47 questions. Try each one, then open the answer to check yourself and read the explanation.
If U = {1, 2, 3, 4, 5, 6}, P = {3, 4, 5}, Q = {2, 4, 6} and R = {1, 2, 3, 4}, list the elements of (P ∪ Q)′ ∩ R.
Answer: C ({1})
P ∪ Q = {2, 3, 4, 5, 6}, so its complement is {1}. {1} ∩ R = {1}.
A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain, 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group?
Answer: A (25)
Fill a Venn diagram from the centre: all three 3; plantain and maize only 2; yam and maize only 1; yam and plantain only 2. Then yam only 4, plantain only 7, maize only 6. Total = 3 + 2 + 1 + 2 + 4 + 7 + 6 = 25.
If log810 = x, evaluate log85 in terms of x.
Answer: C (x − ⅓)
log₈5 = log₈(10/2) = log₈10 − log₈2. Since 8^(1/3) = 2, log₈2 = 1/3, so log₈5 = x − 1/3.
Find the value of x if √2/(x + √2) = 1/(x − √2)
Answer: A (3√2 + 4)
Cross-multiply: √2x − 2 = x + √2, so x(√2 − 1) = 2 + √2. Multiply top and bottom by √2 + 1: x = (2 + √2)(√2 + 1) = 4 + 3√2.
A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss.
Answer: B (25% gain)
Cost = 20 × N1.20 = N24. He sells 80 oranges: 20 × N1.50 = N30. Gain = N6, and 6/24 × 100 = 25% gain.
Divide 24346 by 426.
Answer: B (356)
2434₆ = 2×216 + 4×36 + 3×6 + 4 = 598 and 42₆ = 26. 598 ÷ 26 = 23, and 23 = 3×6 + 5 = 35₆.
If 29 × (Y3)9 = 35 × (Y3)5, find the value of Y.
Answer: D (1)
In base ten: 2(9Y + 3) = 3(5Y + 3), so 18Y + 6 = 15Y + 9 and Y = 1.
Simplify √((0.0023 × 750)/(0.00345 × 1.25))
Answer: B (20)
0.0023 × 750 = 1.725 and 0.00345 × 1.25 = 0.0043125. Their quotient is 400, and √400 = 20.
If m∗n = m/n − n/m for m, n ∈ ℜ, evaluate −3 ∗ 4
Answer: C (7/12)
−3 ∗ 4 = −3/4 − 4/(−3) = −3/4 + 4/3 = (−9 + 16)/12 = 7/12.
The sum of two numbers is twice their difference. If the difference of the numbers is P, find the larger of the two numbers.
Answer: B (3P/2)
Let the numbers be x > y. x + y = 2(x − y) gives x = 3y. With x − y = P, 2y = P so y = P/2 and x = 3P/2.
Find the matrix T if ST = I where S = (−1 1 / 1 −2) and I is the identity matrix. (Rows are separated by /.)
Answer: B ((−2 −1 / −1 −1))
T is the inverse of S. det S = 2 − 1 = 1, so S⁻¹ = (−2 −1 / −1 −1): swap the diagonal entries and change the signs of the others.
The shaded portion in the graph above is represented by
Answer: A (y + x − x3 ≥ 0, y − x ≤ 0)
The curve is y = x³ − x and the line is y = x (45°). The shaded part is above the curve (y + x − x³ ≥ 0) and below the line (y − x ≤ 0).
A binary operation ∗ is defined by a∗b = ab + a + b for any real number a and b. If the identity element is zero, find the inverse of 2 under this operation.
Answer: D (−2/3)
The inverse b satisfies 2∗b = 0: 2b + 2 + b = 0, so 3b = −2 and b = −2/3.
Given that Q = (6 0 / 4 5) and Q + P = (7 −2 / 6 8), evaluate |Q + 2P|. (Rows are separated by /.)
Answer: D (120)
P = (1 −2 / 2 3), so Q + 2P = (8 −4 / 8 11). Its determinant is 8 × 11 − (−4)(8) = 88 + 32 = 120.
Factorize completely x2 + 2xy + y2 + 3x + 3y − 18
Answer: A ((x + y + 6)(x + y − 3))
Let u = x + y: the expression is u² + 3u − 18 = (u + 6)(u − 3) = (x + y + 6)(x + y − 3).
Tope bought x oranges at N5.00 each and some mangoes at N4.00 each. If she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of the values of x.
Answer: C (5 ≤ x ≤ 10)
She spends 5x + 4(2x) = 13x naira. 65 ≤ 13x ≤ 130 gives 5 ≤ x ≤ 10.
Three consecutive positive integers k, l and m are such that l2 = 3(k + m). Find the value of m.
