59 questions. Try each one, then open the answer to check yourself and read the explanation.
S = {1, 2, 3, 4, 5, 6}, T = {2, 4, 5, 7} and R = {1, 4, 5}, find (S∩T)∪R
Answer: C ({1, 2, 4, 5})
S∩T = {2, 4, 5}. Joining R = {1, 4, 5} gives {1, 2, 4, 5}.
Simplify: 3/4 ÷ 1¼ × (1½ − 2/3)
Answer: D (½)
1½ − 2/3 = 5/6. Then 3/4 ÷ 5/4 = 3/5, and 3/5 × 5/6 = 1/2.
Solve the inequality 3m + 3 > 9
Answer: A (m > 2)
Subtract 3: 3m > 6. Divide by 3: m > 2.
Convert 89ten to a number in base two.
Answer: B (1011001)
89 = 64 + 16 + 8 + 1 = 26 + 24 + 23 + 20, which is 1011001 in base two.
A stick of length 1.75m was measured by a boy as 1.80m. Find the percentage error in his measurement
Answer: B (2 6/7%)
Error = 1.80 − 1.75 = 0.05m. Percentage error = 0.05/1.75 × 100 = 20/7 = 2 6/7%.
The nth term of a sequence is given by (−1)n−2 2n−1. Find the sum of the second and third terms.
Answer: A (−2)
n = 2: (−1)0 × 21 = 2. n = 3: (−1)1 × 22 = −4. Sum = 2 + (−4) = −2.
Simplify: (4−½ × 163/4)/41/2
Answer: D (2)
4−½ = ½, 163/4 = 23 = 8 and 41/2 = 2. So (½ × 8) ÷ 2 = 2.
Simplify: log√27 / log 81
Answer: B (3/8)
√27 = 33/2 and 81 = 34, so the ratio is (3/2 log 3) ÷ (4 log 3) = 3/8.
Factorize the expression 2s2 − 3st − 2t2.
Answer: B ((2s + t)(s − 2t))
Split −3st as −4st + st: 2s(s − 2t) + t(s − 2t) = (2s + t)(s − 2t).
Solve the equation x2 − 2x − 3 = 0
Answer: E ((−1, 3))
x2 − 2x − 3 = (x − 3)(x + 1) = 0, so x = −1 or x = 3.
Write as a single fraction 5/(6r) − 3/(4r)
Answer: E (1/(12r))
Common denominator 12r: 10/(12r) − 9/(12r) = 1/(12r).
Factorize 2x2 − 21x + 45
Answer: D ((2x − 15)(x − 3))
Find two numbers with product 2 × 45 = 90 and sum −21: −15 and −6. 2x2 − 6x − 15x + 45 = 2x(x − 3) − 15(x − 3) = (2x − 15)(x − 3).
Solve the simultaneous equations y = 3x; 4y − 5x = 14
Answer: B (2, 6)
Substitute y = 3x: 12x − 5x = 14, so 7x = 14, x = 2 and y = 6.
A sector of a circle of radius 9cm subtends angle 120° at the centre of the circle. Find the area of the sector to the nearest cm2 (Take π = 22/7).
Answer: C (85cm2)
Area = (120/360) × 22/7 × 9 × 9 = 84.86, which is 85 cm2 to the nearest cm2.
Calculate the total surface area of a cone of height 12cm and base radius 5cm. (Take π = 22/7)
Answer: D (282 6/7cm2)
Slant height l = √(122 + 52) = 13 cm. Total surface area = πr(r + l) = 22/7 × 5 × 18 = 1980/7 = 282 6/7 cm2.
A cone is 14cm deep and the base radius is 4½cm. Calculate the volume of water that is exactly half the volume of the cone. (Take π = 22/7)
Answer: C (148.5cm3)
Volume = ⅓ × 22/7 × 4.52 × 14 = 297 cm3. Half of it is 148.5 cm3.
The area and a diagonal of a rhombus are 60cm2 and 12cm respectively. Calculate the length of the other diagonal.
Answer: C (10cm)
Area of a rhombus = ½ × d1 × d2, so 60 = ½ × 12 × d and d = 10 cm.
The angle of a sector of a circle radius 10.5cm is 120°. Find the perimeter of the sector (Take π = 22/7)
Answer: C (43cm)
Arc = (120/360) × 2 × 22/7 × 10.5 = 22 cm. Perimeter = 22 + 10.5 + 10.5 = 43 cm.
