62 questions. Try each one, then open the answer to check yourself and read the explanation.
Express 302.10495 correct to five significant figures.
Answer: A (302.10)
The first five significant figures are 3, 0, 2, 1, 0 and the next digit is 4, so round down: 302.10.
Simplify: (3√5 × 4√6)/(2√2 × 3√3)
Answer: D (2√5)
Numerator = 12√30 and denominator = 6√6. 12√30/(6√6) = 2√5.
In 1995, the enrolments of two schools X and Y were 1,050 and 1,190 respectively. Find the ratio of the enrolments of X to Y.
Answer: B (15 : 17)
Divide both by 70: 1050 : 1190 = 15 : 17.
Convert 3510 to a number in base 2.
Answer: C (100011)
35 = 32 + 2 + 1, so it is 1000112.
The nth term of a sequence is Tn = 5 + (n − 1)2. Evaluate T4 − T6.
Answer: C (−16)
T4 = 5 + 9 = 14 and T6 = 5 + 25 = 30, so T4 − T6 = −16.
Mr. Manu travelled from Accra to Pamfokrom a distance of 720 km in 8 hours. What will be his speed in m/s?
Answer: A (25 m/s)
Speed = 720,000 m ÷ (8 × 3600 s) = 25 m/s.
If ₦2,500.00 amounted to ₦3,500.00 in 4 years at simple interest, find the rate at which the interest was charged.
Answer: D (10%)
Interest = ₦1,000. R = (100 × 1000)/(2500 × 4) = 10%.
Simplify: (54k2 − 6)/(3k + 1).
Answer: C (6(3k − 1))
54k2 − 6 = 6(9k2 − 1) = 6(3k − 1)(3k + 1). Cancel (3k + 1) to get 6(3k − 1).
Represent the inequality −7 < 4x + 9 ≤ 13 on a number line.
Answer: B (Diagram B)
Subtract 9: −16 < 4x ≤ 4, so −4 < x ≤ 1. That is an open circle at −4 and a filled dot at 1, as in B.
Make p the subject of the relation: q = 3p/r + s/2.
Answer: D (p = (2qr − rs)/6)
Multiply by 2r: 2qr = 6p + rs, so 6p = 2qr − rs and p = (2qr − rs)/6.
If x + y = 2y − x + 1 = 5, find the value of x.
Answer: B (2)
x + y = 5 and 2y − x + 1 = 5 (so 2y − x = 4). Substituting y = 5 − x: 10 − 2x − x = 4, so x = 2.
The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation.
Answer: D (5n − 33 = 0)
12 + n/3 = 2n + 1. Multiply by 3: 36 + n = 6n + 3, so 5n − 33 = 0.
Solve the inequality: −m/2 − 5/4 ≤ 5m/12 − 7/6.
Answer: C (m ≥ −1/11)
Multiply by 12: −6m − 15 ≤ 5m − 14, so −11m ≤ 1. Dividing by −11 reverses the sign: m ≥ −1/11.
The curved surface area of a cylindrical tin is 704 cm2. If the radius of its base is 8 cm, find the height. [Take π = 22/7]
Answer: A (14 cm)
2πrh = 704: h = 704 ÷ (2 × 22/7 × 8) = 704 ÷ 50.29 = 14 cm.
The lengths of the minor and major arcs of a circle are 54 cm and 126 cm respectively. Calculate the angle of the major sector.
Answer: B (252°)
Circumference = 180 cm. Major angle = 126/180 × 360° = 252°.
A sector of a circle which subtends 172° at the centre of the circle has a perimeter of 600 cm. Find, correct to the nearest cm, the radius of the circle. [Take π = 22/7]
Answer: A (120 cm)
Arc = 172/360 × 2 × 22/7 × r ≈ 3.003r. Perimeter = arc + 2r = 5.003r = 600, so r ≈ 120 cm.
In the diagram, |QR| = 10 m, |SR| = 8 m, ∠QPS = 30°, ∠QRP = 90° and |PS| = x. Find x.
Answer: C (9.32 m)
tan 30° = 10/(x + 8), so x + 8 = 10/0.5774 = 17.32 and x = 9.32 m.
In ΔXYZ, |XY| = 8cm, |YZ| = 10 cm and |XZ| = 6 cm. Which of these relations is true?
Answer: C (|XZ|2 = |YZ|2 − |XY|2)
62 + 82 = 100 = 102, so |XZ|2 + |XY|2 = |YZ|2, i.e. |XZ|2 = |YZ|2 − |XY|2.
In the diagram, O is the centre of the circle PQRS and ∠PSR = 86°. If ∠POR = x°, find x.
