61 questions. Try each one, then open the answer to check yourself and read the explanation.
Correct 0.04945 to two significant figures.
Answer: B (0.049)
The first two significant figures are 4 and 9; the next digit is 4, so round down: 0.049.
Simplify: 5/√3 − 3/√2
Answer: D (1/6 (10√3 − 9√2))
5/√3 = 5√3/3 = 10√3/6 and 3/√2 = 3√2/2 = 9√2/6, so the difference is (10√3 − 9√2)/6.
Evaluate, correct to the nearest whole number: 7 1/2 − (2 1/2 + 3) ÷ 33/2
Answer: C (7)
Brackets: 5 1/2. Division: 11/2 ÷ 33/2 = 1/3. Then 7 1/2 − 1/3 = 7 1/6 ≈ 7.
Simplify the expression log1018 − log102.88 + log1016.
Answer: C (2)
log(18 × 16 ÷ 2.88) = log 100 = 2.
Find the equation whose roots are 2 and −3 1/2.
Answer: D (2x2 + 3x − 14 = 0)
(x − 2)(2x + 7) = 2x2 + 3x − 14 = 0.
A man bought 220 mangoes at N5x. He sold each for 3x kobo and made a gain of N8. Find the value of x.
Answer: B (5)
He sold them for 220 × 3x kobo = N6.60x. Gain = 6.6x − 5x = 1.6x = 8, so x = 5.
From a point P, R is 5 km due west and 12 km due south. Find the distance between P and R.
Answer: C (13 km)
The two movements are at right angles: √(52 + 122) = √169 = 13 km.
A fair die is tossed once, what is the probability of obtaining neither 5 nor 2?
Answer: B (2/3)
Four of the six faces (1, 3, 4, 6) are neither 5 nor 2: 4/6 = 2/3.
In the diagram, KL//MN, ∠LKP = 30° and ∠NMP = 45°. Find the size of the reflex ∠KPM.
Answer: A (285°)
The non-reflex angle KPM = 30° + 45° = 75° (alternate angles with a line through P parallel to KL), so the reflex angle = 360 − 75 = 285°.
In the diagram, PR is a diameter, ∠PRQ = (3x − 8)° and ∠RPQ = (2y − 7)°. Express x in terms of y.
Answer: C (x = (105 − 2y)/3)
∠PQR = 90° (angle in a semicircle), so (3x − 8) + (2y − 7) = 90, giving 3x = 105 − 2y and x = (105 − 2y)/3.
Convert 101101two to a number in base ten.
Answer: C (45)
32 + 8 + 4 + 1 = 45.
Solve the equation 27 = 85−x.
Answer: B (8/3)
85−x = 215−3x, so 7 = 15 − 3x and x = 8/3.
Expand (2x − 3y)(x − 5y)
Answer: D (2x2 − 13xy + 15y2)
2x2 − 10xy − 3xy + 15y2 = 2x2 − 13xy + 15y2.
Make f the subject of the relation v = u + ft.
Answer: A ((v − u)/t)
ft = v − u, so f = (v − u)/t.
In the diagram, PQ and MN are straight lines. Find the value of x.
Answer: B (17°)
Angles on the straight line MN add to 180°: 2(x + 30) + 86 = 180, so 2x = 34 and x = 17°.
In the diagram, PQRS is a parallelogram and ∠QRT = 30°. Find x.
Answer: D (150°)
SRT is a straight line, so ∠QRS = 180 − 30 = 150°. Opposite angles of a parallelogram are equal, so x = ∠SPQ = 150°.
From a point R, 300 m north of P, man walks eastwards to a place Q which is 600 m from P. Find the bearing of P from Q, correct to the nearest degree.
Answer: D (240°)
cos ∠RPQ = 300/600, so ∠RPQ = 60° and the bearing of Q from P is 060°. The bearing of P from Q is 060° + 180° = 240°.
The lengths of the adjacent sides of a right-angled triangle are x cm, (x − 1) cm. If the length of the hypotenuse is √13 cm, find the value of x.
Answer: B (3)
x2 + (x − 1)2 = 13 gives x2 − x − 6 = 0, (x − 3)(x + 2) = 0, so x = 3.
What is the diameter of a circle of area 77cm2? (Take π = 22/7)
Answer: D (7√2 cm)
r2 = 77 × 7/22 = 49/2, so r = 7/√2 and the diameter = 14/√2 = 7√2 cm.
