55 questions. Try each one, then open the answer to check yourself and read the explanation.
A student approximated 1.2476 to two significant figures instead of two decimal places. Find the percentage error
Answer: C (4%)
To 2 s.f. it is 1.2; to 2 d.p. it is 1.25. Error = 0.05, and 0.05/1.25 × 100% = 4%.
In the diagram, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find ∠QPR
Answer: B (52°)
RP = RS gives ∠RSP = 32°, so the exterior angle PRQ = 64°. PQ = PR gives ∠PQR = 64°, so ∠QPR = 180° − 128° = 52°.
In the diagram, P, Q, R and S are points on the circle with centre O and reflex ∠POR = 220°. Find the angle marked x
Answer: D (110°)
The angle at S stands on the arc subtended by the reflex angle at the centre, so x = ½ × 220° = 110°.
Write 6.02517 to three significant figures
Answer: B (6.03)
The first three significant figures are 6.02 and the next digit is 5, so round up to 6.03.
Given that x : 3/4 = 1/2 : 9/8, find the value of x
Answer: B (1/3)
x ÷ ¾ = ½ ÷ 9/8 = 4/9, so x = ¾ × 4/9 = ⅓.
Given that (x − 2) is a factor of 8 − 2qx − x2, find the value of q
Answer: B (1)
By the factor theorem, x = 2 gives 0: 8 − 4q − 4 = 0, so q = 1.
A point Q is 6 km due north of P and R is 8 km due east of P. A bus travels on a straight route from Q to R at an average speed of 40 km/h. How long does it take to travel at that speed?
Answer: B (15 minutes)
QR = √(6² + 8²) = 10 km. Time = 10/40 h = ¼ h = 15 minutes.
Given that Sn = 2n2 − 3n, find S6
Answer: A (54)
S6 = 2(36) − 18 = 54.
If 110x = 1234 − 1112, find the value of x
Answer: C (4)
1234 = 27 and 1112 = 7, so x² + x = 20, giving (x + 5)(x − 4) = 0 and x = 4.
A generator costs ₦82,000.00. It depreciates by 10% the first year and 5% the second year. Find the value after the second year
Answer: D (₦70,110.00)
82 000 × 0.9 = 73 800 after one year, then 73 800 × 0.95 = ₦70,110.
A right rectangular prism has a height 9 cm. The length of the base is x cm and it is 7 cm more than the width of the base. Write a function of the volume (V) of the prism in terms of the length of the base.
Answer: A (V(x) = 9x(x − 7))
The width is x − 7, so V = length × width × height = 9x(x − 7).
The result of increasing a quantity x by 400% is 60. Find the value of x
Answer: D (12)
Increasing by 400% gives 500% of x: 5x = 60, so x = 12.
Factorize completely: (h2 − k2) − p(h + k)
Answer: B ((h + k)(h − k − p))
h² − k² = (h + k)(h − k), so the expression is (h + k)[(h − k) − p] = (h + k)(h − k − p).
It takes y years for ₦87,300.00 to amount to ₦99,522.00 at 7% per annum simple interest. Find the value of y
Answer: A (2)
Interest = 12 222. Each year earns 87 300 × 0.07 = 6111, so y = 12 222 ÷ 6111 = 2.
A wire is bent to form a square of side 22 cm. If the wire is re-bent to form a circle, find the radius [Take π = 22/7]
Answer: B (14 cm)
Wire length = 4 × 22 = 88 cm = 2πr, so r = 88 × 7 ÷ 44 = 14 cm.
A pilot flies from Lisbon to Kuala Lumpur on a bearing of 163°. If the pilot flew from Kuala Lumpur to Lisbon on the same route, find the bearing
Answer: A (343°)
The return bearing is 163° + 180° = 343°.
The average speed of a car is 63 km/h. Express the speed in ms−1.
Answer: B (17.5 ms−1)
63 km/h = 63 000 m ÷ 3600 s = 17.5 m/s.
A cuboid of length 25 cm and breadth 8 cm holds exactly 1 litre of water. Find the height of the cuboid
Answer: D (5 cm)
1 litre = 1000 cm³, so height = 1000 ÷ (25 × 8) = 5 cm.
Given that cos p = 7/25, 0° < p < 90°, find the value of sin p
Answer: B (24/25)
The opposite side is √(25² − 7²) = 24, so sin p = 24/25.
