52 questions. Try each one, then open the answer to check yourself and read the explanation.
If (√5 − 3)/(2√5 + 3) ≡ x + y√5, find the value of (x + y).
Answer: B (10/11)
Multiply top and bottom by (2√5 − 3): (√5 − 3)(2√5 − 3)/(20 − 9) = (19 − 9√5)/11. So x = 19/11, y = −9/11 and x + y = 10/11.
If (92x−1 × 32x)/27(x−1) = 1, find the value of x.
Answer: B (−1/3)
In powers of 3: 3^(4x − 2) × 3^(2x) ÷ 3^(3x − 3) = 3^(3x + 1) = 3⁰. So 3x + 1 = 0 and x = −1/3.
If logx 16 = 2, find the value of x
Answer: A (4)
log_x 16 = 2 means x² = 16; the base is positive, so x = 4.
Consider the statements:
p: the land is fertile
q: the farmer harvests big cassava tubers
r: the sun is harsh
Which of these represents the statement, "If the land is fertile and the sun is not harsh then the farmer harvests big cassava tubers"?
Answer: D ((p ∧ ~r) ⇒ q)
'The land is fertile and the sun is not harsh' is p ∧ ~r; 'if … then the farmer harvests' makes it (p ∧ ~r) ⇒ q.
Find the truth set of x for which (2x + 3)/(4x3 − x) is undefined
Answer: C ({−½, 0, ½})
The fraction is undefined when 4x³ − x = x(2x − 1)(2x + 1) = 0, i.e. x = −½, 0 or ½.
Find the equation of the line passing through points (1, 4) and (−2, 6).
Answer: D (2x + 3y = 14)
Gradient = (6 − 4)/(−2 − 1) = −2/3. y − 4 = −⅔(x − 1) gives 3y − 12 = −2x + 2, i.e. 2x + 3y = 14.
Given that g: x → 2x + 8, f: x → x + 2 and p: x → 7 − x, find [g o (p o f)](x)
Answer: C (18 − 2x)
p(f(x)) = 7 − (x + 2) = 5 − x. Then g(5 − x) = 2(5 − x) + 8 = 18 − 2x.
Find the reminder when f(x) = 2x3 − 4x2 + x − 7 is divided by (x + 1).
Answer: A (−14)
By the remainder theorem the remainder is f(−1) = −2 − 4 − 1 − 7 = −14.
The third and seventh terms of an exponential sequence (G.P) are 9 and 1/9 respectively. Find the first term
Answer: A (81)
ar² = 9 and ar⁶ = 1/9, so r⁴ = 1/81 and r² = 1/9. Then a = 9 ÷ (1/9) = 81.
If P = {x : 0 < x ≤ 6} and Q = {x : 1 ≤ x ≤ 7}, where x is an integer, find P ∩ Q.
Answer: A (P)
P = {1, 2, 3, 4, 5, 6} and Q = {1, …, 7}. Every member of P is in Q, so P ∩ Q = P.
Given that 11, p, q, r, s, t, 29 are consecutive terms of a linear sequence (A.P), find the value of (p + q + r + s + t)
Answer: D (100)
29 is the 7th term: 11 + 6d = 29, so d = 3. The terms are 14, 17, 20, 23, 26, which add up to 100.
If y = (2 − x)/(3 + 5x), find dy/dx
Answer: B (−13/(5x + 3)2)
Quotient rule: dy/dx = [(3 + 5x)(−1) − (2 − x)(5)]/(3 + 5x)² = (−3 − 5x − 10 + 5x)/(5x + 3)² = −13/(5x + 3)².
The operation * is defined on the set of real numbers, R, by x * y = (√x − √y)/(2√x + 3√y), where x, y > 0. Evaluate 7 * 4
Answer: B (−13/4 + (5/4)√7)
7 * 4 = (√7 − 2)/(2√7 + 6). Multiply top and bottom by (2√7 − 6): (14 − 10√7 + 12)/(28 − 36) = (26 − 10√7)/(−8) = −13/4 + (5/4)√7.
If m and n are the roots of x2 + kx + 4 = 0 and m2 + n2 = 8, find the positive value of k.
Answer: D (4)
m + n = −k and mn = 4. m² + n² = (m + n)² − 2mn = k² − 8 = 8, so k² = 16 and the positive value is k = 4.
Given that d2y/dx2 = x − 1, y′(1) = 5 and y(0) = 6, find y(x).
