50 questions. Try each one, then open the answer to check yourself and read the explanation.
Simplify (4¾ − 6¼)/(4⅕ of 1¼)
Answer: B (−2/7)
The top is 4¾ − 6¼ = −3/2 and the bottom is 21/5 × 5/4 = 21/4. So −3/2 ÷ 21/4 = −3/2 × 4/21 = −2/7.
The H.C.F of a2bx + ab2x and a2b − b3 is
Answer: C (b(a + b))
a2bx + ab2x = abx(a + b) and a2b − b3 = b(a − b)(a + b). The factors common to both are b and (a + b), so the H.C.F. is b(a + b).
Correct 241.34 (3 × 10−3)2 to 4 significant figures
Answer: D (0.002172)
(3 × 10−3)2 = 9 × 10−6, and 241.34 × 9 × 10−6 = 0.00217206, which is 0.002172 to 4 significant figures.
At what rate would a sum of ₦100.00 deposited for 5 years raise an interest of ₦7.50?
Answer: A (1½%)
Simple interest I = PRT/100, so 7.50 = 100 × R × 5/100 = 5R. Hence R = 1.5%.
Three children shared a basket of mangoes in such a way that the first child took ¼ of the mangoes and the second ¾ of the remainder. What fraction of the mangoes did the third child take?
Answer: A (3/16)
After the first child, ¾ is left. The second takes ¾ of ¾ = 9/16, so the third gets ¾ − 9/16 = 3/16.
Simplify and express in standard form (0.00275 × 0.0064)/(0.025 × 0.08)
Answer: C (8.8 × 10−3)
Top: 0.00275 × 0.0064 = 0.0000176. Bottom: 0.025 × 0.08 = 0.002. Dividing gives 0.0088 = 8.8 × 10−3.
Three brothers in a business deal shared the profit. The first received ⅓ of the profit and the second ⅔ of the remainder. If the third received the remaining N12,000.00, how much profit did they share?
Answer: B (₦54,000.00)
After the first takes ⅓, ⅔ is left. The second takes ⅔ of ⅔ = 4/9, so the third gets ⅔ − 4/9 = 2/9 of the profit. 2/9 of the profit = ₦12,000, so the profit is ₦54,000.
Simplify √(160r2 + √(71r4 + √(100r8)))
Answer: C (13r)
Work from the inside: √(100r8) = 10r4, then √(71r4 + 10r4) = √(81r4) = 9r2, then √(160r2 + 9r2) = √(169r2) = 13r.
Simplify √27 + 3/√3
Answer: A (4√3)
√27 = 3√3 and 3/√3 = √3, so the sum is 4√3.
Simplify 3log69 + log612 + log664 − log672
Answer: A (5)
Combine the logs: log6(93 × 12 × 64 ÷ 72) = log67776. Since 65 = 7776, the answer is 5.
Simplify (1/x−1 ÷ 1/y−1)−1
Answer: C (y/x)
1/x−1 = x and 1/y−1 = y, so the bracket is x/y. Its reciprocal is y/x.
If a = 2, b = −2 and c = −½, evaluate (ab2 − bc2)(a2c − abc)
Answer: D (−34)
ab2 − bc2 = 8 − (−½) = 8½ and a2c − abc = −2 − 2 = −4. The product is 8½ × (−4) = −34.
Y varies inversely as X2 and X varies directly as Z2. Find the relationship between Y and Z, if C is a constant.
Answer: D (Z4Y = C)
Y = k/X2 and X = mZ2, so Y = k/(m2Z4). Hence Z4Y is a constant: Z4Y = C.
Express r in terms of p and q in the following equation: p/√2 = √(r/(r + q))
Answer: A (r = p2q/(2 − p2))
Square both sides: p2/2 = r/(r + q), so p2r + p2q = 2r. Then r(2 − p2) = p2q, giving r = p2q/(2 − p2).
If f(x − 4) = x2 + 2x + 3, find f(2)
Answer: D (51)
f(2) means x − 4 = 2, so x = 6. Then 62 + 2(6) + 3 = 51.
Factorize 9(x + y)2 − 4(x − y)2
Answer: C ((x + 5y)(5x + y))
It is a difference of two squares: [3(x + y) − 2(x − y)][3(x + y) + 2(x − y)] = (x + 5y)(5x + y).
If a2 + b2 = 16 and 2ab = 7, find all the possible values of (a − b).
