55 questions. Try each one, then open the answer to check yourself and read the explanation.
In the diagram, ∠PQU = 36°, ∠QRT = 29°, PQ||RS and UQ||RT. Find ∠PQR.
Answer: B (65°)
∠PQR = ∠QRS (alternate angles, PQ ∥ RS). ∠TRS = ∠PQU = 36° (the arms are parallel), so ∠QRS = 29° + 36° = 65°.
The variance of a given distribution is 25. What is the standard deviation?
Answer: D (5)
Standard deviation = √variance = √25 = 5.
Given that P = {b, d, e, f} and Q = {a, c, f, g} are subsets of the universal set U = {a, b, c, d, e, f, g}. Find P'∩Q.
Answer: D ({a, c, g})
P' = {a, c, g}. All three are in Q, so P'∩Q = {a, c, g}.
Find the value of x in 0.5x + 2.6 = 5x + 0.35
Answer: A (0.5)
2.6 − 0.35 = 5x − 0.5x, so 2.25 = 4.5x and x = 0.5.
Find the value of x in the diagram
Answer: B (35°)
The marked angles are the exterior angles of a hexagon, which add to 360°: (2x + 5) + 2x + (x − 20) + x + (3x + 10) + (x + 15) = 10x + 10 = 360, so x = 35°.
If M5ten = 1001011two, find the value of M.
Answer: C (7)
1001011two = 64 + 8 + 2 + 1 = 75, so the digit M is 7.
The diagram is the graph of y = 6 + x − x2. The graph intercepts the x-axis at P and R and the y-axis at Q. What is the value of y at Q?
Answer: B (6)
At the y-axis x = 0, so y = 6 + 0 − 0 = 6.
The diagram is the graph of y = 6 + x − x2. The graph intercepts the x-axis at P and R and the y-axis at Q. What is the length of RP?
Answer: B (5)
6 + x − x2 = −(x − 3)(x + 2), so the curve cuts the x-axis at −2 and 3. RP = 3 − (−2) = 5.
The diagram is the graph of y = 6 + x − x2. When y = 3 1/2, what is the positive value of x?
Answer: B (2 1/5)
Solve 6 + x − x2 = 3.5: x2 − x − 2.5 = 0 gives x ≈ 2.16, which reads as about 2 1/5 on the graph.
The lengths, in cm, of the sides of a right-angled triangle are x, (x + 2) and (x + 1) where x > 0. Find, in cm, the length of its hypotenuse.
Answer: B (5)
x2 + (x + 1)2 = (x + 2)2 gives x2 − 2x − 3 = 0, so x = 3 and the hypotenuse x + 2 = 5 cm.
If log 2 = 0.3010 and log 2y = 1.8062, find, correct to the nearest whole number, the value of y.
Answer: A (6)
y log 2 = 1.8062, so y = 1.8062 ÷ 0.3010 ≈ 6.
Points X, Y and Z are located in the same horizontal plane such that Y is 12 km north of X and Z is on a bearing of 270° from X. If |XZ| = 6km, calculate, correct to one decimal place, |YZ|
Answer: B (13.4km)
Z is due west of X and Y due north, so the angle at X is 90°: YZ = √(122 + 62) = √180 ≈ 13.4 km.
The ratio of the number of men to the number of women in a 20-member committee is 3:1. How many women must be added to the 20-member committee so as to make the ratio of men to women 3:2?
Answer: B (5)
There are 15 men and 5 women. For 15 : w = 3 : 2 we need w = 10, so 5 women must be added.
If (x + 3) varies directly as y and x = 3 when y = 12, what is the value of x when y = 8?
Answer: A (1)
x + 3 = ky: 6 = 12k, so k = 1/2. When y = 8, x + 3 = 4 and x = 1.
From the diagram, find the bearing of Q from P.
Answer: A (236°)
From P, the line PO points due west (270°) and Q lies 34° below it, towards the south: 270° − 34° = 236°.
Find the nth term of the sequence 4, 10, 16,...
Answer: A (2(3n − 1))
It is an A.P. with a = 4 and d = 6: Tn = 4 + 6(n − 1) = 6n − 2 = 2(3n − 1).
If (−3, −4) is a point on the line y = mx + 2, find the value of m.