In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ∠SQR = 68°, find ∠PQT
Answer: C (56°)
By the alternate segment theorem ∠QTS = ∠SQR = 68°. Triangle QTS is isosceles (QT = ST), so ∠TQS = (180° − 68°)/2 = 56°. On the straight line PQR: ∠PQT = 180° − 56° − 68° = 56°.
The sum of the interior angles of a regular polygon is 30 right angles. How many sides has the polygon?
Answer: D (17 sides)
Sum of interior angles = (2n − 4) right angles, so 2n − 4 = 30 and n = 17.
In the diagram below, |QR| = 12cm and |QS| = 10cm. If ∠PRQ = 90°, ∠RSQ = 90° and PSQ is a straight line, find |PS|
Answer: B (4.4cm)
In right triangle RSQ, RS2 = 122 − 102 = 44. In a right triangle the altitude to the hypotenuse gives RS2 = PS × QS, so PS = 44/10 = 4.4 cm.
In the diagram below, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30° and ∠PQR = 80°. Find ∠SPQ.
Answer: B (25°)
Join PR. In isosceles triangle PSR, ∠SPR = (180° − 30°)/2 = 75°. In isosceles triangle PQR, ∠QPR = (180° − 80°)/2 = 50°. So ∠SPQ = 75° − 50° = 25°.
Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31° and ∠PKR = 42°, find ∠KQR
Answer: C (107°)
In triangle KPS, ∠KSP = 180° − 31° − 42° = 107°. In cyclic quadrilateral PQRS, the exterior angle ∠KQR equals the interior opposite angle ∠PSR, so ∠KQR = 107°.
In the diagram below, QRS is a straight line, QP//RT, ∠PRQ = 56°, ∠QPR = 84° and ∠TRS = x°. Find x.
Answer: B (40)
In triangle PQR, ∠PQR = 180° − 56° − 84° = 40°. Since QP is parallel to RT, ∠TRS = ∠PQR = 40° (corresponding angles).
Simplify 2.25/0.015 leaving your answer in standard form.
Answer: E (1.5 × 102)
2.25 ÷ 0.015 = 2250 ÷ 15 = 150 = 1.5 × 102.
State the fifth and seventh terms of the sequence −2, −3, −4½ …………
Answer: A (−81/8, −729/32)
This is a G.P. with a = −2 and r = 3/2. T5 = −2 × (3/2)4 = −81/8 and T7 = −2 × (3/2)6 = −729/32.
A plane flies 90km on a bearing 030° and then flies 150km due east. How far east of the starting point is the plane?
Answer: C (195km)
On bearing 030° the eastward part is 90 sin 30° = 45 km. Add the 150 km due east: 45 + 150 = 195 km.
If sin x = cos 50°, then x equals
Answer: A (40°)
sin x = cos(90° − x), so cos(90° − x) = cos 50° and x = 40°.
From a point on the edge of the sea, one ship is 24km away on a bearing S 50°E and another ship is 7km away on a bearing S40°W. How far apart are the ships?
Answer: C (25km)
The angle between S50°E and S40°W is 50° + 40° = 90°. Distance = √(242 + 72) = √625 = 25 km.
Given that tan x = 5/12, what is the value of sin x + cos x?
Answer: D (17/13)
Opposite 5, adjacent 12, hypotenuse 13: sin x = 5/13 and cos x = 12/13, so the sum is 17/13.
Cos 75° has the same value as
Answer: C (cos 285°)
285° = 360° − 75°, and cosine is positive in the fourth quadrant, so cos 285° = cos 75°.
If sin x = 12/13, where 0° < x < 90°, find the value of 1 − cos2x.
Answer: D (144/169)
1 − cos2x = sin2x = (12/13)2 = 144/169.
The angle of elevation of the top of a tower from a point on the horizontal ground, 40m away from the foot of the tower is 30°. Find the height of the tower.
Answer: B (40√3/3 m)
height = 40 × tan 30° = 40 × 1/√3 = 40√3/3 m.
At a point 500m from the base of a water tank the angle of elevation of the top of the tank is 45°. Find the height of the tank.
Answer: E (500m)
height = 500 × tan 45° = 500 × 1 = 500 m.
Find the median of the following set of numbers: 28, 29, 39, 38, 33, 37, 26, 20, 15, 25
Answer: D (28.5)
In order: 15, 20, 25, 26, 28, 29, 33, 37, 38, 39. With 10 numbers, the median is the mean of the 5th and 6th: (28 + 29)/2 = 28.5.
What is A∩B in the diagram below?
Answer: B ({2, 3, 4})
A∩B is everything inside both circles A and B: 2 and 3, plus 4 in the middle region shared by all three circles.