A bag contains 20 green and a number of red identical balls. The probability of selecting at random a red ball is 3/5. How many balls are there in the bag?
Answer: A (50)
Green balls make up 2/5 of the bag, so 20 = 2/5 of the total and the total is 50.
If 2−n = x, find 2n
Answer: B (1/x)
2n = 1/2−n = 1/x.
A variable Z varies directly as X and inversely as Y. If Z = 12 when X = 4 and Y = 5, find Z when X = 6 and Y = 45.
Answer: B (2)
Z = kX/Y: 12 = 4k/5 gives k = 15. Then Z = 15 × 6 ÷ 45 = 2.
The interior angle of a regular polygon is the same as its exterior angle. Find the sum of the interior angles.
Answer: B (360°)
Interior = exterior = 90°, so the polygon has 360 ÷ 90 = 4 sides and its angles add to (4 − 2) × 180° = 360°.
Given that V = ⅓πr2h, make h the subject
Answer: D (h = 3V/(πr2))
Multiply by 3: 3V = πr²h, then divide by πr²: h = 3V/(πr²).
For what values of m is the expression (m + 1)/(3m2 − 2m − 5) not defined?
Answer: C (−1 and 5/3)
The denominator 3m² − 2m − 5 = (3m − 5)(m + 1) is zero when m = 5/3 or m = −1.
Evaluate 2log4 + 3log5 − log2
Answer: C (3)
2log4 + 3log5 − log2 = log(16 × 125 ÷ 2) = log 1000 = 3.
The mean height of three groups consisting of 20, 16 and 14 students are 1.67 m, 1.50 m and 1.40 m respectively. Find the mean height of all the students.
Answer: A (1.54 m)
Total height = 33.4 + 24 + 19.6 = 77 m for 50 students, so the mean is 1.54 m.
Find the gradient of the graph of y = ⅓(29x + 10) + 5x in the x-y plane
Answer: B (44/3)
y = (29/3)x + 5x + 10/3 = (44/3)x + 10/3, so the gradient is 44/3.
In the diagram, O is the centre of the circle. |OX| = 5 cm, |OZ| = 3 cm and ∠XOZ = ∠YOZ. Find |XY|
Answer: D (8 cm)
OZ bisects the chord at right angles, so XZ = √(5² − 3²) = 4 cm and XY = 8 cm.
A line in the x−y plane passes through the point (−9, 10). If the gradient of the line is ⅓, find the equation of the line.
Answer: A (y = x/3 + 13)
y − 10 = ⅓(x + 9) gives y = x/3 + 3 + 10 = x/3 + 13.
The sum of the ages of 14 learners in a class is 168. A learner was added to the class and the mean age changed to 12.2. Calculate the age of the new learner.
Answer: C (15)
15 learners with mean 12.2 have total age 183, so the new learner is 183 − 168 = 15.
Given that 3a − 2b = 12 and a + 5b = −13, find the value of (4a − b)
Answer: C (11)
From the second equation a = −13 − 5b; substituting gives −39 − 17b = 12, so b = −3 and a = 2. Then 4a − b = 8 + 3 = 11.
The roots of a quadratic equation are −2/3 and −1/4. Find the equation
Answer: D (12x2 + 11x + 2 = 0)
(3x + 2)(4x + 1) = 12x² + 3x + 8x + 2 = 12x² + 11x + 2 = 0.
A cone has a base radius of 12 cm and height of 24 cm. The volume of the cone is equal to the volume of a cylindrical tank with a base radius 6 cm. Calculate the height of the tank. [Take π = 22/7]
Answer: C (32 cm)
⅓π × 144 × 24 = π × 36 × h, so 1152 = 36h and h = 32 cm.
ΔMNP is an isosceles triangle with |MN| = |MP| = 26 cm and |NP| = 20 cm. Find the area of the triangle
Answer: B (240 cm2)
The height from M bisects NP: h = √(26² − 10²) = 24 cm. Area = ½ × 20 × 24 = 240 cm².
In the diagram, P, Q and R are points on a circle with centre O. ∠OPQ = 77° and ∠PQR = 130°. Find ∠ORQ
Answer: D (53°)
∠PQR stands on the major arc PR, so the reflex angle POR = 260° and the angle POR inside OPQR is 100°. The angles of OPQR add to 360°: ∠ORQ = 360° − (77° + 130° + 100°) = 53°.
