58 questions. Try each one, then open the answer to check yourself and read the explanation.
Let U = {1, 2, 3, 4}, P = {2, 3} and Q = {2, 4}. What is (P∩Q)'?
Answer: B ({1, 3, 4})
P∩Q = {2}. Its complement is every element of U except 2, which is {1, 3, 4}.
Simplify (3/4 + 1/3) × 4⅓ ÷ 3¼
Answer: E (13/9)
3/4 + 1/3 = 13/12. Then 13/12 × 13/3 = 169/36, and 169/36 ÷ 13/4 = 169/36 × 4/13 = 13/9.
If x varies over the set of real numbers, which of the following is illustrated in the diagram above?
Answer: B (−3 ≤ x < 2)
The shaded (filled) dot at −3 means −3 is included; the open circle at 2 means 2 is not included. So −3 ≤ x < 2.
Convert 77 to a number in base two
Answer: A (1001101)
77 = 64 + 8 + 4 + 1 = 26 + 23 + 22 + 20, which is 1001101 in base two.
A bricklayer measured the length of a wall and obtained 4.10m. If the actual length of the wall is 4.25m, find his percentage error.
Answer: A (3 9/17%)
Error = 4.25 − 4.10 = 0.15m. Percentage error = 0.15/4.25 × 100 = 300/85 = 3 9/17%.
The nth term of a sequence is given by 3.2n−2. Write down the first three terms of the sequence.
Answer: B (3/2, 3, 6)
The nth term is 3 × 2n−2. n = 1 gives 3 × 2−1 = 3/2; n = 2 gives 3 × 1 = 3; n = 3 gives 3 × 2 = 6.
Evaluate log1025 + log1032 − log108
Answer: B (2)
Combine the logs: log10(25 × 32 ÷ 8) = log10100 = 2.
Factorize the expression 2y2 + xy − 3x2
Answer: D ((2y + 3x)(y − x))
Split xy as 3xy − 2xy: 2y2 − 2xy + 3xy − 3x2 = 2y(y − x) + 3x(y − x) = (2y + 3x)(y − x).
Construct a quadratic equation whose roots are −½ and 2.
Answer: E (2x2 − 3x − 2 = 0)
(x + ½)(x − 2) = 0, or (2x + 1)(x − 2) = 0, which expands to 2x2 − 3x − 2 = 0.
What must be added to the expression x2 − 18x to make it a perfect square?
Answer: E (81)
Add the square of half the coefficient of x: (18/2)2 = 81, giving (x − 9)2.
Solve the equation m/3 + 1/2 = 3/4 + m/4
Answer: D (3)
Multiply through by 12: 4m + 6 = 9 + 3m, so m = 3.
The angle of a sector of a circle of radius 8cm is 240°. This sector is bent to form a cone. Find the radius of the base of the cone.
Answer: A (16/3cm)
The arc becomes the base circumference: (240/360) × 2π × 8 = 2πr, so r = (2/3) × 8 = 16/3 cm.
The curved surface area of a cylinderical tin is 704cm2. Calculate the height when the radius is 8cm. (Take π = 22/7)
Answer: C (14cm)
Curved surface area = 2πrh, so 2 × 22/7 × 8 × h = 704, giving h = 704 × 7 ÷ 352 = 14 cm.
The volume of a cone of height 9cm is 1848cm3. Find its radius. (Take π = 22/7).
Answer: B (14cm)
V = ⅓πr2h: ⅓ × 22/7 × r2 × 9 = 1848, so r2 = 1848 × 7 ÷ 66 = 196 and r = 14 cm.
The area of a parallelogram is 513cm2 and the height is 19cm. Calculate the base.
Answer: C (27cm)
Area = base × height, so base = 513 ÷ 19 = 27 cm.
The diagram below shows the shaded segment of a circle of radius 7cm. If the area of the triangle OXY is 12¼cm2, calculate the area of the segment. (Take π = 22/7)
Answer: B (7/12cm2)
Sector area = (30/360) × 22/7 × 7 × 7 = 77/6 = 12 5/6 cm2. Segment = sector − triangle = 12 5/6 − 12¼ = 7/12 cm2.
In the diagram above PQT is a tangent to the circle QRS at Q. Angle QTR = 48° and ∠QRT = 95°. Find ∠QSR
Answer: C (37°)
In triangle QRT, ∠RQT = 180° − 48° − 95° = 37°. By the alternate segment theorem, ∠QSR = ∠RQT = 37°.
