63 questions. Try each one, then open the answer to check yourself and read the explanation.
Correct, 0.00798516 to three significant figures.
Answer: C (0.00799)
The first three significant figures are 7, 9, 8 and the next digit is 5, so round up: 0.00799.
Simplify: (11two)2.
Answer: A (1001two)
11two = 3 and 32 = 9 = 8 + 1 = 1001two.
Solve: 2√(2x + 1) = 32.
Answer: C (12)
32 = 25, so √(2x + 1) = 5. Squaring, 2x + 1 = 25 and x = 12.
If log102 = m and log103 = n, find log1024 in terms of m and n.
Answer: A (3m + n)
24 = 23 × 3, so log 24 = 3 log 2 + log 3 = 3m + n.
Find the 5th term of the sequence 2, 5, 10, 17, … .
Answer: D (26)
The differences are 3, 5, 7, so the next difference is 9: 17 + 9 = 26. (The terms are n2 + 1.)
If P = {−3 < x < 1} and Q = {−1 < x < 3}, where x is a real number, find P ∩ Q.
Answer: A ({−1 < x < 1})
The overlap of the two intervals starts at −1 and ends at 1, with both ends excluded: −1 < x < 1.
Factorize 6pq − 3rs − 3ps + 6qr.
Answer: B (3(p + r)(2q − s))
Group: 6pq + 6qr − 3ps − 3rs = 6q(p + r) − 3s(p + r) = 3(p + r)(2q − s).
What number should be subtracted from the sum of 2⅙ and 2 7/12 to give 3¼?
Answer: B (1½)
2⅙ + 2 7/12 = 4 9/12 = 4¾. Then 4¾ − 3¼ = 1½.
Mensah is 5 years old and Joyce is thrice as old as Mensah. In how many years will Joyce be twice as old as Mensah?
Answer: C (5 years)
Joyce is 15. In n years: 15 + n = 2(5 + n), so 15 + n = 10 + 2n and n = 5.
If 16 × 2(x + 1) = 4x × 8(1 − x), find the value of x.
Answer: D (−1)
In base 2: 24 + x + 1 = 22x × 23 − 3x, so x + 5 = 3 − x. Then 2x = −2 and x = −1.
The circumference of a circular track is 9 km. A cyclist rides round it a number of times and stops after covering a distance of 302 km. How far is the cyclist from the starting point?
Answer: A (5 km)
302 = 33 × 9 + 5, so after 33 full laps the cyclist has gone a further 5 km along the track.
Simplify: 2√7 − 14/√7 + 7/√21
Answer: C (√21/3)
14/√7 = 2√7, which cancels with 2√7. And 7/√21 = 7√21/21 = √21/3.
If 4x + 2y = 16 and 6x − 2y = 4, find the value of (y − x).
Answer: B (2)
Adding gives 10x = 20, so x = 2. Then 8 + 2y = 16 gives y = 4, and y − x = 2.
Given that R is directly proportional to L and inversely proportional to P, R = 3 when L = 9 and P = 0.8, find R when L = 15 and P = 1.8.
Answer: A (2.2)
R = kL/P: 3 = 9k/0.8 so k = 0.8/3. Then R = (0.8/3) × 15/1.8 ≈ 2.2.
In the diagram, ∠ABC and ∠BCD are right angles, ∠BAD = t and ∠EDF = 70°. Find the value of t.
Answer: D (110°)
∠ADC = ∠EDF = 70° (vertically opposite). AB ∥ CD because both are perpendicular to BC, so t + 70° = 180° and t = 110°.
The sum of the interior angles of a regular polygon with k sides is (3k − 10) right angles. Find the size of the exterior angle.
Answer: A (60°)
(k − 2) × 180° = (3k − 10) × 90°, so 2k − 4 = 3k − 10 and k = 6. Each exterior angle is 360° ÷ 6 = 60°.
Make U the subject of the relation: x = (2U − 3)/(3U + 2).
Answer: C (U = (2x + 3)/(2 − 3x))
3Ux + 2x = 2U − 3, so 3Ux − 2U = −3 − 2x and U(3x − 2) = −(2x + 3). Hence U = (2x + 3)/(2 − 3x).
