60 questions. Try each one, then open the answer to check yourself and read the explanation.
Multiply 3.4 × 10−5 by 7.1 × 108 and leave the answer in standard form.
Answer: C (2.414 × 104)
3.4 × 7.1 = 24.14 and 10−5 × 108 = 103, so the product is 24.14 × 103 = 2.414 × 104.
Given that P = {p: 1 < p < 20}, where p is an integer and R = {r: 0 ≤ r ≤ 25, where r is a multiple of 4}. Find P ∩ R.
Answer: B ({4, 8, 12, 16})
R = {0, 4, 8, 12, 16, 20, 24}. The members that are also between 1 and 20 (not including 20) are 4, 8, 12 and 16.
The first term of an Arithmetic Progression (A.P) is 2 and the last term is 29. If the common difference is 3, how many terms are in the A.P?
Answer: C (10)
29 = 2 + (n − 1) × 3 gives n − 1 = 9, so n = 10.
Express in index form: logax + logay = 3.
Answer: D (xy = a3)
logax + logay = loga(xy) = 3, so xy = a³.
If (3 − 4√2)(1 + 3√2) = a + b√2, find the value of b.
Answer: A (5)
Expanding: 3 + 9√2 − 4√2 − 24 = −21 + 5√2, so b = 5.
Find the time for which $1,250.00 will amount to $2,031.25 at 12.5% per annum simple interest.
Answer: D (5 years)
Interest = 2031.25 − 1250 = 781.25. T = (100 × 781.25) ÷ (1250 × 12.5) = 5 years.
If log3 (2x − 1) = 5, find the value of x.
Answer: D (122)
2x − 1 = 35 = 243, so 2x = 244 and x = 122.
The population of a town increases by 3% every year. In the year 2000, the population was 3,000. Find the population in the year 2003.
Answer: B (3,278)
Three years of 3% growth: 3000 × 1.03³ = 3278.2, about 3,278.
A trader gave a change of ₦540.00 instead of ₦570.00 to a customer. Calculate the percentage error.
Answer: A (55/19%)
Error = ₦30. Percentage error = 30/570 × 100% = 100/19 % = 55/19%.
If 3x − 2y = −5 and x + 2y = 9, find the value of (x − y)/(x + y)
Answer: C (−3/5)
Adding the equations gives 4x = 4, so x = 1 and y = 4. (x − y)/(x + y) = −3/5.
A variable W varies partly as M and partly inversely as P. Which of the following correctly represents the relation with k1 and k2 as constants?
Answer: C (W = k1M + k2/P)
'Partly … and partly …' means a sum of two parts: one proportional to M (k1M) and one inversely proportional to P (k2/P).
A cylindrical metallic barrel of height 2.5 m and radius 0.245 m is closed at one end. Find, correct to one decimal place, the total surface area of the barrel. [Take π = 22/7]
Answer: C (4.0 m2)
Curved surface + one end = 2πrh + πr² = 22/7 × 0.245 × (5 + 0.245) ≈ 4.04, so 4.0 m².
Make R the subject of the relation V = πl(R2 − r2).
Answer: A (R = √(V/(πl) + r2))
Divide by πl: R² − r² = V/(πl). Add r² and take the square root: R = √(V/(πl) + r²).
Consider the following statements:
m: Edna is respectful
n: Edna is brilliant
If m ⇒ n, which of the following is valid?
Answer: C (~n ⇒ ~m)
The contrapositive of m ⇒ n is ~n ⇒ ~m, and it is always equivalent to the original statement.
A number is added to both the numerator and the denominator of the fraction 1/8. If the result is 1/2, find the number.
Answer: D (6)
(1 + x)/(8 + x) = ½ gives 2 + 2x = 8 + x, so x = 6.
Gifty, Justina and Frank shared 60 oranges in the ratio 5 : 3 : 7 respectively. How many oranges did Justina receive?
Answer: A (12)
The ratio has 5 + 3 + 7 = 15 parts. Justina gets 3/15 × 60 = 12.
A piece of rod of length 44 m is cut to form a rectangular shape such that the ratio of the length to the breadth is 7 : 4. Find the breadth.
Answer: A (8 m)
The perimeter is 44 m, so length + breadth = 22 m. Breadth = 4/11 × 22 = 8 m.
In the diagram, MN // KL, ML and KN intersect at X. |MN| = 12 cm, |MX| = 10 cm and |KL| = 9 cm. If the area of ΔMXN is 16 cm2, calculate the area of ΔLXK.
