54 questions. Try each one, then open the answer to check yourself and read the explanation.
Evaluate, correct to four significant figures, (573.06 × 184.25).
Answer: A (105600.00)
573.06 × 184.25 = 105,585.3. The fifth figure is 8, so round up to 105,600.
Change 432five to a number in base three.
Answer: B (11100three)
432five = 4 × 25 + 3 × 5 + 2 = 117. Dividing by 3 repeatedly gives remainders 0, 0, 1, 1, 1, so 117 = 11100three.
Given that A and B are sets such that n(A) = 8, n(B) = 12 and n(A ∩ B) = 3, find n(A ∪ B).
Answer: B (17)
n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 8 + 12 − 3 = 17.
If √24 + √96 − √600 = y√6, find the value of y.
Answer: D (−4)
√24 = 2√6, √96 = 4√6 and √600 = 10√6. So 2 + 4 − 10 = −4.
Evaluate 23 × 54 (mod 7).
Answer: B (3)
23 ≡ 2 and 54 ≡ 5 (mod 7), so the product ≡ 10 ≡ 3 (mod 7).
If 43x = 16x + 1, find the value of x.
Answer: B (2)
16 = 42, so 3x = 2(x + 1) = 2x + 2, giving x = 2.
A weaver bought a bundle of grass for $50.00 from which he made 8 mats. If each mat was sold for $15.00, find the percentage profit.
Answer: B (140%)
He sold the mats for 8 × $15 = $120, a profit of $70 on $50. 70/50 × 100 = 140%.
Find the 17th term of the Arithmetic Progression (A.P.): −6, −1, 4, ….
Answer: C (74)
a = −6 and d = 5. T17 = a + 16d = −6 + 80 = 74.
If a = 3 and b = −7, find the value of (5b + (a + b)2)/(a − b)2
Answer: C (−0.19)
Top: 5(−7) + (−4)2 = −35 + 16 = −19. Bottom: (10)2 = 100. So the value is −0.19.
Three boys shared D10,500.00 in the ratio 6 : 7 : 8. Find the largest share.
Answer: A (D4,000.00)
There are 21 parts. The largest share is 8/21 × 10,500 = D4,000.
The length of a piece of stick is 1.75m. A boy measured it as 1.80m. Find the percentage error.
Answer: B (2 6/7%)
Error = 0.05 m. Percentage error = 0.05/1.75 × 100 = 20/7 = 2 6/7%.
If 5x + 3y = 4 and 5x − 3y = 2, what is the value of (25x2 − 9y2)?
Answer: D (8)
25x2 − 9y2 = (5x + 3y)(5x − 3y) = 4 × 2 = 8.
Mary has $3.00 more than Ben but $5.00 less than Jane. If Mary has $x, how much does Jane and Ben have altogether?
Answer: D ($(2x + 2))
Ben has $(x − 3) and Jane has $(x + 5). Together: $(2x + 2).
Consider the statements: p: Stephen is intelligent; q: Stephen is good at Mathematics. If p ⇒ q, which of the following is a valid conclusion?
Answer: B (If Stephen is not good at Mathematics, then he is not intelligent.)
p ⇒ q always has the same truth as its contrapositive ~q ⇒ ~p: if he is not good at Mathematics, he is not intelligent.
What value of p will make (x2 − 4x + p) a perfect square?
Answer: C (4)
A perfect square needs p = (half of −4)2 = 4, giving (x − 2)2.
Find the value of x such that 1/x + 4/(3x) − 5/(6x) + 1 is zero.
Answer: C (−3/2)
Over 6x: (6 + 8 − 5)/(6x) = −1, so 9/(6x) = −1 and x = −3/2.
In the diagram, |XY| = |YZ| and ∠XYZ = 130°. Find the value of y.
Answer: D (155°)
The base angles are (180° − 130°)/2 = 25°. y is on the straight line at Z: 180° − 25° = 155°.
An exterior angle of a regular polygon is 22.5°. Find the number of sides.
Answer: D (16)
Number of sides = 360° ÷ 22.5° = 16.