A trader paid import duty of 38 kobo in the naira on the cost of an engine. If a total of ₦22,800.00 was paid as import duty, calculate the cost of the engine.
Answer: A (₦60,000.00)
Duty is ₦0.38 for every ₦1, so the cost = 22,800 ÷ 0.38 = ₦60,000.
The height of an equilateral triangle is 10√3 cm. Calculate its perimeter.
Answer: B (60 cm)
Height = side × sin 60° = side × √3/2, so side = 20 cm and the perimeter is 60 cm.
In ΔLMN, |LM| = 6 cm, ∠LMN = 90°, ∠LNM = x and sin x = 3/5. Find the area of ΔLMN.
Answer: C (24 cm2)
LM is opposite x, so LN = 6 ÷ (3/5) = 10 cm and MN = √(100 − 36) = 8 cm. Area = ½ × 6 × 8 = 24 cm2.
Consider the statements: P: All students offering Literature (L) also offer History (H); Q: Students offering History (H) do not offer Geography (G). Which of the venn diagrams correctly illustrates the two statements?
Answer: C (Diagram C)
P puts the L circle inside the H circle, and Q makes the H circle separate from G. Only diagram C shows L inside H with G apart.
Find the quadratic equation whose roots are −2q and 5q.
Answer: D (x2 − 3qx − 10q2 = 0)
(x + 2q)(x − 5q) = x2 − 3qx − 10q2 = 0.
If tan θ = 3/4, 180° < θ < 270°, find the value of cos θ.
Answer: C (−4/5)
From a 3-4-5 triangle, |cos θ| = 4/5. In the third quadrant cosine is negative, so cos θ = −4/5.
If 2/(x − 3) − 3/(x − 2) = P/((x − 3)(x − 2)), find P.
Answer: A ((5 − x))
P = 2(x − 2) − 3(x − 3) = 2x − 4 − 3x + 9 = 5 − x.
The diagonals of a rhombus are 12 cm and 5 cm. Calculate its perimeter.
Answer: A (26 cm)
The diagonals bisect each other at right angles, so each side = √(62 + 2.52) = 6.5 cm. Perimeter = 4 × 6.5 = 26 cm.
In the diagram, ΔXYZ is produced to T. If |XY| = |ZY| and ∠XYT = 40°, find ∠XZT.
Answer: A (110°)
The triangle is isosceles with the 40° angle at Y, so ∠YXZ = ∠YZX = 70°. ∠XZT is on the straight line: 180° − 70° = 110°.
A solid brass cube is melted and recast as a solid cone of height h and base radius r. If the height of the cube is h, find r in terms of h.
Answer: D (r = h√(3/π))
Volumes are equal: h3 = ⅓πr2h, so r2 = 3h2/π and r = h√(3/π).
Which of the following is not an exterior angle of a regular polygon?
Answer: A (66°)
The exterior angle of a regular polygon must divide 360° exactly. 360 ÷ 66 is not a whole number.
From a point T, a man moves 12 km due West and then moves 12 km due South to another point Q. Calculate the bearing of T from Q.
Answer: C (045°)
Q is 12 km west and 12 km south of T, so T is 12 km east and 12 km north of Q. That is exactly north-east: 045°.
In the diagram, O is the centre of circle PQRS, ∠PQR = 72° and OR ∥ PS. Find ∠OPS.
Answer: D (36°)
∠POR = 2 × 72° = 144°, so in isosceles triangle OPR, ∠OPR = ∠ORP = 18°. ∠RPS = ∠ORP = 18° (alternate angles), so ∠OPS = 18° + 18° = 36°.
A trapezium of sides 10 cm and 21 cm and height 8 cm is inscribed in a circle of radius 7cm. Calculate the area of the region not covered by the trapezium. [Take π = 22/7]
Answer: C (30 cm2)
Circle = (22/7) × 49 = 154 cm2. Trapezium = ½(10 + 21) × 8 = 124 cm2. Uncovered = 154 − 124 = 30 cm2.