Answer: C (2)
−4 = −3m + 2, so 3m = 6 and m = 2.
Which of the following is represented by the above sketch?
Answer: B (y = x2 − x − 6)
The curve cuts the x-axis at −2 and 3: y = (x + 2)(x − 3) = x2 − x − 6 (and y = −6 at x = 0).
In the diagram, SQ is a tangent to the circle at P, XP||YQ, ∠XPY = 56° and ∠PXY = 80°. Find angle PQY.
Answer: C (44°)
In triangle PXY, ∠PYX = 180 − 56 − 80 = 44°. The tangent–chord angle ∠SPX equals ∠PYX = 44°, and XP ∥ YQ makes ∠PQY = ∠SPX = 44° (corresponding angles).
For what values of x is the expression (3x − 2)/(4x2 + 9x − 9) undefined?
Answer: D (3/4 or −3)
The denominator 4x2 + 9x − 9 = (4x − 3)(x + 3) is zero when x = 3/4 or x = −3.
The base diameter of a cone is 14cm, and its volume is 462 cm3. Find its height. [Take π = 22/7]
Answer: D (9cm)
r = 7 cm. 462 = ⅓ × 22/7 × 49 × h = 154h/3, so h = 9 cm.
A number is selected at random from the set Y = {18, 19, 20,...28, 29}. Find the probability that the number is prime
Answer: A (1/4)
The set has 12 numbers; the primes are 19, 23 and 29. Probability = 3/12 = 1/4.
In the diagram O and O' are the centres of the circles radii 15cm and 8cm respectively. If PQ = 12cm, find |OO'|.
Answer: B (19.04cm)
OO' cuts PQ at right angles at its midpoint, 6 cm from P. OO' = √(152 − 62) + √(82 − 62) = 13.75 + 5.29 ≈ 19.04 cm.
In the diagram. O is the centre of the circle. Which of the following is/are not true? I. a = b II. b + c = 180° III. a + b = c
Answer: D (III only)
An angle in a semicircle is 90°, so a = b = c = 90°. Then a = b and b + c = 180° are true, but a + b = 180° ≠ c, so only III is not true.
Simplify (21/2 × 81/2)/4
Answer: A (1)
21/2 × 81/2 = √16 = 4, and 4 ÷ 4 = 1.
In the diagram, |SR| = |RQ| and ∠PRQ = 58°, ∠VQT = 19°. PQT, SQV and PSR are straight lines. Find ∠QPS.
Answer: A (42°)
SR = RQ, so ∠RSQ = (180 − 58)/2 = 61°. ∠PQS = ∠VQT = 19° (vertically opposite). ∠RSQ is an exterior angle of triangle PQS, so ∠QPS = 61 − 19 = 42°.
The angle of depression of a point Q from a vertical tower PR, 30m high, is 40°. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, |PQ|.
Answer: A (35.75m)
tan 40° = 30/PQ, so PQ = 30 ÷ 0.8391 ≈ 35.75 m.
The volume of a cylinder of radius 14cm is 210cm3. What is the curved surface area of the cylinder?
Answer: B (30cm2)
Curved surface 2πrh = 2 × (πr2h)/r = 2 × 210/14 = 30 cm2.
Simplify 3√12 + 10√3 − 6/√3
Answer: C (14√3)
3√12 = 6√3 and 6/√3 = 2√3, so 6√3 + 10√3 − 2√3 = 14√3.
Factorize m(2a − b) − 2n(b − 2a).
Answer: C ((2a − b)(m + 2n))
−2n(b − 2a) = +2n(2a − b), so the expression is (2a − b)(m + 2n).
If q oranges are sold for t Naira, how many oranges can be bought for p Naira?
Answer: D (pq/t)
q oranges cost t naira, so 1 naira buys q/t oranges and p naira buy pq/t oranges.
In the diagram, QRT is a straight line. If angle PTR = 90°, angle PRT = 60°, angle PQR = 30° and |PQ| = 6√3 cm, calculate |RT|.
Answer: C (3cm)
PT = 6√3 sin 30° = 3√3 cm. In triangle PRT, tan 60° = PT/RT, so RT = 3√3/√3 = 3 cm.
In the diagram, calculate the value of x.
