45 questions. Try each one, then open the answer to check yourself and read the explanation.
Find the principal which amounts to N5,500 at simple interest in 5 years at 2% per annum
Answer: A (N5,000)
In 5 years at 2% the interest is 10% of the principal, so the amount is 1.1P. Then 1.1P = 5500, giving P = N5,000.
A car dealer bought a second-hand car for N250,000.00 and spent N70,000.00 refurbishing it. He then sold the car for N400,000.00. What is the percentage gain?
Answer: B (25%)
Total cost = 250 000 + 70 000 = N320 000, so the gain is N80 000. As a percentage of cost: 80 000/320 000 × 100 = 25%.
Evaluate 21.05347 − 1.6324 × 0.43, to 3 decimal places.
Answer: B (20.352)
Multiply first: 1.6324 × 0.43 = 0.701932. Then 21.05347 − 0.701932 = 20.351538, which is 20.352 to 3 decimal places.
Given that p = 1 + √2 and q = 1 − √2, evaluate (p2 − q2)/2pq
Answer: C (−2√2)
p2 − q2 = (p + q)(p − q) = 2 × 2√2 = 4√2, and pq = 1 − 2 = −1, so 2pq = −2. The value is 4√2 ÷ (−2) = −2√2.
Simplify (∛(64a3))−1
Answer: C (1/(4a))
The cube root of 64a3 is 4a. Raising to the power −1 gives 1/(4a).
Factorize 4x2 − 9y2 + 20x + 25
Answer: D ((2x − 3y + 5)(2x + 3y + 5))
Group as (4x2 + 20x + 25) − 9y2 = (2x + 5)2 − (3y)2. As a difference of two squares this is (2x + 5 − 3y)(2x + 5 + 3y).
If two graphs y = px2 + q and y = 2x2 − 1 intersect at x = 2, find the value of p in terms of q.
Answer: D ((7 − q)/4)
At x = 2 the second graph gives y = 2(4) − 1 = 7. The first must give the same: 4p + q = 7, so p = (7 − q)/4.
Solve the equations; m2 + n2 = 29, m + n = 7
Answer: D ((2,5) and (5,2))
Substitute m = 7 − n: (7 − n)2 + n2 = 29 gives n2 − 7n + 10 = 0, so n = 2 or 5. The pairs are (5, 2) and (2, 5).
Divide a3x − 26a2x + 156ax − 216 by a2x − 24ax + 108
Answer: C (ax − 2)
Let u = ax. Then (u − 2)(u2 − 24u + 108) = u3 − 26u2 + 156u − 216, so the quotient is ax − 2.
Find the integral values of x and y satisfying the inequality 3y + 5x ≤ 15, given that y > 0, y < 3 and x > 0.
Answer: C ((1,1), (1,2), (2,1))
y can only be 1 or 2. For x = 1, 3y ≤ 10 allows y = 1, 2; for x = 2, 3y ≤ 5 allows y = 1; for x = 3, no positive y works. So the points are (1,1), (1,2), (2,1).
Triangle SPT is the solution of the linear inequalities
Answer: A (2y − x − 2 ≤ 0, y + 2x + 2 ≤ 0, y ≥ 0, x ≤ 0)
The triangle lies above the x-axis (y ≥ 0), below the line 2y − x − 2 = 0 and to the left of y + 2x + 2 = 0. Testing the point (−1.4, 0.1) confirms both line expressions are negative there, matching option A.
The sixth term of an arithmetical progression is half of its twelfth term. The first term is equal to
Answer: C (the common difference)
a + 5d = ½(a + 11d), so 2a + 10d = a + 11d, giving a = d.
A man saves N100.00 in his first year of work and each year saves N20.00 more than in the preceeding year. In how many years will he save N5,800.00?
Answer: A (20 years)
This is an A.P. with a = 100, d = 20. Setting n/2[200 + 20(n − 1)] = 5800 gives n2 + 9n − 580 = 0, so n = 20.
An operation is defined on the set of real numbers by a * b = a + b + 1. If the identity element is −1, find the inverse of the element 2 under *.
