59 questions. Try each one, then open the answer to check yourself and read the explanation.
Which of the following correctly expresses 48 as a product of prime factors?
Answer: D (2 × 2 × 2 × 2 × 3.)
Only D uses prime numbers alone: 2 × 2 × 2 × 2 × 3 = 48. 4 and 8 are not prime.
Evaluate (20three)2 − (11three)2 in base three
Answer: C (202)
20three = 6 and 11three = 4. 36 − 16 = 20 = 2×9 + 0×3 + 2 = 202three.
Evaluate log105 + log1020
Answer: A (2)
log10(5 × 20) = log10100 = 2.
If N varies directly as M and N = 8 when M = 20, find M when N = 7.
Answer: C (17½)
N = kM with k = 8/20 = 2/5. Then 7 = 2M/5, so M = 35/2 = 17½.
In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there altogether?
Answer: A (81)
4 parts = 36, so 1 part = 9 and the total of 9 parts = 81.
A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate.
Answer: D (5%)
Error = 10; as a percentage of the actual fare, 10/200 × 100% = 5%.
Evaluate 5 2/5 × (2/3)2 ÷ (1 1/2)−1
Answer: C (3 3/5)
27/5 × 4/9 = 12/5, and (3/2)−1 = 2/3. 12/5 ÷ 2/3 = 12/5 × 3/2 = 18/5 = 3 3/5.
The nth term of a sequence is 22n−1. Which term of the sequence is 29?
Answer: C (5th)
2n − 1 = 9 gives n = 5, the 5th term.
Evaluate 1/2 + 3/4 of 2/5 ÷ 1 3/5
Answer: B (11/16)
'Of' first: 3/4 × 2/5 = 3/10. Then ÷ 8/5: 3/10 × 5/8 = 3/16. Finally 1/2 + 3/16 = 11/16.
A man is four times as old as his son. The difference between their ages is 36 years. Find the sum of their ages.
Answer: C (60 years)
4x − x = 36 gives x = 12. The father is 48, so the sum is 60 years.
Given that y = px + q and y = 5 when x = 3, while y = 4 when x = 2, find the values of p and q.
Answer: B (p = 1, q = 2)
3p + q = 5 and 2p + q = 4. Subtracting gives p = 1, then q = 2.
Evaluate (x2 + x − 2)/(2x2 + x − 3) when x = −1
Answer: D (1)
Numerator: 1 − 1 − 2 = −2. Denominator: 2 − 1 − 3 = −2. −2 ÷ −2 = 1.
Factorize 6x2 + 7x − 20
Answer: D ((3x − 4)(2x + 5))
(3x − 4)(2x + 5) = 6x2 + 15x − 8x − 20 = 6x2 + 7x − 20.
Simplify (2x − 1)/3 − (x + 3)/2
Answer: C ((x − 11)/6)
Over 6: 2(2x − 1) − 3(x + 3) = 4x − 2 − 3x − 9 = x − 11.
If (y − 3)/2 < (2y − 1)/3, which of the following is true?
Answer: C (y > −7)
Multiply by 6: 3y − 9 < 4y − 2, so −7 < y, i.e. y > −7.
If (4m + 3n)/(4m − 3n) = 5/2, find the ratio m:n
Answer: A (7 : 4)
8m + 6n = 20m − 15n, so 21n = 12m and m : n = 21 : 12 = 7 : 4.
The number line represents
Answer: C (−7 < x ≤ −1)
The circle at −7 is open (not included) and the circle at −1 is shaded (included): −7 < x ≤ −1.
If 2x2 + Kx − 14 = (x + 2)(2x − 7), find the value of K.
Answer: A (−3)
(x + 2)(2x − 7) = 2x2 − 3x − 14, so K = −3.
Which of the following is not a quadratic expression?
Answer: D (y = 5(x − 1))
5(x − 1) = 5x − 5 has no x2 term, so it is linear.
The graph of the curve y = 2x2 − 5x − 1 and a straight line PQ were drawn to solve the equation 2x2 − 5x + 2 = 0. What is the equation of the line PQ?
Answer: D (y = −3)
2x2 − 5x + 2 = 0 is the same as 2x2 − 5x − 1 = −3, so the line is y = −3.
Subtract (−y + 3x + 5z) from (4y − x − 2z).