Answer: B (172)
The angle at the centre is twice the angle at the circumference on the same arc: x = 2 × 86 = 172.
The diagram is a circle centre O. If ∠SPR = 2m and ∠SQR = n, express m in terms of n.
Answer: A (m = n/2)
∠SPR and ∠SQR stand on the same arc SR, so they are equal: 2m = n and m = n/2.
In the diagram, MQ||RS, ∠TUV = 70° and ∠RLV = 30°. Find the value of x.
Answer: C (100°)
RS makes 30° with UV, so the parallel line MQ also makes 30° with UV (corresponding angles). In the triangle formed by UV, UT and MQ, the exterior angle x at X equals the sum of the opposite interior angles: 70° + 30° = 100°.
In the diagram, MN, PQ, and RS are three intersecting straight lines. Which of the following statement(s) is/are true?
I. t = y
II. x + y + z + m = 180°
III. x + m + n = 180°
IV. x + n = m + z
Answer: C (III only)
n = y (vertically opposite), and m + y + x = 180° (angles on the straight line MN). So x + m + n = 180° (III). t equals x, not y, so I is false; II and IV are also false.
If cos (x + 40)° = 0.0872, what is the value of x?
Answer: D (45°)
cos−1(0.0872) = 85°, so x + 40 = 85 and x = 45°.
A kite flies on a taut string of length 50 m inclined at an angle of 54° to the horizontal ground. The height of the kite above the ground is
Answer: B (50 sin 54°)
The string is the hypotenuse and the height is opposite the 54° angle: h = 50 sin 54°.
The positions of three ships P, Q and R at sea are illustrated in the diagram. The arrows indicate the North direction. The bearing of Q from P is 050° and ∠PQR = 72°. Calculate the bearing of R from Q.
Answer: B (158°)
The bearing of P from Q is 050° + 180° = 230°. R is 72° anticlockwise from that direction: 230° − 72° = 158°.
Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores.
Answer: C (5)
Squared deviations from 21: 36, 0, 16, 25, 9, 64, total 150. Variance = 150/6 = 25, so SD = 5.
A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours.
Answer: A (8/15)
P(red then black) + P(black then red) = 4/10 × 6/9 + 6/10 × 4/9 = 48/90 = 8/15.
The bar chart shows the frequency distribution of marks scored by students in a class test. How many students are in the class?
Answer: C (25)
Add the bar heights for marks 1, 2, 3, 4: 6 + 8 + 6 + 5 = 25.
The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.
Answer: C (2.4)
Σfx = 1(6) + 2(8) + 3(6) + 4(5) = 60. Mean = 60/25 = 2.4.
The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?
Answer: A (2)
With 25 students the median is the 13th mark. Cumulative frequencies 6, 14, … so the 13th student scored 2.
Which of these statements about y = 8√m is correct?
Answer: D (log y = 3 log 2 + ½ log m)
log y = log 8 + log m½ = 3 log 2 + ½ log m.
If x + 0.4y = 3 and y = ½x, find the value of (x + y).
Answer: C (3¾)
x + 0.2x = 3, so x = 2.5 and y = 1.25. x + y = 3.75 = 3¾.
Express 3 − ((x − y)/y) as a single fraction.
Answer: D ((4y − x)/y)
(3y − (x − y))/y = (3y − x + y)/y = (4y − x)/y.
Find the coefficient of m in the expansion of (m/2 − 1½)(m + 2/3).
Answer: D (−1⅙)
The m-terms are (m/2)(2/3) = m/3 and (−3/2)(m) = −3m/2. Coefficient = 1/3 − 3/2 = −7/6 = −1⅙.
In the diagram, MN//PO, ∠PMN = 112°, ∠PNO = 129°, ∠NOP = 37° and ∠MPN = y. Find the value of y.
Answer: B (54°)
MN ∥ PO, so ∠MPO = 180° − 112° = 68°. In ΔPNO, ∠NPO = 180° − 129° − 37° = 14°. y = 68° − 14° = 54°.
If P = {prime factors of 210} and Q = {prime numbers less than 10}, find P∩Q.
Answer: D ({2, 3, 5, 7})
210 = 2 × 3 × 5 × 7, so P = {2, 3, 5, 7} = Q, and P∩Q = {2, 3, 5, 7}.
Alfred spent ¼ of his money on food, ⅓ on clothing and saved the rest. If he saved ₦72,000.00, how much did he spend on food?
Answer: A (₦43,200.00)
He saved 1 − ¼ − ⅓ = 5/12 of his money, so the total is 72,000 × 12/5 = ₦172,800. Food = ¼ of that = ₦43,200.