Solve for y in the equations: 2x − 3y = 22; 3x + 2y = 7
Answer: B (−4)
Eliminate y: 4x − 6y = 44 and 9x + 6y = 21 add to 13x = 65, x = 5. Then 15 + 2y = 7, y = −4.
Solve the equation (2y − 1)/3 − (3y − 1)/4 = 1
Answer: B (−13)
Multiply by 12: 8y − 4 − 9y + 3 = 12, so −y = 13 and y = −13.
If log9x = 1.5, find x.
Answer: B (27)
x = 91.5 = (√9)3 = 27.
A sequence is given by 2 1/2, 5, 7 1/2, ....... If the nth term is 25, find n.
Answer: B (10)
a = 2½ and d = 2½, so Tn = 2½n = 25 gives n = 10.
Given that one of the roots of the equation 2x2 + (k + 2)x + k = 0 is 2, find the value of k.
Answer: A (−4)
Substitute x = 2: 8 + 2(k + 2) + k = 0, so 3k = −12 and k = −4.
The figure shows a quadrilateral PQRS having equal sides and opposite sides parallel. The diagonals PR and QS intersect perpendicularly at O, which of the following statements cannot be correct?
Answer: B (PQR is an equilateral triangle)
PQRS is a rhombus: its diagonals bisect each other, it is a parallelogram, and it has two lines of symmetry. Triangle PQR is isosceles but not, in general, equilateral.
Find the mean of the numbers 1, 3, 4, 8, 8, 4 and 7.
Answer: B (5)
Sum = 35 and there are 7 numbers: 35 ÷ 7 = 5.
What is the total surface area of a closed cylinder of height 10 cm and diameter 7 cm? (Take π = 22/7)
Answer: C (297 cm2)
2πr(r + h) = 2 × 22/7 × 3.5 × 13.5 = 297 cm2.
An arc of a circle of radius 14 cm subtends angle 300° at the centre. Find the perimeter of the sector formed by the arc. (Take π = 22/7)
Answer: C (101.33 cm)
Arc = 300/360 × 2 × 22/7 × 14 = 73.33 cm. Perimeter = 73.33 + 14 + 14 = 101.33 cm.
The figure is the graph of a quadratic equation. What are the roots of the equation?
Answer: C (1 and 4)
The curve crosses the x-axis at x = 1 and x = 4.
The figure is the graph of a quadratic equation. What is the equation of the curve?
Answer: C (y = x2 − 5x + 4)
Roots 1 and 4 give y = (x − 1)(x − 4) = x2 − 5x + 4 (it also passes through (0, 4)).
Which of the following statements describes the locus of a point R which moves in a plane such that it is equidistant from two intersecting lines?
Answer: A (The bisector of the angles formed by the lines)
Points the same distance from two intersecting lines lie on the bisectors of the angles between them.
P = {3, 9, 11, 13} and Q = {3, 7, 9, 15} are subsets of the universal set ε = {1, 3, 7, 9, 11, 13, 15}. Find P'∩Q'.
Answer: C ({1})
P' = {1, 7, 15} and Q' = {1, 11, 13}; only 1 is in both.
A train moving at a uniform speed, covers 36 km in 21 minutes. How long does it take to cover 60 km?
Answer: A (35 minutes)
Time is proportional to distance: 21 × 60/36 = 35 minutes.
The salary of a man was increased in the ratio 40:47. Calculate the percentage increase in the salary.
Answer: C (17.5%)
The increase is 7 parts on 40: 7/40 × 100% = 17.5%.
A regular polygon has 9 sides. What is the size of one of its exterior angles?
Answer: B (40°)
Exterior angle = 360° ÷ 9 = 40°.
Simplify: 2/(3xy) − 3/(4yz)
Answer: D ((8z − 9x)/(12xyz))
Over 12xyz: 2 × 4z − 3 × 3x = 8z − 9x.
The total surface area of the walls of a room 7 m long, 5 m wide and x m high is 96 m2. Find the value of x.
Answer: C (4)
The four walls have area 2(7x) + 2(5x) = 24x = 96, so x = 4.
The bar chart shows the pocket money received by students in a class in a week. Find the total number of students in the class.
Answer: D (34)
Add the bars: 4 + 7 + 6 + 2 + 3 + 11 + 1 = 34.
The bar chart shows the pocket money received by students in a class in a week. How many students received N300 or N500 for a week?
Answer: C (18)
7 students received N300 and 11 received N500: 7 + 11 = 18.
In the diagram PQRS is a circle, |PT| = |QT| and ∠QPT = 70°. What is the size of ∠PRS?
