60 questions. Try each one, then open the answer to check yourself and read the explanation.
Express 0.0462 in standard form.
Answer: D (4.62 × 10−2)
Standard form needs one non-zero digit before the point: 0.0462 = 4.62 × 10−2.
The population of a village is 5846. Express this number to three significant figures
Answer: A (5850)
Keep 5, 8, 4; the next digit 6 rounds the 4 up to 5, and the place value is kept with a zero: 5850.
Simplify: (1/4)−1½
Answer: E (8)
(1/4)−3/2 = 43/2 = (√4)3 = 23 = 8.
For what value of y is the expression (y + 2)/(y2 − 3y − 10) undefined?
Answer: D (y = 5)
It is undefined when the denominator is zero: y2 − 3y − 10 = (y − 5)(y + 2) = 0, so y = 5 or y = −2; only y = 5 is offered.
Simplify: log6 + log2 − log12
Answer: C (0)
log6 + log2 − log12 = log(6 × 2 ÷ 12) = log 1 = 0.
Find the number whose logarithm to base 10 is 2.6025
Answer: A (400.4)
Antilog of 0.6025 is 4.004; the characteristic 2 means 3 digits before the point: 400.4.
Factorize: 3a2 − 11a + 6
Answer: A ((3a − 2)(a − 3))
(3a − 2)(a − 3) = 3a2 − 9a − 2a + 6 = 3a2 − 11a + 6.
Solve the equation: 3a + 10 = a2
Answer: D (a = 5 or a = −2)
a2 − 3a − 10 = 0 gives (a − 5)(a + 2) = 0, so a = 5 or a = −2.
Simplify (3/x − 15/(2y)) ÷ 6/(xy)
Answer: A ((2y − 5x)/4)
3/x − 15/(2y) = (6y − 15x)/(2xy). Dividing by 6/(xy) means multiplying by xy/6: (6y − 15x)/12 = 3(2y − 5x)/12 = (2y − 5x)/4.
Simplify: ¼(2n − 2n+2)
Answer: A (2n−2 − 2n)
¼ = 2−2, so ¼(2n − 2n+2) = 2n−2 − 2n.
Find the equation whose roots are −2/3 and −1/4
Answer: A (12x2 + 11x + 2 = 0)
(x + 2/3)(x + 1/4) = 0; multiply by 12: (3x + 2)(4x + 1) = 12x2 + 11x + 2 = 0.
Solve: 6(x − 4) + 3(x + 7) = 3
Answer: B (2/3)
6x − 24 + 3x + 21 = 3 gives 9x − 3 = 3, so 9x = 6 and x = 2/3.
If 2x + y = 7 and 3x − 2y = 3, by how much is 7x greater than 10?
Answer: C (7)
Double the first equation: 4x + 2y = 14. Add the second: 7x = 17, which is 7 more than 10.
A cylinder of base radius 4cm is open at one end. If the ratio of the area of its base to that of its curved surface is 1:4, calculate the height of the cylinder.
Answer: D (8cm)
πr2 : 2πrh = r : 2h = 1 : 4, so 2h = 4r and h = 2r = 8cm.
A headmaster contributes 7% of his income into a fund and his wife contributes 4% of her income. If the husband earns N5,500 per annum (p.a) and the wife earns N4,000 (p.a), find the sum of their annual contribution to the fund.
Answer: C (N545)
7% of N5,500 = N385 and 4% of N4,000 = N160; together N545.
A car is travelling at an average speed of 80km/hr. Its speed in metres per second (m/s) is
Answer: B (22.2m/s)
80km/h = 80 000m ÷ 3600s = 22.2m/s.
Mrs. Kofi sold an article for C7.50 instead of C12.75. Calculate her percentage error, correct to one decimal place.
Answer: A (41.2%)
Error = 12.75 − 7.50 = 5.25. Percentage error = 5.25/12.75 × 100 = 41.2%.
In the diagram below, PXR and PYO are two semi-circles with diameters 14cm and 7cm respectively. (Take π = 22/7)
Find the area of the enclosed region, PXROY correct to the nearest whole number.
Answer: A (96cm2)
Large semicircle: ½ × 22/7 × 72 = 77cm2. Small semicircle: ½ × 22/7 × 3.52 = 19.25cm2. Total = 96.25 ≈ 96cm2.