Solve: (27/125)−1/3 × (4/9)1/2
Answer: A (10/9)
(27/125)−1/3 = 5/3 and (4/9)1/2 = 2/3. Product = 10/9.
The sum of the interior angles of a regular polygon is 1800°. How many sides has the polygon?
Answer: B (12)
(n − 2) × 180° = 1800°, so n − 2 = 10 and n = 12.
The diagram is a circle with centre O. PRST are points on the circle. Find the value of ∠PRS.
Answer: A (144°)
∠PTS = x stands on the arc PRS, so x = ½(360° − 8x), giving x = 36°. ∠PRS stands on the arc PTS, so ∠PRS = ½ × 8x = 4 × 36° = 144°.
The diagram is a circle of radius |OQ| = 4 cm. TR is a tangent to the circle at R. If TP̂O = 120°, find |PQ|.
Answer: C (0.62 cm)
∠OPR = 180° − 120° = 60° and OR = 4 cm is perpendicular to TR. OP = 4/sin 60° ≈ 4.62 cm, so PQ = OP − OQ ≈ 0.62 cm.
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
Answer: B (y = k/x)
In inverse variation, y = k/x: as x increases, y decreases.
In the diagram, |SR| = |QR|, ∠SRP = 65° and ∠RPQ = 48°, find ∠PRQ.
Answer: D (19°)
∠RSQ = ∠RPQ = 48° (same segment), and ΔSRQ is isosceles, so ∠RQS = 48° and ∠SRQ = 84°. ∠PRQ = 84° − 65° = 19°.
The graph is that of y = 2x2 − 5x − 3. For what values of x will y be negative?
Answer: C (−½ < x < 3)
y = (2x + 1)(x − 3) is zero at x = −½ and 3 and below the x-axis between them: −½ < x < 3.
The graph is that of y = 2x2 − 5x − 3. What is the gradient of y = 2x2 − 5x − 3 at the point x = 4?
Answer: A (11.1)
The gradient is dy/dx = 4x − 5, which is 11 at x = 4. (A tangent drawn on the graph gives about 11.)
The diagram is a polygon. Find the largest of its interior angles.
Answer: D (150°)
The marked angles 4x, x, 3x, 2x, 2x are exterior angles, which add to 360°: 12x = 360°, x = 30°. The smallest exterior angle (x = 30°) gives the largest interior angle: 180° − 30° = 150°.
The volume of a cuboid is 54 cm3. If the length, width and height of the cuboid are in the ratio 2 : 1 : 1 respectively, find its total surface area.
Answer: B (90 cm2)
2k × k × k = 54 gives k = 3: dimensions 6, 3, 3. Surface area = 2(18 + 18 + 9) = 90 cm2.
A side and a diagonal of a rhombus are 10 cm and 12 cm respectively. Find its area.
Answer: D (96 cm2)
The diagonals bisect at right angles: half of the other diagonal = √(102 − 62) = 8, so it is 16 cm. Area = ½ × 12 × 16 = 96 cm2.
Factorize completely: 32x2y − 48x3y3.
Answer: A (16x2y(2 − 3xy2))
The HCF is 16x2y: 32x2y − 48x3y3 = 16x2y(2 − 3xy2).
(a) Simplify: (1¼ + 7/9)/(1 4/9 − 2⅔ × 9/64)
(b) Given that sin x = 2/3, evaluate, leaving your answer in surd form and without using tables or calculator, tan x − cos x.
(a) Numerator: 5/4 + 7/9 = (45 + 28)/36 = 73/36.
Denominator: 8/3 × 9/64 = 3/8; 13/9 − 3/8 = (104 − 27)/72 = 77/72.
73/36 × 72/77 = 146/77 = 1 69/77.
(b) Opposite 2, hypotenuse 3, adjacent √(9 − 4) = √5.
tan x = 2/√5, cos x = √5/3.
tan x − cos x = 2/√5 − √5/3 = (6 − 5)/(3√5) = 1/(3√5) = √5/15.
Do the multiplication in the denominator before subtracting.
Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?
Let Wale be y, Sonny 2y. Four years ago: 2y − 4 = 4(y − 4), so 2y = 12 and y = 6. Sonny is 12.
In n years: (12 + n) + (6 + n) = 66, so 2n = 48 and n = 24.
The sum of their ages will be 66 in 24 years' time.
Both ages increase by n, so the sum increases by 2n.
(a) In the diagram, TU is a tangent to the circle. RV̂U = 100° and ∠URS = 36°. Calculate the value of angle STU.
(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.