The table below shows the scores of a group of 40 students in a Physics test.
Score: 1, 2, 3, 4, 5, 6, 7, 8, 9
Freq.: 2, 3, 6, 7, 9, 6, 2, 2, 3
If the mode is m and the median is n, then (m,n) is
Answer: A ((5,5))
The highest frequency (9) is at score 5, so m = 5. Cumulative totals 2, 5, 11, 18, 27 show the 20th and 21st scores are both 5, so n = 5.
The table below shows the scores of a group of 40 students in a Physics test.
Score: 1, 2, 3, 4, 5, 6, 7, 8, 9
Freq.: 2, 3, 6, 7, 9, 6, 2, 2, 3
What is the mean of the distribution?
Answer: C (4.8)
Σfx = 2 + 6 + 18 + 28 + 45 + 36 + 14 + 16 + 27 = 192. Mean = 192 ÷ 40 = 4.8.
Find the median of the set of numbers 12, 15, 13, 14, 12, 12.
Answer: B (12.5)
In order: 12, 12, 12, 13, 14, 15. The median is the mean of the 3rd and 4th: (12 + 13)/2 = 12.5.
The distribution by state of 840 students in the Faculty of Science of a Nigerian University in a certain session is as follows:
Bendel 45
kwara 410
Ogun 105
Ondo 126
Oyo 154
In a pie chart drawn to represent this distribution, the angle subtended by Ondo is
Answer: E (54°)
Angle = 126/840 × 360° = 54°.
The diagram below shows the distribution of marks for a class test. How many students took the examination?
Answer: E (24)
Add the bar heights for marks 3 to 9: 2 + 4 + 3 + 6 + 6 + 2 + 1 = 24.
The diagram below shows the distribution of marks for a class test. How many students scored less than 7 marks?
Answer: B (15)
Marks 3, 4, 5 and 6 have 2 + 4 + 3 + 6 = 15 students.
What is the probability that an integer selected from the set of integers (20, 21, ........ 30) is a prime number?
Answer: A (2/11)
There are 11 integers from 20 to 30, and only 23 and 29 are prime: probability 2/11.
A fair die is rolled once. What is the probability of obtaining a number less than 3?
Answer: B (1/3)
Only 1 and 2 are less than 3: 2 out of 6 = 1/3.
What is the probability that a total sum of seven would appear in two tosses of a fair die?
Answer: B (1/6)
Six of the 36 equally likely pairs add to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Probability = 6/36 = 1/6.
A die is rolled 200 times. The outcomes obtained are shown in the table below.
Number: 1, 2, 3, 4, 5, 6
No. of times: 25, 30, 45, 28, 40, 32
Find the probability of obtaining 2.
Answer: C (0.15)
2 came up 30 times out of 200: 30/200 = 0.15.
A die is rolled 200 times. The outcomes obtained are shown in the table below.
Number: 1, 2, 3, 4, 5, 6
No. of times: 25, 30, 45, 28, 40, 32
What is the probability of obtaining a number less than 3?
Answer: C (0.275)
1 or 2 came up 25 + 30 = 55 times out of 200: 55/200 = 0.275.
(a) Simplify, without using Mathematical tables: log10(30/16) − 2log10(5/9) + log10(400/243)
(b) Without using Mathematical tables, calculate √(P/Q) where P = 3.6 × 10−3 and Q = 2.25 × 106, leaving your answer in standard form.
(a) 2log(5/9) = log(25/81)
Expression = log[(30/16) ÷ (25/81) × (400/243)] = log[(30/16) × (81/25) × (400/243)]
= log1010 = 1
(b) P/Q = 3.6 × 10−3 ÷ (2.25 × 106) = 1.6 × 10−9 = 16 × 10−10
√(P/Q) = 4 × 10−5
Make the power of ten even before taking the square root.
The Universal set is the set of all integers and the subjects P, Q, R of ε are given by
P = {x: x < 0}
Q = {…, −5, −3, −1, 1, 3, 5, …..}
R = {x: −2 ≤ x < 7}
(a) Find Q∩R. (b) Find R' where R' is the complement of R with respect to ε. (c) Find P'∩R'. (d) List the members of (P∩Q)'.
R = {−2, −1, 0, 1, 2, 3, 4, 5, 6}; Q is the set of odd integers
(a) Q∩R = {−1, 1, 3, 5}
(b) R' = {x: x < −2 or x ≥ 7} = {…, −5, −4, −3, 7, 8, 9, …}
(c) P' = {0, 1, 2, 3, …}, so P'∩R' = {7, 8, 9, …}
(d) P∩Q = {…, −5, −3, −1} (negative odd integers)
(P∩Q)' = {…, −6, −4, −2, 0, 1, 2, 3, 4, …}: all even integers together with the non-negative odd integers
Since ε is all integers, a complement contains every integer not in the set.