In the diagram below, PXR and PYO are two semi-circles with diameters 14cm and 7cm respectively. (Take π = 22/7)
Find the perimeter of the region
Answer: C (40cm)
Arc PXR = ½ × 22/7 × 14 = 22cm, arc PYO = ½ × 22/7 × 7 = 11cm, plus the straight edge OR = 7cm: 22 + 11 + 7 = 40cm.
Use the diagram below. Find the coordinates of point B.
Answer: E ((1½, 0))
B is where 4x + 3y = 6 meets the x-axis: put y = 0, so 4x = 6 and x = 1½. B = (1½, 0).
The first term a of an A.P. is equal to twice the common difference d. Find, in terms of d, the 5th term of the A.P.
Answer: C (6d)
5th term = a + 4d = 2d + 4d = 6d.
The common ratio of a G.P. is 2. If the 5th term is greater than the 1st term by 45, find the 5th term,
Answer: D (48)
ar4 − a = 16a − a = 15a = 45, so a = 3 and the 5th term = 3 × 16 = 48.
Two towns, P and Q, are on (4°N 40°W) and (4°N 20°E) respectively. What is the distance between them, along their line of latitude? (Give your answer in terms of π and R, the radius of the earth).
Answer: D ((πR/3) cos 4°)
Difference in longitude = 40° + 20° = 60°. Distance = 60/360 × 2πR cos 4° = (πR/3) cos 4°.
The graph above is a sketch of
Answer: D (y = 2sin x)
The curve starts at 0, crosses the x-axis at π and 2π (period 2π) and reaches −2, so it is a sine curve with amplitude 2: y = 2 sin x.
Given that sin θ = −0.9063, where 0 ≤ θ° ≤ 270°, find θ.
Answer: D (245°)
sin 65° = 0.9063. Sine is negative in the third quadrant within the range, so θ = 180° + 65° = 245°.
From the top of a cliff, the angle of depression of a boat on the sea is 60°, if the top of the cliff is 25m above the sea level, calculate the horizontal distance from the bottom of the cliff to the boat.
Answer: C (25√3m/3)
tan 60° = 25/d, so d = 25/√3 = 25√3/3 m.
The angle of elevation of the top of a tree 39m away from a point on the ground is 30°. Find the height of the tree.
Answer: B (13√3m)
h = 39 tan 30° = 39/√3 = 13√3 m.
In the diagram above, ATR is a tangent at the point T; to the circle centre O, if ∠TOB = 145°, find ∠TAO
Answer: D (55°)
AOB is a straight line, so ∠AOT = 180° − 145° = 35°. The radius OT is perpendicular to the tangent, so ∠TAO = 180° − 90° − 35° = 55°.
P = {2, 1, 3, 9, ½}, Q = {1, 2½, 3, 7} and R = {5, 4, 2½}
Find P∪Q∪R.
Answer: E ({½, 1, 2, 2½, 3, 4, 5, 7, 9})
The union lists every element that is in at least one set: ½, 1, 2, 2½, 3, 4, 5, 7, 9.
P = {2, 1, 3, 9, ½}, Q = {1, 2½, 3, 7} and R = {5, 4, 2½}
Find P∩Q∩R.
Answer: B (∅)
No element is in all three sets (R has none of P's elements), so the intersection is empty.
Which of the following is not a measure of dispersion?
Answer: A (Mode)
The mode is a measure of central tendency (an average); the others measure spread.
The mean of 30 observations recorded in an experiment is 5. If the observed largest value of 34 is deleted, find the mean of the remaining obsvervations
Answer: B (4)
Total = 30 × 5 = 150. Remove 34: 116 ÷ 29 = 4.
The table below shows the frequency distribution of a number of chairs in each of 40 rooms of various houses.
No. of chairs: 1, 2, 3, 4, 5, 6, 7
Frequency: 2, 7, 5, 4, 9, 7, 6
Find the mean of the distribution.
Answer: C (4.4)
Σfx = 2 + 14 + 15 + 16 + 45 + 42 + 42 = 176; mean = 176 ÷ 40 = 4.4.
The table below shows the frequency distribution of a number of chairs in each of 40 rooms of various houses.
No. of chairs: 1, 2, 3, 4, 5, 6, 7
Frequency: 2, 7, 5, 4, 9, 7, 6
Find the mode of the distribution.
Answer: D (5)
The highest frequency is 9, for 5 chairs, so the mode is 5.
The table below shows the frequency distribution of a number of chairs in each of 40 rooms of various houses.