(a) RVUS is cyclic, so the exterior angle ∠UST = ∠RVU = 100°.
Alternate segment: ∠SUT = ∠URS = 36°.
In ΔSUT: ∠STU = 180° − 100° − 36° = 44°.
(b) PQ ∥ YZ, so XQ/XZ = XP/XY = 2/5. XQ = 2/5 × 6 = 2.4 cm.
|QZ| = 6 − 2.4 = 3.6 cm.
The exterior angle of a cyclic quadrilateral equals the interior opposite angle.
(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is ¼, calculate the number of: (i) white discs; (ii) red discs that should be added such that the probability of picking a red disc will be ⅓.
(b) A salesman bought some plates at ₦50.00 each. If he sold all of them for ₦600.00 and made a profit of 20% on the transaction, how many plates did he buy?
(a) Red = ¼ × 40 = 10.
(i) White = 40 − 10 = 30.
(ii) (10 + r)/(40 + r) = ⅓ → 30 + 3r = 40 + r → r = 5 red discs.
(b) ₦600 is 120% of the cost, so cost = 600 × 100/120 = ₦500. Number of plates = 500 ÷ 50 = 10.
Adding red discs increases both the red count and the total.
In the diagram, O is the centre of the circle and XY is a chord. If the radius is 5 cm and |XY| = 6 cm, calculate, correct to 2 decimal places, the: (a) angle which XY subtends at the centre O; (b) area of the shaded portion.
(a) Half the chord is 3 cm: sin(½θ) = 3/5 = 0.6, so ½θ = 36.87° and θ = 73.74°.
(b) Sector = 73.74/360 × 22/7 × 52 ≈ 16.09 cm2.
Triangle OXY = ½ × 6 × 4 = 12 cm2 (height from O = √(25 − 9) = 4).
Shaded segment ≈ 16.09 − 12 = 4.09 cm2.
The shaded minor segment is the sector minus the triangle.
(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister's share was equal to ⅓ of M, find the value of M.
(b) A number of tourists were interviewed on their choice of means of travel. Two-thirds said that they travelled by road, 13/30 by air and 4/15 by both air and road. If 20 tourists did not travel by either air or road, (i) represent the information on a Venn diagram; (ii) how many tourists (A) were interviewed; (B) travelled by air only?
(a) (M − 15)/2 = M/3 → 3M − 45 = 2M → M = 45.
(b)(i) Two overlapping circles, Road and Air: road only 2/3 − 4/15 = 2/5, both 4/15, air only 13/30 − 4/15 = 1/6, outside 20 tourists.
(ii)(A) Road or air = 2/5 + 4/15 + 1/6 = 5/6, so neither = 1/6 of the total = 20. Total = 120.
(B) Air only = 1/6 × 120 = 20.
Write each region as a fraction of the total, then use the 20 who used neither.
(a)(i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of y − 3x/4 = 3 and y + 2x = 6. (ii) From your graph, find the coordinates of the point of intersection of the two graphs. (iii) Show, on the graph sheet, the region satisfied by the inequality y − 3x/4 ≥ 3.
(b) Given that x2 + bx + 18 is factorized as (x + 2)(x + c). Find the values of c and b.
(a)(i) y = 3x/4 + 3 passes through (0, 3) and (−4, 0); y = 6 − 2x passes through (0, 6) and (3, 0). Draw both lines.
(ii) 3x/4 + 3 = 6 − 2x → 11x/4 = 3 → x ≈ 1.1, y ≈ 3.8: about (1.1, 3.8).
(iii) The region on and above the line y = 3x/4 + 3 (shade it, or shade the unwanted side, as instructed).
(b) (x + 2)(x + c) = x2 + (2 + c)x + 2c. 2c = 18 → c = 9; b = 2 + 9 = 11.
Compare coefficients after expanding the brackets.
A point H is 20 m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point (P) on the tower and the top (T) of the tower are 30° and 50° respectively. Calculate, correct to 3 significant figures: (a) |PT|; (b) the distance between H and the top of the tower; (c) the position of H if the angle of depression of H from the top of the tower is to be 40°.
(a) Height of T = 20 tan 50° = 23.84 m; height of P = 20 tan 30° = 11.55 m. |PT| = 12.3 m.
(b) |HT| = 20/cos 50° ≈ 31.1 m.
(c) New distance from the foot = 23.84/tan 40° ≈ 28.4 m, i.e. H must move about 8.41 m further away from the tower.
The angle of depression from T equals the angle of elevation from H.