Answer: B (70°)
PT = QT, so ∠PQT = ∠QPT = 70°, i.e. ∠PQS = 70°. ∠PRS = ∠PQS (angles on the same arc PS) = 70°.
The angle of elevation of the top of a cliff 15 meters high from a landmark is 60°. How far is the landmark from the foot of the cliff? Leave your answer in surd form.
Answer: D (5√3m)
Distance = 15/tan 60° = 15/√3 = 5√3 m.
For what values of y is the expression (6y − 1)/(y2 − y − 6) not defined?
Answer: B (3, −2)
The denominator y2 − y − 6 = (y − 3)(y + 2) is zero when y = 3 or −2.
XOY is a sector of a circle centre O of radius 3.5 cm which subtends an angle of 144° at the centre. Calculate, in terms of π, the area of the sector.
Answer: C (4.9π cm2)
Area = 144/360 × π × 3.52 = 0.4 × 12.25π = 4.9π cm2.
Represent the solution of 3x + 15 > 0 on a number line.
Answer: D (Number line D)
3x > −15, so x > −5. Line D has an open circle at −5 with the arrow pointing right.
A rectangular packet has inner dimension 16 cm by 12 cm by 6 cm. How many cubes of sugar of side 2 cm can be neatly packed into the packet?
Answer: B (144)
8 × 6 × 3 = 144 cubes fit along the three edges.
If y ∝ 1/x2 and y = 1 1/4 when x = 4, find the value of y when x = 1/2.
Answer: D (80)
k = yx2 = 5/4 × 16 = 20. When x = 1/2, y = 20 ÷ 1/4 = 80.
Ladi sold a car for N84,000 at a loss of 4%. How much did Ladi buy the car?
Answer: D (N87,500)
N84,000 is 96% of the cost price: 84,000 ÷ 0.96 = N87,500.
The monthly salary of a man increased from N2,700 to N3,200. Find the percentage increase.
Answer: D (18.5%)
Increase = N500; 500/2700 × 100% ≈ 18.5%.
(a) Evaluate without using the mathematical table or calculator. log10√30 − log10√6 + log10√2.
(b) U = {1, 2, 3, ........... 10}, A = {1, 2, 3, 4, 5}, B = {2, 3, 5} and C = {6, 8, 10}. (i) Given that the Venn diagram represents the sets above, copy and fill in the elements. (ii) Find A∩C. (iii) Find A∩B'.
(a) log √(30 × 2/6) = log √10 = ½ log 10 = ½.
(b)(i) B inside A: put 2, 3, 5 in B; 1 and 4 in A outside B; 6, 8, 10 in C (separate from A); 7 and 9 outside all sets.
(ii) A∩C = ∅ (no common elements).
(iii) A∩B' = {1, 4}.
Combine the logs first, then use log √10 = ½.
(a) Evaluate: 2 ÷ (64/125)−2/3
(b) The lines y = 3x + 5 and y = −4x − 1 intersect at a point k. Find the coordinates of k.
(a) Dividing by a negative power is multiplying by the positive power: 2 × (64/125)2/3 = 2 × (4/5)2 = 2 × 16/25 = 32/25 = 1 7/25.
(b) 3x + 5 = −4x − 1, so 7x = −6 and x = −6/7.
y = 3(−6/7) + 5 = 17/7 = 2 3/7. K = (−6/7, 2 3/7).
(64/125)^(1/3) = 4/5; then square it.
(a) Simplify 5/8 of 2 1/2 − 3/4 ÷ 3/5
(b) A cone and a right pyramid have equal heights and volumes. If the area of the base of the pyramid is 154 cm2, find the base radius of the cone (Take π = 22/7)
(a) 5/8 × 5/2 = 25/16 and 3/4 ÷ 3/5 = 5/4 = 20/16. 25/16 − 20/16 = 5/16.
(b) ⅓πr2h = ⅓ × 154 × h, so πr2 = 154.
r2 = 154 × 7/22 = 49, r = 7 cm.
Equal heights and volumes mean the base areas are equal.
(a) Two fair dice are thrown once. Find the probability of getting: (i) the same digit (ii) a total score greater than 5.
(b) Given that x = cos 30° and y = sin 30°, evaluate without using a mathematical table or calculator: (x2 + y2)/(y2 − x2).
(a)(i) Doubles (1,1)…(6,6): 6/36 = 1/6.