Answer: A (−4)
The inverse b satisfies 2 * b = −1 (the identity). So 2 + b + 1 = −1, giving b = −4.
⊗ | k | l | m
k | l | m | k
l | m | k | l
m | k | l | m
The identity element with respect to the multiplication shown in the table above is
Answer: C (m)
The row for m reads k, l, m, the same as the column headings, so combining m with any element leaves it unchanged. m is the identity.
Given the matrix K = [2, 1; 3, 4], the matrix K2 + K + 1, where I is the 2 × 2 identity matrix, is
Answer: B ([10, 7; 21, 24])
K2 = [7, 6; 18, 19]. Adding K = [2, 1; 3, 4] and I = [1, 0; 0, 1] gives [10, 7; 21, 24].
Evaluate the determinant |−1, −1, −1; 3, 1, −1; 1, 2, 1|
Answer: C (−4)
Expanding along the first row: −1(1 + 2) − (−1)(3 + 1) + (−1)(6 − 1) = −3 + 4 − 5 = −4.
If P = [3, −2, 4; 5, 0, 6; 7, 5, −1], then −2p is
Answer: D ([−6, 4, −8; −10, 0, −12; −14, −10, 2])
Multiply every entry of P by −2: the rows become (−6, 4, −8), (−10, 0, −12) and (−14, −10, 2).
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
Answer: C (6)
Interior + exterior = 180° and interior = 2 × exterior, so the exterior angle is 60°. The polygon has 360 ÷ 60 = 6 sides.
A cylindrical tank has a capacity of 3080m3. What is the depth of the tank if the diameter of its base is 14m?
Answer: A (20m)
The radius is 7 m, so the base area is 22/7 × 49 = 154 m2. Depth = 3080 ÷ 154 = 20 m.
A sector of a circle of radius 7.2cm which subtends an angle of 300° at the centre is used to form a cone. What is the radius of the base of the cone?
Answer: A (6cm)
The arc length becomes the base circumference: (300/360) × 2π × 7.2 = 2πr. So r = 7.2 × 300/360 = 6 cm.
The chord ST of a circle is equal to the radius, r, of the circle. Find the length of arc ST.
Answer: B (πr/3)
A chord equal to the radius forms an equilateral triangle with the centre, so it subtends 60°. Arc length = (60/360) × 2πr = πr/3.
A point P moves such that it is equidistant from points Q and R. Find QR when PR = 8cm and ∠PRQ = 30°.
Answer: D (8√3cm)
PQ = PR = 8 cm, so triangle PQR is isosceles with base angles 30°. Half of QR is 8 cos 30° = 4√3, so QR = 8√3 cm.
Find the locus of a point which moves such that its distance from the line y = 4 is a constant, k.
Answer: D (y = 4 ± k)
Points at distance k from y = 4 lie on two parallel lines, one k above and one k below: y = 4 ± k.
A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.
Answer: A (√3y = x + 5√3)
The gradient is tan 30° = 1/√3 and the intercept is 5, so y = x/√3 + 5. Multiplying by √3 gives √3y = x + 5√3.
P(−6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.
Answer: B (6.5 units)
PQ = √(122 + 52) = √169 = 13, which is the diameter. The radius is 6.5 units.
Find the value of p if the line joining (p, 4) and (6, −2) is perpendicular to the line joining (2, p) and (−1, 3).
Answer: C (4)
The gradients are −6/(6 − p) and (p − 3)/3. Their product must be −1: 6(p − 3) = 3(6 − p), so 9p = 36 and p = 4.
The bearings of P and Q from a common point N are 020° and 300° respectively. If P and Q are also equidistant from N, find the bearing of P from Q.
Answer: C (070°)
∠PNQ = 80° and NP = NQ, so the base angles are 50°. The bearing of N from Q is 120°; turning 50° towards P gives 120° − 50° = 070°.
Find the value of θ in the diagram above.
Answer: D (120°)
The triangle has two sides l and base √3 l. Half the base is (√3/2)l, so each base angle has cosine √3/2 and is 30°. Then θ = 180° − 60° = 120°.