Answer: B (9 cm2)
Because MN // KL, triangles MXN and LXK are similar with scale factor 9/12 = 3/4. Areas scale by (3/4)² = 9/16, so the area is 16 × 9/16 = 9 cm².
A ladder 15 m long leans against a vertical pole, making an angle of 72° with the horizontal. Calculate, correct to one decimal place, the distance between the foot of the ladder and the pole.
Answer: D (4.6 m)
The distance along the ground is the adjacent side: 15 cos 72° = 15 × 0.309 ≈ 4.6 m.
In the diagram, O is the centre of the circle. If |OA| = 25 cm and |AB| = 40 cm, find |OH|.
Answer: A (15 cm)
The perpendicular from the centre bisects the chord, so AH = 20 cm. OH = √(25² − 20²) = √225 = 15 cm.
Given that P is 25 m on a bearing of 330° from Q, how far south of P is Q?
Answer: B (21.7 m)
The bearing 330° is 30° west of north, so the north–south distance is 25 cos 30° ≈ 21.7 m.
A car valued at $600,000.00 depreciates by 10% each year. What will be the value of the car at the end of two years?
Answer: C ($486,000.00)
Each year the value is multiplied by 0.9: 600 000 × 0.9 × 0.9 = $486,000.
The length and breadth of a cuboid are 15 cm and 8 cm respectively. If the volume of the cuboid is 1,560 cm3, calculate the total surface area.
Answer: B (838 cm2)
Height = 1560 ÷ (15 × 8) = 13 cm. Total surface area = 2(15×8 + 15×13 + 8×13) = 2(120 + 195 + 104) = 838 cm².
The number 1621 was subtracted from 6244 in base x. If the result was 4323, find x.
Answer: A (Seven)
Check by adding 4323 + 1621 = 6244: in the tens column 2 + 2 = 4, but in the hundreds column 3 + 6 = 9 must give 2 carry 1, so x = 9 − 2 = 7.
Factorize completely: 27x2 − 48y2
Answer: B (3(3x + 4y)(3x − 4y))
Take out 3: 3(9x² − 16y²), then use the difference of two squares: 3(3x + 4y)(3x − 4y).
For what values of x is (x − 3)/4 + (x + 1)/8 ≥ 2 ?
Answer: C (x ≥ 7)
Multiply by 8: 2(x − 3) + (x + 1) ≥ 16, so 3x − 5 ≥ 16 and x ≥ 7.
In the diagram, ∠SQR = 52° and ∠PRT = 16°. Find the value of the angle marked y.
Answer: C (112°)
PQST is cyclic, so the exterior angle SQR equals the interior opposite angle PTS = 52°. In triangle PRT, y = 180° − 52° − 16° = 112°.
In the diagram, JKL is a tangent to the circle GHIK at K. ∠LKG = 38° and ∠HIK = 87°. Calculate the value of the angle marked x.
Answer: B (55°)
The exterior angle KGL of the cyclic quadrilateral GHIK equals the opposite interior angle HIK = 87°. In triangle GKL, x = 180° − 87° − 38° = 55°.
A cone and a cylinder are of equal volume. The base radius of the cone is twice the radius of the cylinder. What is the ratio of the height of the cylinder to that of the cone?
Answer: B (4 : 3)
With cylinder radius r: (1/3)π(2r)²hcone = πr²hcyl, so hcyl = (4/3)hcone and the ratio is 4 : 3.
Find, correct to the nearest whole number, the value of h in the diagram.
Answer: C (23 m)
The sloping 20 m side spans 19 m horizontally, so it rises √(20² − 19²) = √39 ≈ 6.2 m. h = 17 + 6.2 ≈ 23 m.
The gradient of the line joining the points P(2, −8) and Q(1, y) is −4. Find the value of y.
Answer: C (−4)
(y + 8)/(1 − 2) = −4, so y + 8 = 4 and y = −4.
In the diagram, PQ // RS, ∠WYZ = 44° and ∠WXZ = 50°. Find ∠WTX.
Answer: C (86°)
∠XWY = ∠WYZ = 44° (alternate angles). In triangle WTX, ∠WTX = 180° − 44° − 50° = 86°.
The perimeter of a rectangular garden is 90 m. If the width is 7 m less than the length, find the length of the garden.
Answer: D (26 m)
Length + width = 45 and width = length − 7, so 2 × length − 7 = 45 and length = 26 m.
Four of the angles of a hexagon sum up to 420°. If the remaining angles are equal, find the value of each of the angles.
Answer: D (150°)
A hexagon's angles add to (6 − 2) × 180° = 720°. The other two angles share 720° − 420° = 300°, so each is 150°.
