57 questions. Try each one, then open the answer to check yourself and read the explanation.
Express, correct to three significant figures, 0.003597.
Answer: C (0.00360)
The first three significant figures are 3, 5, 9 and the next digit is 7, so round 359 up to 360: 0.00360. The final zero must be kept to show three significant figures.
Evaluate: (0.064)−1/3
Answer: A (5/2)
0.064 = 64/1000, whose cube root is 4/10 = 2/5. The negative power turns it upside down: 5/2.
Solve: (y + 1)/2 − (2y − 1)/3 = 4.
Answer: B (y = −19)
Multiply by 6: 3(y + 1) − 2(2y − 1) = 24, so 3y + 3 − 4y + 2 = 24, giving −y = 19 and y = −19.
Simplify, correct to three significant figures, (27.63)2 − (12.37)2.
Answer: D (610)
Use the difference of two squares: (27.63 + 12.37)(27.63 − 12.37) = 40 × 15.26 = 610.4, which is 610 to 3 s.f.
If 7 + y = 4 (mod 8), find the least value of y, 10 ≤ y ≤ 30.
Answer: B (13)
y = 4 − 7 = −3 ≡ 5 (mod 8), so y can be 5, 13, 21, 29, … The least value from 10 upwards is 13.
Evaluate: (log39 − log28)/log39
Answer: D (−1/2)
log39 = 2 and log28 = 3, so the value is (2 − 3)/2 = −1/2.
If 23y = 1111two, find the value of y.
Answer: C (6)
1111two = 8 + 4 + 2 + 1 = 15. So 2y + 3 = 15, giving y = 6.
If 6, P and 14 are consecutive terms in an Arithmetic Progression (A.P), find the value of P.
Answer: B (10)
The middle term of an A.P. is the average of its neighbours: P = (6 + 14)/2 = 10.
Evaluate: 2√28 − 3√50 + √72
Answer: C (4√7 − 9√2)
2√28 = 4√7, 3√50 = 15√2 and √72 = 6√2. So the result is 4√7 − 15√2 + 6√2 = 4√7 − 9√2.
If m : n = 2 : 1, evaluate (3m2 − 2n2)/(m2 + mn)
Answer: B (5/3)
Put m = 2 and n = 1: (12 − 2)/(4 + 2) = 10/6 = 5/3.
H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y.
Answer: C (H = p/(2y2))
H = kp/y2. Then 1 = 8k/4, so k = 1/2 and H = p/(2y2).
Solve 4x2 − 16x + 15 = 0.
Answer: B (x = 1½ or x = 2½)
4x2 − 16x + 15 = (2x − 3)(2x − 5) = 0, so x = 3/2 = 1½ or x = 5/2 = 2½.
Simplify: log106 − 3log103 + (2/3)log1027
Answer: B (log102)
Write each term as one log: log 6 − log 27 + log 9 = log(6 × 9/27) = log102.
Bala sold an article for ₦6,900.00 and made a profit of 15%. Calculate his percentage profit if he had sold it for ₦6,600.00.
Answer: B (10%)
Cost price = 6,900 ÷ 1.15 = ₦6,000. Selling at ₦6,600 gives a profit of ₦600, which is 600/6,000 × 100 = 10%.
If 3p = 4q and 9p = 8q − 12, find the value of pq.
Answer: A (12)
8q = 2(4q) = 6p, so 9p = 6p − 12, giving p = −4. Then q = 3p/4 = −3, and pq = 12.
If (0.25)y = 32, find the value of y.
Answer: A (y = −5/2)
0.25 = 2−2 and 32 = 25, so −2y = 5 and y = −5/2.
Simplify: (x2 − 5x − 14)/(x2 − 9x + 14)
Answer: D ((x + 2)/(x − 2))
Factorise: (x − 7)(x + 2) over (x − 7)(x − 2). Cancel (x − 7) to get (x + 2)/(x − 2).
The total surface area of a solid cylinder 165 cm2. If the base diameter is 7 cm, calculate its height. [Take π = 22/7]
Answer: C (4.0 cm)
Total surface area = 2πr2 + 2πrh with r = 3.5: 165 = 77 + 22h, so 22h = 88 and h = 4 cm.
If 2a = √64 and b/a = 3, evaluate a2 + b2.