Answer: A (5y − 4x − 7z)
(4y − x − 2z) − (−y + 3x + 5z) = 4y + y − x − 3x − 2z − 5z = 5y − 4x − 7z.
If y ∝ 1/√x and x = 16 when y = 2, find x when y = 24
Answer: A (1/9)
y = k/√x: 2 = k/4, so k = 8. Then 24 = 8/√x gives √x = 1/3 and x = 1/9.
If 2x : (x + 1) = 3:2, what is the value of x?
Answer: D (3)
2x/(x + 1) = 3/2, so 4x = 3x + 3 and x = 3.
The bearing S40°E is the same as
Answer: D (140°)
From south (180°) turn 40° towards east: 180 − 40 = 140°.
Given that sin p = 5/13, where p is acute, find the value of cos p − tan p.
Answer: A (79/156)
The adjacent side is 12, so cos p = 12/13 and tan p = 5/12. 12/13 − 5/12 = (144 − 65)/156 = 79/156.
Three observation posts P, Q and R are such that Q is due east of P and R is due north of Q. If |PQ| = 5km and |PR| = 10 km, find |QR|.
Answer: D (8.7 km)
The angle at Q is 90°, so QR = √(102 − 52) = √75 ≈ 8.7 km.
Express 25° 45' in decimal (Hint 1° = 60')
Answer: A (25.75°)
45' = 45/60 of a degree = 0.75°, so 25°45' = 25.75°.
A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that the two balls are red?
Answer: A (5/33)
There are 12 balls. P(red, red) = 5/12 × 4/11 = 20/132 = 5/33.
A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that one is green the other is blue?
Answer: A (2/11)
Green then blue or blue then green: 3/12 × 4/11 + 4/12 × 3/11 = 24/132 = 2/11.
If the probability that an event will occur is p and the probability that it will not occur is q, which of the following is true?
Answer: C (p + q = 1)
An event either happens or it does not, so the two probabilities add up to 1.
The histogram shows the distribution of a group of students according to their ages. What is the range of their ages?
Answer: B (20 years)
The ages run from 10.5 to 30.5 years on the histogram, so the range is 30.5 − 10.5 = 20 years.
Which of the following is not the size of an exterior angle of a regular polygon?
Answer: D (21°)
An exterior angle must divide 360° exactly. 360 ÷ 21 is not a whole number, so 21° is impossible.
In the diagram, PQR is a triangle. |PQ| = |PR|, |PS| = |SQ| and ∠PRS = 50°. What is the size of ∠PSQ?
Answer: B (80°)
PQ = PR, so ∠PQR = 50°. PS = SQ, so ∠SPQ = 50° too, and ∠PSQ = 180 − 50 − 50 = 80°.
PQRS is a rhombus of side 16 cm. The diagonal |QS| = 20 cm. Calculate, correct to the nearest degree, ∠PQR.
Answer: D (103°)
Half of QS is 10 cm, so half of ∠QPS has sine 10/16, giving ∠QPS ≈ 77.4°. Adjacent angles of a rhombus add to 180°, so ∠PQR ≈ 103°.
In the diagram, PQ is a diameter of the circle and ∠PRS = 58°. Find ∠STQ.
Answer: B (32°)
∠PQS = ∠PRS = 58° (same arc PS) and ∠PSQ = 90° (semicircle), so ∠SPQ = 32°. ∠STQ = ∠SPQ (same arc SQ) = 32°.
Which of the following pairs of triangles is/are similar?
Answer: B (III only)
In III every side of the larger triangle is 4 times the smaller (5.2, 7.2, 8.4 against 1.3, 1.8, 2.1). In I, 4.2 is not twice 2.6, and the sides in II are not in one ratio.
In the diagram, O is the centre of the circle. Reflex angle XOY = 210° and the length of the minor arc is 5.5m. Find, correct to the nearest metre the length of the major arc.
Answer: A (8 m)
The minor arc has angle 150°. Arc lengths are in proportion to their angles: 5.5 × 210/150 = 7.7 ≈ 8 m.
A right pyramid is on a square base of side 4 cm. The slanting side of the pyramid is 2√3 cm. Calculate the volume of the pyramid.
Answer: B (10 2/3 cm3)
Half the base diagonal = 2√2, so height = √(12 − 8) = 2 cm. V = ⅓ × 16 × 2 = 10 2/3 cm3.
The height of a right circular cone is 4 cm. The radius of its base is 3 cm. Find its curved surface area.
