56 questions. Try each one, then open the answer to check yourself and read the explanation.
The table below gives the distribution of marks obtained by a number of pupils in a class test.
Marks: 0, 1, 2, 3, 4, 5
Frequency: 4, 7, 12, 18, 11, 8
The mode of the distribution is
Answer: A (3)
The mode is the mark with the highest frequency: 3 marks (18 pupils).
The table below gives the distribution of marks obtained by a number of pupils in a class test.
Marks: 0, 1, 2, 3, 4, 5
Frequency: 4, 7, 12, 18, 11, 8
Find the median of the distribution
Answer: C (3)
There are 60 pupils, so the median is between the 30th and 31st. The cumulative frequencies are 4, 11, 23, 41, so both have 3 marks: median = 3.
The table below gives the distribution of marks obtained by a number of pupils in a class test.
Marks: 0, 1, 2, 3, 4, 5
Frequency: 4, 7, 12, 18, 11, 8
How many pupils scored at least 2 marks?
Answer: D (49)
At least 2 means 2, 3, 4 or 5 marks: 12 + 18 + 11 + 8 = 49.
Interprete the inequality represented on the number line
Answer: A (−4 < x ≤ 5)
The circle at −4 is open (−4 not included) and the dot at 5 is shaded (5 included): −4 < x ≤ 5.
A student bought 3 notebooks and 1 pen for N35. After misplacing these items, she again bought 2 notebooks and 2 pens, all of the same type for N30. What is the cost of a pen?
Answer: A (N5.00)
3n + p = 35 and 2n + 2p = 30 (n + p = 15). Subtracting: 2n = 20, n = 10, so p = 5.
Simplify 4/(a − 3) − 1/(a + 2)
Answer: D ((3a + 11)/((a − 3)(a + 2)))
4(a + 2) − (a − 3) = 3a + 11 over the common denominator (a − 3)(a + 2).
Simplify 4√18/√8
Answer: C (6)
√18/√8 = √(9/4) = 3/2, so 4 × 3/2 = 6.
In the diagram, PQ||RT, |RS| = |QS|, ∠QST = 70° and ∠PQW = 55°. Find the size of reflex ∠WQS
Answer: A (235°)
∠PQS = ∠QST = 70° (alternate angles, PQ ∥ RT). The non-reflex ∠WQS = 55° + 70° = 125°, so the reflex angle = 360° − 125° = 235°.
Three men, Bedu, Bakre and Kofi shared N500 in the ratio 3:2: x respectively. If Bedu's share is N150, find the value of x
Answer: C (5)
3/(5 + x) × 500 = 150, so 5 + x = 10 and x = 5.
Simplify (1 2/3)2 − (2/3)2
Answer: A (2 1/3)
(5/3)2 − (2/3)2 = 25/9 − 4/9 = 21/9 = 2 1/3.
Find the value of x which satisfies the equation 5(x − 7) = 7 − 2x
Answer: C (x = 6)
5x − 35 = 7 − 2x, so 7x = 42 and x = 6.
A pyramid of volume 120cm3 has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid
Answer: C (12cm)
V = ⅓ × 30 × h = 10h = 120, so h = 12 cm.
PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?
Answer: C (105°)
Opposite angles of a cyclic quadrilateral add to 180°: 180 − 75 = 105°.
In the diagram, ZM is a straight line. Calculate the value of x.
Answer: C (35°)
The exterior angle at Y equals the sum of the opposite interior angles: 3x + 14 = 2x + (2x − 21), so x = 35°.
Find the value of x in the equation 3x2 − 8x − 3 = 0
Answer: C (−1/3, 3)
3x2 − 8x − 3 = (3x + 1)(x − 3) = 0, so x = −1/3 or 3.
In ΔPQR, ∠PQR = 84°, ∠QPR = 43° and |PQ| = 5cm. Find |QR| in cm, correct to 1 decimal place.
Answer: B (4.3)
∠R = 180 − 84 − 43 = 53°. Sine rule: QR = 5 sin 43°/sin 53° = 5 × 0.6820/0.7986 ≈ 4.3 cm.
Given, that 4P45 = 11910, find the value of P
Answer: C (3)
4 × 25 + 5P + 4 = 119, so 5P = 15 and P = 3.
A ladder, 6m long, leans against a vertical wall at an angle 53° to the horizontal. How high up the wall does the ladder reach?
Answer: C (4.792m)
Height = 6 sin 53° = 6 × 0.7986 ≈ 4.792 m.
What is the size of angle x in the diagram?