(ii) Totals of 5 or less: 1 + 2 + 3 + 4 = 10 outcomes, so P(total > 5) = 1 − 10/36 = 26/36 = 13/18.
(b) x2 + y2 = 3/4 + 1/4 = 1; y2 − x2 = 1/4 − 3/4 = −1/2. Value = 1 ÷ (−1/2) = −2.
Count the outcomes that are not wanted when that is easier, then subtract from 1.
(a) In the diagram, A, B, C and D are points on the circumference of a circle. XY is a tangent at A. Find: (i) ∠CAX; (ii) ∠ABY
(b) If (m + 1) and (m − 3) are factors of m2 − km + c, find the values of k and c.
(a)(i) ∠ACB = ∠ADB = 20° (same segment). C, B and Y are in a straight line, so in triangle ACY the exterior angle ∠CAX = ∠ACY + ∠AYC = 20° + 60° = 80°.
(ii) ∠BAY = ∠ADB = 20° (angle between tangent and chord equals the angle in the alternate segment). In triangle ABY, ∠ABY = 180 − 60 − 20 = 100°.
(b) (m + 1)(m − 3) = m2 − 2m − 3, so k = 2 and c = −3.
Use the tangent–chord angle and the exterior angle of a triangle.
(a) Without using mathematical table or calculator, evaluate: √((0.18 × 12.5)/(0.05 × 0.2)).
(b) Simplify: (8 − 4√18)/√50 leaving your answer in the a + b√n where a and b are rational numbers and n is an integer.
(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.
(a) 0.18 × 12.5 = 2.25 and 0.05 × 0.2 = 0.01, so √225 = 15.
(b) √18 = 3√2 and √50 = 5√2: (8 − 12√2)/(5√2). Multiply by √2/√2: (8√2 − 24)/10 = −12/5 + (4/5)√2.
(c) x = ky3/z2: 108 = 27k/16, so k = 64.
4000 = 64 × 1000/z2, so z2 = 16 and z = 4.
Rationalise the denominator by multiplying by √2/√2.
(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find: (i) the value of n; (ii) the sum of all the interior angles.
(b) A boy walks 6 km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13 km, on a bearing of 146°. (i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometer, the distance PR.
(a) I − E = 120 and I + E = 180, so E = 30° and I = 150°.
(i) n = 360 ÷ 30 = 12.
(ii) Sum = (12 − 2) × 180° = 1800°.
(b)(i) Sketch north lines at P and Q, PQ at 065° and QR at 146°.
(ii) The bearing of P from Q is 245°, so ∠PQR = 245 − 146 = 99°.
PR2 = 62 + 132 − 2(6)(13) cos 99° = 205 + 24.4 = 229.4, PR ≈ 15.1 ≈ 15 km.
Find the angle at Q from the back bearing, then use the cosine rule.
A plane flies due East from A (lat. 53°N, long. 25°E) to a point B (lat. 53°N, long. 85°E) at an average speed of 400 km/h. The plane then flies South from B to a point C 2000 km away. Calculate, correct to the nearest whole number: (a) the distance between A and B; (b) the time the plane takes to reach point B; (c) the latitude of C. [Take the radius of the earth to be 6400 km and π = 22/7].
(a) Difference in longitude = 85 − 25 = 60°. AB = 60/360 × 2 × 22/7 × 6400 × cos 53° = 6704.8 × 0.6018 ≈ 4035 km.
(b) Time = 4035 ÷ 400 ≈ 10.09 h ≈ 10 hours.
(c) 2000 = θ/360 × 2 × 22/7 × 6400, so θ ≈ 17.9°. Latitude of C = 53 − 17.9 = 35.1 ≈ 35°N.
Along a latitude use R cos(latitude); along a meridian use R.
(a) Find the smallest integer that satisfies the inequality x + 8 < 4x − 15.
(b) A sales girl is paid a monthly salary of N2,500.00 in addition to a commission of 5 kobo in the naira on all sales made by her during the month. If her sales for a month amounts to N200,000.00, calculate her income for that month.
(c) The diagram shows a window consisting of a rectangular and a semi-circular parts. The radius of the semi-circular part is 35 cm and the height of the rectangular part is 50cm. Find the area of the window. (Take π = 22/7)
(a) 23 < 3x, so x > 7⅔; the smallest integer is 8.
(b) Commission = 5% of 200,000 = N10,000. Income = 2,500 + 10,000 = N12,500.00.
(c) The rectangle is 70 cm wide (the diameter) and 50 cm high: 3500 cm2.