Answer: B (15π cm2)
Slant height = √(32 + 42) = 5 cm. Curved surface = πrl = π × 3 × 5 = 15π cm2.
A chord of length 30 cm is 8 cm away from the centre of the circle. What is the radius of the circle?
Answer: B (17 cm)
r = √(152 + 82) = √289 = 17 cm.
A sector of a circle radius 14 cm subtends an angle 135° at the centre of the circle. What is the perimeter of the sector? [Take π = 22/7].
Answer: B (61 cm)
Arc = 135/360 × 2 × 22/7 × 14 = 33 cm. Perimeter = 33 + 14 + 14 = 61 cm.
In the diagram, MN||PQ, |LM| = 3 cm and |LP| = 4cm. If the area of ΔLMN is 18 cm2, find the area of quadrilateral MPQN.
Answer: C (14 cm2)
ΔLMN and ΔLPQ are similar with sides in ratio 3 : 4, so their areas are in ratio 9 : 16. Area LPQ = 32 cm2, and MPQN = 32 − 18 = 14 cm2.
The diagram shows a triangular prism of length 7 cm. The right-angled triangle PQR is a cross-section of the prism. |PR| = 5 cm and |RQ| = 3 cm. What is the area of the cross-section?
Answer: B (6 cm2)
PR is the hypotenuse, so PQ = √(25 − 9) = 4 cm. Area = ½ × 4 × 3 = 6 cm2.
The diagram shows a triangular prism of length 7 cm. The right-angled triangle PQR is a cross-section of the prism. |PR| = 5 cm and |RQ| = 3 cm. What is the volume of the prism?
Answer: B (42 cm3)
Volume = cross-section × length = 6 × 7 = 42 cm3.
A solid cylinder of radius 7 cm is 10 cm long. Find its total surface area.
Answer: D (238π cm2)
2πrh + 2πr2 = 2π × 7 × 10 + 2π × 49 = 140π + 98π = 238π cm2.
PQRS is a trapezium in which |PS| = 9 cm, |QR| = 15 cm, |PQ| = 2√3, ∠PQR = 90° and ∠QRS = 30°. Calculate the area of the trapezium.
Answer: A (24√3 cm2)
PQ is the height. Area = ½(9 + 15) × 2√3 = 24√3 cm2.
Given that log102 = 0.3010 and log103 = 0.4771, calculate without using mathematical tables or calculator, the value of: (a) log1054, (b) log100.24.
(a) 54 = 2 × 33: log 54 = log 2 + 3 log 3 = 0.3010 + 1.4313 = 1.7323.
(b) 0.24 = 23 × 3 ÷ 100: log 0.24 = 3(0.3010) + 0.4771 − 2 = 1.3801 − 2 = −0.6199 (= 1̄.3801).
Write each number as a product of powers of 2, 3 and 10.
(a) Simplify (1/35n) × 9n−1 × 27n+1
(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years time, the ratio of their ages will be 7:2. Find their present ages.
(a) Write everything as powers of 3: 3−5n × 32n−2 × 33n+3 = 3−5n + 2n − 2 + 3n + 3 = 31 = 3.
(b) w + d = 46 and (w + 4)/(d + 4) = 7/2, so 2w + 8 = 7d + 28, i.e. 2w − 7d = 20.
Substitute w = 46 − d: 92 − 9d = 20, d = 8 and w = 38.
The woman is 38 and the daughter is 8.
Change 9 and 27 to powers of 3 and add the indices.
The sides of a rectangular floor are xm and (x + 7)m. The diagonal is (x + 8)m, calculate, in metres: (a) the value of x; (b) the area of the floor.
(a) Pythagoras: x2 + (x + 7)2 = (x + 8)2, so 2x2 + 14x + 49 = x2 + 16x + 64.
x2 − 2x − 15 = 0, (x − 5)(x + 3) = 0, x = 5 (a length is positive).
(b) Area = 5 × 12 = 60 m2.
The diagonal is the hypotenuse of the right-angled triangle formed by two sides.
The diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992. (a) How many vehicles were produced in February? (b) What fraction of the vehicles manufactured in February were cars? (c) How many buses were produced altogether from January to March, 1992? (d) What is the ratio in lowest term of the number of lorries produced in February to that in March?
Readings from the chart (car, lorry, bus): January 10, 15, 20; February 40, 25, 30; March 5, 35, 50.