In the diagram, ∠POQ = 150° and the radius of the circle PSQR is 4.2cm. [Take π = 22/7] Find the length of the minor arc PRQ.
Answer: A (11.00cm)
Arc = (150/360) × 2 × (22/7) × 4.2 = 11.00 cm.
In the diagram, ∠POQ = 150° and the radius of the circle PSQR is 4.2cm. [Take π = 22/7] Find the area of the sector OPSQ.
Answer: D (32.34cm2)
Sector OPSQ is the major sector, with angle 360° − 150° = 210°. Area = (210/360) × (22/7) × 4.22 = 32.34 cm2.
A cylinder, opened at one end, has a radius of 3.5cm and height 8cm. Calculate the total surface area. [Take π = 22/7]
Answer: D (214.5cm2)
Curved surface 2πrh = 176 cm2 plus one base πr2 = 38.5 cm2: 214.5 cm2.
In the diagram, ∠WZY and ∠WYX are right angles. Find the perimeter of WXYZ.
Answer: B (32cm)
WY = √(42 + 32) = 5 cm and WX = √(52 + 122) = 13 cm. Perimeter = 4 + 3 + 12 + 13 = 32 cm.
The length of a rectangle is 10cm. If its perimeter is 28cm, find the area.
Answer: B (40cm2)
Length + width = 14, so width = 4 cm and area = 10 × 4 = 40 cm2.
In the diagram, MRW and MNST are straight lines. |MN| = |NR|, ∠MNR = 110° and ∠WRS = 86°. Find the value of x.
Answer: C (51°)
Triangle MNR is isosceles, so ∠NMR = 35°. ∠MRS = 180° − 86° = 94°. In triangle MRS, x = 180° − 35° − 94° = 51°.
A boy 1.4m tall, stood 10m away from a tree of height 12m. Calculate, correct to the nearest degree, the angle of elevation of the top of the tree from the boy's eyes.
Answer: B (47°)
Above eye level the tree rises 12 − 1.4 = 10.6 m. tan θ = 10.6/10 = 1.06, so θ ≈ 47°.
In the diagram, MNR is a tangent to the circle centre O at N and ∠NOS = 108°. Find ∠OSN.
Answer: C (36°)
OS = ON (radii), so triangle ONS is isosceles: ∠OSN = (180° − 108°)/2 = 36°.
In the diagram, MNR is a tangent to the circle centre O at N and ∠NOS = 108°. Find ∠SNR.
Answer: C (54°)
The radius ON is perpendicular to the tangent, so ∠SNR = 90° − ∠ONS = 90° − 36° = 54°.
Mrs. Gabriel is pregnant. The probability that she will give birth to a girl is ½ and the probability that the baby will have blue eyes is ¼. What is the probability that she will give birth to a girl with blue eyes?
Answer: C (1/8)
The two events are independent, so multiply: ½ × ¼ = 1/8.
The mean of a set of 10 numbers is 56. If the mean of the first nine numbers is 55, find the 10th number.
Answer: B (65)
Total of 10 numbers = 560 and of the first nine = 495. The 10th number = 560 − 495 = 65.
The diagram shows triangle PQR inscribed in a circle. PS is a tangent to the circle at P. Find ∠PRQ.
Answer: A (49°)
By the alternate segment theorem ∠PQR = ∠RPS = 73°. With ∠QPR = 58°, ∠PRQ = 180° − 58° − 73° = 49°.
In the diagram, triangle MNR is inscribed in circle MNR and PQ is a straight line. If ∠MRN = 41° and ∠PMR = 141°, find ∠QNR.
Answer: B (80°)
∠RMN = 180° − 141° = 39°. ∠QNR is an exterior angle of triangle MNR, so it equals 39° + 41° = 80°.
Solve: (y + 2)/4 − (y − 1)/3 > 1.
Answer: B (y < −2)
Multiply by 12: 3y + 6 − 4y + 4 > 12, so −y > 2 and y < −2.
The ages (years) of some members in a singing group are: 12, 47, 49, 15, 43, 41, 13, 39, 43, 41 and 36. Find the lower quartile.