Semicircle = ½ × 22/7 × 352 = 1925 cm2. Window = 3500 + 1925 = 5425 cm2.
5 kobo in the naira is 5%.
The sketch shows a plot of land (|AB| = 85 m, ∠CAB = 40°, ∠ABC = 95°). (a) Using a scale of 1 cm to 10 m draw an accurate diagram of the plot. (b) Construct: (i) the locus l1 of points equidistant from AC and BC (ii) the locus l2 of points 60 m from A. (c) A tree T inside the plot, is on both l1 and l2. Locate T and find |TC| in metres. (d) A flag pole, P is to be placed such that it is nearer AC than BC and more than 60 m from A. Shade the region where P can be located.
(a) Draw AB = 8.5 cm, angle 40° at A and 95° at B; the arms meet at C (AC ≈ 12.0 cm, BC ≈ 7.7 cm, ∠C = 45°).
(b)(i) l1: bisector of ∠ACB.
(ii) l2: arc of a circle centre A, radius 6 cm.
(c) T is where the bisector meets the arc inside the plot; TC ≈ 7.2 cm on the drawing, i.e. about 72 m.
(d) Shade the part of the plot on the AC side of the bisector of ∠C and outside the arc of radius 6 cm about A.
Equidistant from two lines: angle bisector. A fixed distance from a point: circle.
The table shows the age distributions of the members of a club.
Age (years): 10-14, 15-19, 20-24, 25-29, 30-34, 35-39
Frequency: 7, 18, 25, 17, 9, 4
(a) Calculate, correct to one decimal place, the mean age. (b)(i) Draw a histogram to illustrate the information (ii) Use the histogram to estimate the modal age. (c) If a member is selected at random, what is the probability that he/she is in the modal class?
(a) Class marks 12, 17, 22, 27, 32, 37; Σf = 80; Σfx = 84 + 306 + 550 + 459 + 288 + 148 = 1835. Mean = 1835/80 ≈ 22.9 years.
(b)(i) Bars on boundaries 9.5, 14.5, …, 39.5 with heights 7, 18, 25, 17, 9, 4.
(ii) In the tallest bar join its top corners to the neighbouring bars' tops; they cross at about 19.5 + 7/15 × 5 ≈ 21.8 years.
(c) Modal class 20–24 has 25 members: P = 25/80 = 5/16.
Use the class mid-points for x when finding the mean.
(a) In a diagram, ΔABD is right-angled at B. |AB| = 3 cm, |AD| = 5cm, ∠ACB = 61° and ∠DAC = x°. Calculate, correct to one decimal place, the value of x.
(b) In the diagram, OABCD is a pyramid with a square base of side 2 cm and a slant height of 4cm. Calculate, correct to three significant figures: (i) the vertical height of the pyramid; (ii) the volume of the pyramid.
(a) tan ∠DAB = BD/AB = 4/3, so ∠DAB = 53.13°. In right-angled triangle ABC, ∠CAB = 90 − 61 = 29°.
x = 53.13 − 29 = 24.1°.
(b) The diagram marks the slant edge as 4 cm. Half the base diagonal = √2 cm.
(i) Height = √(42 − 2) = √14 ≈ 3.74 cm.
(ii) Volume = ⅓ × 22 × 3.742 ≈ 4.99 cm3.
(If 4 cm is taken as the height of a triangular face instead, h = √15 ≈ 3.87 cm and V ≈ 5.16 cm3.)
First find BD = 4 cm with Pythagoras (3-4-5 triangle).
(a) Copy and complete the table. y = x2 − 2x − 2 for −4 ≤ x ≤ 4.
x: −4, −3, −2, −1, 0, 1, 2, 3, 4
y: 22, , , , −2, , , 1, 6
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = x2 − 2x − 2.
(c) Use your graph to find: (i) the roots of the equation x2 − 2x − 2 = 0; (ii) the values of x for which x2 − 2x − 4½ = 0 (iii) the equation of the line of symmetry of the curve.
(a) x = −3: 13; −2: 6; −1: 1; 1: −3; 2: −2.
(b) Plot and join with a smooth U-shaped curve (lowest point (1, −3)).
(c)(i) The curve crosses y = 0 at x ≈ −0.7 and x ≈ 2.7.
(ii) x2 − 2x − 4½ = 0 means x2 − 2x − 2 = 2½; draw y = 2.5: x ≈ −1.3 and x ≈ 3.3.
(iii) x = 1.
The line of symmetry passes through the lowest point of the curve.