A cylindrical tin of diameter 13.2 cm and height 20 cm is half-filled with water. Find the volume of the water in the tin
Answer: C (435.6π cm3)
r = 6.6 cm and the water is 10 cm deep: V = π × 6.6² × 10 = 435.6π cm³.
The point M is on the same horizontal as the foot of a vertical pole. The angle of elevation of the top of the pole from M is 55°. If M is 10 m from the foot of the pole, calculate, correct to the nearest metre, the height of the pole.
Answer: C (14.0 m)
h = 10 tan 55° = 10 × 1.428 ≈ 14 m.
In the diagram, Q, R and S are points on the circle centre O. If ∠QRS = 62°, find ∠SQR
Answer: B (28°)
QR passes through O, so it is a diameter and ∠QSR = 90°. Then ∠SQR = 180° − 90° − 62° = 28°.
The first term of an Arithmetic progression is 20. If the fifth term is four-fifth of the third term, find the common difference.
Answer: C (−5/3)
20 + 4d = ⅘(20 + 2d) gives 100 + 20d = 80 + 8d, so 12d = −20 and d = −5/3.
The expression x2 + 10x + 18 can be written as (x + p)2 + q, where p and q are integers. Find the values of p and q
Answer: C (5 and −7)
(x + 5)² = x² + 10x + 25, so x² + 10x + 18 = (x + 5)² − 7: p = 5, q = −7.
Solve 2x − 3 ≤ 4x + 5 < 3x + 7
Answer: B (−4 ≤ x < 2)
2x − 3 ≤ 4x + 5 gives x ≥ −4, and 4x + 5 < 3x + 7 gives x < 2, so −4 ≤ x < 2.
A fun club donates to children's home annually. The amount given is increased by 8% of its yearly value. If the donation in 2010 was $7,000.00, find the: (a) donation given in 2012 (b) total amount of money donated from 2010 to 2013
(a) 2011: 7000 × 1.08 = $7,560; 2012: 7560 × 1.08 = $8,164.80
(b) 2013: 8164.80 × 1.08 = $8,817.98
Total = 7000 + 7560 + 8164.80 + 8817.98 = $31,542.78
Each year's donation is 1.08 times the year before; part (b) asks for the sum of four years.
Two cars X and Y, start from a common point and travel in opposite directions. X travels at an average speed of 60 km/h and Y at 80 km/h on the same course.
(a) Determine, how far apart X and Y will be after 3 hours
(b) If the speed of X is increased by 20 km/h within the 3 hours, what would be the effect on the distance between X and Y? (c) State the importance of considering speed when calculating total distance over the same time frame
(a) X: 60 × 3 = 180 km; Y: 80 × 3 = 240 km; apart = 420 km
(b) X: 80 × 3 = 240 km; apart = 240 + 240 = 480 km, an increase of 60 km
(c) Distance = speed × time, so for the same time a higher speed gives a greater distance (the separation grows with the combined speed)
Moving in opposite directions, add the distances.
Concrete blocks of length 14/25 m were laid end to end along the four sides of a rectangular lawn 28 m long and 21 m wide.
(a) How many block were required to complete one layer?
(b) If each block costs GH¢ 1.80 and the cost of labor is one-third the cost of a block, find the total cost of laying the blocks.
(a) Perimeter = 2(28 + 21) = 98 m; blocks = 98 ÷ 14/25 = 175
(b) Blocks: 175 × 1.80 = GH¢315; labour = ⅓ × 315 = GH¢105
Total cost = GH¢420.00
The blocks run round the perimeter.
In the diagram, MNPQ is a circle with centre O. ∠MNQ = y°, ∠NEP = x°, ∠NMP = (100 − x)° and ∠NFP = (5x)°. Find the values of x and y.
∠NQP = ∠NMP = 100 − x (same segment), so arc NP subtends 2(100 − x) at the centre
Angle between secants: ∠NEP = ½(arc NP − arc MQ) ⇒ arc MQ = 200 − 2x − 2x = 200 − 4x
Angle between chords: ∠NFP = ½(arc NP + arc MQ) = ½(400 − 6x) = 200 − 3x = 5x ⇒ x = 25
y = ∠MNQ = ½ arc MQ = 100 − 2(25) = 50
Angles in the same segment are equal; use the angle sum in the triangles at F and E.
