60 questions. Try each one, then open the answer to check yourself and read the explanation.
Evaluate (101.5)2 − (100.5)2
Answer: D (202)
Use the difference of two squares: (101.5 − 100.5)(101.5 + 100.5) = 1 × 202 = 202.
Express the product of 0.06 and 0.09 in standard form
Answer: A (5.4 × 10−3)
0.06 × 0.09 = 0.0054, and 0.0054 = 5.4 × 10−3.
Simplify 36½ × 64−⅓ × 50
Answer: D (1½)
36½ = 6, 64−⅓ = 1/4 and 50 = 1, so the product is 6 × 1/4 × 1 = 1½.
Find the 8th term of the A.P. −3, −1, 1, ...
Answer: B (11)
First term a = −3 and common difference d = 2, so T8 = a + 7d = −3 + 14 = 11.
A group of students took a test and the following frequency table shows the scores:
Score: 0, 1, 2, 3, 4, 5
Frequency: 2, 3, 4, 2, 7, 2
Find the mode
Answer: C (4)
The mode is the score with the highest frequency. Score 4 occurs 7 times, more than any other.
A group of students took a test and the following frequency table shows the scores:
Score: 0, 1, 2, 3, 4, 5
Frequency: 2, 3, 4, 2, 7, 2
Find the mean score
Answer: D (2.75)
Σfx = 0 + 3 + 8 + 6 + 28 + 10 = 55 and Σf = 20, so the mean = 55 ÷ 20 = 2.75.
A group of students took a test and the following frequency table shows the scores:
Score: 0, 1, 2, 3, 4, 5
Frequency: 2, 3, 4, 2, 7, 2
The median score is
Answer: C (3)
There are 20 scores, so the median is the average of the 10th and 11th. Cumulative frequencies are 2, 5, 9, 11, …, so both the 10th and 11th scores are 3.
If cos 60° = 1/2, which of the following angles has a cosine of 1/2?
Answer: E (300°)
Cosine is positive in the first and fourth quadrants, so cos 300° = cos(360° − 60°) = cos 60° = 1/2.
In the diagram below, find the ratio |PQ|/|PR|
Answer: B (3/5)
The parallel lines cut the upper line into 5 equal parts, so they also cut PR into 5 equal parts (intercept theorem). PQ covers 3 of them, so |PQ|/|PR| = 3/5.
A ladder 9m long against a vertical wall, making an angle of 64° with the horizontal ground. Calculate, correct to one decimal place, how far the foot of the ladder is from the wall.
Answer: A (4.0m)
Distance from wall = 9 cos 64° = 9 × 0.4384 = 3.95 ≈ 4.0 m.
What is the difference in longitude between P (lat. 50°N, long. 50°W) and Q (lat. 50°N, long. 150°W)?
Answer: D (100°)
Both longitudes are west, so subtract: 150° − 50° = 100°.
The number 186 047 was corrected to 186 000. Which of the following can correctly describe the degree of approximation made? I. to the nearest hundred II. to the nearest thousand III. to 3 significant figures IV. to 4 significant figures
Answer: E (All of them.)
186 047 rounds to 186 000 to the nearest hundred (047 < 50), to the nearest thousand, to 3 s.f. (186|047) and to 4 s.f. (1860|47), so all four descriptions are correct.
Points P and Q are respectively 24m north and 7m east of point R. Calculate |PQ| in metres
Answer: C (25)
PRQ is a right angle at R, so |PQ| = √(24² + 7²) = √625 = 25 m.
Points P and Q are respectively 24m north and 7m east of point R. What is the bearing of Q from P to the nearest whole degree?
Answer: E (164°)
From P, Q lies to the south-east. tan θ = 7/24 gives θ ≈ 16° east of south, so the bearing is 180° − 16° = 164°.
If sin θ = 3/5 find tan θ for 0 < θ < 90°.
Answer: B (3/4)
Opposite = 3, hypotenuse = 5, so adjacent = √(25 − 9) = 4 and tan θ = 3/4.