Answer: D (7)
Write l = k + 1 and m = k + 2: (k + 1)² = 3(2k + 2), so k² − 4k − 5 = 0 and k = 5 (positive). Then m = 7.
Express 1/(x3 − 1) in partial fractions.
Answer: B (⅓(1/(x − 1) − (x + 2)/(x2 + x + 1)))
x³ − 1 = (x − 1)(x² + x + 1). Write 1 = A(x² + x + 1) + (Bx + C)(x − 1): x = 1 gives A = 1/3, then B = −1/3 and C = −2/3.
The first term of a geometrical progression is twice its common ratio. Find the sum of the first two terms of the progression if its sum to infinity is 8.
Answer: C (72/25)
a = 2r and a/(1 − r) = 8 give 2r = 8 − 8r, so r = 4/5 and a = 8/5. The first two terms add to 8/5 + 32/25 = 72/25.
Divide 4x3 − 3x + 1 by 2x − 1
Answer: D (2x2 + x − 1)
Check by multiplying: (2x − 1)(2x² + x − 1) = 4x³ + 2x² − 2x − 2x² − x + 1 = 4x³ − 3x + 1.
Find the length XZ in the triangle above.
Answer: A (√7 m)
By the cosine rule with XY = 2 m, YZ = 1 m and ∠Y = 120°: XZ² = 4 + 1 − 2(2)(1)cos 120° = 5 + 2 = 7, so XZ = √7 m.
Find a positive value of α if the coordinate of the centre of a circle x2 + y2 − 2αx + 4y − α = 0 is (α, −2) and the radius is 4 units.
Answer: C (3)
For this circle r² = α² + 2² + α = 16, so α² + α − 12 = 0, i.e. (α + 4)(α − 3) = 0. The positive value is 3.
A man 1.7m tall observes a bird on top of a tree at an angle of 30°. If the distance between the man's head and the bird is 25m, what is the height of the tree?
Answer: B (14.2m)
The 25 m line of sight is the hypotenuse, so the height above the man's eye is 25 sin 30° = 12.5 m. Adding his height: 12.5 + 1.7 = 14.2 m.
In ΔMNO, MN = 6 units, MO = 4 units and NO = 12 units. If the bisector of angle M meets NO at P, calculate NP.
Answer: B (7.2 units)
The angle bisector divides NO in the ratio of the other two sides: NP : PO = 6 : 4. So NP = 6/10 × 12 = 7.2 units.
Find the tangent of the acute angle between the lines 2x + y = 3 and 3x − 2y = 5.
Answer: C (7/4)
The gradients are −2 and 3/2. tan θ = |(−2 − 3/2)/(1 + (−2)(3/2))| = (7/2)/2 = 7/4, positive because the angle is acute.
In the figure above, PQRS is a circle with ST||RQ. Find the value of x if PT = PS.
Answer: B (55°)
Opposite angles of cyclic quadrilateral PQRS add to 180°, so ∠SPT = 180° − 110° = 70°. Triangle PST is isosceles, so ∠PTS = (180° − 70°)/2 = 55°. Since ST ∥ RQ, x = ∠PTS = 55° (corresponding angles).
In the diagram above, EFGH is a cyclic quadrilateral in which EH||FG, EG and FH are chords. If ∠FHG = 42° and ∠EFH = 34°, calculate ∠HEG.
Answer: C (52°)
∠EGH = ∠EFH = 34° (same segment). Let ∠HEG = y; then ∠HFG = y (same segment) and ∠EHF = y (alternate angles, EH ∥ FG). In triangle EHG: y + (y + 42°) + 34° = 180°, so y = 52°.
From a point P, the bearings of two points Q and R are N67°W and N23°E respectively. If the bearing of R from Q is N68°E and PQ = 150m, calculate PR.
Answer: C (150m)
∠QPR = 67° + 23° = 90°. At Q the direction to P is S67°E and to R is N68°E, so ∠PQR = 180° − 67° − 68° = 45°. The right-angled triangle is isosceles, so PR = PQ = 150 m.
Find the equation of the locus of a point P(x, y) such that PV = PW, when V = (1, 1) and W = (3, 5).
Answer: D (x + 2y = 8)
P lies on the perpendicular bisector of VW. The midpoint is (2, 3) and VW has gradient 2, so the bisector has gradient −1/2: y − 3 = −½(x − 2), i.e. x + 2y = 8.
In the figure above, TZ is tangent of the circle QPZ. Find x if TZ = 6 units and PQ = 9 units.
Answer: A (3)
Tangent–secant rule: TZ² = TP × TQ, so 36 = x(x + 9). Then x² + 9x − 36 = 0, (x + 12)(x − 3) = 0 and x = 3.