A bowl contains red, green and black marbles all of the same size. The number of red marbles are 3 more than the number of green marbles and the number of black marbles is 1 less than the number of green marbles. If the probability of drawing at random a black marble is 1/4, find the: (a) Number of each color of marble (b) Probability that a marble drawn at random from the bowl is not black
(a) Green = g, red = g + 3, black = g − 1; total = 3g + 2
(g − 1)/(3g + 2) = ¼ ⇒ 4g − 4 = 3g + 2 ⇒ g = 6
Green 6, red 9, black 5 (total 20)
(b) P(not black) = 1 − 5/20 = 15/20 = ¾
Express every colour in terms of the number of green marbles.
In a remedial school of 13 teachers, 8 teach Government (G), 7 teach Geography (P) and 7 teach History (H). 2 teach Government and Geography only, 2 teach Government and History only and 3 teach Geography and History only.
(a) Represent the information in a Venn diagram
(b) Find: (i) n(G ∩ P ∩ H) (ii) n(P ∩ H) (iii) n[(P ∪ H)′ ∩ G]
(a) Three overlapping circles G, P, H in U (13); G∩P only = 2, G∩H only = 2, P∩H only = 3, all three = x; G only = 4 − x, P only = 2 − x, H only = 2 − x
(b)(i) (4 − x) + (2 − x) + (2 − x) + 2 + 2 + 3 + x = 13 ⇒ 15 − 2x = 13 ⇒ x = 1
(ii) n(P ∩ H) = 3 + 1 = 4
(iii) n[(P ∪ H)′ ∩ G] = Government only = 4 − 1 = 3
Every teacher teaches at least one subject, so all regions add up to 13.
(a) On a certain day, Mabel rode her bicycle for 45 minutes and then walked for 1 hour, covering a total distance of 10.4 km. The average cycling speed is three times the walking speed.
(i) Find, correct to two significant figures, the walking speed
(ii) If Mabel increases the average cycling speed by 20% but maintains the walking speed for 1 hour, find, in minutes, the new cycling time
(b) A line, L passing through K(−3, 5) is parallel to the equation 3x − 2y + 5 = 0. Find the equation of L.
(a)(i) Walking speed w: ¾ × 3w + w = 10.4 ⇒ 3.25w = 10.4 ⇒ w = 3.2 km/h
(ii) New cycling speed = 9.6 × 1.2 = 11.52 km/h; walking covers 3.2 km, so cycling distance = 7.2 km
Time = 7.2 ÷ 11.52 = 0.625 h = 37.5 minutes
(b) 2y = 3x + 5 ⇒ gradient 3/2; y − 5 = 3/2(x + 3) ⇒ 3x − 2y + 19 = 0
Parallel lines have the same gradient.
(a) In the diagram, A, B, C and D are points on the circle with centre O. ∠DAB = 64° and |AD| = |DC|. Find the size of the angle marked x.
(b) Two consecutive even numbers are such that one-third of the larger exceeds one-fourth of the smaller by 7. Find the numbers.
(a) ∠DCB = 180° − 64° = 116° (cyclic quadrilateral); ∠ACB = 90° (AB is a diameter) ⇒ ∠ACD = 26°
AD = DC ⇒ ∠DAC = ∠DCA = 26°; x = ∠ADC = 180° − 52° = 128°
(b) Numbers n and n + 2: (n + 2)/3 − n/4 = 7 ⇒ 4n + 8 − 3n = 84 ⇒ n = 76
The numbers are 76 and 78
AB passes through O, so the angle in the semicircle is 90°.
(a) The angle of elevation of the top of a 75 m tower from a ship and a boat are 26.4° and 57.4° respectively. Given that, the ship, boat and the foot of the tower are on the same horizontal: (i) Illustrate the information in a diagram (ii) Calculate, correct to three significant figures, the distance between the ship and the boat.
(b) A sector of a circle has a radius of 18 cm and subtends an angle of 120° at the centre of the circle. Find, correct to five significant figures, the area of the sector. [Take π = 22/7]
(a)(i) Vertical tower 75 m; ship and boat on the ground line, ship further away (26.4°), boat nearer (57.4°)
(ii) Ship to tower = 75 ÷ tan 26.4° ≈ 151.09 m; boat to tower = 75 ÷ tan 57.4° ≈ 47.96 m
Same side of the tower: distance ≈ 151.09 − 47.96 ≈ 103 m (if on opposite sides: 151.09 + 47.96 ≈ 199 m)
(b) Area = 120/360 × 22/7 × 18² = 339.43 cm²
The question does not say which side; the usual reading is that both are on the same side of the tower.