Find, correct to two decimal places, the mean of 1½, 2⅔, 3¾, 4⅘ and 5⅚.
Answer: A (3.71)
Sum = 1.5 + 2.667 + 3.75 + 4.8 + 5.833 = 18.55. Mean = 18.55 ÷ 5 = 3.71.
A cyclist moved at a speed of X km/h for 2 hours. He then increased his speed by 2 km/h for the next 3 hours. If the total distance covered is 36 km, calculate his intial speed, X.
Answer: D (6 km/h)
Distance = 2X + 3(X + 2) = 5X + 6 = 36, so X = 6 km/h.
Find the value of (x + y) in the diagram.
Answer: A (215°)
Using the parallel lines, y = 180° − 110° = 70° and x = 180° − 35° = 145°. So x + y = 215°.
In the diagram, MP is a tangent to the circle NQR, ∠PNQ = 64° and |RQ| = |RN|. Find the angle marked t.
Answer: C (58°)
By the alternate segment theorem ∠NRQ = 64°. Triangle RQN is isosceles, so ∠RQN = (180° − 64°)/2 = 58°, and t = ∠RQN = 58° (alternate segment again).
Find the first quartile of 7, 8, 7, 9, 11, 8, 7, 9, 6, and 8.
Answer: B (7.0)
In order: 6, 7, 7, 7, 8, 8, 8, 9, 9, 11. The lower quartile is the middle of the lower half (6, 7, 7, 7, 8), which is 7.
In the diagram, PQRS is a circle. Find the value of x.
Answer: A (50°)
Opposite angles of a cyclic quadrilateral add to 180°: x + (x + y) = 180° and x + (2y − 30°) = 180°. So 2x + y = 180 and x + 2y = 210, giving x = 50°.
A cone has a base radius of 8 cm and height 11 cm. Calculate, correct to two decimal places, the curved surface area. [Take π = 22/7]
Answer: A (341.98 cm2)
Slant height l = √(82 + 112) = √185 ≈ 13.601 cm. Curved surface = πrl = (22/7) × 8 × 13.601 ≈ 341.98 cm2.
Given that sin x = 3/5, 0° ≤ x ≤ 90°, evaluate (tan x + 2 cos x).
Answer: B (2 7/20)
From a 3-4-5 triangle, tan x = 3/4 and cos x = 4/5. So 3/4 + 8/5 = 47/20 = 2 7/20.
In the diagram, EC is a diameter of the circle ABCDE. If ∠ABC = 158°, find ∠ADE.
Answer: C (68°)
ABCE is cyclic, so ∠AEC = 180° − 158° = 22°. ∠EAC = 90° (angle in a semicircle), so ∠ACE = 68°, and ∠ADE = ∠ACE = 68° (same segment).
The table shows the heights of thirty-seven players of a basketball team. Calculate, correct to one decimal place, the mean height of the players.
Height (cm): 160, 161, 162, 163, 164, 165
No of Players: 4, 6, 3, 7, 8, 9
Answer: A (163.0)
Σfx = 640 + 966 + 486 + 1141 + 1312 + 1485 = 6030. Mean = 6030 ÷ 37 ≈ 162.97, which is 163.0.
XY is a line segment with the coordinates X(−8, −12) and Y(p, q). If the midpoint of XY is (−4, −2), find the coordinates of Y.
Answer: B ((0, 8))
(−8 + p)/2 = −4 gives p = 0, and (−12 + q)/2 = −2 gives q = 8. So Y is (0, 8).
Five hundred tickets were sold for a concert. Tickets for adults and children were sold at $4.50 and $3.00 respectively. If the total receipts for the concert was $1,987.50, how many tickets for adults were sold?
Answer: A (325)
If all 500 were child tickets: $1,500. The extra $487.50 comes from $1.50 more per adult ticket, so 487.50 ÷ 1.50 = 325 adults.
The distance (d) between two villages is more than 18 km but not more than 23 km. Which of these inequalities represents the statement?