A simple measuring device is used at points X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to be 5,200m above sea level, |XY| = 4,000m and the measurements of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (Take tan 15° = 0.3 and tan 35° = 0.7)
Let h be the height of P above the level of X and Y, and O the point on that level directly below P. X, Y and O are in a line, Y nearer the mountain.
XO = h/tan 15° = h/0.3; YO = h/tan 35° = h/0.7
XO − YO = 4000: h(1/0.3 − 1/0.7) = 4000, h × 40/21 = 4000
h = 2100 m
Height of mountain above sea level = 5200 + 2100 = 7300 m
(If X and Y are taken on opposite sides of the mountain, XO + YO = 4000 gives h(1/0.3 + 1/0.7) = 4000, h = 840 m and the height is 6040 m.)
Express both horizontal distances in terms of h, then use the 4,000 m between X and Y.
(a) Simplify 3/(m + 2n) − 2/(m − 3n)
(b) A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.
(a) = [3(m − 3n) − 2(m + 2n)] / [(m + 2n)(m − 3n)]
= (3m − 9n − 2m − 4n)/[(m + 2n)(m − 3n)] = (m − 13n)/[(m + 2n)(m − 3n)]
(b) Let the number be 10t + u. Then t + u = 11
Interchanged: (10u + t) − (10t + u) = 9, so 9u − 9t = 9 and u − t = 1
Solving: u = 6, t = 5
The number is 56 (check: 65 − 56 = 9)
A two-digit number with tens digit t and units digit u has value 10t + u.
A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that: (a) three are red; (b) the first is blue and the other two are red; (c) two are white and one is blue.
Total = 36 balls
(a) P(3 red) = 12/36 × 11/35 × 10/34 = 11/357
(b) P(blue, red, red) = 8/36 × 12/35 × 11/34 = 44/1785
(c) One order (W, W, B): 16/36 × 15/35 × 8/34 = 32/714 = 16/357
The blue ball can be 1st, 2nd or 3rd, so P = 3 × 16/357 = 16/119
(If the order white, white, blue is intended, the answer is 16/357.)
Without replacement, each draw has one ball fewer in the box.
(a)(i) Given that log105 = 0.699 and log103 = 0.477, find log1045, without using mathematical tables.
(ii) Hence, solve x0.8265 = 45
(b) Use mathematical tables to evaluate √(2.067/(0.0348 × 0.538))
(a)(i) log 45 = log(9 × 5) = 2 log 3 + log 5 = 0.954 + 0.699 = 1.653
(ii) Take logs: 0.8265 log x = log 45 = 1.653
log x = 1.653 ÷ 0.8265 = 2, so x = 102 = 100
(b) log 2.067 = 0.3154; log 0.0348 = 2̄.5416; log 0.538 = 1̄.7308
Denominator: 2̄.5416 + 1̄.7308 = 2̄.2724
0.3154 − 2̄.2724 = 2.0430; ÷ 2 = 1.0215
antilog 1.0215 ≈ 10.51
log 45 = log 9 + log 5, and log 9 = 2 log 3.
(a) In the diagram below, BA is parallel to DE. Find the value of x.
(b) Illustrate graphically and shade the region in which the inequalities y − 2x < 5, 2y + x ≥ 4
(a) The interior angle at D is 360° − 312° = 48°.
Draw a line through C parallel to BA and DE: it splits ∠BCD into 52° (alternate to ∠ABC) and 48° (alternate to the 48° at D), so ∠BCD = 100°.
x is the reflex angle at C: x = 360° − 100° = 260°
(b) Draw y − 2x = 5 as a broken line (through (0, 5) and (−2.5, 0)) and x + 2y = 4 as a solid line (through (0, 2) and (4, 0)).
The required region is below y = 2x + 5 and on or above x + 2y = 4 (it contains e.g. (2, 2)); the lines meet at (−1.2, 2.6). Shade (or leave unshaded, as instructed) and label the region.
Through the bend point, draw a line parallel to the two parallel lines and use alternate angles.
(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) In the diagram below, O is the centre of the circle ACDB. If ∠CAO = 26° and ∠AOB = 130°, calculate: (i) ∠OBC; (ii) ∠COB
(a) Given: arc AB subtends ∠AOB at the centre O and ∠ACB at C on the remaining arc. To prove: ∠AOB = 2∠ACB.