Differentiate (2x + 5)2(x − 4) with respect to x.
Answer: A ((2x + 5)(6x − 11))
By the product rule: (2x + 5)2 + (x − 4) × 4(2x + 5) = (2x + 5)(2x + 5 + 4x − 16) = (2x + 5)(6x − 11).
If y = x sin x, find dy/dx when x = π/2.
Answer: B (1)
dy/dx = sin x + x cos x. At x = π/2 this is 1 + (π/2)(0) = 1.
If the gradient of the curve y = 2kx2 + x + 1 at x = 1 is 9, find k.
Answer: B (2)
dy/dx = 4kx + 1. At x = 1, 4k + 1 = 9, so k = 2.
Find the rate of change of the volume v of a sphere with respect to its radius r when r = 1.
Answer: A (4π)
v = (4/3)πr3, so dv/dr = 4πr2. At r = 1 this is 4π.
Find the dimensions of the rectangle of greatest area which has a fixed perimeter p.
Answer: A (Square of sides p/4)
With sides x and p/2 − x, the area x(p/2 − x) is greatest when x = p/4. Both sides are then p/4, a square.
Evaluate ∫2(2x − 3)2/3 dx.
Answer: D ((3/5)(2x − 3)5/3 + k)
Raise the power by one to 5/3 and divide by 5/3 and by the inner derivative 2: 2 × (2x − 3)5/3 ÷ (5/3 × 2) = (3/5)(2x − 3)5/3 + k.
Find the area bounded by the curves y = 4 − x2 and y = 2x + 1.
Answer: B (10⅔ sq. units)
The curves meet where x2 + 2x − 3 = 0, i.e. x = −3 and 1. The area is ∫(3 − 2x − x2)dx from −3 to 1 = 5/3 − (−9) = 32/3 = 10⅔.
Score: 4, 7, 8, 11, 13, 8
Frequency: 3, 5, 2, 7, 2, 1
Use the table above. Find the square of the mode.
Answer: D (121)
The score with the highest frequency (7) is 11, so the mode is 11. Its square is 121.
Score: 4, 7, 8, 11, 13, 8
Frequency: 3, 5, 2, 7, 2, 1
Use the table above. The mean score is
Answer: C (8.7)
Σfx = 12 + 35 + 16 + 77 + 26 + 8 = 174 and Σf = 20. The mean is 174 ÷ 20 = 8.7.
Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3.
Answer: A (4/3)
Range = largest − smallest. The largest is 3/2 and the smallest is 1/6, so the range is 9/6 − 1/6 = 8/6 = 4/3.
Find the variance of 2, 6, 8, 6, 2 and 6.
Answer: C (5)
The mean is 30 ÷ 6 = 5. The squared deviations are 9, 1, 9, 1, 9, 1, which add to 30. Variance = 30 ÷ 6 = 5.
The graph above shows the cumulative frequency of the distribution of masses of fertilizer for 48 workers in one institution. Which of the following gives the interquartile range?
Answer: A (Q3 − Q1)
The interquartile range is the upper quartile minus the lower quartile, Q3 − Q1. Half of it would be the semi-interquartile range.
Find the number of ways of selecting 8 subjects from 12 subjects for an examination.
Answer: C (495)
This is a selection, so use combinations: 12C8 = 12C4 = (12 × 11 × 10 × 9)/24 = 495.
If 6Pr = 6, find the value of 6Pr + 1.
Answer: B (30)
6P1 = 6, so r = 1. Then 6P2 = 6 × 5 = 30.
Colour: Blue, Black, Yellow, White, Brown
No. of beads: 1, 2, 4, 5, 3
The distribution of colours of beads in a bowl is given above. What is the probability that a bead selected at random will be blue or white?
Answer: C (2/5)
There are 15 beads, of which 1 + 5 = 6 are blue or white. The probability is 6/15 = 2/5.
Teams P and Q are involved in a game of football. What is the probability that the game ends in a draw?
Answer: B (⅓)
There are three possible results: P wins, Q wins, or a draw. Taking them as equally likely, the chance of a draw is 1/3.