Answer: C (90)
√64 = 8 = 23, so a = 3 and b = 9. Then a2 + b2 = 9 + 81 = 90.
In XYZ, |YZ| = 32 cm, ∠YXZ = 52° and ∠XZY = 90°. Find, correct to the nearest centimetre, |XZ|.
Answer: B (25 cm)
YZ is opposite the 52° angle and XZ is next to it, so tan 52° = 32/|XZ|. |XZ| = 32 ÷ 1.2799 ≈ 25 cm.
If logx2 = 0.3, evaluate logx8.
Answer: C (0.9)
logx8 = logx23 = 3 logx2 = 3 × 0.3 = 0.9.
An arc subtends an angle of 72° at the centre of a circle. Find the length of the arc if the radius of the circle is 3.5 cm. [Take π = 22/7]
Answer: C (4.4 cm)
Arc length = (72/360) × 2 × (22/7) × 3.5 = (1/5) × 22 = 4.4 cm.
Make b the subject of the relation lb = ½(a + b)h.
Answer: A (ah/(2l − h))
Multiply by 2: 2lb = ah + bh. Collect b: b(2l − h) = ah, so b = ah/(2l − h).
Eric sold his house through an agent who charged 8% commission on the selling price. If Eric received $117,760.00 after the sale, what was the selling price of the house?
Answer: B ($128,000.00)
Eric kept 92% of the selling price, so the price = 117,760 ÷ 0.92 = $128,000.
Find the angle which an arc of length 22 cm subtends at the centre of a circle of radius 15 cm. [Take π = 22/7]
Answer: B (84°)
22 = (θ/360) × 2 × (22/7) × 15, so θ = 22 × 360 × 7/(44 × 15) = 84°.
A rectangular board has length 15 cm and width x cm. If its sides are doubled, find its new area.
Answer: A (60x cm2)
The new sides are 30 cm and 2x cm, so the area is 30 × 2x = 60x cm2.
Factorize completely: (2x + 2y)(x − y) + (2x − 2y)(x + y).
Answer: A (4(x − y)(x + y))
Each part is 2(x + y)(x − y), so the sum is 4(x − y)(x + y).
The interior angles of a polygon are 3x°, 2x°, 4x°, 3x° and 6x°. Find the size of the smallest angle of the polygon.
Answer: B (60°)
A 5-sided polygon has angles adding to 540°. 18x = 540, so x = 30 and the smallest angle is 2x = 60°.
A box contains 2 white and 3 blue identical balls. If two balls are picked at random from the box, one after the other with replacement, what is the probability that they are of different colours?
Answer: D (12/25)
P(white then blue) + P(blue then white) = (2/5)(3/5) + (3/5)(2/5) = 12/25.
Find the equation of a straight line passing through the point (1, −5) and having gradient of 3/4.
Answer: D (3x − 4y − 23 = 0)
y + 5 = (3/4)(x − 1). Multiply by 4: 4y + 20 = 3x − 3, so 3x − 4y − 23 = 0.
The foot of a ladder is 6 m from the base of an electric pole. The top of the ladder rest against the pole at a point 8 m above the ground. How long is the ladder?
Answer: C (10 m)
The ladder is the hypotenuse: √(62 + 82) = √100 = 10 m.
If tan x = 3/4, 0 < x < 90°, evaluate cos x/(2 sin x).
Answer: D (2/3)
With sides 3, 4, 5: sin x = 3/5 and cos x = 4/5. So (4/5) ÷ (6/5) = 2/3.
From the top of a vertical cliff 20 m high a boat at sea can be sighted 75 m away and on the same horizontal position as the foot of the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.
Answer: D (15°)
tan θ = 20/75 = 0.267, so θ ≈ 14.9°, which is 15° to the nearest degree.
In the diagram, O is the centre of the circle of radius 18 cm. If ∠ZXY = 70°, calculate the length of arc ZY. [Take π = 22/7]
Answer: C (44 cm)
The angle at the centre is twice the angle at the circumference: ∠ZOY = 140°. Arc = (140/360) × 2 × (22/7) × 18 = 44 cm.
In the diagram, RT is a tangent to the circle at R, ∠PQR = 70°, ∠QRT = 52°, ∠QSR = y and ∠PRQ = x. Find the value of y.
