56 questions. Try each one, then open the answer to check yourself and read the explanation.
Express 0.0000407, correct to 2 significant figures.
Answer: C (0.000041)
The first two significant figures are 4 and 0, and the next digit (7) rounds the 0 up to 1: 0.000041.
Evaluate: (3¼ × 1⅗)/(11⅓ − 5⅓)
Answer: B (13/15)
Numerator = 13/4 × 8/5 = 26/5; denominator = 6. 26/5 ÷ 6 = 26/30 = 13/15.
Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n ⊗ 4) ⊕ 3 = 0(mod 5).
Answer: C (3)
n ⊗ 4 must be 2 (since 2 ⊕ 3 = 0). From the multiplication table, 3 ⊗ 4 = 2, so n = 3.
The ages of Tunde and Ola are in the ratio 1:2. If the ratio of Ola's age to Musa's age is 4:5, what is the ratio of Tunde's age to Musa's age?
Answer: C (2:5)
Tunde : Ola = 4 : 8 and Ola : Musa = 8 : 10, so Tunde : Musa = 4 : 10 = 2 : 5.
If M = {x: 3 ≤ x < 8} and N = {x: 8 < x ≤ 12}, which of the following is true? I. 8 ∈ M ∩ N II. 8 ∈ M ∪ N III. M ∩ N = ∅
Answer: A (III only)
8 is in neither set (M stops below 8 and N starts above 8), so I and II are false; the sets share no element, so III is true.
Given that a = log 7 and b = log 2, express log 35 in terms of a and b.
Answer: C (a − b + 1)
log 35 = log(7 × 10/2) = log 7 + log 10 − log 2 = a + 1 − b.
If x = ⅔ and y = −6, evaluate xy − y/x.
Answer: B (5)
xy = −4 and y/x = −6 ÷ ⅔ = −9, so xy − y/x = −4 + 9 = 5.
Solve the equation: 1/5x + 1/x = 3.
Answer: B (⅖)
1/(5x) + 5/(5x) = 6/(5x) = 3, so 15x = 6 and x = ⅖.
A sum of N18,100.00 was shared among 5 boys and 4 girls with each boy taking N20.00 more than each girl. Find a boy's share.
Answer: C (N2,020.00)
If a girl gets g, then 5(g + 20) + 4g = 18 100, so 9g = 18 000, g = 2000 and a boy gets N2,020.
One factor of 7x2 + 33x − 10 is
Answer: C (7x − 2)
7x² + 33x − 10 = 7x² − 2x + 35x − 10 = (7x − 2)(x + 5).
Solve: −¼ < ¾(3x − 2) < ½
Answer: A (5/9 < x < 8/9)
Multiply by 4/3: −⅓ < 3x − 2 < ⅔, so 5/3 < 3x < 8/3 and 5/9 < x < 8/9.
Simplify: 3x − (p − x) − (r − p)
Answer: C (4x − r)
3x − p + x − r + p = 4x − r.
An arc of a circle of radius 7.5 cm is 7.5 cm long. Find, correct to the nearest degree, the angle which the arc subtends at the centre of the circle. [Take π = 22/7]
Answer: B (57°)
θ/360 × 2 × 22/7 × 7.5 = 7.5 gives θ = 360 × 7/44 ≈ 57.3°, i.e. 57°.
Water flows out of a pipe at a rate of 40π cm3 per second into an empty cylindrical container of base radius 4 cm. Find the height of water in the container after 4 seconds.
Answer: A (10 cm)
Volume after 4 s = 160π cm³. π × 4² × h = 160π, so h = 10 cm.
The dimensions of a water tank are 13 cm, 10 cm and 70 cm. If it is half-filled with water, calculate the volume of water in litres.
Answer: A (4.55 litres)
Full volume = 13 × 10 × 70 = 9100 cm³; half is 4550 cm³ = 4.55 litres (1000 cm³ = 1 litre).
If the total surface area of a solid hemisphere is equal to its volume, find the radius.
Answer: B (4.5 cm)
Total surface area = 3πr² and volume = ⅔πr³. 3πr² = ⅔πr³ gives r = 4.5 cm.
Which of the following is true about parallelograms?
Answer: C (Opposite angles are equal)
In a parallelogram opposite angles are equal (it is adjacent angles that add to 180°).
The diagram shows a circle centre O. If ∠STR = 29° and ∠RST = 46°, Calculate the value of ∠STO.