Answer: B (30°)
The 5 cm side is opposite x and the hypotenuse is 10 cm: sin x = 5/10 = 0.5, so x = 30°.
The four interior angles of a quadrilateral are (x + 20)°, (x + 10)° (2x − 45)° and (x − 25)°. Find the value of x
Answer: B (80)
The angles add to 360°: 5x − 40 = 360, so x = 80.
Calculate the value of y in the diagram
Answer: A (17)
Through the 136° corner draw a line parallel to the two arrowed lines; the two parts of 136° are 5y and 3y (alternate angles), so 8y = 136 and y = 17.
Out of 60 members of an Association, 15 are Doctors and 9 are Lawyers. If a member is selected at random from the Association, what is the probability that the member is neither a Doctor nor a Lawyer?
Answer: A (3/5)
Assuming no one is both, 60 − 24 = 36 members are neither: 36/60 = 3/5.
Find the values of x for which 1/(2x2 − 13x + 15) is not defined,
Answer: A (5 or 3/2)
The denominator 2x2 − 13x + 15 = (2x − 3)(x − 5) is zero when x = 3/2 or 5.
In the diagram, ∠PQS = 65°, ∠RPS = 40° and ∠QSR = 20°. Find ∠PSQ.
Answer: C (55°)
∠QPR = ∠QSR = 20° (same arc QR), so ∠QPS = 20 + 40 = 60°. In triangle PQS, ∠PSQ = 180 − 65 − 60 = 55°.
Evaluate (111two)2 − (101two)2
Answer: D (11000two)
111two = 7 and 101two = 5. 49 − 25 = 24 = 11000two.
The graphs above are for the relations y = ax2 + bx + c and y = d − x. The equation y = ax2 + bx + c is equivalent to
Answer: C (y = 2 + x − x2)
The curve cuts the x-axis at −1 and 2 and opens downwards, with y = 2 at x = 0: y = −(x + 1)(x − 2) = 2 + x − x2.
The graphs above are for the relations y = ax2 + bx + c and y = d − x. What are the values of x for which ax2 + bx + c = 0?
Answer: D (−1, 2)
The roots are where the curve crosses the x-axis: x = −1 and x = 2.
If 3(1−n)/9−2n = 1/9, find n
Answer: C (−1)
9−2n = 3−4n, so 31 − n + 4n = 3−2: 1 + 3n = −2 and n = −1.
Which of the following is/are not the interior angle(s) of a regular polygon? I. 108° II. 116° III. 120°
Answer: B (II only)
The exterior angles would be 72°, 64° and 60°. 360 ÷ 64 is not a whole number, so 116° is impossible; 108° (pentagon) and 120° (hexagon) are possible.
In a ΔXYZ, |YZ| = 6cm ∠YXZ = 60° and ∠XYZ is a right angle. Calculate |XZ| in cm, leaving your answer in surd form
Answer: B (4√3)
YZ is opposite the 60° angle and XZ is the hypotenuse: XZ = 6/sin 60° = 6 ÷ (√3/2) = 12/√3 = 4√3.
Use the Venn diagram. How many elements are in P∩Q?
Answer: B (2)
Q lies inside P, so P∩Q = Q = {f, e}: 2 elements.
Use the Venn diagram. Find Q'∩R
Answer: C ((c, g, h))
R = {e, h, c, g}. Remove e, which is in Q: Q'∩R = {c, g, h}.
The square root of a number is 2k. What is half of the number
Answer: D (2k2)
The number is (2k)2 = 4k2, and half of it is 2k2.
Given that p varies as the square of q and q varies inversely as the square root of r. How does p vary with r?
Answer: D (p varies inversely as r)
p = kq2 and q = c/√r, so p = kc2/r: p varies inversely as r.
The probabilities of a boy passing English and Mathematics test are x and y respectively. Find the probability of the boy failing both tests
Answer: C (1 − (x + y) + xy)
P(fail both) = (1 − x)(1 − y) = 1 − x − y + xy = 1 − (x + y) + xy.
The locus of points equidistant from two intersecting straight lines PQ and PR is
Answer: D (the bisector of angle QPR)
Any point on the bisector of angle QPR is the same perpendicular distance from line PQ and line PR.
Find the equation whose roots are −2/3 and 3
Answer: D (3x2 − 7x − 6 = 0)
(3x + 2)(x − 3) = 3x2 − 7x − 6 = 0.
