50 questions. Try each one, then open the answer to check yourself and read the explanation.
Convert 241 in base 5 to base 8
Answer: B (1078)
2415 = 2×25 + 4×5 + 1 = 71 in base 10. Dividing 71 by 8 gives 8 remainder 7, then 1 remainder 0, then 0 remainder 1, so it is 1078.
Find the length of a rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length
Answer: B (240cm)
The rod must be a multiple of both 40 and 48. The smallest such length is the LCM of 40 and 48, which is 240cm.
A rectangular lawn has an area of 1815 square yards. If its length is 50 metres, find its width in metres, given that 1 metre equals 1.1 yards.
Answer: D (30.00)
1 m = 1.1 yd, so 1 m² = 1.1 × 1.1 = 1.21 sq yd. The area is 1815 ÷ 1.21 = 1500 m², so the width is 1500 ÷ 50 = 30 m.
Reduce each number to 2 significant figures and then evaluate
(0.021741 × 1.2047)/0.023789
Answer: C (1.1)
To 2 s.f. the numbers are 0.022, 1.2 and 0.024. Then 0.022 × 1.2 = 0.0264 and 0.0264 ÷ 0.024 = 1.1.
A train moves from P to Q at an average speed of 90km/h and immediately returns from Q to P through the same route at an average speed of 45km/h. Find the average speed for the entire journey.
Answer: B (60.00km/h)
Let PQ = d. Time taken = d/90 + d/45 = 3d/90 = d/30. Average speed = total distance ÷ total time = 2d ÷ (d/30) = 60 km/h.
If the length of a square is increased by 20% while its width is decreased by 20% to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?
Answer: D (24:25)
If the side is 1, the rectangle is 1.2 by 0.8, with area 0.96. The ratio is 0.96 : 1 = 24 : 25.
Two brothers invested a total sum of N5,000.00 on a farm project. The farm yield was sold for N15,000.00 at the end of the season. If the profit was shared in the ratio 2:3, what is the difference in the amount of profit received by the brothers?
Answer: A (N2,000.00)
Profit = 15,000 − 5,000 = N10,000. The shares are 2/5 × 10,000 = N4,000 and 3/5 × 10,000 = N6,000, a difference of N2,000.
Peter's weekly wages are N20.00 for the first 20 weeks and N36.00 for the next 24 weeks. Find his average weekly wage for the remaining 8 weeks of the year, if his average weekly wage for the whole year is N30.00.
Answer: A (N37.00)
For 52 weeks he earns 52 × 30 = N1,560. The first 44 weeks give 400 + 864 = N1,264, leaving N296 for 8 weeks, which is N37 a week.
A man invests a sum of money at 4% per annum simple interest. After 3 years the principal amounts to N7,000.00. Find the sum invested.
Answer: B (N6,250.00)
Amount = P(1 + 3 × 0.04) = 1.12P. So 1.12P = 7,000 and P = N6,250.
By selling 20 oranges for N1.35 a trader makes a profit of 8%. What is his percentage gain or loss if he sells the same 20 oranges for N1.10?
Answer: C (12%)
N1.35 is 108% of the cost, so the cost is 1.35 ÷ 1.08 = N1.25. Selling for N1.10 is a loss of N0.15, and 0.15/1.25 × 100 = 12%.
Four boys and ten girls can cut a field in 5 hours. If the boys work at 5/4 hrs rate of which the girls work, how many boys will be needed to cut the field in 3 hours?
Answer: D (20)
A boy does 5/4 of a girl's work, so 10 girls equal 8 boys and the team is 12 boys working 5 hours (60 boy-hours). To finish in 3 hours needs 60 ÷ 3 = 20 boys.
Evaluate without using tables
(0.008)−1/3 × (0.16)−3/2
Answer: A (625/8)
0.008 = 1/125, so (0.008)−1/3 = 5. 0.16 = 4/25, so (0.16)−3/2 = (25/4)3/2 = 125/8. The product is 625/8.
Instead of writing 35/6 as a decimal correct to 3 significant figures, a student wrote it correct to 3 places of decimals. Find his error in standard form.
Answer: B (3.0 × 10−3)
35/6 = 5.8333…, which is 5.83 to 3 s.f. and 5.833 to 3 d.p. The error is 0.003, which in standard form is 3.0 × 10−3.
Simplify without using tables
(log26 − log23)/(log28 − 2log2½)
Answer: A (1/5)
The top is log2(6/3) = log22 = 1. The bottom is 3 − 2(−1) = 5. So the value is 1/5.
Simplify without using tables
(2√14 × 3√21)/(7√24 × 2√98)
Answer: D (3√2/28)
Top: 6√294 = 6 × 7√6 = 42√6. Bottom: 7 × 2√6 × 2 × 7√2 = 196√12 = 392√3. So 42√6/(392√3) = 42√2/392 = 3√2/28.