Answer: C (15)
In order: 12, 13, 15, 36, 39, 41, 41, 43, 43, 47, 49. The lower quartile is the (11 + 1)/4 = 3rd value, 15.
The ages (years) of some members in a singing group are: 12, 47, 49, 15, 43, 41, 13, 39, 43, 41 and 36. Find the mean.
Answer: C (34.45)
The ages add up to 379, so the mean is 379 ÷ 11 ≈ 34.45.
Find, correct to two decimal places, the volume of a sphere whose radius is 3cm. [Take π = 22/7]
Answer: D (113.14cm3)
V = (4/3) × (22/7) × 27 = 792/7 ≈ 113.14 cm3.
A local community has two newspapers: the Morning Times and the Evening Dispatch. The Morning Times is read by 45% of households and the Evening Dispatch by 60%. If 20% of the households read both papers, find the probability that a particular household reads at least one paper.
Answer: C (0.85)
P(at least one) = 0.45 + 0.60 − 0.20 = 0.85.
A rectangle has width ¾ cm and an area 3⅜ cm2. Find the length.
Answer: B (4½ cm)
Length = area ÷ width = 27/8 ÷ 3/4 = 27/8 × 4/3 = 9/2 = 4½ cm.
The mean of two numbers x and y is 4. Find the mean of the four numbers x, 2x, y and 2y.
Answer: C (6)
x + y = 8, so x + 2x + y + 2y = 3(x + y) = 24 and the mean is 24 ÷ 4 = 6.
The straight line y = mx − 4 passes through the point (−4, 16). Calculate the gradient of the line.
Answer: A (−5)
16 = −4m − 4, so −4m = 20 and m = −5.
If the equations x2 − 5x + 6 = 0 and x2 + px + 6 = 0 have common roots, find the value of p.
Answer: D (−5)
Both equations have the same roots and the same constant 6, so they are the same equation: p = −5 (roots 2 and 3).
Given that log327 = 2x + 1, find the value of x.
Answer: B (1)
log327 = 3, so 2x + 1 = 3 and x = 1.
Solve: 6x2 = 5x − 1.
Answer: C (x = 1/2, 1/3)
6x2 − 5x + 1 = 0 factorises as (3x − 1)(2x − 1) = 0, so x = 1/3 or 1/2.
(a) Given that (7 − 2x), 9, (5x + 17) are consecutive terms of a Geometric Progression (G.P.) with common ratio, r > 0, find the values of x.
(b) Two positive numbers are in the ratio 3 : 4. The sum of thrice the first number and twice the second is 68. Find the smaller number.
(a) Middle term squared = product of neighbours: 81 = (7 − 2x)(5x + 17)
81 = 35x + 119 − 10x2 − 34x → 10x2 − x − 38 = 0
(10x + 19)(x − 2) = 0 → x = 2 or x = −19/10
Check r > 0: x = 2 gives 3, 9, 27 (r = 3); x = −1.9 gives 10.8, 9, 7.5 (r = 5/6). Both are valid
(b) Numbers 3k and 4k: 9k + 8k = 68 → k = 4
Smaller number = 12
In a G.P., T2/T1 = T3/T2.
Given that y = [pr/m − p2r]−3/2, (a) make r the subject; (b) find the value of r when y = −8, m = 1 and p = 3.
(a) Raise both sides to the power −2/3: y−2/3 = pr/m − p2r = r(p − mp2)/m
r = my−2/3/(p − mp2) = m/(y2/3p(1 − mp))
(b) y2/3 = ((−8)1/3)2 = (−2)2 = 4
r = 1/(4 × 3 × (1 − 3)) = −1/24
Undo the power −3/2 by raising both sides to −2/3.
A chord subtends an angle of 72° at the centre of a circle of radius 24.5m. Calculate the perimeter of the minor segment. [Take π = 22/7]
Chord = 2r sin(72°/2) = 49 sin 36° ≈ 28.80 m
Minor arc = 72/360 × 2 × (22/7) × 24.5 = 30.8 m
Perimeter = 28.80 + 30.80 ≈ 59.6 m
The minor segment is bounded by the chord and the minor arc.