No. of chairs: 1, 2, 3, 4, 5, 6, 7
Frequency: 2, 7, 5, 4, 9, 7, 6
Find the median of the distribution.
Answer: C (5)
The median is between the 20th and 21st rooms. Cumulative frequencies 2, 9, 14, 18, 27 show both are in the 5-chair group.
If events X and Y are mutually exclusive, P(X) = 1/3 and P(Y) = 2/5,
P(X∩Y) is
Answer: A (0)
Mutually exclusive events cannot happen together, so P(X∩Y) = 0.
If events X and Y are mutually exclusive, P(X) = 1/3 and P(Y) = 2/5,
P(X∪Y) is
Answer: D (11/15)
For mutually exclusive events P(X∪Y) = P(X) + P(Y) = 1/3 + 2/5 = 11/15.
A box contains 2 white and 3 blue identical marbles. If two marbles are picked at random, one after the other without replacement, what is the probability of picking two marbles of different colours?
Answer: B (3/5)
P(white then blue) + P(blue then white) = 2/5 × 3/4 + 3/5 × 2/4 = 6/20 + 6/20 = 3/5.
Mrs. Jones is expecting a baby. The probability that it will be a boy is ½ and probability that the baby will have blue eyes is ¼. What is the probability that she will have a blue-eyed boy?
Answer: A (1/8)
Multiply the independent probabilities: ½ × ¼ = 1/8.
Which of the following about a rhombus may not be true?
Answer: A (The diagonals are equal)
The diagonals of a rhombus are equal only when it is a square; all the other statements are always true.
The angles of a pentagon are x°, 2x°, (x + 60)°, (x + 10)°, (x − 10)°. Find the value of x.
Answer: D (80)
The angles of a pentagon add up to 540°: 6x + 60 = 540, so x = 80.
In the diagram below, ∠PTQ = ∠URP = 25° and XPU = 4URP. Calculate ∠USQ.
Answer: D (130°)
∠XPU = 100°, so ∠UPQ = 80°. In triangle PTQ, ∠PQT = 180° − 80° − 25° = 75°, so ∠SQR = 105°. In triangle SQR, ∠QSR = 180° − 105° − 25° = 50°, so ∠USQ = 180° − 50° = 130°.
In a given regular polygon, the ratio of the exterior angle to the interior angles is 1 : 3. How many side has the polygon?
Answer: D (8)
Exterior + interior = 180°, so the exterior angle = 180° ÷ 4 = 45°. Number of sides = 360° ÷ 45° = 8.
In the diagram below, O is the centre of the circle, |SQ| = |QR| and ∠PQR = 68°. Calculate ∠PRS.
Answer: A (34°)
PQ is a diameter, so ∠PRQ = 90°. Triangle SQR is isosceles with apex angle 68°, so ∠QRS = (180° − 68°)/2 = 56°. ∠PRS = 90° − 56° = 34°.
Find the area of an equilateral triangle of side 16cm.
Answer: A (64√3cm2)
Area = ½ × 16 × 16 × sin 60° = 128 × √3/2 = 64√3cm2.
In the diagram below, PQ and XY are two concentric arcs; centre O. If the ratio of the lengths of the two arcs is 1 : 3, find the ratio of the areas of the two sectors OPQ and OXY.
Answer: C (1 : 9)
Both sectors have the same angle, so the arc lengths are in the ratio of the radii, 1 : 3. Areas go as the square of the radius: 1 : 9.
In the diagram below, |AD| = 10cm, |DC| = 8cm and |CF| = 15cm.
If the area of triangle DCF = 24cm2, find the area of the quadrilateral ABCD.
Answer: B (48cm2)
DCFE is a parallelogram, so its area is twice triangle DCF: 48cm2. Parallelogram ABCD has the same base DC and lies between the same parallels, so its area is also 48cm2.
In the diagram below, |AD| = 10cm, |DC| = 8cm and |CF| = 15cm.
Which of the following is correct?
Answer: E (Area DABO = Area CFEO)
Parallelograms ABCD and DCFE have equal areas. Removing the common triangle DOC from each leaves DABO and CFEO, so these are equal.
Simplify: (i) 2 2/3 − (2½ − 1 4/5) (ii) (3.25 − 1.64)/(2.47 − 2.01)
(i) 2½ − 1⅘ = 5/2 − 9/5 = 7/10
2⅔ − 7/10 = 80/30 − 21/30 = 59/30 = 1 29/30
(ii) 1.61 ÷ 0.46 = 161/46 = 3.5
Work out the bracket first.