In the diagram below, O is the centre of the circle and ∠POR = 126°. Find ∠PQR.
Answer: B (117°)
The 126° angle faces Q, so the arc PR not containing Q has central angle 360° − 126° = 234°. ∠PQR is half of that: 117°.
In the diagram below, TR = TS and ∠TRS = 60°. Find the value of x.
Answer: C (60)
TR = TS with ∠TRS = 60° makes triangle RST equilateral, so ∠RST = 60° and ∠PSR = 180° − 60° = 120°. PSRQ is cyclic, so x = 180° − 120° = 60°.
In the diagram below, O is the centre of the circle PQRS and ∠QPS = 36°. Find ∠QOS
Answer: B (72°)
The angle at the centre is twice the angle at the circumference on the same arc: ∠QOS = 2 × 36° = 72°.
In the diagram below, O is the centre of the circle. QOR is a diameter and ∠PSR is 37°. Find ∠PRQ.
Answer: B (53°)
∠PQR = ∠PSR = 37° (angles in the same segment) and ∠QPR = 90° (angle in a semicircle). So ∠PRQ = 180° − 90° − 37° = 53°.
In the diagram below, ML//PQ and NP//QR. If ∠LMN = 40° and ∠MNP = 55°. Find ∠QPR
Answer: A (15°)
Draw a line through N parallel to ML and PQ: it splits the 55° angle into 40° (alternate to ∠LMN) and 15°, so NP makes 15° with PQ. Since QR is parallel to NP, the angle between QR and PQ is also 15°.
The angles marked in the diagram below are measured in degrees. Find x
Answer: B (24°)
The marked angles are the exterior angles of the pentagon, which add up to 360°: 2x + 3x + 5x + 3x + 2x = 15x = 360°, so x = 24°.
Find the value of t in the diagram below.
Answer: A (5.6cm)
The line parallel to QR divides the two sides in the same ratio: t/4 = 7/5, so t = 28/5 = 5.6 cm.
A regular polygon has each of its exterior angles as 18°. How many sides has the polygon?
Answer: D (20)
Exterior angles of a polygon add up to 360°, so the number of sides = 360° ÷ 18° = 20.
In the diagram above, PQT is an isosceles triangle. |PQ| = |QT|, ∠SRQ = 75°, ∠QPT = 25° and PQR is a straight line. Find ∠RST.
Answer: C (55°)
PQ = QT, so ∠QTP = 25° and the exterior angle ∠TQR = 25° + 25° = 50°. In triangle QRS, ∠RST = 180° − 50° − 75° = 55°.
Sin 60° has the same value as I. sin 120° II. cos 240° III. −sin 150° IV. cos 210° V. sin 240°
Answer: A (I only)
sin 60° = √3/2. sin 120° = sin 60° = √3/2; cos 240° = −½, −sin 150° = −½, cos 210° = −√3/2 and sin 240° = −√3/2. Only I matches.
Which of the following is the graph of cos θ?
Answer: C (Graph C)
cos θ equals 1 at θ = 0, 0 at π/2, −1 at π and 0 again at 3π/2. Graph C is the only one with this shape.
The value of tan 315° is
Answer: D (−1)
315° is in the fourth quadrant where tangent is negative, and its reference angle is 45°, so tan 315° = −tan 45° = −1.
If cos θ = 5/13, what is the value of tan θ for 0 < θ < 90°?
Answer: D (12/5)
With adjacent side 5 and hypotenuse 13, the opposite side is √(169 − 25) = 12, so tan θ = 12/5.
The value of sin 210° is
Answer: A (−½)
210° is in the third quadrant where sine is negative, with reference angle 30°: sin 210° = −sin 30° = −½.
Town P is on bearing 315° from town Q while town R is south of town P and west of town Q. If town R is 60km away from Q, how far is R from P?
Answer: D (60km)
R, P and Q form a right angle at R, and the bearing 315° (north-west) makes the angle at Q 45°. The triangle is isosceles, so RP = RQ = 60 km.
A ladder leans against a vertical wall at an angle 60° to the wall. If the foot of the ladder is 5 metres away from the wall, calculate the length of the ladder.