In the diagram BCDE, is a circle with centre A. ∠BCD = (2x + 40°), ∠BAD = (5x − 35)°, ∠BED = (2y + 10°) and ∠ADC = 40°. Find: (a) the values of x and y; (b) ∠ABC
(a) Angle at centre = twice angle at circumference: 5x − 35 = 2(2y + 10) → 5x − 4y = 55
BCDE is cyclic: (2x + 40) + (2y + 10) = 180 → x + y = 65
Solve: 9x = 315 → x = 35, y = 30
(b) ∠BAD = 140°, ∠BCD = 110°
Quadrilateral ABCD: ∠ABC = 360° − (140° + 110° + 40°) = 70°
Form two equations: one from the angle at the centre, one from the cyclic quadrilateral.
(a) Given that m = tan 30° and n = tan 45°, simplify, without using calculator, (m − n)/(mn), leaving the answer in the form p + √q.
(b) There are 20 women in a bus. 15 of them wear glasses and 10 wear wrist watches. If a woman is chosen at random from the bus, find the probability that she wears both glasses and wrist watch.
(a) m = 1/√3, n = 1
(m − n)/(mn) = (1/√3 − 1) ÷ (1/√3) = 1 − √3
So p = 1 and the surd part is −√3
(b) Assume every woman wears at least one: 15 + 10 − x = 20 → x = 5
P(both) = 5/20 = 1/4
Dividing by 1/√3 is the same as multiplying by √3.
(a) A man purchased 180 copies of a book at ₦250.00 each. He sold y copies at ₦300.00 each and the rest at the discount of 5 kobo in the Naira of the cost price. If he made a profit of ₦7,125.00, find the value of y.
(b) A trader bought x bags of rice at a cost, c = 24x + 103 and sold them at a price, s = 33x − x2/20. (i) Find the expression for the profit. (ii) If 20 bags of rice were sold, calculate the percentage profit.
(a) Cost = 180 × 250 = ₦45,000; total sales = 45,000 + 7,125 = ₦52,125
Discounted price = 250 × 0.95 = ₦237.50
300y + 237.50(180 − y) = 52,125 → 42,750 + 62.5y = 52,125
y = 150
(b)(i) Profit = s − c = 9x − x2/20 − 103
(ii) x = 20: profit = 180 − 20 − 103 = 57; cost = 24(20) + 103 = 583
Percentage profit = 57/583 × 100 ≈ 9.8%
5 kobo in the naira is a 5% discount on the cost price.
The table shows the monthly expenditure (in percentage) of Mr. Okafor's salary.
Item: Food and drinks, Fuel, Rent, Building Project, Education, Savings
Percentage (%): 35, 7.5, 10, 15, 17.5, x
(a) Calculate the percentage of Mr. Okafor's salary that was put into savings.
(b) Illustrate the information on a pie chart.
(c) If Mr Okafor's annual gross salary is $28,800.00 and he pays tax of 12%, calculate: (i) his monthly tax; (ii) amount saved each month.
(a) x = 100 − (35 + 7.5 + 10 + 15 + 17.5) = 15%
(b) Angles (× 3.6°): food 126°, fuel 27°, rent 36°, building 54°, education 63°, savings 54°; draw and label the sectors
(c)(i) Monthly salary = 28,800 ÷ 12 = $2,400; tax = 12% of 2,400 = $288
(ii) Savings = 15% of $2,400 = $360 a month
Percentages of a whole always add up to 100.
(a) Copy and complete the table of values for y = 3 sin x + 7 cos x for 0° ≤ x ≤ 180°.
x: 0°, 20°, 40°, 60°, 80°, 100°, 120°, 140°, 160°, 180°
y: 7.0, __, __, __, 4.2, __, −0.9, __, __, __
(b) Using a scale of 2cm to 20° on the x-axis and 2cm to 2 units on the y-axis, draw the graph of y = 3 sin x + 7 cos x for 0° ≤ x ≤ 180°.
(c) Using the graph, find the: (i) value of y when x = 150°; (ii) range of values of x for which y > 0.