Find the value of l in the frustrum above.
Answer: A (5 cm)
The radii differ by 6 − 3 = 3 cm and the height is 4 cm, so the slant height l = √(4² + 3²) = 5 cm.
Find the area bounded by the curve y = x(2 − x), the x-axis, x = 0 and x = 2.
Answer: C (1⅓ sq units)
Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3 = 1⅓ square units.
Evaluate ∫0π/4 (sin x − cos x) dx.
Answer: D (1 − √2)
The integral is [−cos x − sin x] from 0 to π/4 = (−√2/2 − √2/2) − (−1) = 1 − √2.
Find the volume of solid generated when the area enclosed by y = 0, y = 2x and x = 3 is rotated about the x-axis.
Answer: B (36π cubic units)
V = π∫₀³ y² dx = π∫₀³ 4x² dx = π × 4 × 27/3 = 36π cubic units (a cone of radius 6 and height 3).
What is the derivation of t2 sin(3t − 5) with respect to the variable t?
Answer: C (2t sin (3t − 5) + 3t2 cos (3t − 5))
Product rule: 2t × sin(3t − 5) + t² × 3cos(3t − 5).
Evaluate ∫−21 (x − 1)2dx
Answer: C (9)
∫(x − 1)² dx = (x − 1)³/3. From −2 to 1: 0 − (−27/3) = 9.
Find the value of x for which the function y = x3 − x has a minimum value.
Answer: C (√3/3)
dy/dx = 3x² − 1 = 0 gives x = ±1/√3. d²y/dx² = 6x is positive at x = 1/√3 = √3/3, so that is the minimum.
If the maximum value of y = 1 + hx − 3x2 is 13, find h.
Answer: B (12)
dy/dx = h − 6x = 0 at x = h/6. Then y = 1 + h²/6 − h²/12 = 1 + h²/12 = 13, so h² = 144 and h = 12 (taking the positive value offered).
How many two-digit numbers can be formed from the digits 0, 1, 2, 3 if a digit can be repeated and no number may begin with 0?
Answer: B (12)
The first digit can be 1, 2 or 3 (3 ways) and the second any of the 4 digits, so 3 × 4 = 12.
Three boys play a game of luck in which their respective chances of winning are ½, ⅓ and ¼. What is the probability that one and only one of the boys wins the game?
Answer: C (11/24)
P(exactly one) = ½·⅔·¾ + ½·⅓·¾ + ½·⅔·¼ = 6/24 + 3/24 + 2/24 = 11/24.
The grades of 36 students in a class test are as shown in the pie chart above. How many students had Excellent?
Answer: A (7)
Very Good is a right angle (90°), so Excellent = 360° − 120° − 80° − 90° = 70°. 70/360 × 36 = 7 students.
A number is selected at random from 0 to 20. What is the probability that the numbers is an odd prime?
Answer: B (1/3)
0 to 20 is 21 numbers. The odd primes are 3, 5, 7, 11, 13, 17 and 19, which is 7 numbers. 7/21 = 1/3.
A student calculated the mean of 5 numbers as 45.3. While rechecking his working, he discovered that his total was short by 20.5. What is the correct mean of the 5 numbers?
Answer: C (49.4)
His total was 5 × 45.3 = 226.5. The correct total is 226.5 + 20.5 = 247, so the correct mean is 247 ÷ 5 = 49.4.
The sectorial allocations to various ministries in a state budget are as follows: Agriculture − N25 000 000.00, Education − N20 000 000.00, Women Affairs − N35 000 000.00, Commerce and Industries − N20 000 000.00. In a pie chart to represent this information the corresponding angle to Agriculture is
Answer: D (90°)
Total budget = 25 + 20 + 35 + 20 = 100 million naira. Agriculture's angle = 25/100 × 360° = 90°.
The mean of four numbers is 5 and the mean deviation is 3. Find the fourth number if the mean deviation of the first three numbers is 2.
Answer: C (11)
The four deviations from 5 add to 4 × 3 = 12 and the first three add to 3 × 2 = 6, so the fourth number is 6 away from 5. So it is 11 (the other possibility, −1, is not an option).
Marks: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
No. of Students: 2, 2, 11, 10, 16, 51, 40, 10, 25, 15, 20
The marks scored by students in a test are given in the table above. Find the median.
Answer: B (6)
There are 202 students, so the median lies between the 101st and 102nd. The cumulative frequency is 92 up to mark 5 and 132 up to mark 6, so the median mark is 6.
If 6Cr/6Pr = 1/6, find the value of r
Answer: B (3)
ⁿCᵣ = ⁿPᵣ/r!, so the ratio is 1/r! = 1/6. Since 3! = 6, r = 3.