Evaluate Cos 45° Cos 30° − Sin 45° Sin 30° leaving the answer in surd form
Answer: D ((√6 − √2)/4)
(√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
In constructing an angle, Olu draws line OX. With centre O and a convenient radius, he draws an arc intersecting OX at P. With centres P and the same radius, he draws an arc intersecting the first arc at Q and finally joins OQ. What is the size of angle POQ so constructed?
Answer: C (60°)
OP = OQ = PQ (all the same radius), so triangle OPQ is equilateral and ∠POQ = 60°.
In the diagram |PS| = 9 cm, |QR| = 5cm, ∠PSR = 63° and ∠SPQ = ∠PQR = 90°. Find, correct to the nearest whole number, the area of the trapezium.
Answer: A (55cm2)
Drop a line from R across to PS: it cuts off 9 − 5 = 4 cm, so the height PQ = 4 tan 63° ≈ 7.85 cm. Area = ½(9 + 5) × 7.85 ≈ 55 cm2.
A right circular cone is such that its radius r is twice its height h. Find its volume in terms of h
Answer: D (4/3 πh3)
V = ⅓π(2h)2h = 4/3 πh3.
A man made a loss of 15% by selling an article for N595. Find the cost price of the article
Answer: C (N700.00)
N595 is 85% of the cost price: 595 ÷ 0.85 = N700.
The bearing of P from Q is x, where 270° < x < 360°. Find the bearing of Q from P
Answer: D ((x − 180)°)
A back bearing differs by 180°; since x is more than 180°, the bearing of Q from P is x − 180°.
(a) If p varies directly as r2 and p = 3.2 when r = 4, find the value of p when r = 6.5
(b) Solve the simultaneous equations x/2 + y/4 = 1; x/3 − y/4 = −1/6
(a) p = kr2: 3.2 = 16k, k = 0.2. p = 0.2 × 6.52 = 8.45.
(b) Add the equations: x/2 + x/3 = 1 − 1/6, so 5x/6 = 5/6 and x = 1.
Then 1/2 + y/4 = 1, y/4 = 1/2, y = 2.
Adding removes the y terms because they have opposite signs.
Without using Mathematical Tables or a Calculator, simplify
(a) √50 − 3√2(2√2 − 5) − 5√32
(b) ½ log10(25/4) − 2log10(4/5) + log10(320/125)
(a) √50 = 5√2; 3√2(2√2 − 5) = 12 − 15√2; 5√32 = 20√2.
5√2 − 12 + 15√2 − 20√2 = −12.
(b) ½ log(25/4) = log(5/2); 2 log(4/5) = log(16/25).
log[(5/2) × (25/16) × (320/125)] = log 10 = 1.
Turn coefficients into powers, then combine into one log.
(a) Simplify: 6253/8 × 51/2 ÷ 25
(b) Solve the following equations, correct to one decimal place. (i) tan (θ + 25°) = 5.145 (ii) 5 cos θ − 1 = 0, where 0 ≤ θ ≤ 90°
(a) 6253/8 = (54)3/8 = 53/2. 53/2 × 51/2 ÷ 52 = 50 = 1.
(b)(i) θ + 25° = tan−15.145 = 79.0°, so θ = 54.0°.
(ii) cos θ = 0.2, so θ = 78.5°.
Write 625 and 25 as powers of 5.
In the diagram, PQRS is a circle with centre O and radius 7cm. SQ and PR intersect at K and ∠SKR = 90°. If the length of the arc SR is four times that of arc PQ, find the length of arc SR. [Take π = 22/7]
The angle between two chords equals half the sum of the arcs they cut off: ∠SKR = ½(arc SR + arc PQ) in degrees.
So arc SR + arc PQ = 180°. With arc SR = 4 × arc PQ: 5 × arc PQ = 180°, arc PQ = 36°, arc SR = 144°.
Length of arc SR = 144/360 × 2 × 22/7 × 7 = 17.6 cm.
When two chords cross, the angle between them is half the sum of the two arcs they cut off.
The Table shows the number of suitcases possessed by a group of travellers
No. of suitcases: 0, 1, 2, 3, 4, 5
Travellers: 2, 7, 7, 2, 3, 9
(a) Calculate the (i) median (ii) mean, correct to the nearest whole number (b) Draw a bar chart to represent the information
There are 30 travellers.
(a)(i) The median is between the 15th and 16th; cumulative frequencies 2, 9, 16…, so both have 2 suitcases: median = 2.
(ii) Σfx = 0 + 7 + 14 + 6 + 12 + 45 = 84; mean = 84/30 = 2.8 ≈ 3.