Simplify (27⅓)2
Answer: C (9)
27⅓ = 3 (the cube root of 27), and 32 = 9.
A student is told to draw the graph of y = x2 + 4x − 6. He is then told to draw a linear graph on the same axis such that the intersection of the two graphs will give the solutions to the equation x2 + 4x − 7 = 0. What is the equation of the linear graph he needs to draw?
Answer: C (y = 1)
x2 + 4x − 7 = 0 is the same as x2 + 4x − 6 = 1, so the line needed is y = 1.
Calculate the value of x in the diagram below
Answer: A (18°)
Each 3x angle is vertically opposite an angle of the triangle, so the triangle's angles are 3x, 3x and 4x. Then 10x = 180°, so x = 18°.
Find the quadratic equation whose roots are x = −2 or x = 7
Answer: D (x2 − 5x − 14 = 0)
(x + 2)(x − 7) = 0 expands to x2 − 5x − 14 = 0.
A sales girl gave a change of N1.15 to a customer instead of N1.25. Calculate her percentage error.
Answer: C (8.0%)
Error = 1.25 − 1.15 = 0.10. Percentage error = 0.10/1.25 × 100 = 8.0%.
What is the probability of having an odd number in a single toss of a fair die?
Answer: C (1/2)
The odd numbers are 1, 3 and 5: 3 out of 6 outcomes, so the probability is 1/2.
A water tank is made with a right-angled trapezium-shaped vertical cross-section so that when it is placed on sloping ground the top of the tank is level. The dimensions are as given in the diagram. By what number will you multiply your answer in 22 above to determine the capacity of the tank in litres?
Answer: E (1000.)
1 m³ = 1000 litres, so multiply the volume in cubic metres by 1000.
Evaluate log106 + log1045 − log1027 without using logarithm tables.
Answer: B (1)
log 6 + log 45 − log 27 = log(6 × 45 ÷ 27) = log 10 = 1.
Find the roots of the equation 2x2 − 3x − 2 = 0
Answer: E (x = −½ or 2.)
2x2 − 3x − 2 = (2x + 1)(x − 2) = 0, so x = −½ or x = 2.
What value of k makes the given expression a perfect square? m2 − 8m + k
Answer: D (16)
For a perfect square, k = (half the coefficient of m)2 = (−4)2 = 16, giving (m − 4)2.
If log10q = 2.7078, what is q?
Answer: C (510.2)
q = antilog of 2.7078. The mantissa 0.7078 gives 5.102 and the characteristic 2 means multiply by 102: q = 510.2.
In the diagram below, O is the centre of the circle, BOC is a diameter and angle ADC is 37°. What is ACB?
Answer: D (53°)
∠ABC = ∠ADC = 37° (angles on the same arc AC). ∠BAC = 90° (angle in a semicircle), so ∠ACB = 180° − 90° − 37° = 53°.
Factorize 5y2 + 2ay − 3a2
Answer: E ((y + a)(5y − 3a).)
We need two terms multiplying to −15a2y2 and adding to 2ay: 5ay and −3ay. So 5y2 + 5ay − 3ay − 3a2 = (y + a)(5y − 3a).
Cos x is negative and sin x is negative. Which of the following is true of x?
Answer: C (180° < x < 270°)
Sine and cosine are both negative only in the third quadrant, 180° < x < 270°.
Simplify 0.63954 ÷ 0.003 giving your answer correct to two significant figures.
Answer: D (210)
0.63954 ÷ 0.003 = 213.18, which is 210 to two significant figures.
Which of the following is a correct method for constructing an angle of 60° at Q?
Answer: B (II only)
In II, an arc centred at the vertex cuts the line, and an arc of the same radius from that point cuts the first arc; joining gives an equilateral triangle, so 60°. Figure I bisects a right angle (45°).
If log10a = 4; what is a?
Answer: E (10,000.)
log10a = 4 means a = 104 = 10,000.
The student measured the length of a room and obtained the measurement of 3.99m. If the percentage error of his measurement was 5% and his own measurement was smaller than the length, what is the length of the room?