Answer: A (3, −3)
(a − b)2 = a2 + b2 − 2ab = 16 − 7 = 9. So a − b = 3 or −3.
Divide x3 − 2x2 − 5x + 6 by (x − 1).
Answer: A (x2 − x − 6)
Long (or synthetic) division by x − 1 leaves no remainder and gives x2 − x − 6. Check: (x − 1)(x2 − x − 6) = x3 − 2x2 − 5x + 6.
If x + 1/x = 4, find x2 + 1/x2
Answer: B (14)
Square x + 1/x = 4: x2 + 2 + 1/x2 = 16, so x2 + 1/x2 = 14.
What must be added to 4x2 − 4 to make it a perfect square?
Answer: B (1/x2)
(2x − 1/x)2 = 4x2 − 4 + 1/x2. So adding 1/x2 to 4x2 − 4 makes a perfect square.
Find the solution of the equation x − 8√x + 15 = 0
Answer: C (9, 25)
Let u = √x: u2 − 8u + 15 = 0 gives u = 3 or 5. So x = 9 or 25.
The lengths of the sides of a right-angled triangle are x cm, (3x − 1)cm and (3x + 1) cm. Find x.
Answer: D (12)
The longest side is 3x + 1, so (3x + 1)2 = x2 + (3x − 1)2. This simplifies to 12x = x2, so x = 12.
The perimeter of a rectangular lawn is 24m. If the area of the lawn is 35m3, how wide is the lawn?
Answer: A (5m)
Length + width = 12 and length × width = 35, so the sides are 7m and 5m. The width is the shorter side, 5m.
A car painter charges N40.00 per day for himself and N10.00 per day for his assistant. If a fleet of cars were painted for N2,000.00 and the painter worked 10 days more than his assistant, how much did the assistant received?
Answer: B (N320.00)
If the assistant worked d days, 40(d + 10) + 10d = 2000, so 50d = 1600 and d = 32. The assistant received 32 × N10 = N320.
Simplify x/(x + y) + y/(x − y) − x2/(x2 − y2)
Answer: B (y2/(x2 − y2))
Over the common denominator x2 − y2, the top is x(x − y) + y(x + y) − x2 = y2. So the answer is y2/(x2 − y2).
Given that x2 + y2 + z2 = 194, calculate z if x = 7 and √y = 3.
Answer: B (8)
√y = 3 means y = 9. Then 49 + 81 + z2 = 194, so z2 = 64 and z = 8.
Find the sum of the first twenty terms of the arithmetic progression log a, log a2, log a3, .......
Answer: D (log a210)
The terms are log a, 2 log a, 3 log a, … so the sum is (1 + 2 + … + 20) log a = 210 log a = log a210.
Find the sum of the first 18 terms of the progression 3, 6, 12,...
Answer: D (3(218 − 1))
This is a G.P. with a = 3 and r = 2, so S18 = 3(218 − 1)/(2 − 1) = 3(218 − 1).
What is the equation of the quadratic function represented by the graph above?
Answer: D (y = −x2 + x + 2)
The curve cuts the x-axis at −1 and 2 and opens downwards, so y = −(x + 1)(x − 2) = −x2 + x + 2.
At what value of x is the function x2 + x + 1 minimum?
Answer: B (−½)
dy/dx = 2x + 1 = 0 gives x = −½, and the curve opens upwards, so this is the minimum.
In the diagram above, the area of PQRS is 73.5cm2 and its height is 10.5cm. Find the length of PS if QR is one-third of PS.
Answer: D (10½cm)
Let PS = x, so QR = x/3. Area = ½(x + x/3) × 10.5 = 7x = 73.5, so x = 10.5cm.
The angle of a sector of a circle, radius 10.5cm, is 48°. Calculate the perimeter of the sector
Answer: D (29.8cm)
Arc length = 48/360 × 2 × 22/7 × 10.5 = 8.8cm. Perimeter = arc + two radii = 8.8 + 21 = 29.8cm.
In the figure above, PS = QS = RS and QŜR = 100°, find QP̂R.
Answer: B (50°)
S is the same distance from P, Q and R, so it is the centre of a circle through them. The angle QSR at the centre is twice the angle QPR at the circumference, so QPR = 50°.
In triangle XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY
Answer: B (6cm)
Triangles XPQ and XZY are similar, with XY = 5 + 3 = 8cm matching XP = 4cm. So ZY/PQ = 8/4, giving ZY = 2 × 3 = 6cm.