Answer: B (15°)
∠ROT = 2 × ∠RST = 92° (angle at centre). Triangle ROT is isosceles, so ∠RTO = (180° − 92°)/2 = 44°. ∠STO = 44° − 29° = 15°.
In the diagram, XY is a straight line, ∠POX = ∠POQ and ∠ROY = ∠QOR. Find the value of ∠POQ + ∠ROY.
Answer: B (90°)
The four angles on line XY add to 180°: 2∠POQ + 2∠ROY = 180°, so ∠POQ + ∠ROY = 90°.
The diagram shows a circle centre O. If ∠ZYW = 33°, find ∠ZWX.
Answer: B (57°)
WX is a diameter so ∠WZX = 90°; ∠ZXW = ∠ZYW = 33° (same segment). ∠ZWX = 180° − 90° − 33° = 57°.
In the diagram, PQ and PS are tangents to the circle centre O. If ∠PSQ = m, ∠SPQ = n and ∠SQR = 33°, find the value of (m + n)
Answer: B (123°)
QR is a diameter, so the tangent PQ ⟂ QR and ∠PQS = 90° − 33° = 57°. Tangents from P are equal, so m = ∠PSQ = 57° and n = 180° − 114° = 66°. m + n = 123°.
Calculate the gradient (slope) of the line joining points (−1, 1) and (2, −2).
Answer: A (−1)
(−2 − 1)/(2 − (−1)) = −3/3 = −1.
If P(2, 3) and Q(2, 5) are points on a graph, calculate the length PQ.
Answer: D (2 units)
Both points have x = 2, so PQ = 5 − 3 = 2 units.
A bearing of 320° expressed as a compass bearing is
Answer: B (N 40° W)
320° is 40° short of North going clockwise, i.e. 40° west of north: N 40° W.
A stationary boat is observed from a height of 100 m. If the horizontal distance between the observer and the boat is 80 m, calculate, correct to two decimal places, the angle of depression of the boat from the point of observation.
Answer: C (51.34°)
tan θ = 100/80 = 1.25, so θ ≈ 51.34°.
The average age of a group 25 girls is 10 years. If one girl, aged 12 years and 4 months joins the group, find, correct to one decimal place, the new average age of the group.
Answer: A (10.1 years)
Total age = 250 + 12⅓ = 262⅓ years for 26 girls; 262⅓ ÷ 26 ≈ 10.1 years.
In a class of 45 students, 28 offer chemistry and 25 offer Biology. If each student offers at least one of the two subjects, calculate the probability that a student selected at random from the class offers Chemistry only.
Answer: B (4/9)
Both = 28 + 25 − 45 = 8, so Chemistry only = 20. Probability = 20/45 = 4/9.
In what number base was the addition 1 + nn = 100, where n > 0, done?
Answer: C (n + 1)
In base b: 1 + nb + n = b², so n(b + 1) = (b − 1)(b + 1) and b = n + 1.
Three exterior angles of a polygon are 30°, 40° and 60°. If the remaining exterior angles are 46° each, name the polygon.
Answer: C (Octagon)
Exterior angles add to 360°. 360° − 130° = 230° = 5 × 46°, so there are 3 + 5 = 8 sides: an octagon.
In the diagram, NQ//TS, ∠RTS = 50° and ∠PRT = 100°. Find the value of ∠NPR.
Answer: B (130°)
∠PRQ = 180° − 100° = 80° and ∠PQR = ∠RTS = 50° (alternate angles). So ∠QPR = 50° and ∠NPR = 180° − 50° = 130°.
Simplify the expression (a2b4 − b2a4)/(ab(a + b)).
Answer: D (ab2 − a2b)
Numerator = a²b²(b² − a²) = a²b²(b − a)(b + a). Dividing by ab(a + b) leaves ab(b − a) = ab² − a²b.
Find the 6th term of the sequence: ⅔, 7/15, 4/15,…
Answer: A (−⅓)
The terms are 10/15, 7/15, 4/15, so d = −3/15. T₆ = 10/15 + 5 × (−3/15) = −5/15 = −⅓.
The diagonal of a square is 60 cm. Calculate its perimeter.
Answer: D (120√2)
Side = 60/√2 = 30√2 cm, so the perimeter = 4 × 30√2 = 120√2 cm.
The roots of a quadratic equation are −½ and ⅔. Find the equation.