Find the value of x in the diagram.
Answer: C (120°)
In triangle NMP, ∠M = 180° − 65° − 55° = 60°. QR // MP, so ∠NQR = 60° (corresponding angles), and x = 180° − 60° = 120° (angles on a straight line).
The following are the masses (in kg) of members in a club: 59, 44, 53, 57, 49 40, 48 and 50. Calculate the mean mass.
Answer: C (50 kg)
The eight masses add up to 400 kg, so the mean is 400 ÷ 8 = 50 kg.
The following are the masses (in kg) of members in a club: 59, 44, 53, 57, 49 40, 48 and 50. Calculate the variance of the distribution
Answer: A (35)
Deviations from 50 are 9, −6, 3, 7, −1, −10, −2, 0; their squares add to 280. Variance = 280 ÷ 8 = 35.
Two opposite sides of a rectangle are (5x + 3) m and (2x + 9) m. If an adjacent side is (6x − 7) m, find in m2, the area of the rectangle.
Answer: B (65)
Opposite sides are equal: 5x + 3 = 2x + 9 gives x = 2. The sides are 13 m and 6(2) − 7 = 5 m, so the area is 65 m².
A die is tossed once. Find the probability of getting a prime number.
Answer: A (1/2)
The primes on a die are 2, 3 and 5: 3 out of 6 outcomes, so ½.
The area of a sector of a circle with radius 7 cm is 51.3 cm2. Calculate, correct to the nearest whole number, the angle of the sector. [Take π = 22/7]
Answer: B (120°)
θ/360 × 22/7 × 49 = 51.3, so θ = 51.3 × 360 ÷ 154 ≈ 119.9 ≈ 120°.
A cliff on the bank of a river is 87 m high. A boat on the river is 22 m away from 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: A (76°)
The angle of depression equals the angle of elevation from the boat: tan α = 87/22 ≈ 3.95, so α ≈ 76°.
In the diagram TU is a tangent to the circle SPQR at P. If ∠PTS = 44° and ∠SQP = 35°, find ∠PST.
Answer: A (101°)
By the alternate segment theorem, ∠SPT = ∠SQP = 35°. In triangle PST, ∠PST = 180° − 44° − 35° = 101°.
The probability that Amaka will pass an examination is 3/7 and that Bala will pass is 4/9. Find the probability that both will pass the examination.
Answer: B (4/21)
Independent events: 3/7 × 4/9 = 12/63 = 4/21.
Which of the following points lies on the line 3x − 8y = 11?
Answer: B ((1, −1))
For (1, −1): 3(1) − 8(−1) = 3 + 8 = 11, so it lies on the line.
Find the range of the following set of numbers: 28, 29, 39, 38, 37, 26, 20, 15 and 25.
Answer: B (24)
Range = largest − smallest = 39 − 15 = 24.
The fourth and eighth terms of an Arithmetic Progression are 16 and 40 respectively. Find the common difference.
Answer: B (6)
From the 4th to the 8th term there are 4 steps: 4d = 40 − 16 = 24, so d = 6.
For what values of y is (y + 2)/(8y2 − 10y + 3) not defined
Answer: D (3/4, 1/2)
The fraction is undefined when 8y² − 10y + 3 = (4y − 3)(2y − 1) = 0, i.e. y = 3/4 or 1/2.
The time (t) taken to buy fuel at a filling station varies directly as the number of vehicles (V) in a queue and varies inversely as the number of pumps (P), available at the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find the: (a) relationship between t, P and V; (b) time it takes to fuel 50 vechicles at a station with 2 pumps; (c) number of pumps required to fuel 40 vehicles in 20 minutes.
(a) t = kV/P; 10 = 20k/5 ⇒ k = 2.5, so t = 2.5V/P (= 5V/2P)
(b) t = 2.5 × 50 ÷ 2 = 62.5 minutes
(c) 20 = 2.5 × 40 ÷ P ⇒ P = 5 pumps
Joint variation: V goes on top, P underneath.
A car travelled a distance of (2x + 13) km at 67.5 km/h and (5x − 20) km at 72 km/h. If the total time for the entire journey was 90 minutes, find the velue of x.
Time = distance ÷ speed, and 90 minutes = 1.5 h
(2x + 13)/67.5 + (5x − 20)/72 = 1.5
Multiply by 1080: 16(2x + 13) + 15(5x − 20) = 1620
107x − 92 = 1620 ⇒ 107x = 1712 ⇒ x = 16
Change 90 minutes to hours before forming the equation.