Answer: C (x3/6 − x2/2 + 11x/2 + 6)
dy/dx = x²/2 − x + c; y′(1) = 5 gives c = 11/2. Then y = x³/6 − x²/2 + 11x/2 + d, and y(0) = 6 gives d = 6.
Find the coefficient of the 6th term in ascending powers of x in the binomial expansion of (1 − 3x/4)9
Answer: A (−15309/512)
The 6th term is ⁹C₅(−3x/4)⁵ = 126 × (−243/1024)x⁵, and 126 × 243/1024 = 15309/512, so the coefficient is −15309/512.
Find the value of x for which |2x −3 / x −5| = 14. (Determinant written row by row; / separates the rows.)
Answer: D (−2)
2x(−5) − (−3)(x) = −10x + 3x = −7x = 14, so x = −2.
Calculate the standard deviation of 2, 7, 10, 11, 16 and 20
Answer: A (5.83)
Mean = 66/6 = 11. Squared deviations: 81, 16, 1, 0, 25, 81, total 204. SD = √(204/6) = √34 ≈ 5.83.
The dot product of p and q is 1 where p = (3/5)i + xj, q = (−5/3)i + (4/7)j. Find the value of x
Answer: A (7/2)
p·q = (3/5)(−5/3) + (4/7)x = −1 + 4x/7 = 1, so 4x/7 = 2 and x = 7/2.
If u = 3i + 4j and v = 2i − 3j, find the direction of (u + v)
Answer: A (079°)
u + v = 5i + j, which makes an angle tan⁻¹(1/5) ≈ 11.3° with the east (i) direction. As a bearing that is 90° − 11.3° ≈ 079°.
The gradient of point Q on the curve y = 7x2 + 2x + 1 is 30. Find the coordinate of Q.
Answer: A ((2, 33))
dy/dx = 14x + 2 = 30 gives x = 2, and y = 28 + 4 + 1 = 33.
The ages (in years) of 10 children in a school are 4, 4.5, 7.8, 9.5, 6, 9, 12.5, 11, 5 and 5.5. Calculate, correct to two decimal places, the mean deviation.
Answer: B (2.48)
Mean = 74.8/10 = 7.48. The absolute deviations add up to 24.8, so the mean deviation = 24.8/10 = 2.48.
The distance S in metres covered by a particle in t seconds is S = (7/2)t2 + 5t + 18. Find the acceleration of the particle.
Answer: D (7 ms−2)
v = dS/dt = 7t + 5 and a = dv/dt = 7 ms⁻².
Given that g(x) = 2x + 5, find the image of −3 under g−1(x).
Answer: A (−4)
g⁻¹(x) = (x − 5)/2, so g⁻¹(−3) = −8/2 = −4.
Find the coefficient of x3 in the binomial expansion of (m − 5x)7
Answer: A (−4375m4)
The x³ term is ⁷C₃ m⁴(−5x)³ = 35 × (−125)m⁴x³ = −4375m⁴x³.
A particle of mass 4 kg moves from point P with velocity (3i + 10j) ms−1 and reached point Q with velocity (2i + 3j) ms−1. Find the change in momentum of the particle from P to Q.
Answer: B (−4i − 28j)
Change in momentum = m(v − u) = 4[(2i + 3j) − (3i + 10j)] = 4(−i − 7j) = −4i − 28j.
If the lines 3x + 2y = 4 and 2x − ky = 1 are perpendicular, find the value of constant k.
Answer: A (3)
The gradients are −3/2 and 2/k. Perpendicular lines need (−3/2)(2/k) = −1, so −3/k = −1 and k = 3.
In a dart firing contest, the probabilities that Ade and John hit the target are 1/6 and 3/5 respectively. What is the probability that only one of them will hit the target?
Answer: A (17/30)
P(only Ade) = (1/6)(2/5) = 2/30 and P(only John) = (5/6)(3/5) = 15/30. Total = 17/30.
Two forces F1 (8N, 060°) and F2 (8N, 330°) act on a particle. Calculate the magnitude of the force that will keep the particle in equilibrium.
Answer: D (11.31 N)
The bearings 060° and 330° are 90° apart, so the resultant is √(8² + 8²) = √128 ≈ 11.31 N. The equilibrant has the same size.
Given that tan p = 3/4 and tan q = 8/15, evaluate tan (p + q)
Answer: C (77/36)
tan(p + q) = (3/4 + 8/15)/(1 − (3/4)(8/15)) = (77/60)/(36/60) = 77/36.