(a) y = 7.0, 7.6, 7.3, 6.1, 4.2, 1.7, −0.9, −3.4, −5.6, −7.0
(b) Plot the points and join with a smooth curve
(c)(i) At x = 150°, y ≈ −4.6
(ii) The curve is above the x-axis from x = 0° to where it crosses at about 113°: 0° ≤ x < 113°
Use your calculator in degree mode for each value of x.
The table shows the distribution of ages of a number of children in a school.
Age (years): 3, 4, 5, 6, 7, 8, 9, 10
Number of Children: 2, 6, 5, x, 6, 9, 8, 5
If the mean of the distribiution is 7, find the: (a) value of x; (b) standard deviation of their ages.
(a) Σf = 41 + x; Σfx = 6 + 24 + 25 + 6x + 42 + 72 + 72 + 50 = 291 + 6x
(291 + 6x)/(41 + x) = 7 → 287 + 7x = 291 + 6x → x = 4
(b) Σf = 45; deviations from 7: −4, −3, −2, −1, 0, 1, 2, 3
Σf(x − 7)2 = 32 + 54 + 20 + 4 + 0 + 9 + 32 + 45 = 196
Variance = 196/45 ≈ 4.36; standard deviation ≈ 2.09
Use the mean of 7 to find x first, then the deviations from 7.
(a) The exterior angles of a polygon are 42°, 38°, 57°, x°, (x + y)°, (2x − 15)° and (3x − y)°. If x is 7° less than y, find the values of x and y.
(b) In the diagram, O is the centre of the circle XYZ, ∠ZXO = 34° and ∠XOY = 146°. Find ∠OYZ.
(a) y = x + 7; exterior angles add to 360°
137 + x + (2x + 7) + (2x − 15) + (3x − x − 7) = 360 → 122 + 7x = 360
x = 34, y = 41
(b) ∠XZY = ½ × 146° = 73° (angle at centre)
OX = OZ → ∠OZX = 34°, so ∠OZY = 73° − 34° = 39°
OY = OZ → ∠OYZ = 39°
Exterior angles of any polygon add up to 360°.
(a) The probability that an athlete will not win any of the three races is ¼. If the athlete runs in all the races, what is the probability that the athlete will win: (i) only the second race; (ii) all the three races; (iii) only two of the races?
(b) A cone with perpendicular height 24cm has a volume of 24cm3. Find the volume of a cone with same base radius and height 84cm. [Take π = 22/7]
(a) Taking P(lose a race) = ¼ and P(win a race) = ¾ for each race:
(i) P(L W L) = ¼ × ¾ × ¼ = 3/64
(ii) P(W W W) = (¾)3 = 27/64
(iii) P(LWW) + P(WLW) + P(WWL) = 3 × (¾ × ¾ × ¼) = 27/64
(b) Same base radius, so volume is proportional to height
V = 24 × 84/24 = 84 cm3
For (b), keep r fixed: V = ⅓πr2h grows in step with h.
(a) The diameter of a cylinder closed at both ends is 7cm. If the total surface area is 209 cm2, calculate the height. [Take π = 22/7]
(b) The points X and Y, 19m apart are on the same side of a tree. The angles of elevation of the top, T, of the tree from X and Y on the horizontal ground with the foot of the tree are 43° and 38° respectively. (i) Illustrate the information in a diagram. (ii) Find, correct to one decimal place, the height of the tree.
(a) r = 3.5 cm; TSA = 2πr(r + h) = 2 × (22/7) × 3.5 × (3.5 + h) = 22(3.5 + h)
22(3.5 + h) = 209 → 3.5 + h = 9.5 → h = 6 cm
(b)(i) Tree TB vertical; X nearer (43°), Y 19 m further away (38°), all on a line with B
(ii) Let XB = d, height h: h = d tan 43° = (d + 19) tan 38°
d(tan 43° − tan 38°) = 19 tan 38° → d ≈ 98.2 m
h = 98.2 × tan 43° ≈ 91.5 m
The larger angle of elevation belongs to the point nearer the tree.