Three towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A Health Centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y will always travel equal distance to access the Health Centre while patients from Z will travel exactly 10 km to reach the Health Centre. (a) Using a scale of 1 cm to 2 km, find by construction, using a pair of compasses and ruler only, the possible positions the Health Centre can be located. (b) In how many possible locations can the Health Centre be built? (c) Measure and record the distances of the locations from town X. (d) Which of these locations would be convenient for all the three towns?
(a) Draw triangle XYZ with XY = 10 cm, YZ = 11 cm, XZ = 9 cm. Construct the perpendicular bisector of XY. With centre Z and radius 5 cm (10 km) draw a circle. The Health Centre is where they meet.
(b) Two possible locations.
(c) About 12.7 km and 28.0 km from X (6.3 cm and 14.0 cm on the drawing).
(d) The location about 12.7 km from X (inside the triangle), since it is closest to all three towns.
Points equidistant from X and Y lie on the perpendicular bisector of XY; points 10 km from Z lie on a circle.
The table shows the distribution of marks scored by students in an examination.
Class Interval: 60 − 64, 65 − 69, 70 − 74, 75 − 79, 80 − 84, 85 − 89, 90 − 94, 95 − 99
Frequency: 2, 3, 6, 11, 8, 7, 2, 1
Calculate, correct to 2 decimal places, the (a) mean; (b) standard deviation of the distribution.
Class marks: 62, 67, 72, 77, 82, 87, 92, 97; Σf = 40.
(a) Σfx = 124 + 201 + 432 + 847 + 656 + 609 + 184 + 97 = 3150. Mean = 3150/40 = 78.75.
(b) Σf(x − 78.75)2 = 561.13 + 414.19 + 273.38 + 33.69 + 84.50 + 476.44 + 351.13 + 333.06 = 2527.5.
SD = √(2527.5/40) = √63.19 ≈ 7.95.
Use the class mid-points as x.
(a) In the diagram, ABCD is a rectangular garden (3n − 1)m long and (2n + 1) m wide. A wire-mesh 135 m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.
(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take π = 22/7]
(a) The wire runs along 3 lines of length (3n − 1) (top, middle, bottom) and 5 lines of length (2n + 1) (two ends and three dividers).
3(3n − 1) + 5(2n + 1) = 135 → 19n + 2 = 135 → n = 7.
(b) 22/7 × 142 × h = 223 = 10648, so h = 10648/616 ≈ 17.29 cm.
Cylinder TSA = 2πr(r + h) = 2 × 22/7 × 14 × (14 + 17.29) ≈ 2753.1 cm2. Cube = 6 × 222 = 2904 cm2.
Ratio ≈ 2753.1 : 2904 = 73 : 77 (about 0.95 : 1).
Count every length of wire in the grid, including the inside dividers.
(a) Copy and complete the table of values for y = 1 − 4 cos x.
x: 0°, 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°
y: −3.0, __, __, 1.0, __, __, __, 4.5, __, __, −1.0
(b) Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of y = 1 − 4 cos x for 0° ≤ x ≤ 300°.
(c) Use the graph to: (i) solve the equation 1 − 4 cos x = 0; (ii) find the value of y when x = 105°; (iii) find x when y = 1.5.
(a) 30°: −2.5; 60°: −1.0; 120°: 3.0; 150°: 4.5; 180°: 5.0; 240°: 3.0; 270°: 1.0.
(b) Plot the points and draw a smooth curve.
(c)(i) Where the curve crosses the x-axis: x ≈ 76° and x ≈ 284°.
(ii) At x = 105°, y ≈ 2.0.
(iii) Where y = 1.5: x ≈ 97° and x ≈ 263°.
(Accept graph readings within about ±2°.)
For example at 30°: 1 − 4(0.866) = −2.46 ≈ −2.5.
(a) In the diagram, |PQ| = 6 cm, |QR| = 13 cm, |RS| = 5 cm and ∠RSQ is a right angle. Calculate, correct to one decimal place, |PS|.
(b) The diagram shows a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take π = 22/7]
(a) |SQ| = √(132 − 52) = 12 cm. cos ∠SQR = 12/13. P, Q, R are on a straight line, so cos ∠PQS = −12/13.
|PS|2 = 62 + 122 − 2(6)(12)(−12/13) = 180 + 132.9 = 312.9, so |PS| ≈ 17.7 cm.
(b) Slant height = √(242 + 72) = 25 cm.
Cone curved surface = 22/7 × 7 × 25 = 550 cm2; hemisphere curved surface = 2 × 22/7 × 72 = 308 cm2.
Surface area = 858 cm2.
Only the curved surfaces are on the outside; the flat circle is hidden where the two parts join.