A circular floor of a building is to be tiled with ceramic tiles each of side 40 cm × 40 cm. If the perimeter of the floor is 66 m. Calculate, correct to the nearest whole number, the number of tiles required to completely tile the floor. [Take π = 22/7]
2πr = 66 ⇒ r = 66 × 7 ÷ 44 = 10.5 m
Area of floor = 22/7 × 10.5² = 346.5 m²
Area of one tile = 0.4 × 0.4 = 0.16 m²
Number of tiles = 346.5 ÷ 0.16 ≈ 2165.6 ≈ 2166 tiles
Work in metres throughout: 40 cm = 0.4 m.
In the diagram, ABCD is a circle O. The quadrilateral OBCD is a rhombus such that ∠ADO = ∠OBA = y and ∠BAD = t. Find (a) the value of t; (b) the value of y; (c) ∠ADC.
OB = OC = OD (radii) and OBCD is a rhombus, so OB = BC = CD = DO = OC: triangles OBC and OCD are equilateral
∠BOD (through C) = 60° + 60° = 120°
(a) t = ∠BAD = ½ × 120° = 60° (angle at centre is twice angle at circumference)
(b) OA = OB = OD, so ∠OAB = ∠OBA = y and ∠OAD = ∠ODA = y; t = 2y ⇒ y = 30°
(c) ∠ADC = ∠ADO + ∠ODC = 30° + 60° = 90°
Every side of the rhombus equals the radius, so look for equilateral triangles.
A basket contains 3 gold-plated marbles, 4 diamond marbles and some silver marbles, all of the same size and shape. Two marbles were drawn from the basket at random one after the other without replacement. If the probability that the two marbles were all silver is 1/15, find the number of silver marbles.
Let n = number of silver marbles; total = n + 7
n/(n + 7) × (n − 1)/(n + 6) = 1/15
15n² − 15n = n² + 13n + 42 ⇒ 14n² − 28n − 42 = 0 ⇒ n² − 2n − 3 = 0
(n − 3)(n + 1) = 0 ⇒ n = 3 silver marbles
Without replacement, the second fraction has one less on top and bottom.
Given that (x + 2), (4x + 3) and (7x + 24) are consecutive terms of a Geometric Progression (G.P.), find the: (a) value of x; (b) common ratio.
(a) (4x + 3)/(x + 2) = (7x + 24)/(4x + 3) ⇒ (4x + 3)² = (x + 2)(7x + 24)
16x² + 24x + 9 = 7x² + 38x + 48 ⇒ 9x² − 14x − 39 = 0 ⇒ (x − 3)(9x + 13) = 0
x = 3 (or x = −13/9)
(b) With x = 3 the terms are 5, 15, 45, so the common ratio = 3 (with x = −13/9 the terms are 5/9, −25/9, 125/9 and the ratio is −5)
In a G.P. the middle term squared equals the product of its neighbours.
The table shows the ages of 20 children in a household.
Age: 5, 6, 7, 8, 9, 10
Frequency: 2, 2x − 1, y + 2, 4, 2, y − 1
Given that x : y = 1 : 2, find the (a) values of x and y; (b) mean age of the children.
(a) Σf = 8 + 2x + 2y = 20 ⇒ x + y = 6; y = 2x ⇒ 3x = 6 ⇒ x = 2, y = 4
(b) Frequencies 2, 3, 6, 4, 2, 3; Σfx = 10 + 18 + 42 + 32 + 18 + 30 = 150
Mean age = 150/20 = 7.5 years
Use the total frequency of 20 to form the first equation.
Two observers Abu and Badu, 46 m apart, observe a bird on a vertical pole from the same side of the bird. The angles of elevation of the bird from Abu's and Badu's eye are 40° and 48° respectively. If at the foot of the pole, Abu and Badu are on the same horizontal: (a) illustrate the information in a diagram; (b) calculate, correct to one decimal place, the height of the pole.
(a) Vertical pole with the bird on top; Abu and Badu on one horizontal line on the same side, Badu nearer the pole, 46 m apart; elevations 40° at Abu and 48° at Badu
(b) Let Badu be d m from the pole and h the height: h = d tan 48° = (d + 46) tan 40°
d(1.1106 − 0.8391) = 46 × 0.8391 ⇒ d ≈ 142.2 m
h = 142.16 × 1.1106 ≈ 157.9 m
The observer with the larger angle is nearer the pole.