(a) February: 40 + 25 + 30 = 95 vehicles.
(b) Cars in February: 40/95 = 8/19.
(c) Buses: 20 + 30 + 50 = 100.
(d) Lorries February : March = 25 : 35 = 5 : 7.
Read each bar against the vertical scale, then add or compare as asked.
(a) In the diagram, MN||ST, NP||QT and ∠STQ = 70°. Find x.
(b) In the diagram, AC is a straight line, |BC| = |BD|, ∠BCD = 50° and ∠BAD = 55°. Find ∠BDA.
(a) MN ∥ ST and NP ∥ QT, but NP points the opposite way to TQ, so ∠MNP and ∠STQ add up to 180°.
x = 180° − 70° = 110°.
(b) BC = BD, so ∠BDC = ∠BCD = 50° and ∠CBD = 180 − 100 = 80°.
∠CBD is an exterior angle of triangle ABD: 80° = 55° + ∠BDA, so ∠BDA = 25°.
An exterior angle of a triangle equals the sum of the two opposite interior angles.
(a) Draw the table of values for the relation y = x2 for the interval −3 ≤ x ≤ 4. (b) Using a scale of 2 cm to 1 unit on the x-axis and 2cm to 2 units on the y-axis, draw the graphs of: (i) y = x2 (ii) y = 2x + 3 for −3 ≤ x ≤ 4. (c) Use your graph to find: (i) the roots of the equation x2 = 2x + 3; (ii) the gradient of y = x2 at x = −2.
(a) x: −3, −2, −1, 0, 1, 2, 3, 4; y: 9, 4, 1, 0, 1, 4, 9, 16.
(b) Plot the parabola and the straight line through (0, 3) and (3, 9).
(c)(i) The graphs meet at x = −1 and x = 3.
(ii) Draw the tangent at (−2, 4); its gradient is about −4.
To find a gradient from a graph, draw a tangent and use rise ÷ run.
The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows:
50 99 81 86 69 85 93 63 92 65
77 74 76 71 90 74 81 94 67 75
95 81 68 105 99 68 75 75 76 73
79 74 80 69 74 62 74 80 79 68
79 75 75 71 83 75 80 85 81 82
(a) Construct a frequency distribution table, using class intervals 45–54, 55–64, etc. (b) Draw the histogram for the distribution. (c) Use your histogram to estimate the mode. (d) Calculate the mean number of births.
(a) 45–54: 1; 55–64: 2; 65–74: 15; 75–84: 21; 85–94: 7; 95–104: 3; 105–114: 1 (total 50).
(b) Bars on class boundaries 44.5, 54.5, … 114.5 with heights equal to the frequencies.
(c) In the tallest bar (74.5–84.5) join its top corners to the neighbouring bars' tops; the lines cross at about 77.5, so the mode ≈ 77.5.
(d) Class marks 49.5, 59.5, … 109.5: Σfx = 49.5 + 119 + 1042.5 + 1669.5 + 626.5 + 298.5 + 109.5 = 3915. Mean = 3915/50 = 78.3.
Tally the data one class at a time and check the frequencies add to 50.
(a) In a market survey, 100 traders sell fruits. 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three fruits. Each of the 100 traders sells at least one of the three fruits. (i) Represent the information in a Venn diagram. (ii) Find the number that sell oranges and mangoes only.
(b) Find the value of x for which 312four + 52x = 96ten.
(a)(i) Let y sell oranges and mangoes only. Regions: all three 10; apples and oranges only 4; apples and mangoes only 5; apples only 40 − 19 = 21; oranges only 46 − 14 − y = 32 − y; mangoes only 50 − 15 − y = 35 − y.
(ii) Total: 21 + 4 + 5 + 10 + y + (32 − y) + (35 − y) = 100, so 107 − y = 100 and y = 7.
(b) 312four = 48 + 4 + 2 = 54, so 52x = 42: 5x + 2 = 42 and x = 8.
Fill the Venn diagram from the centre (all three) outwards.
Using a ruler and a pair of compasses only, (a) Construct a quadrilateral MNOP with |OP| = 11 cm, |PM| = 6cm, |NO| = 5 cm, |MN| = 8 cm and |MO| = 9cm. (b) Measure ∠MPO. (c) Locate the point Y which is equidistant from the straight lines ON and OP and also equidistant from points M and N. (d) Draw YP and find the perimeter of ΔOYP.