If P = 2/3 (1 − r2)/n2, find n when r = √(1/3) and p = 1
Answer: D (2/3)
With r² = 1/3: 1 = (2/3)(1 − 1/3)/n² = (2/3)(2/3)/n² = (4/9)/n². So n² = 4/9 and n = 2/3.
If a = u2 − 3v2 and b = 2uv + v2, evaluate (2a − b)(a − b2), when u = 1 and v = −1
Answer: A (9)
a = 1 − 3 = −2 and b = −2 + 1 = −1. Then 2a − b = −3 and a − b² = −2 − 1 = −3, so the product is 9.
The formula Q = 1.5 + 0.5n, gives the cost Q (in naira) of feeding n people for a week. Find in kobo the extra cost of feeding one additional person.
Answer: D (50k)
Increasing n by 1 increases Q by 0.5 naira, the coefficient of n. That is 50 kobo.
If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2
Answer: B (PR2 = 98)
P = k/V and V = cR², so P = K/R². With R = 7 and P = 2, K = 2 × 49 = 98, giving PR² = 98.
Make y the subject of the formula Z = x2 + 1/y3
Answer: C (y = 1/(Z − x2)1/3)
1/y³ = Z − x², so y³ = 1/(Z − x²). Taking cube roots, y = 1/(Z − x²)1/3.
Find the values of m which make the following quadratic function a perfect square.
x2 + 2(m + 1)x + m + 3
Answer: C (1, −2)
A perfect square needs (half the x-coefficient)² = constant: (m + 1)² = m + 3. This gives m² + m − 2 = 0, so (m + 2)(m − 1) = 0 and m = 1 or −2.
Factorize 62x + 1 + 7(6x) − 5
Answer: B ({3(6x) + 5}{2(6x) − 1})
Let y = 6x; then 62x+1 = 6y². So the expression is 6y² + 7y − 5 = (3y + 5)(2y − 1).
Find two values of y which satisfy the simultaneous equations x + y = 5, x2 − 2y2 = 1
Answer: C (−12, 2)
Put x = 5 − y into the second equation: (5 − y)² − 2y² = 1, which becomes y² + 10y − 24 = 0. So (y + 12)(y − 2) = 0 and y = −12 or 2.
An (n − 2)2-sided figure has n diagonals. Find the number n of diagonals for a 25-sided figure.
Answer: A (7)
Set (n − 2)² = 25, so n − 2 = 5 and n = 7.
A cubic function f(x) is specified by the graph shown above. The values of the independent variable for which the function vanishes are
Answer: A (−1, 0, 1)
"Vanishes" means f(x) = 0. The graph crosses the x-axis at x = −1, x = 0 and x = 1.
Solve the inequality x − 1 > 4(x + 2)
Answer: B (x < −3)
x − 1 > 4x + 8 gives −3x > 9. Dividing by −3 reverses the sign: x < −3.
Simplify
(x2 − y2)/(2x2 + xy − y2)
Answer: C ((x − y)/(2x − y))
x² − y² = (x − y)(x + y) and 2x² + xy − y² = (2x − y)(x + y). Cancelling (x + y) leaves (x − y)/(2x − y).
The minimum value of y in the equation y = x2 − 6x + 8 is
Answer: D (−1)
dy/dx = 2x − 6 = 0 at x = 3. Then y = 9 − 18 + 8 = −1.
Find the sum of the first 21 terms of the progression −10, −8, −6, …
Answer: D (210)
a = −10, d = 2, n = 21. S = 21/2 × (2(−10) + 20 × 2) = 21/2 × 20 = 210.
Find the eleventh term of the progression 4, 8, 16, …
Answer: B (212)
This is a GP with a = 4 and r = 2. The 11th term is 4 × 210 = 22 × 210 = 212.
In the diagram above, POQ is a diameter, O is the centre of the circle and TP is a tangent. Find the value of x.
Answer: A (30°)
OP is perpendicular to the tangent TP, so in triangle OPT angle POT = 180° − 90° − 30° = 60°. The angle x at Q stands on the same arc PR, so it is half of 60°, i.e. 30°.
In the diagram above, QR||TS, QR:TS = 2:3. Find the ratio of the area of triangle PQR to the area of the trapezium QRST.
Answer: B (4:5)
Triangles PQR and PTS are similar with sides in ratio 2:3, so their areas are in ratio 4:9. The trapezium is 9 − 4 = 5 parts, so the ratio is 4:5.
Three angles of a nonagon are equal and the sum of six other angles is 1110°. Calculate the size of one of the equal angles.
Answer: D (50°)
The angles of a nonagon add up to (9 − 2) × 180° = 1260°. The three equal angles share 1260° − 1110° = 150°, so each is 50°.
In the figure above, XŶZ = YT̂Z = 90°, XT = 9cm and TZ = 16cm. Find YZ
Answer: B (20cm)
In a right-angled triangle with the altitude YT, YZ² = TZ × XZ. So YZ² = 16 × 25 = 400 and YZ = 20cm.