(a) Using logarithm table, evaluate ∛1.376/∜0.007 correct to three s.f.
(b) Without using Mathematical tables, find the value of (log 81)/(log 1/3)
(a) log 1.376 = 0.1386; ÷ 3 = 0.0462
log 0.007 = 3̄.8451; ÷ 4 = (4̄ + 1.8451) ÷ 4 = 1̄.4613
0.0462 − 1̄.4613 = 0.5849
antilog 0.5849 = 3.845 ≈ 3.85
(b) log 81 = 4 log 3 and log 1/3 = −log 3, so the value is 4 log 3/(−log 3) = −4
To divide a negative characteristic by 4, first make it exactly divisible: 3̄.8451 = 4̄ + 1.8451.
In a certain class, 22 pupils take one or more of Chemistry, Economics and Government. 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 pupils take Economics and Government.
(a)(i) Using set notation and the letters indicated above, write down the two statements in the last sentence. (ii) Draw a Venn diagram to illustrate the information.
(b) How many pupils take (i) both Chemistry and Government? (ii) Government only?
(a)(i) E ∩ C = ∅ (or n(E ∩ C) = 0); n(E ∩ G) = 4
(ii) Circles E and G overlap (4 in the overlap); circle C overlaps G only and does not touch E; let x be in C ∩ G
E only = 8, G only = 4 − x, C only = 7 − x
(b)(i) 8 + 4 + (4 − x) + x + (7 − x) = 22, so 23 − x = 22 and x = 1
(ii) Government only = 8 − 4 − 1 = 3
Draw the C circle so that it does not overlap E, because nobody takes both.
A man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.
(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?
(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?
Red balls = 9 − 3 − 2 = 4
(a)(i) P(not blue) = 1 − 2/9 = 7/9 (ii) P(not red) = 1 − 4/9 = 5/9
(b)(i) 3/9 × 2/8 = 1/12
(ii) 2/9 × 2/9 = 4/81
Without replacement, both the number of black balls and the total drop by one for the second draw.
ABCDE is a regular pentagon and a rectangle. AXYE is drawn on the side AE such that the vertices X and Y lie on the sides BC and CD respectively. Calculate the size of (i) an interior angle of the pentagon, (ii) ∠BXA.
(i) Sum of interior angles = (5 − 2) × 180° = 540°; each angle = 540° ÷ 5 = 108°
(ii) ∠XAE = 90° (rectangle), so ∠BAX = 108° − 90° = 18°
In triangle ABX: ∠BXA = 180° − 108° − 18° = 54°
Draw the figure: the rectangle shares side AE with the pentagon.
(a) Solve the equation, correct to two decimal places. 2x2 + 7x − 11 = 0
(b) Using the substitution p = 1/x, q = 1/y Solve the simultaneous equation 2/x + 1/y = 3, 1/x − 5/y = 7
(a) x = (−7 ± √(49 + 88))/4 = (−7 ± √137)/4 = (−7 ± 11.705)/4
x = 1.18 or x = −4.68
(b) 2p + q = 3 and p − 5q = 7
From the first, q = 3 − 2p; then p − 15 + 10p = 7, so 11p = 22, p = 2 and q = −1
x = 1/p = ½, y = 1/q = −1
Solve for p and q first, then turn them back into x and y.
A man bought 5 reams of duplicating paper, each of which was supposed to contain 480 sheets. The actual number of sheets in the packets were 435, 420, 405, 415 and 440.
(a) Calculate, correct to the nearest whole number the average percentage error for the packets of paper
(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.
(a) Expected = 5 × 480 = 2400 sheets; actual = 2115; shortfall = 285
Average percentage error = 285/2400 × 100 = 11.875% ≈ 12%
(b) Cost of one sheet = N35 ÷ 480; amount cheated = 285 × 35/480 = N20.78 ≈ N21
The average of the five percentage errors gives the same 11.875% because every ream should have 480 sheets.
Using a scale of 2cm to 1unit on the x-axis and 1cm to 1unit on the y-axis, draw on the same axes the graphs of y = 3 + 2x − x2 y = 2x − 3 for −3 ≤ x ≤ 4.
Using your graph: (i) Solve the equation 6 − x2 = 0. (ii) find the maximum value of 3 + 2x − x2, (iii) find the range of x for which 3 + 2x − x2 ≤ 1, expressing all your answers correct to one decimal place.