Answer: D (18 < d ≤ 23)
'More than 18' means d > 18 and 'not more than 23' means d ≤ 23: 18 < d ≤ 23.
The pie chart represents the distribution of fruits on display in a shop. If there are 60 apples on display, how many oranges are there?
Answer: D (90)
The orange sector is 360° − (80° + 60° + 100°) = 120°. 80° stands for 60 apples, so 120° stands for 60 × 120/80 = 90 oranges.
A box contains 40 identical balls of which 10 are red and 12 are blue. If a ball is selected at random from the box, what is the probability that it is neither red nor blue?
Answer: A (9/20)
40 − 10 − 12 = 18 balls are neither colour, so the probability is 18/40 = 9/20.
A fair die is tossed twice. What is the probability of getting a sum of at least 10?
Answer: D (1/6)
Sums of 10 or more: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) – 6 of the 36 outcomes, so 6/36 = 1/6.
A man will be (x + 10) years old in 8 years time. If 2 years ago, he was 63 years, find the value of x.
Answer: B (63)
He is 63 + 2 = 65 now, so in 8 years he will be 73. x + 10 = 73 gives x = 63.
The equation of a line is given as 3x − 5y = 7. Find its gradient (slope).
Answer: B (3/5)
Rearrange: 5y = 3x − 7, so y = (3/5)x − 7/5 and the gradient is 3/5.
For what value of x is (4 − 2x)/(x + 1) undefined?
Answer: B (−1)
The fraction is undefined when the denominator is zero: x + 1 = 0, so x = −1.
(a) Mr Sarto borrowed $25,000 from Afiak Financial Services at 21% simple interest per annum for 3 years. If he was able to pay back the loan in 2 years at equal yearly instalments, how much did he pay each year?
(b) Two consecutive numbers are such that the sum of thrice the smaller and twice the larger is 17. Find, correct to three significant figures, the smaller number as a percentage of the sum of the two numbers.
(a) Interest = 25,000 × 21 × 3/100 = $15,750
Total to repay = $40,750
Each year = 40,750 ÷ 2 = $20,375
(b) Let the numbers be n and n + 1: 3n + 2(n + 1) = 17 → 5n = 15 → n = 3
Numbers 3 and 4, sum 7
Percentage = 3/7 × 100 ≈ 42.9%
Work out the full amount owed for 3 years first, then split it into 2 payments.
A man left town M at 10:00 a.m. and travelled by car to town N at an average speed of 72 km/h. He spent 2 hours for a meeting and returned to town M by bus at an average speed of 40 km/h. If the distance covered by the bus was 2 km longer than that of the car and he arrived at town M at 1:55 p.m., calculate the distance from M to N.
Total time 10:00 to 1:55 = 3 h 55 min; travelling time = 1 h 55 min = 23/12 h
Let MN = x km: x/72 + (x + 2)/40 = 23/12
Multiply by 360: 5x + 9(x + 2) = 690 → 14x = 672
x = 48 km
Take away the 2-hour meeting before setting up the time equation.
The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270°. Calculate, correct to: (a) two significant figures, |YZ|; (b) the nearest degree, the bearing of Y from Z.
Sketch: Z is 20 km due west of X, Y is 15 km due south of X; ∠ZXY = 90°
(a) |YZ| = √(202 + 152) = √625 = 25 km
(b) At Z: tan θ = 15/20 → θ = 36.87° below east
Bearing of Y from Z = 90° + 36.87° ≈ 127°
The right angle is at X, between the west and south directions.
In the diagram, AD is a diameter of a circle with centre O. If ABD is a triangle in a semi-circle and ∠OAB = 34°, find: (a) ∠OBD; (b) ∠OCB.
(a) OA = OB (radii) → ∠OBA = 34°
∠ABD = 90° (angle in a semicircle)
∠OBD = 90° − 34° = 56°
(b) BC touches the circle at B, so ∠OBC = 90° (radius ⊥ tangent)
∠BOC = 34° + 34° = 68° (exterior angle of ΔOAB)
∠OCB = 180° − 90° − 68° = 22°
Look for the isosceles triangle made by two radii.