Answer: D (4.20m)
His reading is 5% less than the true length L, so 0.95L = 3.99 and L = 3.99 ÷ 0.95 = 4.20 m.
When an aeroplane is 800m above the ground, its angle of elevation from a point P on the ground is 30°. How far is the plane from P by line of sight?
Answer: D (1600m)
sin 30° = 800/d, so d = 800 ÷ 0.5 = 1600 m.
In the diagram below, LK//PQ, reflex angle KLM = 241° and ∠QPM = 89°. What is the value of ∠LMP?
Answer: A (30°)
Draw a line through M parallel to KL. ∠KLM = 360° − 241° = 119°, so the angle between ML and that line is 180° − 119° = 61°. The angle between MP and that line is 180° − 89° = 91° (co-interior with PQ). So ∠LMP = 91° − 61° = 30°.
Abidjan is 4° west of Accra and on the same circle of latitude. If the radius of this circle of latitude is 6370km, how far is Abidjan west of Accra, correct to the nearest kilometer? (Take π = 22/7)
Answer: B (445km)
Distance = 4/360 × 2 × 22/7 × 6370 = 444.9 ≈ 445 km (the radius given is already the radius of the latitude circle).
Find the product xy if x, 3/2, 6/7, y are in G.P.
Answer: C (9/7)
Common ratio r = (6/7) ÷ (3/2) = 4/7. Then x = (3/2) ÷ (4/7) = 21/8 and y = 6/7 × 4/7 = 24/49, so xy = 21/8 × 24/49 = 9/7.
Which of the regions P, Q, R, S and T satisfies the following inequalities: 0 < y < 1, y < x + 2, x < 0
Answer: E (T)
The region must be between the x-axis and y = 1, to the left of the y-axis, and below/right of the line y = x + 2. Only region T lies in all three.
A cone is made from a sector of a circle of radius 14cm and angle of 90°. What is the area of the curved surface of the cone? (Take π = 22/7)
Answer: D (154cm2)
The curved surface of the cone is the sector itself: 90/360 × 22/7 × 142 = 154 cm².
In the diagram below, O is the centre of the circle through points L, M and N, if ∠MLN = 74° and ∠MNL = 39°, calculate ∠LON.
Answer: E (134°)
∠LMN = 180° − 74° − 39° = 67°. The angle at the centre is twice the angle at the circumference on the same arc, so ∠LON = 2 × 67° = 134°.
A rope of length 18m is used to form a sector of a circle of radius 3.5m, on a school playing field. What is the size of the angle of the sector, correct to the nearest degree?
Answer: D (180°)
The rope forms two radii and the arc: arc = 18 − 2(3.5) = 11 m. The full circumference is 2 × 22/7 × 3.5 = 22 m, so the angle = 11/22 × 360° = 180°.
In the diagram, PR is the perpendicular from P to QS, PQ = 2cm, QR = 1cm and PR = RS. What is the size of the angle QPS?
Answer: D (75°)
In triangle PQR, sin ∠QPR = 1/2, so ∠QPR = 30°. Triangle PRS is right-angled and isosceles, so ∠RPS = 45°. ∠QPS = 30° + 45° = 75°.
Cotonou and Niamey are on the same line of longitude and Niamey is 7° north of Cotonou. If the radius of the earth is 6400km, how far is Niamey north of Cotonou along the line of longitude, correct to the nearest kilometer? (Take π = 22/7)
Answer: C (782km)
Distance along a great circle = 7/360 × 2 × 22/7 × 6400 = 782.2 ≈ 782 km.
If Q = (all perfect squares less than 30) and P = (all odd numbers from 1 to 10). Find Q∩P.
Answer: D ((1,9))
Q = {1, 4, 9, 16, 25} and P = {1, 3, 5, 7, 9}. The common elements are 1 and 9.
The histogram shows the number of candidates, in thousands, obtaining given ranges of marks in a State Examination. Find the total number of candidates that sat for the examination
Answer: E (270,000.)