Answer: C (52°)
The angle between a tangent and a chord equals the angle in the alternate segment, so ∠QSR = ∠QRT = 52°.
In the diagram, RT is a tangent to the circle at R, ∠PQR = 70°, ∠QRT = 52°, ∠QSR = y and ∠PRQ = x. Calculate the value of x.
Answer: B (58°)
∠QPR = ∠QRT = 52° (alternate segment). In triangle PQR, x = 180° − 70° − 52° = 58°.
The fourth term of an Arithmetic Progression (A.P.) is 37 and the first term is −20. Find the common difference.
Answer: C (19)
T4 = a + 3d, so −20 + 3d = 37, 3d = 57 and d = 19.
In the diagram, PQ is parallel to RS, ∠QFG = 105° and ∠FEG = 50°. Find the value of m.
Answer: D (55°)
∠QFG is an exterior angle of triangle EFG, so it equals the two opposite interior angles: 50° + m = 105°, giving m = 55°.
In the diagram, PQ is parallel to RS, ∠QFG = 105° and ∠FEG = 50°. Find the value of n.
Answer: C (75°)
∠FGS and ∠QFG are co-interior angles between parallel lines, so ∠FGS = 180° − 105° = 75°. Angle n is vertically opposite ∠FGS, so n = 75°.
A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?
Answer: C (1/3)
There are 18 pencils and 6 are green: 6/18 = 1/3.
The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?
Answer: A (90)
The orange sector is 360° − (60° + 100° + 120°) = 80°. So 80° stands for 60 fruits and the apple sector of 120° stands for 60 × 120/80 = 90 apples.
The following are scores obtained by some students in a test: 8, 18, 10, 14, 18, 11, 13, 14, 13, 17, 15, 8, 16, 13. Find the mode of the distribution.
Answer: C (13)
13 appears three times, more than any other score, so the mode is 13.
The following are scores obtained by some students in a test: 8, 18, 10, 14, 18, 11, 13, 14, 13, 17, 15, 8, 16, 13. Find the median score.
Answer: C (13.5)
In order the 14 scores are 8, 8, 10, 11, 13, 13, 13, 14, 14, 15, 16, 17, 18, 18. The middle two are 13 and 14, so the median is 13.5.
The following are scores obtained by some students in a test: 8, 18, 10, 14, 18, 11, 13, 14, 13, 17, 15, 8, 16, 13. How many students scored above the mean score?
Answer: D (7)
The total is 188, so the mean is 188/14 ≈ 13.4. The scores above it are 14, 14, 15, 16, 17, 18, 18: 7 students.
(a) Given that 110x = 40five, find the value of x.
(b) Simplify 15/√75 + √108 + √432, leaving the answer in the form a√b, where a and b are positive integers.
(a) 110x = x2 + x and 40five = 20
x2 + x − 20 = 0, (x + 5)(x − 4) = 0
A base must be positive, so x = 4
(b) 15/√75 = 15/(5√3) = √3; √108 = 6√3; √432 = 12√3
Sum = 19√3
Change both numbers to base ten; reject the negative root because a base is a positive whole number.
(a) Find the equation of the line which passes through the points A(−2, 7) and B(2, −3).
(b) Given that (5b − a)/(8b + 3a) = 1/5, find, correct to two decimal places, the value of a/b.
(a) Gradient = (−3 − 7)/(2 − (−2)) = −10/4 = −5/2
y − 7 = −5/2(x + 2) → 2y − 14 = −5x − 10
5x + 2y − 4 = 0
(b) Cross-multiply: 5(5b − a) = 8b + 3a → 25b − 5a = 8b + 3a
17b = 8a, so a/b = 17/8 = 2.125 ≈ 2.13
For (b), collect the a terms on one side and the b terms on the other, then divide.
(a) Ali, Musah and Yusif shared ₦420,000.00 in the ratio 3 : 5 : 8 respectively. Find the sum of Ali and Yusif's shares.
(b) Solve: 2(1/8)x = 32x − 1.
(a) Total parts = 16; Ali = 3/16 × 420,000 = ₦78,750; Yusif = 8/16 × 420,000 = ₦210,000
Sum = ₦288,750 (or 11/16 × 420,000)
(b) Write in base 2: 21 × 2−3x = 25(x − 1)
1 − 3x = 5x − 5 → 8x = 6
x = 3/4
Write every number as a power of 2 before equating indices.