(a) Draw OP = 11 cm. Arcs: centre P radius 6 cm and centre O radius 9 cm meet at M. Then arcs centre M radius 8 cm and centre O radius 5 cm meet at N (same side as M). Join MN, NO, PM.
(b) ∠MPO ≈ 55° (cos P = (36 + 121 − 81)/132).
(c) Bisect ∠NOP and construct the perpendicular bisector of MN; Y is where they cross.
(d) OY ≈ 5.2 cm, YP ≈ 8.4 cm, OP = 11 cm: perimeter ≈ 24.6 cm (accept 24–25 cm).
Equidistant from two lines means on the angle bisector; equidistant from two points means on the perpendicular bisector.
(a) The probabilities that three boys pass an examination are 2/3, 5/8 and 3/4 respectively. Find the probability that: (i) all the boys pass; (ii) none of the boys pass; (iii) only two of the boys pass.
(b) A shopkeeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% off the marked price. Calculate the actual percentage profit.
(a)(i) 2/3 × 5/8 × 3/4 = 30/96 = 5/16.
(ii) 1/3 × 3/8 × 1/4 = 3/96 = 1/32.
(iii) Exactly two pass: 2/3 × 5/8 × 1/4 + 2/3 × 3/8 × 3/4 + 1/3 × 5/8 × 3/4 = 10/96 + 18/96 + 15/96 = 43/96.
(b) Cost price = 36,000 ÷ 1.2 = N30,000. Selling price = 95% of 36,000 = N34,200.
Profit = N4,200; actual percentage profit = 4,200/30,000 × 100% = 14%.
For 'only two pass' add the three cases where exactly one boy fails.
(a) The roots of the equation 2x2 + (p + 1)x + q = 0, are 1 and 3, where p and q are constants. Find the values of p and q.
(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth's surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800 km above the surface of the earth. [Take the radius of the earth as 6,400 km]
(a) The equation is 2(x − 1)(x − 3) = 2x2 − 8x + 6 = 0.
Compare: p + 1 = −8, so p = −9; q = 6.
(b) W = k/d2. On the surface d = 6400 km; at 800 km up, d = 7200 km.
W = 80 × (6400/7200)2 = 80 × 64/81 ≈ 63.2, so about 63 kg.
Distance is measured from the centre of the earth, so add the radius to the height.
In the diagram, three points A, B and C are on the same horizontal ground. B is 15 m from A, on a bearing of 053°. C is 18m from B on a bearing of 161°. A vertical pole with top T is erected at B such that angle ATB = 58°. Calculate, correct to three significant figures, (a) the length of AC; (b) the bearing of C from A; (c) the height of the pole BT.
(a) The bearing of A from B is 233°, so ∠ABC = 233 − 161 = 72°.
AC2 = 152 + 182 − 2(15)(18) cos 72° = 549 − 166.9 = 382.1, so AC ≈ 19.5 m.
(b) sin ∠BAC = 18 sin 72°/19.55 = 0.8758, so ∠BAC ≈ 61.1°. Bearing of C from A = 53 + 61.1 ≈ 114°.
(c) In right-angled triangle ABT: tan 58° = AB/BT, so BT = 15/tan 58° = 15/1.600 ≈ 9.37 m.
Find the angle at B from the two bearings first, then use the cosine rule.
(a) A boy blew his rubber balloon to a spherical shape. The balloon burst when its diameter was 15 cm. Calculate, correct to the nearest whole number, the volume of air in the balloon at the point of bursting. [Take π = 22/7].
(b) A point X is on latitude 28°N and longitude 105°W. Y is another point on the same latitude as X but on longitude 35°E. (i) Calculate, correct to 3 significant figures, the distance between X and Y along latitude 28°N. (ii) How far is X from the equator? [Take π = 22/7, and the radius of the Earth as 6,400 km]
(a) r = 7.5 cm: V = 4/3 × 22/7 × 7.53 ≈ 1768 cm3.
(b)(i) Difference in longitude = 105 + 35 = 140°. Radius of the latitude circle = 6400 cos 28°.
Distance = 140/360 × 2 × 22/7 × 6400 × cos 28° ≈ 15644 × 0.8829 ≈ 13,800 km.
(ii) Along the meridian: 28/360 × 2 × 22/7 × 6400 ≈ 3,130 km.
One point is west and the other east of Greenwich, so add the longitudes.