Answer: B (6x2 − x − 2 = 0)
(2x + 1)(3x − 2) = 6x² − 4x + 3x − 2 = 6x² − x − 2.
Make x the subject of the relation d = √(6/x − y/2).
Answer: D (x = 12/(2d2 + y))
d² = 6/x − y/2, so 6/x = (2d² + y)/2 and x = 12/(2d² + y).
Consider the statments: p: it is hot. q: it is raining. Which of the following symbols correctly represents the statement "It is raining if and only if it is cold"?
Answer: D (q ⇔ ~p)
'It is raining' is q and 'it is cold' is 'not hot', ~p, so the statement is q ⇔ ~p.
Given that t = 2−x; find 2x+1 in terms of t.
Answer: A (2/t)
2x = 1/t, so 2x + 1 = 2 × 2x = 2/t.
Find the value of m in the diagram.
Answer: C (44°)
The exterior angle 2x equals the sum of the opposite interior angles: 2x = 68° + x, so x = 68°. Then m = 180° − 2x = 180° − 136° = 44°.
Two bottles are drawn with replacement from a crate containing 8 coke, 12 fanta and 4 sprite bottles. What is the probability that the first is coke and the second is not coke?
Answer: C (2/9)
P(coke) = 8/24 = ⅓ and P(not coke) = 16/24 = ⅔; with replacement the probability is ⅓ × ⅔ = 2/9.
If the simple interest on a certain amount of money saved in a bank for 5 years at 2½% per annum is N500.00, calculate the total amount due after 6 years at the same rate.
Answer: D (N4,600.00)
P × 2.5 × 5/100 = 500 gives P = N4,000. Interest for 6 years = 4000 × 2.5 × 6/100 = N600, so the amount is N4,600.
Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9.
Answer: D (4.2)
Mean = 50/10 = 5. Squared deviations: 9, 4, 4, 1, 0, 0, 0, 4, 4, 16, total 42. Variance = 42/10 = 4.2.
A circular pond of radius 4 m has a path of width 2.5 m round it. Find, correct to two decimal places, the area of the path. [Take π = 22/7]
Answer: D (82.50m2)
Outer radius = 6.5 m. Area = 22/7 × (6.5² − 4²) = 22/7 × 26.25 = 82.50 m².
In the diagram, RP is a diameter of circle RSP, RP is produced to T and TS is a tangent to the circle as S. If ∠PRS = 24°, calculate the value of ∠STR.
Answer: B (42°)
∠RSP = 90° (semicircle) and ∠PST = ∠PRS = 24° (alternate segment), so ∠RST = 114°. In triangle RST, ∠STR = 180° − 24° − 114° = 42°.
(a) If (y − 1) log104 = y log1016, without using mathematical tables or calculator, find the value of y.
(b) When I walk from my house at 4 Km/h, I will get to the office 30 minutes later than when I walk at 5 km/h. Calculate the distance between my house and office.
(a) log 16 = log 4² = 2 log 4
(y − 1) log 4 = 2y log 4 → y − 1 = 2y → y = −1
(b) Let the distance be d km: d/4 − d/5 = ½
d/20 = ½ → d = 10 km
Write log 16 as 2 log 4 and divide both sides by log 4.
(a) Solve the equation: ⅔(3x − 5) − ⅗(2x − 3) = 3
(b) In the diagram, ∠STQ = m, ∠TUQ = 80°, ∠UPQ = r, ∠PQU = n and ∠RQT = 88°. Find the value of (m + n).
(a) Multiply by 15: 10(3x − 5) − 9(2x − 3) = 45 → 12x − 23 = 45 → x = 68/12 = 5⅔
(b) ∠UQT = 180° − n − 88° = 92° − n (angles on straight line PQR)
∠UTQ = 180° − m (angles on straight line PUTS)
In triangle UQT: 80° + (92° − n) + (180° − m) = 180° → m + n = 172°
Express the angles of triangle UQT in terms of m and n.
(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50 m, calculate, correct to the nearest metre, the height of the building.
(b) In the diagram, PT//SU, QS//TR, |SR| = 6cm and |RU| = 10cm. If the area of ΔTRU = 45cm2, calculate the area of trapezium QTUS.