Answer: C (10√3/3 m)
The 5 m distance is opposite the 60° angle at the wall: sin 60° = 5/L, so L = 5 ÷ (√3/2) = 10/√3 = 10√3/3 m.
Points X and Y are respectively 12m North and 5m East of point Z. Calculate XY.
Answer: C (13m)
The angle at Z is 90°, so XY = √(122 + 52) = √169 = 13 m.
Find the median of the following set of scores: 65, 72, 55, 48, 78
Answer: D (65)
In order: 48, 55, 65, 72, 78. The middle score is 65.
The data below shows the frequency distribution of marks scored by a group of students in a class test.
Marks: 2, 3, 4, 5, 6
Frequency: 2, 4, 5, 3, 1
How many students took the test?
Answer: C (15)
Add the frequencies: 2 + 4 + 5 + 3 + 1 = 15.
The data below shows the frequency distribution of marks scored by a group of students in a class test.
Marks: 2, 3, 4, 5, 6
Frequency: 2, 4, 5, 3, 1
What is the modal score.
Answer: C (4)
The mode is the score with the highest frequency: 4 (frequency 5).
The data below shows the frequency distribution of marks scored by a group of students in a class test.
Marks: 2, 3, 4, 5, 6
Frequency: 2, 4, 5, 3, 1
Find the mean mark.
Answer: E (3.8)
Σfx = 4 + 12 + 20 + 15 + 6 = 57 and Σf = 15, so the mean = 57 ÷ 15 = 3.8.
A group of students measured a certain angle (to the nearest degree) and obtained the following results: 75°, 76°, 72°, 73°, 74°, 79°, 72°, 72°, 77°, 72°, 71°, 70°, 78°, 73°. Find the mode
Answer: E (72°)
72° occurs four times, more than any other value, so it is the mode.
The table below shows the scores of 15 students in a Physics test.
Marks: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
No. of Students: 1, 3, 2, 0, 1, 6, 1, 0, 1, 0
How many students scored at least 5?
Answer: D (9)
Add the students scoring 5 or more: 1 + 6 + 1 + 0 + 1 + 0 = 9.
The table below shows the scores of 15 students in a Physics test.
Marks: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
No. of Students: 1, 3, 2, 0, 1, 6, 1, 0, 1, 0
Find the median score
Answer: C (6)
The median is the 8th of the 15 scores. Cumulative totals: 1, 4, 6, 6, 7, 13…, so the 8th score falls in mark 6.
A number is chosen at random from the set {1, 2, 3, ….., 9, 10}. What is the probability that the number is greater than or equal to 7?
Answer: C (2/5)
The numbers 7, 8, 9 and 10 qualify: 4 out of 10, which is 2/5.
What is the probability of throwing a number greater than 2 with a single fair die.
Answer: D (2/3)
3, 4, 5 and 6 are greater than 2: 4 out of 6 = 2/3.
A fair die is rolled once. What is the probability of obtaining 4 or 6?
Answer: C (1/3)
Two favourable outcomes out of six: 2/6 = 1/3.
Three balls are drawn one after the other with replacement, from a bag containing 5 red, 9 white and 4 blue identical balls. What is the probability that they are one red, one white and one blue?
Answer: C (5/162)
There are 18 balls. Taking the draws in the order red, white, blue: 5/18 × 9/18 × 4/18 = 180/5832 = 5/162.
The data below shows the number of workers employed in the various sections of a construction company in Lagos.
Carpenters 24, Plumbers 12, Painters 9, Bricklayers 18, Labourers 27, Plasterers 15, Messengers 3
If one of the workers is absent on a certain day, what is the probability that he is a bricklayer?
Answer: D (1/6)
Total workers = 24 + 12 + 9 + 18 + 27 + 15 + 3 = 108. P(bricklayer) = 18/108 = 1/6.
The data below shows the number of workers employed in the various sections of a construction company in Lagos.
Carpenters 24, Plumbers 12, Painters 9, Bricklayers 18, Labourers 27, Plasterers 15, Messengers 3
If a worker is retrenched, what is the probability that he is a plumber or plasterer.
Answer: D (1/4)
Plumbers + plasterers = 12 + 15 = 27 out of 108 workers, so the probability is 27/108 = 1/4.
(a) If 92x+1 = 81x−2/3x, find x.