In the diagram, PQRS is a quadrilateral, ∠PQR = ∠PRS = 90°, |PQ| = 3 cm, |QR| = 4 cm and |PS| = 13 cm. Find the area of the quadrilateral.
In ΔPQR: |PR| = √(32 + 42) = 5 cm
In ΔPRS: |RS| = √(132 − 52) = √144 = 12 cm
Area = ½ × 3 × 4 + ½ × 5 × 12 = 6 + 30
Area = 36 cm2
Split the quadrilateral along PR into two right-angled triangles.
Three red balls, five green balls and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is 1/6, find: (a) the number of blue balls in the sack; (b) the probability of picking a green ball.
(a) Let the number of blue balls be n; total = 8 + n
3/(8 + n) = 1/6 → 8 + n = 18 → n = 10
(b) Total = 18 balls, 5 green
P(green) = 5/18
Use P(red) = red balls ÷ total balls to find the total first.
(a) The force of attraction, F, between two bodies varies directly as the product of their masses, m1 and m2 and inversely as the square of the distance, d, between them. Given that F = 20 N, when m1 = 25 kg, m2 = 10 kg and d = 5 m, find: (i) an expression for F in terms of m1, m2 and d; (ii) the distance, d when F = 30 N, m1 = 7.5 kg and m2 = 4 kg.
(b) Find the value of x in the diagram.
(a)(i) F = km1m2/d2; 20 = k(25)(10)/25 → k = 2
F = 2m1m2/d2
(ii) 30 = 2(7.5)(4)/d2 = 60/d2 → d2 = 2
d = √2 m ≈ 1.41 m
(b) The figure is a pentagon: angle sum = (5 − 2) × 180° = 540°
x + (x + 20) + (x + 40) + (x + 60) + (x + 80) = 540 → 5x + 200 = 540
x = 68°
Find the constant k first, then use the formula with the new values.
The data show the marks obtained by students in a Biology test.
52 56 25 56 68 73 66 64 56 48
20 39 9 50 46 54 54 40 50 96
36 44 18 97 65 21 60 44 54 32
92 49 37 94 72 88 89 35 59 34
15 88 53 16 84 52 72 46 60 42
(a) Construct a frequency distribution table using the class interval 0 − 9, 10 − 19, 20 − 29, …
(b) Draw a cumulative frequency curve for the distribution.
(c) Use the graph to estimate the: (i) median; (ii) percentage of students who scored at least 66 marks, correct to the nearest whole number.
(a) Frequencies: 0–9: 1; 10–19: 3; 20–29: 3; 30–39: 6; 40–49: 8; 50–59: 12; 60–69: 6; 70–79: 3; 80–89: 4; 90–99: 4 (total 50)
Cumulative: 1, 4, 7, 13, 21, 33, 39, 42, 46, 50
(b) Plot cumulative frequency against upper class boundaries 9.5, 19.5, …, 99.5 and join with a smooth S-shaped curve
(c)(i) Median at the 25th student: about 53 (calculation gives 49.5 + 4/12 × 10 ≈ 52.8)
(ii) About 37 students scored below 65.5, so about 13 scored at least 66
Percentage ≈ 13/50 × 100 ≈ 27% (accept 26–28% from the graph)
Plot at the upper class boundaries, not the class limits.
(a) Solve the inequality (1/3)x − (1/4)(x + 2) ≥ 3x − 1⅓.
(b) In the diagram, ABC is right-angled triangle on a horizontal ground. |AD| is a vertical tower. ∠BAC = 90°, ∠ACB = 35°, ∠ABD = 52° and |BC| = 66 m. Find, correct to two decimal places: (i) the height of the tower; (ii) the angle of elevation of the top of the tower from C.
(a) Multiply by 12: 4x − 3(x + 2) ≥ 36x − 16
x − 6 ≥ 36x − 16 → −35x ≥ −10
x ≤ 2/7 (sign flips when dividing by a negative)
(b)(i) |AB| = 66 sin 35° = 37.86 m; |AD| = |AB| tan 52° = 48.45 m
(ii) |AC| = 66 cos 35° = 54.06 m
tan θ = 48.45/54.06 = 0.8962 → θ = 41.87°
Find AB and AC in the flat triangle first, then use the vertical triangles.