Two chords QR and NP of a circle intersect inside the circle at X. If RQ̂P = 37°, RQ̂N = 49° and QP̂N = 35° find PR̂Q.
Answer: D (59°)
In triangle PQN, angle PQN = 37° + 49° = 86° and angle QPN = 35°, so angle QNP = 180° − 121° = 59°. Angle PRQ stands on the same arc PQ as angle PNQ, so it is also 59°.
In the figure above, find the value of x.
Answer: C (90°)
The two lines are parallel, so the co-interior angles 2y and 2z add to 180°, giving y + z = 90°. In the triangle containing x, x = 180° − (y + z) = 90°.
In the figure above, PQRS is a rectangle. If the shaded area is 72sq.cm, find h.
Answer: D (5cm)
The rectangle is 2h by 3h and the unshaded part inside is (2h − 4) by (3h − 2). Shaded area = 6h² − (2h − 4)(3h − 2) = 16h − 8 = 72, so h = 5cm.
The sine, cosine and tangent of 210° are respectively
Answer: A (−1/2, −√3/2, √3/3)
210° is in the third quadrant with reference angle 30°, where sine and cosine are negative and tangent positive. So sin 210° = −1/2, cos 210° = −√3/2 and tan 210° = 1/√3 = √3/3.
If tan θ = (m2 − n2)/2mn, find sec θ
Answer: B ((m2 + n2)/2mn)
Take a right triangle with opposite side m² − n² and adjacent side 2mn. The hypotenuse is √((m² − n²)² + 4m²n²) = m² + n², so sec θ = hypotenuse/adjacent = (m² + n²)/2mn.
From two points X and Y, 8cm apart, and in line with a pole, the angle of elevation of the top of the pole are 30 and 60 respectively. Find the height of the pole, assuming that X, Y, and the foot of the pole are on the same horizontal plane and X and Y are on the same side of the pole.
Answer: C (4√3 m)
Let Y be a from the foot. Then h = a tan 60° = a√3 and h = (8 + a) tan 30° = (8 + a)/√3. So 3a = 8 + a, a = 4 and h = 4√3.
A room is 12m long, 9m wide and 8cm high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room.
Answer: A (15/17)
The floor diagonal is √(144 + 81) = 15m and the room diagonal is √(225 + 64) = 17m. The cosine of the angle with the floor is 15/17.
What is the circumference of radius of the earth?
Answer: B (2π R cos θ)
The circle of latitude θ° has radius R cos θ, where R is the radius of the earth. Its circumference is 2πR cos θ.
The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that edge is 4√3cm
Answer: C (4cm)
Half the diagonal of the base is 8√2/2 = 4√2. Height² = (4√3)² − (4√2)² = 48 − 32 = 16, so the height is 4cm.
The figure above is an example of the construction of a
Answer: B (perpendicular from a given point to a given line)
An arc centred at the point P cuts the line at Q and R. Equal arcs from Q and R meet below the line, and joining P to that point gives the perpendicular from P to the line.
What is the locus of the mid-points of all chords of length 6cm within a circle of radius 5cm and with centre O?
Answer: A (A circle of radius 4cm and with centre O)
The midpoint of each chord is √(5² − 3²) = 4cm from O. So all the midpoints lie on a circle of radius 4cm centred at O.
Taking the period of daylight on a certain day to be from 5.30 a.m to 7.00 p.m, calculate the periods of daylight and of darkness on that day.
Answer: C (202°30', 157°30')
Daylight lasts 13½ hours out of 24, so its angle is 13.5/24 × 360° = 202.5° = 202°30'. Darkness is 360° − 202°30' = 157°30'.
The goals scored by 40 football teams from three league divisions are recorded below:
Number of goals: 0, 1, 2, 3, 4, 5, 6
Frequency: 4, 3, 15, 16, 1, 0, 1
What is the total number of goals scored by all the teams?
Answer: C (91)
Total goals = Σfx = 0 + 3 + 30 + 48 + 4 + 0 + 6 = 91.
The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If P is the mean and Q the median, then P + Q is
Answer: D (15)
The total is 60, so the mean is 60/8 = 7.5. In order, 2, 3, 5, 7, 8, 9, 12, 14, the median is (7 + 8)/2 = 7.5. P + Q = 15.
Below are the scores of a group of students in a Music test.
Scores: 1, 2, 3, 4, 5, 6, 7, 8, 9
No. of students: 3, 6, 10, 8, 6, 5, 2, 4, 12
If CF(x) is the number of students with scores less than or equal to x, find CF(6).
Answer: B (38)
Add the frequencies for scores 1 to 6: 3 + 6 + 10 + 8 + 6 + 5 = 38.
Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium.
Answer: A (3/5)
The square, rectangle and rhombus are parallelograms; the kite and trapezium are not. So the probability is 3/5.