(a) Angle of elevation of T from P = 23.6°; h = 50 tan 23.6° ≈ 21.8 ≈ 22 m
(b) Area of ΔTRU = ½ × 10 × h = 45 → height h = 9 cm
QSRT is a parallelogram, so |QT| = |SR| = 6 cm; |SU| = 6 + 10 = 16 cm
Area of QTUS = ½(6 + 16) × 9 = 99 cm²
The triangle and the trapezium have the same height.
If the sixth term of an Arithmetic Progression (A. P.) is 37 and the sum of the first six terms is 147, find the: (a) first term; (b) sum of the first fifteen terms.
a + 5d = 37 and S₆ = 3(2a + 5d) = 147 → 2a + 5d = 49
Subtract: a = 12, so d = 5
(a) First term = 12
(b) S₁₅ = 15/2 (2 × 12 + 14 × 5) = 15/2 × 94 = 705
Use Sn = n/2 (2a + (n − 1)d).
Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags; (a) illustrate this information in a diagram; (b) find the number of customers who bought shoes; (c) calculate the probability that a customer selected at random bought bags.
(a) Venn diagram: universal set 120, circles B (bags) and S (shoes) overlapping, intersection 45, nothing outside the circles.
(b) Let n(B) = x, n(S) = x + 11: x + x + 11 − 45 = 120 → 2x = 154 → x = 77
Customers who bought shoes = 88
(c) P(bags) = 77/120
n(B ∪ S) = n(B) + n(S) − n(B ∩ S).
(a) A manufacturing company requires 3 hours of direct labour to process every N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget for direct labour at N18.25 per hour?
(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and N y in bank N at the rate of 8 % simple interest per annum. If a total of N8,000,000.00 was invested in two banks and the investor recieved a total of N2,320,000.00 as interest from two banks after 4 years, calculate the: (i) values of x and y; (ii) interest paid by the second bank.
(a) 30 450 ÷ 87 = 350 lots of N87; hours = 350 × 3 = 1050 h
Budget = 1050 × 18.25 = N19,162.50
(b)(i) x + y = 8 000 000; 0.06 × 4x + 0.08 × 4y = 2 320 000 → 0.24x + 0.32y = 2 320 000
0.24(8 000 000 − y) + 0.32y = 2 320 000 → 0.08y = 400 000 → y = N5,000,000, x = N3,000,000
(ii) Interest from bank N = 0.32 × 5 000 000 = N1,600,000
Set up one equation for the money and one for the interest.
(a) Copy and complete the table of values for the equation y = 2x2 − 7x − 9 for −3 ≤ x ≤ 6.
x: −3, −2, −1, 0, 1, 2, 3, 4, 5, 6
y: __, 13, __, −9, −14, __, −12, __, 6, __
(b) using scales of 2cm to 1 unit on the x-axis and 2cm to 4 units on the y-axis, draw the graph of y = 2x2 − 7x − 9 for −3 ≤ x ≤ 6.
(c) Use the graph to estimate the: (i) roots of equation 2x2 − 7x = 26; (ii) coordinates of the minimum point of y; (iii) range of values for which 2x2 − 7x < 9.
(a) x = −3: 30; x = −1: 0; x = 2: −15; x = 4: −5; x = 6: 21
(b) Plot the points and join with a smooth U-shaped curve.
(c)(i) 2x² − 7x − 9 = 17: draw y = 17; x ≈ −2.3 and x ≈ 5.8
(ii) Minimum point ≈ (1.75, −15.1)
(iii) 2x² − 7x − 9 < 0 where the curve is below the x-axis: −1 < x < 4.5
For (c)(i), rewrite the equation so the left side matches y and draw the horizontal line y = 17.
Marks: 1, 2, 3, 4, 5
Number of students: m + 2, m − 1, 2m − 3, m + 5, 3m − 4
The table shows the distribution of marks scored by some students in a test. (a) If the mean mark is 36/23, find the value of m. (b) Find the: (i) interquartile range; (ii) probability of selecting a student who scored at least 4 marks in the test.
(a) Σf = 8m − 1; Σfx = (m + 2) + 2(m − 1) + 3(2m − 3) + 4(m + 5) + 5(3m − 4) = 28m − 9
(28m − 9)/(8m − 1) = 75/23 → 644m − 207 = 600m − 75 → m = 3
(b) Frequencies 5, 2, 3, 8, 5 (23 students); cumulative 5, 7, 10, 18, 23
(i) Q₁ = 6th mark = 2; Q₃ = 18th mark = 4; interquartile range = 4 − 2 = 2
(ii) At least 4 marks: 8 + 5 = 13 students; probability = 13/23
Write the mean as an improper fraction 75/23 before cross-multiplying.