(a) Copy and complete the table of values for y = 2 cos x + 3 sin x for 0° ≤ x ≤ 360°.
x: 0°, 60°, 120°, 180°, 240°, 300°, 360°
y: 2.0, __, __, __, −3.6, __, __
(b) Using a scale of 2 cm to 60° on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of y = 2 cos x + 3 sin x for 0° ≤ x ≤ 360°.
(c) Using the graph: (i) solve 2 cos x + 3 sin x = −1; (ii) find, correct to one decimal place, the value of y when x = 342°.
(a) y = 2.0, 3.6, 1.6, −2.0, −3.6, −1.6, 2.0
(b) Plot the points with the given scales and join with a smooth curve
(c)(i) Read where the curve meets y = −1: x ≈ 162° and x ≈ 310°
(ii) At x = 342°, y ≈ 1.0
At 180°, cos x = −1 and sin x = 0, so y = −2.0.
A woman bought 130 kg of tomatoes for ₦52,000.00. She sold half of the tomatoes at a profit of 30%. The rest of the tomatoes began to go bad, she then reduced the selling price per kg by 12%. Calculate:
(a) the new selling price per kg;
(b) the percentage profit on the entire sales if she threw away 5 kg of bad tomatoes.
(a) Cost per kg = 52,000/130 = ₦400; first selling price = 400 × 1.3 = ₦520 per kg
New price = 520 × 0.88 = ₦457.60 per kg
(b) First half: 65 × 520 = ₦33,800
Second half: 65 − 5 = 60 kg × 457.60 = ₦27,456
Total sales = ₦61,256; profit = ₦9,256
Percentage profit = 9,256/52,000 × 100 ≈ 17.8%
The 12% cut is on the selling price (₦520), not on the cost price.
(a) The third and sixth terms of a Geometric Progression (G.P.) are 1/4 and 1/32 respectively. Find: (i) the first term and the common ratio; (ii) the seventh term.
(b) Given that 2 and −3 are the roots of the equation ax2 + bx + c = 0, find the values of a, b and c.
(a)(i) ar2 = 1/4, ar5 = 1/32; divide: r3 = 1/8 → r = 1/2
a(1/4) = 1/4 → a = 1
(ii) T7 = ar6 = (1/2)6 = 1/64
(b) (x − 2)(x + 3) = 0 → x2 + x − 6 = 0
a = 1, b = 1, c = −6
Dividing the two term equations removes a.
(a) Given that sin y = 8/17, find the value of tan y/(1 + 2 tan y).
(b) An amount of ₦300,000.00 was shared among Otobo, Ada and Adeola. Otobo received ₦60,000.00, Ada received 5/12 of the remainder, while the rest went to Adeola. In what ratio was the money shared?
(a) Adjacent side = √(172 − 82) = 15, so tan y = 8/15
1 + 2 tan y = 1 + 16/15 = 31/15
tan y/(1 + 2 tan y) = (8/15) ÷ (31/15) = 8/31
(b) Remainder = ₦240,000; Ada = 5/12 × 240,000 = ₦100,000; Adeola = ₦140,000
Ratio 60,000 : 100,000 : 140,000 = 3 : 5 : 7
Draw a right-angled triangle with opposite 8 and hypotenuse 17.
(a) In the diagram, RS and RT are tangent to the circle with centre O. ∠TUS = 68°, ∠SRT = x, and ∠UTO = y. Find the value of x.
(b) Two tanks A and B are filled to capacity with diesel. Tank A holds 600 litres of diesel more than tank B. If 100 litres of diesel was pumped out of each tank, tank A would then contain 3 times as much diesel as tank B. Find the capacity of each tank.
(a) ∠TOS = 2 × 68° = 136° (angle at centre)
∠OTR = ∠OSR = 90° (radius ⊥ tangent)
x = 360° − 90° − 90° − 136° = 44°
(b) Let B = x litres, A = x + 600
x + 600 − 100 = 3(x − 100) → x + 500 = 3x − 300 → x = 400
Tank B = 400 litres, Tank A = 1,000 litres
In (a), the quadrilateral OTRS has two right angles where the tangents touch the circle.