Answer: C (100°)
The right side meets the parallel lines at 180 − 115 = 65° inside the small triangle at the top, so the third angle there is 180 − 35 − 65 = 80°. x is on a straight line with it: x = 100°.
In the diagram, PS and QT are two altitudes of ΔPQR. Which of the following is equal to ∠RQT?
Answer: D (∠SPR)
In right-angled triangles QTR and PSR, ∠RQT = 90° − ∠R and ∠SPR = 90° − ∠R, so they are equal.
In the diagram, |QR| = 10cm, PR ⊥ QS, angle PSR = 30° and angle PQR = 45°. Calculate in metres, |QS|.
Answer: A (10(1 + √3))
Angle Q = 45°, so PR = QR = 10. In triangle PRS, RS = PR/tan 30° = 10√3. QS = 10 + 10√3 = 10(1 + √3).
Calculate, correct to two significant figures, the percentage error in approximating 0.375 to 0.4.
Answer: D (6.7.)
Error = 0.025. 0.025/0.375 × 100% = 6.67% ≈ 6.7%.
To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?
Answer: C (90km/hr)
Scheduled time = 60/72 h = 50 min. Starting 10 min late leaves 40 min = 2/3 h, so speed = 60 ÷ 2/3 = 90 km/h.
In the diagram, QR||TP and ∠WPT = 88°. Find the value of x.
Answer: C (67°)
Through W draw a line parallel to QR. The angle QWP on the left is x + (180 − 88) = x + 92, and the marked reflex angle 3x completes it to 360°: 3x + x + 92 = 360, so x = 67°.
In the diagram, O is the centre of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR.
Answer: D (86°)
OS = OR, so ∠OSR = (180 − 64)/2 = 58° and ∠PSR = 36 + 58 = 94°. Opposite angles of cyclic quadrilateral PQRS add to 180°, so ∠PQR = 86°.
Which of the following represents the solution to the inequality x/3 − (x − 3)/2 ≤ 1?
Answer: D (Number line D)
Multiply by 6: 2x − 3x + 9 ≤ 6, so −x ≤ −3 and x ≥ 3. Line D starts at a shaded dot on 3 and goes right.
Given that p = x − 1/x and q = x2 + 1/x2, express q in terms of p.
Answer: A ((p2 + 2))
p2 = x2 − 2 + 1/x2 = q − 2, so q = p2 + 2.
Which of the following is not a rational number?
Answer: D (√90)
90 is not a perfect square, so √90 cannot be written as a fraction. √4 = 2 is rational.
In the diagram, LMT is a straight line. If O is the centre of circle LMN, ∠OMN = 20°, ∠LTN = 32° and |NM| = |MT|, find ∠LNM.
Answer: B (46°)
NM = MT gives ∠MNT = 32°, so ∠NMT = 116° and ∠LMN = 64°. OM = ON gives ∠MON = 140°, so ∠MLN = 70°. Then ∠LNM = 180 − 64 − 70 = 46°.
(a) Simplify (2a + b)2 − (b − 2a)2
(b) Given that S = K√(m2 + n2); (i) make m the subject of the relation; (ii) if S = 12.2, K = 0.02 and n = 1.1, find, correct to the nearest whole number, the positive value of m.
(a) Difference of two squares: [(2a + b) − (b − 2a)][(2a + b) + (b − 2a)] = 4a × 2b = 8ab.
(b)(i) S/K = √(m2 + n2), so m2 = S2/K2 − n2 and m = √(S2/K2 − n2).
(ii) S/K = 12.2/0.02 = 610. m = √(372100 − 1.21) ≈ 610.
Square both sides to remove the root, then isolate m².
The sets A = {1, 3, 5, 7, 9, 11}, B = {2, 3, 5, 7, 11, 15} and C = {3, 6, 9, 12, 15} are subsets of ε = {1, 2, 3, ...., 15}
(a) Draw a Venn diagram to illustrate the given information. (b) Use your diagram to find: (i) C∩A' (ii) A'∩(B∪C).
(a) Regions: A∩B∩C = {3}; A∩B only = {5, 7, 11}; A∩C only = {9}; B∩C only = {15}; A only = {1}; B only = {2}; C only = {6, 12}; outside all = {4, 8, 10, 13, 14}.
(b)(i) C∩A' = {6, 12, 15}.