(b) Without using Mathematical Tables, evaluate √((0.81 × 10−5)/(2.25 × 107))
(a) Write everything as powers of 3: 32(2x+1) = 34(x−2) ÷ 3x
So 4x + 2 = 4x − 8 − x = 3x − 8
x = −10
(b) (0.81 × 10−5)/(2.25 × 107) = (81/225) × 10−12 = 0.36 × 10−12
Square root: 0.6 × 10−6 = 6 × 10−7
Turn both sides into the same base before comparing indices.
(a) Solve the following pair of simultaneous equations
2x + 5y = 6½
5x − 2y = 9
(b) If log10(2x + 1) − log10(3x − 2) = 1, find x
(a) Double the first equation: 4x + 10y = 13. Multiply the second by 5: 25x − 10y = 45
Add: 29x = 58, so x = 2
Substitute: 4 + 5y = 6½, so y = ½
(b) log10[(2x + 1)/(3x − 2)] = 1 = log1010
(2x + 1)/(3x − 2) = 10, so 2x + 1 = 30x − 20
28x = 21, x = ¾ (check: 3x − 2 = ¼ > 0, so the log exists)
Use the subtraction law of logarithms to combine the left-hand side into one log.
(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector (Take π = 22/7).
(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.
(a) Arc length = (108/360) × 2 × 22/7 × 7 = 0.3 × 44 = 13.2 cm
Perimeter = arc + two radii = 13.2 + 7 + 7 = 27.2 cm
(b) Angle of elevation of the top from the boat = angle of depression = 30°
tan 30° = h/120, so h = 120 × tan 30° = 120 × 0.5774 = 69.28
Height ≈ 69.3 m (3 s.f.)
The perimeter of a sector includes the two radii as well as the arc.
(a) The sides PQ and PR of ΔPQR are produced to T and S respectively, such that ∠TQR = 131° and ∠QRS = 98°. Find ∠QPR.
(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre (Take π = 22/7).
(a) ∠PQR = 180° − 131° = 49° (angles on a straight line)
∠PRQ = 180° − 98° = 82° (angles on a straight line)
∠QPR = 180° − (49° + 82°) = 49° (angles in a triangle)
(b) 2πr = 400, so r = 400 × 7 ÷ (2 × 22) = 63.6
r ≈ 64 m (nearest metre)
Each exterior angle and its interior angle add up to 180°.
(a) In a game, a fair die is rolled once and two unbiased coins are tossed once. What is the probability of obtaining 3 and a tail?
(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting: (i) a red, then a black marble; (ii) two black marbles.
(a) P(3 on the die) = 1/6
Two coins: HH, HT, TH, TT; exactly one tail in 2 of the 4 outcomes, so P(a tail) = 1/2
P(3 and a tail) = 1/6 × 1/2 = 1/12
(If 'a tail' is read as at least one tail, P = 3/4 and the answer is 1/6 × 3/4 = 1/8.)
(b)(i) P(red then black) = 3/10 × 7/9 = 21/90 = 7/30
(ii) P(two black) = 7/10 × 6/9 = 42/90 = 7/15
Without replacement, the second fraction uses one fewer marble in total.
(a) If 17x = 3752 − 3562, find the exact value of x.
(b) If 4x = 21/2 × 8, find x.
(c) The sum of the first 9 terms of an A.P. is 72 and the sum of the next 4 terms is 71. Find the A.P.
(a) Difference of two squares: 3752 − 3562 = (375 − 356)(375 + 356) = 19 × 731 = 13889
x = 13889 ÷ 17 = 817
(b) 22x = 21/2 × 23 = 27/2, so 2x = 7/2 and x = 7/4
(c) S9 = 9/2(2a + 8d) = 72, so a + 4d = 8
S13 = 72 + 71 = 143 = 13/2(2a + 12d), so a + 6d = 11
Subtract: 2d = 3, d = 3/2; then a = 8 − 6 = 2
The A.P. is 2, 3½, 5, 6½, 8, …
The first 13 terms add up to 72 + 71 = 143; use the sum formula twice.
(a) Calculate the area of the shaded segment of the circle shown in the diagram. (Take π = 22/7).
(b) A tin has radius 3cm and height 6cm. Find the: (i) total surface area of tin; (ii) volume in litres of liquid that will fill the tin to capacity, correct to two decimal places. (Take π = 22/7).