(b) Bars of equal width, separated by gaps, with heights 2, 7, 7, 2, 3, 9 above 0–5 suitcases; label both axes.
The median of 30 values is the average of the 15th and 16th.
(a) Simplify (1/3 c2 − 2/3 cd)/(1/2 d2 − 1/4 cd)
(b) In the diagram, YPF is a straight line. ∠XPY = 44°, ∠MPF = 46°, ∠XYP = ∠MFP = 90°, |XY| = 7cm and |MP| = 9cm. (i) Calculate, correct to 3 significant figures, |XM| and |YF|. (ii) Find ∠XMP
(a) Numerator = c(c − 2d)/3; denominator = d(2d − c)/4. Result = 4c(c − 2d)/[3d(2d − c)] = −4c/(3d).
(b)(i) XP = 7/sin 44° = 10.08 cm; YP = 7/tan 44° = 7.249 cm; PF = 9 cos 46° = 6.252 cm.
YF = 7.249 + 6.252 = 13.5 cm.
∠XPM = 180 − 44 − 46 = 90°, so XM = √(10.082 + 92) = √182.5 ≈ 13.5 cm.
(ii) tan ∠XMP = XP/MP = 10.08/9 = 1.120, so ∠XMP ≈ 48.2°.
Notice ∠XPM is a right angle because the three angles at P make a straight line.
The table shows the monthly contributions and expenditure pattern of an employee in 1999.
Pension – 5%
Income Tax – 25%
Food – 40%
Transport – 10%
Rent – 12.5%
Others – 7.5%
(a) Draw a pie chart to illustrate the data (b) If the employee's gross monthly salary was N10,800.00, calculate. (i) the pension contribution of the employee (ii) the income tax paid by the employee (c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take-home pay of the employee.
(a) Angles (percentage × 3.6°): Pension 18°, Income Tax 90°, Food 144°, Transport 36°, Rent 45°, Others 27°. Draw and label the sectors.
(b)(i) 5% of 10,800 = N540.00.
(ii) 25% of 10,800 = N2,700.00.
(c) Take-home pay = 10,800 − 540 − 2,700 = N7,560.00.
1% of the whole is 3.6° on a pie chart.
(a) A = {1,2,5,7} and B = {1,3,6,7} are subsets of the universal set U = {1,2,3,....., 10}. Find: (i) A' (ii) (A∩B)'; (iii) (A∪B)', (iv) the subsets of B each of which has three elements.
(b) Write down the 15th term of the sequence 2/(1 × 3), 3/(2 × 4), 4/(3 × 5), 5/(4 × 6), ..........
(c) An Arithmatic Progression (A.P) has 3 as its first term and 4 as the common difference. (i) Write an expression, in its simplest form for the nth term. (ii) find the least term of the AP that is greater than 100.
(a)(i) A' = {3, 4, 6, 8, 9, 10}.
(ii) A∩B = {1, 7}, so (A∩B)' = {2, 3, 4, 5, 6, 8, 9, 10}.
(iii) A∪B = {1, 2, 3, 5, 6, 7}, so (A∪B)' = {4, 8, 9, 10}.
(iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7}, {3, 6, 7}.
(b) Tn = (n + 1)/[n(n + 2)], so T15 = 16/(15 × 17) = 16/255.
(c)(i) Tn = 3 + 4(n − 1) = 4n − 1.
(ii) 4n − 1 > 100 gives n > 25.25, so n = 26 and the term is 103.
Spot the pattern in the numerator and each factor of the denominator separately.
The marks obtained by 40 students in an examination are as follows:
85 77 87 74 77 78 79 89 85 90
78 73 86 83 91 74 84 81 83 75
77 70 81 69 75 63 76 87 61 78
69 96 65 80 84 80 77 74 88 72
(a) Copy and complete the table for the distribution using the above data.
Class Boundaries: 59.5 – 64.5, 64.5 – 69.5, 69.5 – 74.5, 74.5 – 79.5, 79.5 – 84.5, 84.5 – 89.5, 89.5 – 94.5, 94.5 – 99.5 (with Tally and Frequency columns)
(b) Draw a histogram to represent the distribution (c) Using your histogram, estimate the modal mark (d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.
(a) Frequencies: 59.5–64.5: 2; 64.5–69.5: 3; 69.5–74.5: 6; 74.5–79.5: 11; 79.5–84.5: 8; 84.5–89.5: 7; 89.5–94.5: 2; 94.5–99.5: 1 (total 40).
(b) Bars on the class boundaries with heights equal to the frequencies.