Add the heights of all the bars (in thousands): roughly 5 + 10 + 20 + 57 + 75 + 50 + 30 + 7 + 7 + 7 ≈ 270, i.e. about 270,000 candidates.
The histogram shows the number of candidates, in thousands, obtaining given ranges of marks in a State Examination. How many candidates scored at most 30%?
Answer: D (35,000)
Add the bars for 0–10, 10–20 and 20–30: 5 + 10 + 20 = 35 thousand.
The surnames of 40 children in a class are arranged in alphabetical order. 16 of the surnames begin with O and 9 of the surnames begin with A. 14 of the letters of the alphabet do not appear as the first letter of a surname. What is the probability that the surname of a child picked at random from the class begins with either O or A?
Answer: A (5/8)
16 + 9 = 25 of the 40 surnames begin with O or A, so the probability is 25/40 = 5/8.
The surnames of 40 children in a class are arranged in alphabetical order. 16 of the surnames begin with O and 9 of the surnames begin with A. 14 of the letters of the alphabet do not appear as the first letter of a surname. If more than one surname begins with a letter besides A and O, how many surnames begin with that letter?
Answer: E (6)
26 − 14 = 12 letters are used; besides A and O that leaves 10 letters for 40 − 25 = 15 surnames. Each letter has at least one surname, so 9 letters have one each and the remaining letter has 15 − 9 = 6.
The number of items produced by a company over a five-year period is given below:
Year: 1978, 1979, 1980, 1981, 1982
Number produced: 4100, 2500, 1500, 1800, 9200
(i) Plot a bar chart for this information
(ii) What is the average production for the five-year period?
(i) Years on the horizontal axis, number produced on the vertical axis with a sensible scale (e.g. 1 cm to 1000); five separate bars of equal width with heights 4100, 2500, 1500, 1800 and 9200; title and labelled axes.
(ii) Total = 4100 + 2500 + 1500 + 1800 + 9200 = 19 100
Average = 19 100 ÷ 5 = 3820 items per year.
Bars must have equal widths and gaps; only the heights change.
In the diagram below, ∠PQR = ∠PSQ = 90°, [PS] = 9cm, [SR] = 16cm and [SQ] = xcm.
(a) Find the value of x using a trigonometric ratio.
(b) Calculate: (i) the size of QRS correct to the nearest degree, (ii) PQ.
(a) Let ∠QRS = θ. In △QSR: tan θ = x/16.
∠PQS = 90° − ∠SQR = θ, so in △PQS: tan θ = 9/x.
x/16 = 9/x ⇒ x2 = 144 ⇒ x = 12 cm.
(b)(i) tan ∠QRS = 12/16 = 0.75 ⇒ ∠QRS = 36.87° ≈ 37°.
(ii) PQ2 = PS2 + SQ2 = 81 + 144 = 225 ⇒ PQ = 15 cm (or PQ = PR sin 36.87° = 25 × 0.6 = 15 cm).
Spot that ∠PQS equals ∠QRS; that links the two triangles.
(a) Given that sin α = 0.3907, use tables to find the value of: (i) tan α (ii) cos α
(b) A ladder of length 4.5m leans against a vertical wall making an angle of 50° with the horizontal. If the bottom of a window is 4m above the ground, what is the distance between the top of the ladder and the bottom of the window? (Answer correct to the nearest cm).
(a) sin α = 0.3907 ⇒ α = 23°.
(i) tan 23° = 0.4245 (ii) cos 23° = 0.9205.
(b) Height of top of ladder = 4.5 sin 50° = 4.5 × 0.7660 = 3.447 m.
Distance to bottom of window = 4 − 3.447 = 0.553 m = 55 cm (to the nearest cm).
Find the angle from the sine table first, then read the other ratios.
In a certain school, the lesson periods for each week are as itemised below: English 10, Mathematics 7, Biology 3, Statistics 4, Igbo 3, others 9. Draw a pie chart to illustrate this information.
Total periods = 10 + 7 + 3 + 4 + 3 + 9 = 36.