Find the volume of the solid of revolution generated when the region bounded by y = 3x2, x = 0, x = 2 and x-axis is rotated about the x-axis through 360°. [Take π = 22/7]
Answer: C (181 unit3)
V = π∫₀² (3x²)² dx = π∫₀² 9x⁴ dx = π × 9 × 32/5 = 288π/5 ≈ 181 unit³.
A body initially at rest is acted upon by forces F1 = (6N, 030°), F2 = (8N, 180°) and F3 = (10N, 090°). Find, correct to one decimal place, the magnitude of the resultant force.
Answer: C (13.3 N)
East components: 6 sin 30° + 10 = 13; north components: 6 cos 30° − 8 = −2.80. Resultant = √(13² + 2.80²) = √176.9 ≈ 13.3 N.
Find the locus of points 8 cm from T(0, 3).
Answer: A (x2 + y2 − 6y = 55)
The locus is the circle x² + (y − 3)² = 64, i.e. x² + y² − 6y + 9 = 64, so x² + y² − 6y = 55.
How many three and four digits odd numbers can be formed with the numbers 0, 1, 2, 4 and 5 if repetitions are not allowed?
Answer: A (54)
The last digit must be 1 or 5 (2 ways) and the first can't be 0. Three-digit: 2 × 3 × 3 = 18. Four-digit: 2 × 3 × 3 × 2 = 36. Total 54.
If a fair die is thrown twice, find the probability that the sum is not a prime number.
Answer: D (7/12)
Prime sums: 2 (1 way), 3 (2), 5 (4), 7 (6), 11 (2): 15 of the 36 outcomes. P(not prime) = 21/36 = 7/12.
Given that cos y = −1/2 and tan y > 0, find the value of y.
Answer: D (240°)
cos y < 0 and tan y > 0 put y in the third quadrant. The reference angle is 60°, so y = 180° + 60° = 240°.
A medical group consists of 7 general practitioners and 5 surgeons. In how many ways can an outreach team of 5 medical personnel be selected if the team must have at least three general practitioners?
Answer: D (546)
3 GPs: ⁷C₃ × ⁵C₂ = 350; 4 GPs: ⁷C₄ × ⁵C₁ = 175; 5 GPs: ⁷C₅ = 21. Total 546.
A binary operation * is defined on the set of real numbers, R, by x * y = 2x + 2y − xy/5. Find the: (a) inverse of x under the operation *.
(b) truth set when m * 7 = −2 * m.
(a) Identity: x * e = x gives 2x + 2e − xe/5 = x, so e(10 − x)/5 = −x and e = −5x/(10 − x).
(Note: e depends on x, so strictly the operation has no single identity; following the question:)
x * x⁻¹ = e: 2x + x⁻¹(2 − x/5) = −5x/(10 − x), so x⁻¹(10 − x)/5 = (2x² − 25x)/(10 − x).
x⁻¹ = 5(2x² − 25x)/(10 − x)², x ≠ 10.
(b) m * 7 = 2m + 14 − 7m/5; −2 * m = −4 + 2m + 2m/5.
14 − 7m/5 = −4 + 2m/5, so 18 = 9m/5 and m = 10. Truth set {10}.
Find the identity element first, then use x * x⁻¹ = e.
Given that xC2 + xC3 − 4x = 0, where x is a positive integer, find the value of x.
x(x − 1)/2 + x(x − 1)(x − 2)/6 = 4x.
Divide by x (x ≠ 0) and multiply by 6: 3(x − 1) + (x − 1)(x − 2) = 24.
(x − 1)(x + 1) = 24, so x² = 25.
x = 5 (positive integer).
Write ⁿCᵣ as factorials and cancel.
A function f is defined by f(x) = 72/(px + r), where p and r are constants. Given that f(6) = 12 and f(7) = 9, find: (a) the values of p and r. (b) f−1(5/3)
(a) 72/(6p + r) = 12 gives 6p + r = 6; 72/(7p + r) = 9 gives 7p + r = 8.
Subtracting: p = 2, then r = −6.
(b) f(x) = 72/(2x − 6). f⁻¹(5/3) is the x with 72/(2x − 6) = 5/3:
2x − 6 = 216/5, 2x = 246/5, x = 123/5 (= 24.6).
f⁻¹(k) is the input that gives output k.
Resolve (2x − 7)/(25x − 24 − 6x2) into partial fractions.
25x − 24 − 6x² = −(6x² − 25x + 24) = −(2x − 3)(3x − 8) = (2x − 3)(8 − 3x).
(2x − 7)/((2x − 3)(8 − 3x)) = A/(2x − 3) + B/(8 − 3x); 2x − 7 = A(8 − 3x) + B(2x − 3).
x = 3/2: −4 = A(7/2), A = −8/7. x = 8/3: −5/3 = B(7/3), B = −5/7.