The diameter of a circle centre O is 26 cm. If a chord PQ is drawn such that it is 5 cm from O to the centre of the chord, calculate, correct to the nearest whole number, the: (a) ∠POQ; (b) area of the minor segment formed by the chord PQ. [Take π = 22/7]
(a) Radius = 13 cm; cos(½∠POQ) = 5/13 ⇒ ½∠POQ ≈ 67.38° ⇒ ∠POQ ≈ 135°
(b) Area of sector = 135/360 × 22/7 × 13² ≈ 199.2 cm²
Area of triangle POQ = ½ × 13² × sin 135° ≈ 59.8 cm² (half-chord = 12 cm gives exactly 60 cm²)
Area of minor segment ≈ 139 cm²
The line from the centre to the middle of a chord is perpendicular to it.
(a) In a man's will, he gave 2/5 of the total acres of his cocoa farm to the wife and 1/3 of what is left to the family. The rest of the farm was to be shared amongst his three children in the ratio 3 : 5 : 2. Given that, the child who had the least share received 8 acres, calculate the: (i) total acres the man left; (ii) number of acres the wife received.
(b) The list price of a Television set is $1,600.00. It can be purchased by a deposit of $400.00 and the rest of the amount paid by 12 monthly installment at 25% per annum simple interest. If the Television set is purchased by installment, find the total cost.
(a) Let the farm be x acres. Wife: 2/5 x; left 3/5 x; family: 1/3 × 3/5 x = 1/5 x
Children's part = x − 2/5 x − 1/5 x = 2/5 x
Least share = 2/10 × 2/5 x = 2/25 x = 8 ⇒ (i) x = 100 acres
(ii) Wife = 2/5 × 100 = 40 acres
(b) Balance = 1600 − 400 = $1,200; interest for 1 year at 25% = $300
Total cost = 400 + 1200 + 300 = $1,900
The least child's share is a fraction of the children's part, not of the whole farm.
(a) Find the equation of the line that passes through the origin and the point of intersection of the lines x + 2y = 7 and x − y = 4.
(b) The ratio of an interior angle to an exterior angle of a regular polygon is 4:1. Find the: (i) number of sides; (ii) value of the exterior angle; (iii) sum of the interior angles of the polygon.
(a) Subtracting: 3y = 3 ⇒ y = 1, x = 5; the line through (0, 0) and (5, 1) is y = x/5 (5y = x)
(b) Interior + exterior = 180° and interior = 4 × exterior ⇒ exterior = 36°
(i) Number of sides = 360° ÷ 36° = 10
(ii) Exterior angle = 36°
(iii) Sum of interior angles = (10 − 2) × 180° = 1440°
A line through the origin has equation y = mx.
(a) In the diagram PR is a tangent to the circle centre O at Q. ∠POQ = 56° and PO intersects SQ at V such that ∠SVP = 109°. Calculate: (i) ∠TQP; (ii) ∠QTS.
(b) Simplify: (2n2 − 3n − 2)/(2n2 + 3n + 1) × (n2 − 1)/(n2 − 4)
(a)(i) OT = OQ, so ∠OQT = ∠OTQ = (180° − 56°)/2 = 62°; ∠OQP = 90° (tangent ⊥ radius) ⇒ ∠TQP = 90° − 62° = 28°
(ii) ∠QVO = ∠SVP = 109° (vertically opposite); in triangle QVO, ∠VQO = 180° − 109° − 56° = 15°, so ∠SQP = 90° − 15° = 75°
T lies on the arc on the same side of SQ as P, so ∠QTS = 180° − 75° = 105° (alternate segment theorem)
(b) = (2n + 1)(n − 2)/[(2n + 1)(n + 1)] × (n − 1)(n + 1)/[(n − 2)(n + 2)] = (n − 1)/(n + 2)
Factorise every quadratic before cancelling.
(a) In the diagram ABT is a circle centre O. PQ is a tangent to the circle at T and ABC is a straight line. TC bisects ∠BTQ. ∠BAT = 44° and ∠PTA = 60°. Find ∠ACT.
(b) The circumference of the base of a cyclindrical tank is 11 m. The height of the tank is 3 m more than 6 times the base radius. Calculate the: (i) radius; (ii) height; (iii) volume of the tank. [Take π = 22/7]
(a) ∠BTQ = ∠BAT = 44° (alternate segment) ⇒ ∠BTC = 22°
∠ABT = ∠PTA = 60° (alternate segment); ∠ABT is an exterior angle of triangle BTC ⇒ ∠ACT = 60° − 22° = 38°
(b)(i) 2 × 22/7 × r = 11 ⇒ r = 1.75 m
(ii) h = 6 × 1.75 + 3 = 13.5 m
(iii) V = 22/7 × 1.75² × 13.5 ≈ 129.9 m³
Use the alternate segment theorem twice.