(c) In the tallest bar (74.5–79.5) join its top corners to the tops of the next bars; the lines cross at about 77.6, so the modal mark ≈ 78.
(d) Marks above 79: 8 + 7 + 2 + 1 = 18, so P = 18/40 = 9/20.
Tally the marks one by one and check the frequencies add up to 40.
In the diagram, PQT is a straight line and SQ//RT.
(a) Join QR and show that: (i) ∠RPS = ∠QRT, (ii) ∠PRS = ∠QTR
(b) ABC is a triangle. The sides AB and AC are produced to D and E respectively such that ∠DBC = 132° and ∠ECB = 96°. Show that ΔABC is isosceles.
(a)(i) ∠RPS = ∠RQS (angles in the same segment, on chord RS).
∠RQS = ∠QRT (alternate angles, SQ ∥ RT). Hence ∠RPS = ∠QRT.
(ii) ∠PRS = ∠PQS (angles in the same segment, on chord PS).
∠PQS = ∠QTR (corresponding angles, SQ ∥ RT, PQT a straight line). Hence ∠PRS = ∠QTR.
(b) ∠ABC = 180 − 132 = 48° and ∠ACB = 180 − 96 = 84° (angles on a straight line).
∠BAC = 180 − 48 − 84 = 48°. ∠BAC = ∠ABC, so CA = CB and triangle ABC is isosceles.
Give a reason for every equal pair of angles.
An aeroplane flies due west for 3 hours from P(lat 50°N, long 60°W) to point Q at an average speed of 600kmh−1. The aeroplane then flies due south from Q to a point Y 500 km away. Calculate, correct to 3 significant figures. (a) the longitude of Q (b) the latitude of Y. [Take the radius of the earth to be 6400 km and π = 22/7]
(a) Distance PQ = 600 × 3 = 1800 km along latitude 50°N.
1800 = θ/360 × 2 × 22/7 × 6400 cos 50°, so θ = 1800 × 360/(2 × 22/7 × 6400 × 0.6428) ≈ 25.1°.
Flying west from 60°W: longitude of Q = 60 + 25.1 = 85.1°W.
(b) 500 = α/360 × 2 × 22/7 × 6400, so α ≈ 4.47°.
Flying south from 50°N: latitude of Y = 50 − 4.47 ≈ 45.5°N.
Along a latitude use radius R cos(latitude); along a meridian use the full radius R.
Using ruler and a pair of compasses only. (a) construct a quadrilateral PXYQ such that |PX| = 9.9cm, |QX| = 10.2cm, ∠QPX = 75°, |QY| = 10.4cm and PQ//XY. (b) construct the: (i) locus l1 of points equidistant from X and Y, (ii) locus l2 of point equidistant from QY and YX (c) Locate M, the point of intersection of l1 and l2 (d) Measure |PM|.
(a) Draw PX = 9.9 cm. Construct 75° at P (60° plus half of the 30° between 60° and 90°). With centre X radius 10.2 cm cut this arm at Q (PQ ≈ 6.1 cm). Through X construct a line parallel to PQ; with centre Q radius 10.4 cm cut it at Y. Join QY.
(b)(i) l1: perpendicular bisector of XY.
(ii) l2: bisector of ∠QYX.
(c) M is where l1 and l2 cross.
(d) |PM| ≈ 9.5 cm (accept 9.3–9.7 cm).
Equidistant from two points: perpendicular bisector. Equidistant from two lines: angle bisector.
The table shows the values of the relation y = 11 − 2x − 2x2 for −4 ≤ x ≤ 3
x: −4, −3, −2, −1, 0, 1, 2, 3
y: −13, , , , 11, , ,
(a) Copy and complete the table (b) Using a scale of 2cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 11 − 2x − 2x2. (c) Use your graph to find: (i) the roots of the equation 11 − 2x − 2x2 = 0, (ii) the values of x for which 3 − 2x − 2x2 = 0, (iii) the gradient of the curve at x = 1
(a) x = −3: −1; −2: 7; −1: 11; 1: 7; 2: −1; 3: −13.
(b) Plot the points and join with a smooth upside-down parabola.
(c)(i) The curve crosses y = 0 at x ≈ −2.9 and x ≈ 1.9.
(ii) 3 − 2x − 2x2 = 0 means 11 − 2x − 2x2 = 8; draw y = 8: x ≈ −1.8 and x ≈ 0.8.
(iii) Draw the tangent at (1, 7); its gradient is about −6.
Add 8 to both sides of the new equation so it matches the drawn curve.