Join CO and produce it to D. OA = OC (radii), so ∠OAC = ∠OCA = x; the exterior angle ∠AOD = 2x.
Similarly OB = OC gives ∠OCB = ∠OBC = y and ∠BOD = 2y.
∠AOB = 2x + 2y = 2(x + y) = 2∠ACB. (Diagram, given, to prove, construction and proof required.)
(b)(i) ∠ACB = ½ × 130° = 65° (angle at centre = twice angle at circumference)
OA = OC, so ∠ACO = ∠CAO = 26°; ∠BCO = 65° − 26° = 39°
OB = OC, so ∠OBC = ∠BCO = 39°
(ii) ∠COB = 180° − 39° − 39° = 102°
Equal radii make isosceles triangles, so their base angles are equal.
(a) What is the 25th term of 5, 9, 13……..?
(b) Find the 5th term of 8/9, −4/3, 2 ………
(c) The 3rd and 6th terms of a Geometric Progression (G.P) are 48 and 14 2/9 respectively. Write down the first four terms of the G.P.
(a) A.P. with a = 5, d = 4: T25 = 5 + 24 × 4 = 101
(b) G.P. with a = 8/9, r = (−4/3) ÷ (8/9) = −3/2
T5 = 8/9 × (−3/2)4 = 8/9 × 81/16 = 9/2 = 4½
(c) ar2 = 48 and ar5 = 128/9
Divide: r3 = 128/432 = 8/27, so r = 2/3
a = 48 ÷ 4/9 = 108
First four terms: 108, 72, 48, 32
Dividing two terms of a G.P. removes a and leaves a power of r.
(a) Copy and complete the following table of values for y = 3 sin 2θ − cos θ
θ: 0°, 30°, 60°, 90°, 120°, 150°, 180°
y: −1.0, , , 0, , , 1.0
(b) Using a scale of 2cm to 30° on the θ axis and 2cm to 1 unit on the y axis, draw the graph of y = 3sin 2θ − cos θ for 0° ≤ θ ≤ 180°.
(c) Use your graph to find the: (i) solution of the equation 3 sin 2θ − cos θ = 0, correct to the nearest degree; (ii) maximum value y, correct to one decimal place.
(a) θ = 0° → −1.0; 30° → 1.7; 60° → 2.1; 90° → 0; 120° → −2.1; 150° → −1.7; 180° → 1.0
(b) Plot the points with the given scales and join with a smooth curve.
(c)(i) Read where the curve cuts the θ-axis: θ ≈ 10°, 90° and 170° (exact: cos θ(6 sin θ − 1) = 0 gives 9.6°, 90°, 170.4°)
(ii) Maximum y ≈ 2.3 (at about θ = 49°)
y is zero where the curve crosses the θ-axis.
The table below shows the frequency distribution of the marks scored by fifty students 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: 2, 3, 4, 6, 13, 10, 5, 3, 2, 2
(a) Draw a cumulative frequency curve for the distribution.
(b) Use your curve to estimate the: (i) upper quartile; (ii) pass mark if 60% of the students passed.
(a) Upper class boundaries 9.5, 19.5, …, 99.5 with cumulative frequencies 2, 5, 9, 15, 28, 38, 43, 46, 48, 50. Plot them (start at (−0.5, 0)) and draw a smooth ogive.
(b)(i) Upper quartile at the ¾ × 50 = 37.5th student: about 59 marks (49.5 + 9.5/10 × 10)
(ii) 60% passed, so 40% (20 students) failed. The pass mark is the mark at cumulative frequency 20: about 43 (39.5 + 5/13 × 10 = 43.3)
The pass mark is the score below which the failing 40% lie.
P and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along (a) the parallel of latitude; (b) a great circle. (Take π = 22/7 and radius of the earth = 6400km)
Difference in longitude = 33° + 20° = 53°
(a) Radius of the parallel = 6400 cos 55° = 3671 km
Distance = 53/360 × 2 × 22/7 × 6400 × cos 55° ≈ 3397 km (about 3400 km)
(b) Chord PQ = 2 × 6400 cos 55° × sin 26.5° ≈ 3277 km
Half the angle at the earth's centre: sin(θ/2) = 3277 ÷ (2 × 6400) = 0.2560, so θ ≈ 29.7°
Great-circle distance = 29.7/360 × 2 × 22/7 × 6400 ≈ 3316 km
Along a parallel use the smaller radius R cos(latitude); along a great circle use the full earth radius.