Answer: −8/(7(2x − 3)) − 5/(7(8 − 3x)).
Factorise the denominator first, then substitute the roots.
The table shows the distribution of the weights (Newtons) of 100 blood donors.
Weight (N): 55 − 59, 60 − 65, 66 − 71, 72 − 78, 79 − 86, 87 − 90
Number of donors: 18, 5, 20, 22, 27, 8
Draw a histogram for the distribution.
Class boundaries and widths: 54.5–59.5 (5), 59.5–65.5 (6), 65.5–71.5 (6), 71.5–78.5 (7), 78.5–86.5 (8), 86.5–90.5 (4).
The widths are unequal, so bar heights are frequency densities (frequency ÷ width): 3.6, 0.83, 3.33, 3.14, 3.38, 2.0.
Draw adjacent bars on the class boundaries with these heights; label axes (weight in N; frequency density) and give a scale.
With unequal class widths, the area (not the height) of each bar shows the frequency.
If two fair dice are thrown together twice, find the probability of obtaining a product of six in the first throw and a sum of eight in the second throw.
Product 6: (1,6), (6,1), (2,3), (3,2): P = 4/36 = 1/9.
Sum 8: (2,6), (6,2), (3,5), (5,3), (4,4): P = 5/36.
The throws are independent: P = 1/9 × 5/36 = 5/324.
List the outcomes in a 6 × 6 table.
A light inextensible string passes over a smooth pulley and carries masses 4 kg and 3 kg at its ends. If the masses are released from rest, calculate: (a) their acceleration; (b) their speed after 3 seconds. [Take g = 10 ms−2]
(a) 4 kg: 40 − T = 4a; 3 kg: T − 30 = 3a. Adding: 10 = 7a, a = 10/7 ≈ 1.43 ms⁻².
(b) v = u + at = 0 + (10/7)(3) = 30/7 ≈ 4.29 ms⁻¹.
Write F = ma for each mass separately and add to remove the tension.
A force of F N acts on a body of mass 50 kg and changes its velocity from 30 ms−1 to 40 ms−1 in 6 seconds. Find the: (a) magnitude of the force; (b) acceleration of the body. (c) distance covered in 9 seconds when the initial velocity is 30 ms−1. [Take g = 10 ms−2]
(a) F = m(v − u)/t = 50 × 10/6 = 250/3 ≈ 83.3 N.
(b) a = F/m = (40 − 30)/6 = 5/3 ≈ 1.67 ms⁻².
(c) s = ut + ½at² = 30(9) + ½(5/3)(81) = 270 + 67.5 = 337.5 m.
In (c) use the body's acceleration, not g.
(a) Given that ∫−1m (2x2 − x − 3) dx = −9/2, where m is an integer, find the value of m.
(b) Solve 2(log3 x − 1) = log3 x and y = √x + 1 simultaneously
(a) [2x³/3 − x²/2 − 3x] from −1 to m = (2m³/3 − m²/2 − 3m) − 11/6 = −9/2.
Multiply by 6: 4m³ − 3m² − 18m + 16 = 0. m = 2 is a root (32 − 12 − 36 + 16 = 0); the other roots of 4m² + 5m − 8 = 0 are not integers. m = 2.
(b) 2log₃ x − 2 = log₃ x gives log₃ x = 2, so x = 9.
y = √9 + 1 = 4. Solution: x = 9, y = 4.
In (b) collect the log terms first; x must be positive.
Given that 8Cx, 7Cx and (7/6)(6Cx) form the first three consecutive terms of an exponential sequence (G.P.), find the: (a) value of x; (b) common ratio of the sequence; (c) sum of the first ten terms of the sequence.
(a) (⁷Cₓ)² = ⁸Cₓ × (7/6)⁶Cₓ. Using ⁸Cₓ = 8/(8 − x) × ⁷Cₓ and ⁶Cₓ = (7 − x)/7 × ⁷Cₓ:
1 = 8(7 − x)/(6(8 − x)), so 48 − 6x = 56 − 8x and x = 4.
(b) Terms: ⁸C₄ = 70, ⁷C₄ = 35, (7/6)(15) = 17.5; r = 1/2.
(c) S₁₀ = 70(1 − (½)¹⁰)/(1 − ½) = 140 × 1023/1024 ≈ 139.86.
For a G.P., (middle term)² = first × third.
