D.0.80
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Answer: C (0.78)
Explanation
P(none graduate) = 0.6³ = 0.216, so P(at least one) = 1 − 0.216 = 0.784 ≈ 0.78.
A particle of mass 3 kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t2 + 3t. Find the magnitude of F at time t.
A.0 N
B.2 N
C.3(2t + 3) N
D.6 N
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Answer: D (6 N)
Explanation
v = 2t + 3 and a = 2 ms⁻², so F = ma = 3 × 2 = 6 N.
A particle is acted upon by forces F = (10 N, 060°), P = (15 N, 120°) and Q = (12 N, 200°). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.
A.17.55i + 13.78j
B.17.55i − 13.78j
C.−17.55i + 13.78j
D.−17.55i − 13.78j
Show the answer and explanation
Answer: C (−17.55i + 13.78j)
Explanation
With i east and j north, a force (F, θ) is F sin θ i + F cos θ j. The resultant is 17.55i − 13.78j, so the equilibrant is −17.55i + 13.78j.
cos 2x = 1 − 2sin²x: 3 − 6sin²x − sin x = 0, i.e. 6sin²x + sin x − 3 = 0. sin x = (−1 ± √73)/12 = 0.6287 or −0.7953. sin x = 0.6287: x = 38.95°, 141.05°. sin x = −0.7953: x = 232.69°, 307.31°. x ≈ 39.0°, 141.0°, 232.7°, 307.3°.
Explanation
Write cos 2x in terms of sin x to get a quadratic in sin x.
The probability that Abiola will be late to office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will: (a) be late for only 3 days; (b) not be late in the week; (c) be late throughout the six days.
The table shows the scores obtained by a group of artistes in Vocal (X) and Instrumental (Y) music competion. Vocal (X): 63, 69, 72, 59, 82, 91, 95, 68 Instrumental (Y): 58, 61, 67, 51, 53, 79, 92, 57 Calculate the Spearman's rank correlation coefficient between the scores.
A body of mass 18 kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35°. If the system is in equilibrium, calculate the: (a) value of F; (b) tension in the string. [Take g = 10ms−2]
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Model answer
Resolve: vertically T cos 35° = 180; horizontally T sin 35° = F. (a) F = 180 tan 35° ≈ 126.0 N. (b) T = 180/cos 35° ≈ 219.7 N.
Explanation
Resolve the tension into vertical and horizontal parts.
Given that p = (8 N, 030°) and q = (9 N, 150°), find, in component form, the unit vector along (p − q).
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Model answer
With i east and j north: p = 8 sin 30° i + 8 cos 30° j = 4i + 6.928j; q = 9 sin 150° i + 9 cos 150° j = 4.5i − 7.794j. p − q = −0.5i + 14.722j; |p − q| = √(0.25 + 216.74) ≈ 14.731. Unit vector = (−0.5i + 14.722j)/14.731 ≈ −0.034i + 0.999j.
Explanation
Convert bearings to components, subtract, then divide by the magnitude.
A solid rectangular block has a base which measures 3x cm by 2x cm. The height of the block is y cm and its volume is 72 cm3. (a) Express y in terms of x. (b) Find: (i) an expression for the total surface area of the block in terms of x only; (ii) the value of x for which the total surface area has a stationary value.
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Model answer
(a) 6x²y = 72, so y = 12/x². (b)(i) A = 2(6x²) + 2(3xy) + 2(2xy) = 12x² + 10xy = 12x² + 120/x. (ii) dA/dx = 24x − 120/x² = 0 gives x³ = 5, x = ∛5 ≈ 1.71 cm.
Explanation
Eliminate y using the volume before differentiating.
(a) Find the binomial expansion of (1 + 2x)7 and (1 − 2x)7. (b) Using the result in 11(a), find, correct to three decimal places, the value of (1.2)7 − (0.8)7.
A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers. Two fruits are drawn one after the other without replacement. Calculate the probability that: (a) the first is an orange and the second is an avocado pear; (b) both are of the same fruit; (c) at least one is an apple.
(a) A particle initially at rest moves in a straight line with an acceleration of (10t − 4t2) ms−2. Find the: (i) velocity of the particle after t seconds; (ii) average acceleration of the particle during the 4th second. (b) A load of mass 120 kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards: (i) at a constant velocity; (ii) with an acceleration of 3 ms−2. [Take g = 10ms−2]
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Model answer
(a)(i) v = ∫(10t − 4t²) dt = 5t² − (4/3)t³ (v = 0 at t = 0). (ii) The 4th second is from t = 3 to t = 4: v(3) = 45 − 36 = 9; v(4) = 80 − 256/3 = −16/3. Average acceleration = (v(4) − v(3))/1 = −16/3 − 9 = −43/3 ≈ −14.3 ms⁻². (b)(i) R = mg = 1200 N. (ii) R − 1200 = 120 × 3, R = 1560 N.
The vectors 6i + 8j and 8i − 6j are parallel to OP and OQ respectively. If the magnitude of OP and OQ are 80 units and 120 units respectively, express: (a) OP and OQ in terms of i and j; (b) |PQ|, in the form c√k, where c and k are constants.
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Model answer
(a) |6i + 8j| = 10, so OP = 8(6i + 8j) = 48i + 64j. |8i − 6j| = 10, so OQ = 12(8i − 6j) = 96i − 72j. (b) PQ = OQ − OP = 48i − 136j; |PQ| = √(2304 + 18496) = √20800 = 40√13.
Explanation
Scale each direction vector to the required length.