(a) A man shared his property among his children as follows:
Child's name: Ann, Afia, Kojo, Nuno, Akosua
Percentage share: 5, 15, 10, 45, 25
Represent the information on a pie chart.
(b) A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability that a girl takes away two red beads, one after the other, from the box.
(a) Angles = percentage × 3.6°: Ann 18°, Afia 54°, Kojo 36°, Nuno 162°, Akosua 90° (total 360°)
Draw the circle, measure each sector with a protractor and label it
(b) 12 beads, taken without replacement:
P(red, red) = 5/12 × 4/11 = 20/132 = 5/33
Taking beads away one after the other means without replacement.
(a) In a class of 80 students, ¾ study Biology and ⅗ study Physics. If each student studies at least one of the subjects: (i) draw a venn diagram to represent this information; (ii) how many students study both subjects; (iii) find the fraction of the class that study Biology but not Physics.
(b) Johnson and Jocatol Ltd. owned a business office with floor measuring 15 m by 8 m which was to be carpeted. The cost of carpeting was GH¢ 890.00 per square metre. If a total of GH¢ 216,120.00 was spent on painting and carpeting, how much was the cost of painting?
(a) n(B) = 60, n(P) = 48, n(U) = 80
(i) Two overlapping circles B and P with regions 60 − x, x, 48 − x and nothing outside
(ii) 60 − x + x + 48 − x = 80 → x = 28
(iii) Biology only = 60 − 28 = 32; fraction = 32/80 = 2/5
(b) Floor area = 15 × 8 = 120 m2; carpeting = 120 × 890 = GH¢106,800
Painting = 216,120 − 106,800 = GH¢109,320
Every student is in at least one circle, so nothing goes outside the circles.
(a) Copy and complete the table of values for the relation y = 2x2 − x − 2 for −4 ≤ x ≤ 4.
x: −4, −3, −2, −1, 0, 1, 2, 3, 4
y: __, 19, __, __, −2, __, __, __, 26
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x2 − x − 2 for −4 ≤ x ≤ 4.
(c) On the same axis, draw the graph of y = 2x + 3.
(d) Use the graph to find the: (i) roots of the equation 2x2 − 3x − 5 = 0; (ii) range of values of x for which 2x2 − x − 2 < 0.
(a) y = 34, 19, 8, 1, −2, −1, 4, 13, 26
(b) Plot the points and draw a smooth U-shaped curve
(c) Straight line through e.g. (−4, −5), (0, 3), (4, 11)
(d)(i) 2x2 − 3x − 5 = 0 is where the curve meets y = 2x + 3: x = −1 and x = 2.5
(ii) The curve is below the x-axis between its x-intercepts: about −0.8 < x < 1.3
Subtracting 2x + 3 from the curve's equation gives 2x2 − 3x − 5, so the intersections give its roots.
(a) In ΔPQR, ∠PQR = 90°. If its area is 216 cm2 and |PQ| : |QR| is 3 : 4, find |PR|.
(b) The present ages of a man and his son are 47 years and 17 years respectively. In how many years would the man's age be twice that of his son?
(a) Let PQ = 3k, QR = 4k: ½ × 3k × 4k = 216 → 6k2 = 216 → k = 6
PQ = 18 cm, QR = 24 cm
|PR| = √(182 + 242) = 30 cm
(b) 47 + n = 2(17 + n) → 47 + n = 34 + 2n
n = 13 years
The right angle is at Q, so PQ and QR are the base and height.
In the diagram, PQRS is a trapezium with QR ∥ PS. U and T are points on PS such that |PU| = 5 cm, |QU| = 12 cm and ∠PUQ = ∠STR = 90°. If the area of ΔPQR = 20 cm2, calculate, correct to the nearest whole number, the: (a) perimeter; (b) area; of the trapezium.