Table for y = 3 + 2x − x2: x = −3, −2, −1, 0, 1, 2, 3, 4 gives y = −12, −5, 0, 3, 4, 3, 0, −5
Line y = 2x − 3: e.g. (−3, −9), (0, −3), (4, 5)
(i) 6 − x2 = 0 is where the curve meets the line (3 + 2x − x2 = 2x − 3): x ≈ −2.4 or x ≈ 2.4
(ii) Maximum value = 4 (at x = 1)
(iii) Draw y = 1: the curve is on or below it for −3 ≤ x ≤ −0.7 and 2.7 ≤ x ≤ 4
Rearrange the equation to show it is the curve minus the line equal to zero.
(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) In the diagram above, O is the centre of the circle, ∠OQR = 32° and ∠MPQ = 15°. Calculate (i) ∠QPR; (ii) ∠MQO.
(a) Given a circle centre O, arc BC subtending ∠BOC at O and ∠BAC at A on the circumference. Join AO and produce to N.
OA = OB (radii), so ∠OAB = ∠OBA = x; exterior angle ∠BON = 2x. Similarly ∠CON = 2y where y = ∠OAC.
∠BOC = 2x + 2y = 2(x + y) = 2∠BAC.
(b)(i) OQ = OR, so ∠ORQ = 32° and ∠QOR = 180° − 64° = 116°; ∠QPR = ½ × 116° = 58°
(ii) ∠MOQ = 2 × ∠MPQ = 30°; triangle MOQ is isosceles, so ∠MQO = (180° − 30°)/2 = 75°
Use the angle at the centre twice: once for chord QR and once for chord MQ.
The table below shows the distribution of the waiting times for some customers in a certain petrol station.
Waiting time (in min.): 1.5 − 1.9, 2.0 − 2.4, 2.5 − 2.9, 3.0 − 3.4, 3.5 − 3.9, 4.0 − 4.4
No. of Customers: 3, 10, 18, 10, 7, 2
(a) Write down the class boundaries of the distribution.
(b) Construct a cumulative frequency curve for the data.
(c) Using your graph, estimate: (i) the interquartile range of the distribution; (ii) the proportion of customers who would have waited for more than 3 minutes.
(a) 1.45−1.95, 1.95−2.45, 2.45−2.95, 2.95−3.45, 3.45−3.95, 3.95−4.45
(b) Cumulative frequencies 3, 13, 31, 41, 48, 50 plotted against the upper boundaries 1.95, 2.45, 2.95, 3.45, 3.95, 4.45 (start at (1.45, 0)); join with a smooth S-shaped curve
(c)(i) Q1 (12.5th) ≈ 2.4 min, Q3 (37.5th) ≈ 3.3 min; interquartile range ≈ 0.8−0.9 min
(ii) About 32 customers waited up to 3 minutes, so about 18 of 50 waited longer: 18/50 = 0.36 (36%)
Plot cumulative frequency at the upper class boundary, not at the class mark.
(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm intersecting at K and ∠QKR = 60°.
(b) Construct a rectangle PABS which is equal in area to PQRS in (a) above and on the same side of PS as PQRS. Measure |PA|
(a) Draw PR = 9cm and bisect it to get K. Construct 60° at K for ∠QKR; mark KQ = KS = 3cm on the line through K. Join PQ, QR, RS, SP. Arcs must be shown.
(b) Area of PQRS = ½ × 9 × 6 × sin 60° ≈ 23.4cm2; PS ≈ 4.0cm
Construct perpendiculars to PS at P and S on the side of the parallelogram. The rectangle has the same base PS and lies between the same parallels, so the line QR (extended) cuts the perpendiculars at A and B.
|PA| ≈ 5.9cm (accept 5.8−6.0cm)
A rectangle and a parallelogram on the same base and between the same parallels have equal areas.
From a horizontal distance of 8.5km, a pilot observes that the angles of depression of the top and base of a control tower are 30° and 33° respectively. Calculate, correct to three significant figures: (i) the shortest distance between the pilot and the base of the control tower; (ii) the height of the control tower.
(i) cos 33° = 8.5/d, so d = 8.5/0.8387 = 10.1km
(ii) Height of pilot above the base = 8.5 tan 33° = 5.520km; above the top = 8.5 tan 30° = 4.907km
Height of tower = 5.520 − 4.907 = 0.612km (612m)
Draw one horizontal line from the pilot; both angles of depression are measured from it.