Angles (× 360°/36 = 10° per period): English 100°, Mathematics 70°, Biology 30°, Statistics 40°, Igbo 30°, Others 90° (total 360°).
Draw a circle, measure the sectors with a protractor and label each sector.
Each period is worth 360° ÷ 36 = 10°.
A sector of a circle of radius 7cm subtending an angle of 270° at the centre of the circle, is used to form a cone. (a) Find the base radius of the cone, (b) Calculate the area of the base of the cone correct to the nearest square centimetre. (Take π = 22/7)
(a) Arc length = 270/360 × 2 × 22/7 × 7 = 33 cm.
This arc becomes the base circumference: 2πr = 33 ⇒ r = 33 × 7/44 = 5.25 cm.
(b) Base area = πr2 = 22/7 × 5.252 = 86.625 ≈ 87 cm².
The arc of the sector becomes the circumference of the cone's base.
(a) The subsets A, B and C of a universal set are defined as follows: A = (m,a,p,e) B = (a,e,i,o,u) C = (l,m,n,o,p,q,r,s,t,u)
List the elements of the following sets: (i) A∪B (ii) A∪C (iii) A∪(B∩C)
(b) Out of the 400 students in the final year in a Senior Secondary School, 300 are offering Biology and 190 are offering Chemistry. (i) How many students are offering both Biology and Chemistry, if only 70 students are offering neither Biology nor Chemistry? (ii) How many students are offering at least one of Biology or Chemistry?
(a)(i) A∪B = {a, e, i, m, o, p, u}
(ii) A∪C = {a, e, l, m, n, o, p, q, r, s, t, u}
(iii) B∩C = {o, u}, so A∪(B∩C) = {a, e, m, o, p, u}
(b) Let x offer both. n(B∪C) = 400 − 70 = 330.
(300 − x) + x + (190 − x) = 330 ⇒ 490 − x = 330 ⇒ x = 160.
(i) 160 students offer both.
(ii) 330 students offer at least one of the two.
Draw a Venn diagram; the 70 outside both circles are taken from the 400 first.
Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time.
−x + 5y ≤ 10
3x − 4y ≤ 8
and x > −1
Draw the boundary lines: −x + 5y = 10 (through (−10, 0) and (0, 2)); 3x − 4y = 8 (through (0, −2) and (8/3, 0)); x = −1 (broken line, since the inequality is strict).
−x + 5y ≤ 10: region below the first line (the origin satisfies it).
3x − 4y ≤ 8: region above the second line (the origin satisfies it).
x > −1: region to the right of x = −1.
The required region is the triangle with vertices (−1, 1.8), (−1, −2.75) and about (8.3, 3.5) where the two slanting lines meet; shade it (or shade the unwanted regions, and say which convention is used).
Test the origin in each inequality to decide which side to keep.
The table below is for the relation y = 2 + x − x2
x: −2, −1.5, −1, −0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3
y: −4, −1.75, 0, 1.25, 2, 2.25, 2, 1.25, 0, −1.75, −4
(a) Using a scale of 2cm to 1 unit on each axis, draw the graph of the relation in the interval −2 ≤ x ≤ 3.
(b) From your graph, find the greatest value of y and the value of x for which this occurs,
(c) Using the same scale and axis, draw the graph of y = 1 − x. (d) Use your graph to solve the equation 1 + 2x − x2 = 0.
(a) Plot the 11 points and join with a smooth curve (an upside-down parabola).
(b) Greatest value y = 2.25, at x = 0.5.
(c) Straight line through (0, 1), (1, 0) and (3, −2).
(d) 2 + x − x2 = 1 − x is the same as 1 + 2x − x2 = 0, so the solutions are the x-values where the curve and the line cross: x ≈ −0.4 and x ≈ 2.4.
Subtract the line's equation from the curve's to see why their crossing points solve the equation.