Area ΔPQR = ½ × |QR| × 12 = 20 → |QR| = 10/3 ≈ 3.33 cm
|PQ| = √(52 + 122) = 13 cm; |RT| = 12 cm and ∠RST = 50° (from the diagram)
|RS| = 12/sin 50° ≈ 15.66 cm; |TS| = 12/tan 50° ≈ 10.07 cm
|PS| = 5 + 3.33 + 10.07 ≈ 18.40 cm
(a) Perimeter = 13 + 3.33 + 15.66 + 18.40 ≈ 50 cm
(b) Area = ½(3.33 + 18.40) × 12 ≈ 130 cm2
The height of ΔPQR on base QR is the same 12 cm as the trapezium's height.
(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms, find, correct to: (i) two significant figures, the distance between the two farms; (ii) the nearest degree, the bearing of Manu's farm from Dogbe's.
(b) A ladder 10 m long leaned against a vertical wall x m high. The distance between the wall and the foot of the ladder is 2 m longer than the height of the wall. Calculate the value of x.
(a) From the cottage, Dogbe's farm is on 020° (5 km) and Manu's on 290° (3 km); the angle between them = 290° − 20° = 270°, i.e. 90°
(i) Distance = √(52 + 32) = √34 ≈ 5.8 km
(ii) At Dogbe's farm, tan θ = 3/5 → θ ≈ 31.0°; bearing of Manu's farm = 200° + 31° ≈ 231°
(b) x2 + (x + 2)2 = 102 → 2x2 + 4x − 96 = 0 → x2 + 2x − 48 = 0
(x + 8)(x − 6) = 0 → x = 6 (length must be positive)
Turn each bearing round by 180° to get the direction from the cottage to each farm.
The table shows the distribution of the number of hours per day spent in studying by 50 students.
Number of hours per day: 4, 5, 6, 7, 8, 9, 10, 11
Number of students: 5, 7, 5, 9, 12, 4, 3, 5
Calculate, correct to two decimal places, the: (a) mean; (b) standard deviation.
(a) Σfx = 20 + 35 + 30 + 63 + 96 + 36 + 30 + 55 = 365
Mean = 365/50 = 7.30
(b) Σfx2 = 2873; variance = 2873/50 − 7.32 = 57.46 − 53.29 = 4.17
Standard deviation = √4.17 ≈ 2.04
Use either Σf(x − mean)2 or Σfx2/Σf − mean2; both give 4.17.
(a) In the diagram, PQRS is a circle. |PQ| = |QS|, ∠SPR = 26° and the interior angles of ΔPQS are in the ratio 2 : 3 : 3. Calculate: (i) ∠PQR; (ii) ∠RPQ; (iii) ∠PRQ.
(b) The coordinates of two points P and Q in a plane are (7, 3) and (5, x) respectively, where x is a real number. If |PQ| = √29 units, find the value of x.
(a) Angles of ΔPQS: 2/8 × 180° = 45° at Q; 67.5° at P and at S (equal sides PQ = QS)
(i) ∠SQR = ∠SPR = 26° (same segment); ∠PQR = 45° + 26° = 71°
(ii) ∠RPQ = ∠SPQ − ∠SPR = 67.5° − 26° = 41.5°
(iii) ∠PRQ = 180° − 71° − 41.5° = 67.5°
(b) (5 − 7)2 + (x − 3)2 = 29 → (x − 3)2 = 25
x − 3 = ±5 → x = 8 or x = −2
The angle 45° is at Q, opposite the base PS of the isosceles triangle.
(a) On Sam's first birthday celebration, his grandfather deposited an amount of $1,000.00 in a bank compounded at 4% interest annually. Find how much is in the account if Sam is 4 years old.
(b) In the diagram, ABCD are points on the circle centre O. If |AB| = |BC| and ∠ADC = 50°, find ∠BAD.
(a) From age 1 to age 4 is 3 years
Amount = 1,000 × (1.04)3 = $1,124.86
(b) ∠ABC = 180° − 50° = 130° (opposite angles of cyclic quadrilateral)
AB = BC → ∠BAC = (180° − 130°)/2 = 25°
AD is a diameter → ∠ACD = 90°, so ∠CAD = 180° − 90° − 50° = 40°
∠BAD = 25° + 40° = 65°
Count the years from the first birthday, not from birth.