(ii) A'∩(B∪C) = {2, 6, 12, 15}.
Place the element common to all three sets first, then work outwards.
(a) A manufacturer offers distributors a discount of 20% on any article bought and a further discount of 2½% for prompt payment. (i) If the marked price of an article is N25,000 find the total amount saved by a distributor for paying promptly. (ii) If a distributor pays N11,700 promptly for an article marked Nx, find the value of x.
(b) Factorize 6y2 − 149y − 102, hence solve the equation 6y2 − 149y − 102 = 0
(a)(i) After 20% off: 0.8 × 25,000 = N20,000. A further 2½% off: 0.025 × 20,000 = N500. Total saved = 5,000 + 500 = N5,500.
(ii) He pays 0.8 × 0.975x = 0.78x = 11,700, so x = N15,000.
(b) 6y2 − 153y + 4y − 102 = 3y(2y − 51) + 2(2y − 51) = (3y + 2)(2y − 51).
y = 25½ or y = −2/3.
The second discount is taken off the already-reduced price.
(a) Without using calculator or mathematical tables evaluate: (3/√3)(2/√3 − √12/6)
(b) In the diagram, O is the centre of the circle. The side AB is produced to E, ∠ACB = 49° and ∠CBE = 68°. Calculate: (i) the interior angle AOC; (ii) angle BOC
(a) 3/√3 = √3 and √12/6 = √3/3. √3 × 2/√3 − √3 × √3/3 = 2 − 1 = 1.
(b)(i) ∠ABC = 180 − 68 = 112°. The reflex angle AOC = 2 × 112° = 224°, so the interior angle AOC = 360 − 224 = 136°.
(ii) ∠BAC = ∠CBE − ∠ACB = 68 − 49 = 19° (exterior angle of triangle ABC). ∠BOC = 2 × 19° = 38°.
The angle at the centre is twice the angle at the circumference on the same arc.
The probabilities that Ade, Kujo and Fati will pass an examination are 2/3, 5/8 and 3/4 respectively. Find the probability that: (a) the three (b) none of them; (c) Ade and Kujo only; will pass the examination.
(a) 2/3 × 5/8 × 3/4 = 5/16.
(b) 1/3 × 3/8 × 1/4 = 1/32.
(c) Ade and Kujo pass, Fati fails: 2/3 × 5/8 × 1/4 = 5/48.
'Only' means the third person must fail, so use 1 − 3/4 for Fati.
(a) Simplify ½ log1025 − 2 log103 + log1018
(b) If 123y = 8310, obtain an equation in y, hence find the value of y.
(c) Solve the equation 92x−3/3x+3 = 1
(a) log 5 − log 9 + log 18 = log(5 × 18/9) = log 10 = 1.
(b) y2 + 2y + 3 = 83, so y2 + 2y − 80 = 0, (y + 10)(y − 8) = 0. y = 8 (a base must be positive and bigger than 3).
(c) 34x−6 = 3x+3, so 4x − 6 = x + 3 and x = 3.
Write 9 as 3² so both sides have the same base.
(a) The first term of an Arithmetic Progression (AP) is 3 and the common difference is 4. Find the sum of the first 28 terms.
(b) Given that x = 2m/(1 − m2) and y = 2m/(1 + m), express 2x − y in terms of m in the simplest form.
(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x − 10)°. Find the value of x.
(a) S28 = 28/2 [2(3) + 27(4)] = 14 × 114 = 1596.
(b) 2x − y = 4m/(1 − m2) − 2m(1 − m)/(1 − m2) = (2m + 2m2)/(1 − m2) = 2m(1 + m)/[(1 − m)(1 + m)] = 2m/(1 − m).
(c) Angles of a pentagon add to 540°: 11x − 10 = 540, so x = 50.
Use 1 − m² = (1 − m)(1 + m) as the common denominator.
(a) An open rectangular tank is made of a steel plate of area 1440m2. Its length is twice its width. If the depth of the tank is 4m less than its width, find its length.
(b) A man saved N3,000 in a bank P, whose interest rate was x% p.a. and N2,000 in another bank Q whose interest rate was y% p.a. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.