Two towns K and Q are on the parallel of latitude 46°N. The longitude of town K is 130°W and that of town Q is 103°. A third town P also on latitude 46°N is on longitude 23°E. Calculate: (i) the length of the parallel of latitude 46°N, to the nearest 100km, (ii) the distance K and Q, correct to the nearest 100km, (iii) the distance between Q and P measured along the parallel of latitude, to the nearest 10m. (Take π = 3.142, radius of the earth = 6400km).
Radius of the parallel = R cos 46° = 6400 × 0.6947.
(i) Length = 2 × 3.142 × 6400 × 0.6947 = 27 938 km ≈ 27 900 km.
(ii) K (130°W) and Q (103°W): difference 27°. Distance = 27/360 × 27 938 = 2095 km ≈ 2100 km.
(iii) Q (103°W) and P (23°E): difference 103° + 23° = 126°. Distance = 126/360 × 27 938 = 9778 km ≈ 9780 km.
Distances along a parallel use the radius R cos(latitude); add longitudes on opposite sides of Greenwich.
(a) Prove that the sum of the angles of a triangle is two right angles. (b) In a triangle LMN, the side MN is produced to P and the bisector of ∠LNP meets ML produced at Q. If ∠LMN = 46° and ∠MLN = 80°, calculate ∠LQN, stating clearly your reasons.
(a) Given △ABC. Produce BC to X and draw CY parallel to BA.
∠ACY = ∠BAC (alternate angles); ∠YCX = ∠ABC (corresponding angles).
∠ACB + ∠ACY + ∠YCX = 180° (angles on a straight line), so ∠C + ∠A + ∠B = 180° = two right angles.
(b) ∠LNM = 180° − 46° − 80° = 54° (angle sum of a triangle).
∠LNP = 46° + 80° = 126° (exterior angle = sum of interior opposite angles).
∠LNQ = 126° ÷ 2 = 63° (NQ bisects ∠LNP).
In △MNQ: ∠MNQ = 54° + 63° = 117°, so ∠LQN = ∠MQN = 180° − 46° − 117° = 17°.
Find the exterior angle at N first, then halve it.
Using a ruler and a pair of compasses only. (a) Construct: (i) a △ABC such that |AB| = 5cm, |AC| = 7.5cm and ∠CAB = 120°. (ii) the locus l1 of points equidistant from A and B. (iii) the locus l2 of points equidistant from |AB| and AC, which passes through triangle ABC. (b) Label the point P where l1 and l2 intersect. (c) Measure |CP|.
(a)(i) Draw AB = 5 cm. At A construct 120° (two 60° arcs), mark C on that arm with AC = 7.5 cm, join BC.
(ii) l1 is the perpendicular bisector of AB (arcs from A and B above and below, join the crossings).
(iii) l2 is the bisector of ∠CAB (makes 60° with AB).
(b) Mark P where the two loci meet.
(c) |CP| ≈ 6.6 cm (accept about 6.5–6.8 cm).
Construction arcs must be left visible; the equidistant-from-two-lines locus is an angle bisector.
The table below shows the frequency distribution of the marks of 800 candidates in an examination.
Marks%: 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99
Frequency: 10, 40, 80, 140, 170, 130, 100, 70, 40, 20
(a)(i) Construct a cumulative frequency table, (ii) Draw the ogive: (iii) Use your ogive to determine the 50th percentile: (b) The candidates that scored less than 25% are to be withdrawn from the institution, while those that scored more than 75% are to be awarded scholarship. Estimate the number of candidates that will be retained, but who will not enjoy the award.
(a)(i) Upper class boundaries 9.5, 19.5, …, 99.5 with cumulative frequencies 10, 50, 130, 270, 440, 570, 670, 740, 780, 800.
(ii) Plot cumulative frequency against upper class boundary and join with a smooth S-shaped curve.
(iii) 50th percentile = median = the 400th candidate: about 39.5 + (130/170) × 10 ≈ 47 marks.
(b) From the ogive: about 94 candidates scored below 25%; about 709 scored up to 75%, so about 91 scored above 75%.
Retained without award ≈ 709 − 94 ≈ 615 candidates (accept roughly 600–630 from the graph).
Plot against upper class boundaries, not class midpoints.