(a) Sector area = (63/360) × 22/7 × 102 = 55 cm2
Triangle area = ½ × 10 × 10 × sin 63° = 50 × 0.8910 = 44.55 cm2
Segment area = 55 − 44.55 = 10.45 cm2
(b)(i) Closed tin: total surface area = 2πr2 + 2πrh = 2πr(r + h) = 2 × 22/7 × 3 × 9 = 169.71 cm2
(ii) Volume = πr2h = 22/7 × 9 × 6 = 169.71 cm3
1 litre = 1000 cm3, so volume ≈ 0.17 litres
Segment = sector minus the triangle formed by the two radii and the chord.
(a) Copy and complete the following table for the relation y = 5/2 + x − 4x2
x: −2.0, −1.5, −1.0, −0.5, 0, 0.5, 1, 1.5, 2
y: −15.5, , , 1, 2.5, , , ,
(b) Using a scale of 2cm to 1 unit on the x axis and 2cm to 5 units on the y axis, draw the graph of the relation for −2.0 ≤ x ≤ 2.0
(c) What is the maximum value of y?
(d) From your graph obtain the roots of the equation 8x2 − 2x − 5 = 0.
(a) y values: x = −2.0 → −15.5; −1.5 → −8; −1.0 → −2.5; −0.5 → 1; 0 → 2.5; 0.5 → 2; 1 → −0.5; 1.5 → −5; 2 → −11.5
(b) Plot the points with the given scales and join them with a smooth n-shaped curve.
(c) Maximum y ≈ 2.6 (exact value 2.5625 at x = 0.125)
(d) 8x2 − 2x − 5 = 0 is −2 times (5/2 + x − 4x2) = 0, i.e. y = 0
Read where the curve cuts the x-axis: x ≈ −0.7 and x ≈ 0.9 (exact −0.68 and 0.93)
Dividing 8x² − 2x − 5 = 0 by −2 gives exactly the plotted relation set to zero.
(a) The distribution of junior workers in an illustration is as follows:
Clerks − 78
Drivers − 36
Typists − 44
Messengers − 52
Others − 30
Represent the above information by a pie chart.
(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.
Marks: 4, 5, 6, 7, 8, 9
No. of Students: 5, 8, 5, 6, 4, 2
Find the mean score to the nearest whole number.
(a) Total = 78 + 36 + 44 + 52 + 30 = 240 workers, so each worker is 360°/240 = 1.5°
Clerks 117°, Drivers 54°, Typists 66°, Messengers 78°, Others 45° (total 360°)
Draw the circle, measure the sectors with a protractor and label each one.
(b) Σfx = 20 + 40 + 30 + 42 + 32 + 18 = 182; Σf = 30
Mean = 182 ÷ 30 = 6.07 ≈ 6
Sector angle = (number in group ÷ total) × 360°.
Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the: (a) distance between Q and R; (b) bearing of R and Q.
(a) ∠QPR = 75° + (360° − 310°) = 125°
Cosine rule: QR2 = 502 + 902 − 2 × 50 × 90 × cos 125° = 10600 + 5162 = 15762
QR ≈ 125.5 km
(b) Sine rule: sin ∠PQR = 90 × sin 125° ÷ 125.5 = 0.5873, so ∠PQR ≈ 36.0°
Bearing of P from Q = 075° + 180° = 255°
Bearing of R from Q = 255° + 36° = 291°
Find the angle between the two bearings first, then use the cosine rule.
The table below shows the weekly profit in naira from a Mini-market.
Weekly profit (N): 1-10, 11-20, 21-30, 31-40, 40-50, 51-60
Frequency: 6, 6, 12, 11, 10, 5
(a) Draw the cumulative frequency graph of the data.
(b) From your graph, estimate the: (i) median; (ii) 80th percentile
(c) What is the modal weekly profit?
(a) Upper class boundaries 10.5, 20.5, 30.5, 40.5, 50.5, 60.5 with cumulative frequencies 6, 12, 24, 35, 45, 50. Plot these (start at (0.5, 0)) and draw a smooth ogive.
(b)(i) Median at the 25th value: about N31 (calculation: 30.5 + (1/11) × 10 = 31.4)
(ii) 80th percentile at the 40th value: about N45.5 (40.5 + (5/10) × 10)
(c) The modal class is N21–N30 (frequency 12); estimated mode = 20.5 + 6/(6 + 1) × 10 ≈ N29
Plot cumulative frequency against the upper class boundary, not the class mark.