Find the length of a side of a rhombus whose diagonals are 6cm and 8cm.
Answer: B (5cm)
The diagonals of a rhombus bisect each other at right angles, giving a right triangle with legs 3cm and 4cm. The side is √(32 + 42) = 5cm.
Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?
Answer: A (9)
Each exterior angle is 180° − 140° = 40°. The exterior angles add up to 360°, so there are 360 ÷ 40 = 9 sides.
In the figure above, PQRS is a circle. If PQT and SRT are straight lines, find the value of x.
Answer: C (103°)
In triangle PST, angle S = 180° − 81° − 22° = 77°. PQRS is cyclic, so x (angle PQR) = 180° − 77° = 103°.
In a regular pentagon, PQRST, PR intersects QS at O. Calculate RÔS.
Answer: B (72°)
Each interior angle is 108°, and the diagonals make 36° angles with the sides, so in triangle QOR the angles at Q and R are 36° each and angle QOR = 108°. RÔS is on a straight line with QOR, so it is 180° − 108° = 72°.
If cos θ = 12/13, find 1 + cot2θ
Answer: A (169/25)
With cos θ = 12/13, the opposite side is 5, so sin θ = 5/13. 1 + cot2θ = cosec2θ = (13/5)2 = 169/25.
In the figure above, YX̂Z = 30°, XŶZ = 105° and XY = 8cm. Calculate YZ.
Answer: C (4√2cm)
Angle Z = 180° − 30° − 105° = 45°. By the sine rule, YZ/sin 30° = 8/sin 45°, so YZ = 8 × ½ ÷ (√2/2) = 4√2cm.
In the figure above, PQR is a semicircle. Calculate the area of the shaded region.
Answer: A (125 2/7cm2)
OS = √(62 + 82) = 10cm, which is also the diameter PR. Shaded area = rectangle 11 × 10 + semicircle ½ × 22/7 × 52 − triangle ½ × 6 × 8 = 110 + 39 2/7 − 24 = 125 2/7cm2.
A cylindrical pipe, made of metal, is 3cm thick, if the internal radius of the pipe is 10cm, find the volume of metal used in making 3m of the pipe.
Answer: D (20,700πcm3)
The outer radius is 13cm and the length is 300cm. Volume of metal = π(132 − 102) × 300 = 69 × 300π = 20,700πcm3.
If the heights of two circular cylinders are in the ratio 2 : 3 and their base radii are in the ratio 9 : 8, what is the ratio of their volumes?
Answer: A (27 : 32)
Volume is πr2h, so the ratio is (92 × 2) : (82 × 3) = 162 : 192 = 27 : 32.
Find the curved surface area of the frustrum in the figure above.
Answer: B (20π√10cm2)
The radii are 4cm and 6cm and the height is 6cm, so the slant height is √(62 + 22) = 2√10cm. Curved surface = π(4 + 6) × 2√10 = 20π√10cm2.
The locus of a point which moves so that it is equidistant from two intersecting straight lines is the
Answer: B (angle bisector of the two lines)
Points the same distance from two crossing lines lie on the lines that bisect the angles between them.
4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the sectors representing all numbers equal to or greater than 16
Answer: D (276°)
The total is 120. The numbers 16 or more are 16, 30, 20 and 26, which add up to 92. Their angles add up to 92/120 × 360° = 276°.
The mean of 10 positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number
Answer: D (38)
The first 10 numbers add up to 160. The 11 numbers add up to 11 × 18 = 198, so the new number is 198 − 160 = 38.
Below are the scores of a group of students in a test.
Scores: 1, 2, 3, 4, 5, 6
No. of Students: 1, 4, 5, 6, x, 2
If the average score is 3.5, find the value of x.
Answer: B (2)
Total score = 1 + 8 + 15 + 24 + 5x + 12 = 60 + 5x and the number of students is 18 + x. (60 + 5x)/(18 + x) = 3.5 gives 1.5x = 3, so x = 2.
Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?
Answer: C (⅓)
There are 6 possible pairs. Only 1 + 3 and 2 + 4 give even sums, so the probability is 2/6 = ⅓.
Find the probability that a number selected at random from 41 to 56 is a multiple of 9
Answer: A (⅛)
There are 16 numbers from 41 to 56, and only 45 and 54 are multiples of 9. Probability = 2/16 = ⅛.