(a) Width b, length 2b, depth b − 4. Open top: base + four sides = 2b2 + 2(2b)(b − 4) + 2b(b − 4) = 8b2 − 24b = 1440.
b2 − 3b − 180 = 0, (b − 15)(b + 12) = 0, b = 15. Length = 30 m.
(b) 30x + 20y = 640 and 20x + 30y = 660, i.e. 3x + 2y = 64 and 2x + 3y = 66.
Solving: x = 12, y = 14.
An open tank has no top, so count the base and four sides only.
The diagram is a portion of a right circular solid cylinder of radius 7cm and height 15cm. The centre of the base of the cylinder is Q, while that of the top is B, where ∠ABC = ∠PQR = 120°. Calculate, correct to one decimal place; (a) the volume (b) the total surface area of the solid. [Take π = 22/7]
A 120° wedge has been removed, so the solid is 240/360 = 2/3 of the cylinder.
(a) V = 2/3 × 22/7 × 72 × 15 = 1540.0 cm3.
(b) Curved surface = 2/3 × 2 × 22/7 × 7 × 15 = 440 cm2.
Two flat rectangles = 2 × 7 × 15 = 210 cm2.
Top and bottom = 2 × 2/3 × 22/7 × 49 = 205.3 cm2.
Total = 440 + 210 + 205.3 = 855.3 cm2.
Remember the two flat rectangular faces left by the cut.
Using ruler and a pair of compasses only, (a) construct: (i) triangle XYZ with |XY| = 8cm, ∠YXZ = 60° and ∠XYZ = 30°; (ii) the perpendicular ZT to meet XY in T; (iii) the locus l1 of points equidistant from ZY and XY. (b) If l1 and ZT intersect at S, measure |ST|.
(a)(i) Draw XY = 8 cm. Construct 60° at X and 30° at Y (bisect a 60° angle); the arms meet at Z (∠Z = 90°).
(ii) From Z drop the perpendicular to XY, meeting it at T.
(iii) Bisect ∠XYZ; this bisector is l1.
(b) ST ≈ 1.6 cm (by calculation: XZ = 4, XT = 2, TY = 6, ST = 6 tan 15° ≈ 1.61 cm).
Points equidistant from two lines lie on the bisector of the angle between them.
In the diagram, |PQ| = 8m, |QR| = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°. (a) Calculate, correct to 3 significant figures, (i) |PR| (ii) the bearing of R from P. (b) Calculate the shortest distance between point Q and PR, hence the area of triangle PQR.
(a)(i) ∠PQR = 50° + (180° − 130°) = 100°.
PR2 = 82 + 132 − 2(8)(13) cos 100° = 233 + 36.1 = 269.1, so PR ≈ 16.4 m.
(ii) sin ∠QPR = 13 sin 100°/16.40 = 0.7805, so ∠QPR ≈ 51.3°. Bearing of R from P = 50 + 51.3 ≈ 101°.
(b) Shortest distance = perpendicular from Q to PR = 8 sin 51.3° ≈ 6.24 m.
Area = ½ × 16.4 × 6.24 ≈ 51.2 m2 (check: ½ × 8 × 13 × sin 100° = 51.2).
Use the north lines at P and Q to find the angle at Q first.
Below is the graph of the relation y = ax2 + bx + c, where a, b, c, are constants. Use the graph to: (a) Find the roots of the quadratic equation ax2 + bx + c = 0; (b) Determine the values of constants a, b, and c in the relation using the values of the coordinates P, Q and hence write down the relation illustrated in the graph; (c) Find the maximum value of y and the corresponding value of x at this point (d) Find the values of x when y = 2.
(a) The curve cuts the x-axis at P(−1.5, 0) and Q(2, 0): roots x = −1.5 and x = 2.
(b) y = a(x + 1.5)(x − 2). The curve meets the y-axis at (0, 6), so 6 = a(1.5)(−2) and a = −2.
y = −2(x + 1.5)(x − 2) = −2x2 + x + 6: a = −2, b = 1, c = 6.
(c) Maximum at x = −b/2a = 0.25: y = 6.125 (about 6.1 from the graph).
(d) −2x2 + x + 6 = 2 gives 2x2 − x − 4 = 0, x ≈ 1.7 or x ≈ −1.2.
Known roots let you write y = a(x − r₁)(x − r₂); one more point gives a.