(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, ∠TPS = 34° and ∠TSQ = 65°. (i) illustrate the information in a diagram. (ii) Find the value of: (A) ∠RTS; (B) ∠SRP
(b) In the diagram, |VZ| = |YZ|, ∠YXZ = 20° and ∠ZVY = 52°. Calculate the size of ∠WYZ.
(a)(i) Circle through R, S, T; tangent PSQ at S; P outside the circle with P, R, T on one straight line (R between P and T).
(ii)(A) ∠PST = 180° − 65° = 115°; in triangle PST, ∠RTS = 180° − 34° − 115° = 31°
(B) ∠SRT = ∠TSQ = 65° (alternate segment), so ∠SRP = 180° − 65° = 115°
(b) ∠ZYV = ∠ZVY = 52° (|VZ| = |YZ|)
In triangle XYZ: ∠XZY = 180° − 20° − 52° = 108°
∠YWZ = ∠YVZ = 52° (same segment)
In triangle WYZ: ∠WYZ = 180° − 108° − 52° = 20°
Use the alternate segment theorem for the tangent at S.
(a) Given that sin x = 5/13, 0° < x < 90°, find (cos x − 2 sin x)/(2 tan x).
(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6 metres above the ground. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the: (i) length of ladder LA; (ii) angle which LA makes with the ground.
(a) cos x = 12/13, tan x = 5/12
(12/13 − 10/13) ÷ (10/12) = 2/13 × 12/10 = 12/65
(b) Foot of pole to B = √(12² − 9.6²) = √51.84 = 7.2 m; so foot to A = 10 − 7.2 = 2.8 m
(i) LA = √(9.6² + 2.8²) = √100 = 10 m
(ii) tan θ = 9.6/2.8 → θ ≈ 73.7° ≈ 74°
Use a right-angled triangle with sides 5, 12, 13 in (a).
(a) It takes 8 students two-thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one-third of an hour at the same rate?
(b) A chord, 20 cm long, is 12cm from the centre of circle. Calculate, correct to one decimal place, the: (i) angle subtended by the chord at the centre of the circle; (ii) perimeter of minor segment cut off by the chord. [Take π = 3.142]
(a) Half the students for half the time: 12 × ½ × ½ = 3 tanks
(b)(i) Half-angle: tan θ = 10/12 → θ = 39.8°; angle at centre ≈ 79.6°
(ii) Radius = √(10² + 12²) = √244 ≈ 15.62 cm
Arc = 79.61/360 × 2 × 3.142 × 15.62 ≈ 21.7 cm
Perimeter of minor segment ≈ 21.7 + 20 = 41.7 cm
The perpendicular from the centre bisects the chord.
(a) Using completing the square method, solve, correct to 2 decimal places, the equation 3y2 − 5y + 2 = 0.
(b) Given that M = (1 2 / 4 3), N = (m x / n y) and MN = (2 1 / 3 4), find the matrix N.
(a) y² − (5/3)y = −⅔ → (y − 5/6)² = 25/36 − 24/36 = 1/36
y − 5/6 = ±1/6 → y = 1.00 or y = 0.67
(b) MN = (m + 2n x + 2y / 4m + 3n 4x + 3y) = (2 1 / 3 4)
m + 2n = 2 and 4m + 3n = 3 → n = 1, m = 0
x + 2y = 1 and 4x + 3y = 4 → y = 0, x = 1
N = (0 1 / 1 0)
Matrices are written row by row: (a b / c d) means top row a b, bottom row c d.
(a) The operation (*) is defined on the set of real numbers, R, by x(*)y = (x + y)/2, x, y ∈ R. (i) Evaluate 3(*)⅖. (ii) If 8(*)y = 8¼, find the value of y.
(b) In ΔABC, AB = (−4 / 6) and AC = (3 / −8). If P is the mid point of AB, express CP as a column.
(a)(i) 3(*)⅖ = (3 + ⅖)/2 = 17/10 = 1.7
(ii) (8 + y)/2 = 8¼ → 8 + y = 16½ → y = 8½
(b) AP = ½AB = (−2 / 3)
CP = CA + AP = −AC + AP = (−3 / 8) + (−2 / 3) = (−5 / 11)
Go from C back to A and then along to P.