Two equal chords AB and BC are each 15 cm long. They meet at a point B on a circle with centre O, such that ∠ABC = 120°.
(a) Illustrate the information in a diagram
(b) Calculate, correct to the nearest whole number the: (i) radius of the circle (ii) perimeter of the major segment cut off by the chord AC. [Take π = 22/7]
(a) Circle with B on it, chords BA = BC = 15 cm, ∠ABC = 120°, chord AC drawn
(b)(i) AC = 2 × 15 sin 60° ≈ 25.98 cm; sine rule: 2r = AC ÷ sin 120° = 30 ⇒ r = 15 cm (triangles OAB and OBC are equilateral)
(ii) The major arc AC (the arc not containing B) subtends 2 × 120° = 240° at the centre
Major arc = 240/360 × 2 × 22/7 × 15 ≈ 62.86 cm; perimeter = 62.86 + 25.98 ≈ 89 cm
The angle at B on the circumference is half the angle subtended at the centre by the opposite arc.
(a) The profit, P, made by a retailer for selling y number of articles in a day is given by P = (3y2 − 9)/(y − 2) − 18/(y − 2)
(i) If $16.00 profit was made in a particular day, determine the number of articles sold.
(ii) On a particular day, no profit was made for sales of articles. Find the number of articles sold.
(b) Solve (5 + x)/6 + (4x − 1)/3 ≥ (3x + 2)/5
(a)(i) (3y² − 27)/(y − 2) = 16 ⇒ 3y² − 16y + 5 = 0 ⇒ (3y − 1)(y − 5) = 0 ⇒ y = 5 articles (y must be a whole number)
(ii) 3y² − 27 = 0 ⇒ y² = 9 ⇒ y = 3 articles
(b) Multiply by 30: 5(5 + x) + 10(4x − 1) ≥ 6(3x + 2) ⇒ 45x + 15 ≥ 18x + 12 ⇒ 27x ≥ −3 ⇒ x ≥ −1/9
Combine the two fractions first; they share the denominator y − 2.
(a) A study on average daily temperature in °C over a period of 12 months are: 24, 25, 23, 27, 26, 28, 29, 30, 27, 26, 25 and 24
(i) Find the: (I) mean (II) median (III) mode of the data
(ii) State, with explanation, which of the measures in 12(a)(i) best represents the data?
(b) Two towns lie on the same meridian and have latitudes of 15°N and 36°N. (i) Illustrate the information in a diagram (ii) Calculate, correct to three significant figures, the distance between the towns [Take π = 22/7 and R = 6400 km]
(a)(i)(I) Mean = 314 ÷ 12 ≈ 26.17 °C
(II) Ordered: 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 29, 30; median = (26 + 26)/2 = 26 °C
(III) Modes: 24, 25, 26 and 27 (each appears twice)
(ii) The median (or the mean): the data has no extreme values, both are close to the centre; the mode is not useful because there are four modes
(b)(i) Sphere with the equator, one meridian and the two towns at 15°N and 36°N on it
(ii) Angle = 36° − 15° = 21°; distance = 21/360 × 2 × 22/7 × 6400 ≈ 2,350 km
Both towns are north, so subtract the latitudes.
(a) Using a ruler and a pair of compasses only, construct ΔMNR in which |MN| = 8.7 cm, |MR| = 10.5 cm and ∠MNR = 75°.
(b) Locate by construction, a point S such that RS is parallel to MN and S is equidistant from points M and R.
(c) Measure: (i) |NR|; (ii) ∠NRS
(d) What shape is SMNR?
(a) Draw MN = 8.7 cm; at N construct 75° (60° + half of 30°); with centre M and radius 10.5 cm cut the arm at R; join MR
(b) Construct the line through R parallel to MN and the perpendicular bisector of MR; S is where they meet
(c)(i) |NR| ≈ 8.5 cm (ii) ∠NRS ≈ 105° (allow ±1)
(d) SMNR is a trapezium (RS ∥ MN)
RS ∥ MN, so ∠NRS and ∠MNR are co-interior angles adding to 180°.