(a) Express [(4 + √7)/(4 − √7)]2 − [(4 − √7)/(4 + √7)]2 in the form p + q√7, where p and q are real numbers
(b) Two functions f and g are defined on the set of real numbers by f: x → √((16 − x2)/(x2 + 4)) and g: x → (x2 − 4)/(8 − x2), x ≠ ±2√2. Find: (i) the domain of f; (ii) g o f(x).
(a) (4 + √7)/(4 − √7) = (4 + √7)²/9 = (23 + 8√7)/9; its reciprocal is (23 − 8√7)/9.
Difference of squares: (a + b)(a − b) = (46/9)(16√7/9) = 736√7/81, so p = 0, q = 736/81.
(b)(i) Need 16 − x² ≥ 0 (x² + 4 > 0 always): domain {x : −4 ≤ x ≤ 4}.
(ii) [f(x)]² = (16 − x²)/(x² + 4). g(f(x)) = ([f]² − 4)/(8 − [f]²) = (−5x²/(x² + 4)) ÷ ((9x² + 16)/(x² + 4)) = −5x²/(9x² + 16).
Spot that the two fractions are reciprocals and use a² − b² = (a + b)(a − b).
If the mean and standard deviation of the numbers x, 2.5, 4, y and 7 are 4 and √(31/10) respectively. Find the values of x and y where x < y.
Mean: (x + 2.5 + 4 + y + 7)/5 = 4, so x + y = 6.5.
Variance 3.1: Σ(value − 4)² = 15.5; (2.5 − 4)² + 0 + 3² = 11.25, so (x − 4)² + (y − 4)² = 4.25.
With y = 6.5 − x: 2x² − 13x + 18 = 0, (2x − 9)(x − 2) = 0.
x = 2, y = 4.5 (since x < y).
Use the mean to link x and y, then the variance to get a quadratic.
An association is made up of 10 journalists and 9 engineers. If a committee of 4 members is to be formed from the association, find the probability that it will consist of: (a) at least two engineers; (b) 75% journalists; (c) odd number of engineers.
Total committees: ¹⁹C₄ = 3876.
(a) 2E: ⁹C₂ × ¹⁰C₂ = 1620; 3E: ⁹C₃ × ¹⁰C₁ = 840; 4E: ⁹C₄ = 126. P = 2586/3876 = 431/646 ≈ 0.667.
(b) 3 journalists and 1 engineer: ¹⁰C₃ × 9 = 1080. P = 1080/3876 = 90/323 ≈ 0.279.
(c) 1E: 1080; 3E: 840. P = 1920/3876 = 160/323 ≈ 0.495.
Count the favourable committees, then divide by ¹⁹C₄.
(a) Five forces acting on a particle are represented by (2i + 3j), (4i − 7j), (−5i + 8j), (i + j) and (pi + qj), where p, q ∈ R. If the particle is in equilibrium, find the values of p and q. (b) WXYZ is a parallelogram whose vertices are W(−1, 1), X(m, n), Y(5, −7) and Z(2, 4). Find the: (i) values of m and n; (ii) unit vector in the direction of XY.
(a) Sum of forces = 0: i: 2 + 4 − 5 + 1 + p = 0, p = −2; j: 3 − 7 + 8 + 1 + q = 0, q = −5.
(b)(i) Diagonals of a parallelogram bisect each other: W + Y = X + Z, so (4, −6) = (m + 2, n + 4); m = 2, n = −10.
(ii) XY = (5 − 2)i + (−7 + 10)j = 3i + 3j; unit vector = (3i + 3j)/(3√2) = (i + j)/√2.
In equilibrium the i and j parts each add up to zero.
An object is thrown vertically upwards from the top of a building 20 m high. If the object passes the point it was thrown after 4 seconds on its way down, find the: (a) velocity at which the object was thrown; (b) time taken when the object is 10 m above the level it was thrown; (c) time taken when the object is 10 m below the level it was thrown; (d) velocity with which the object hits the ground. [Take g = 10 ms−2]
Take up as positive: s = ut − 5t².
(a) s = 0 at t = 4: 4u − 80 = 0, u = 20 ms⁻¹.
(b) 20t − 5t² = 10: t² − 4t + 2 = 0, t = 2 ± √2 = 0.59 s (going up) and 3.41 s (coming down).
(c) 20t − 5t² = −10: t² − 4t − 2 = 0, t = 2 + √6 ≈ 4.45 s.
(d) v² = u² + 2as = 400 + 2(10)(20) = 800, v ≈ 28.3 ms⁻¹ downwards.
Measure displacement from the point of throwing; below it is negative.