60 questions. Try each one, then open the answer to check yourself and read the explanation.
Evaluate, correct to three decimal places, (4.314 × 0.000056)/0.0067
Answer: B (0.036)
4.314 × 0.000056 = 0.00024158, and 0.00024158 ÷ 0.0067 = 0.03606, which is 0.036 to three decimal places.
There are 30 students in a class. 15 study woodwork and 13 study metalwork, 6 study neither of the two subjects. How many students study woodwork but not metalwork?
Answer: C (11)
30 − 6 = 24 study at least one subject, so 15 + 13 − 24 = 4 study both. Woodwork only = 15 − 4 = 11.
Solve: 25x ÷ 2x = 5√(210)
Answer: C (1/2)
The left side is 24x and the right side is 210/5 = 22, so 4x = 2 and x = ½.
Solve: 1 + ∛(x − 3) = 4.
Answer: C (30)
∛(x − 3) = 3, so x − 3 = 27 and x = 30.
Express 4137 in base 5.
Answer: B (13115)
4137 = 4 × 49 + 1 × 7 + 3 = 206. Dividing 206 repeatedly by 5 gives remainders 1, 1, 3, 1, so 206 = 13115.
Solve: log3 x + log3 (x − 8) = 2.
Answer: D (9)
log3[x(x − 8)] = 2 gives x² − 8x = 9, so (x − 9)(x + 1) = 0. Logs need x > 8, so x = 9.
Mr. Manu is 4 times as old as his son, Adu. 7 years ago, the sum of their ages was 76 years. How old is Adu?
Answer: C (18 years)
Let Adu be a, so Manu is 4a. Seven years ago: (4a − 7) + (a − 7) = 76, so 5a = 90 and a = 18.
Factorize completely: x2 − (y + z)2.
Answer: B ((x + y + z)(x − y − z))
Difference of two squares: x² − (y + z)² = [x + (y + z)][x − (y + z)] = (x + y + z)(x − y − z).
Find the roots of the quadratic equation: 3m2 − 2m − 65 = 0.
Answer: C ((−13/3, 5))
3m² − 2m − 65 = (3m + 13)(m − 5), so m = −13/3 or m = 5.
M varies jointly as the square of n and the square root of q. If M = 24 when n = 2 and q = 4, find M when n = 5 and q = 9.
Answer: D (225)
M = kn²√q: 24 = k × 4 × 2 gives k = 3. Then M = 3 × 25 × 3 = 225.
If m : n = 2⅓ : 1⅕ and n : q = 1½ : 1⅓, find q : m.
Answer: A (16 : 35)
m : n = 7/3 : 6/5 = 35 : 18 and n : q = 3/2 : 4/3 = 9 : 8 = 18 : 16. So m : n : q = 35 : 18 : 16 and q : m = 16 : 35.
One-third the sum of two numbers is 12. Twice their difference is 12. Find the numbers
Answer: A (21 and 15)
x + y = 36 and x − y = 6. Adding gives 2x = 42, so x = 21 and y = 15.
Find the quadratic equation whose roots are 2/3 and −3/4.
Answer: C (12y2 + y − 6 = 0)
(3y − 2)(4y + 3) = 12y² + 9y − 8y − 6 = 12y² + y − 6 = 0.
Make x the subject of the relation: y = (ax3 − b)/3z
Answer: B (x = ∛((3yz + b)/a))
Multiply by 3z: 3yz = ax³ − b. Then ax³ = 3yz + b, x³ = (3yz + b)/a, and take the cube root.
The price of a shoe was decreased by 22%. If the new price was $27.30, what was the original price?
Answer: D ($35.00)
The new price is 78% of the old: 0.78P = 27.30, so P = 27.30 ÷ 0.78 = $35.00.
The radius and height of a solid cylinder are 8 cm and 14 cm respectively. Find, correct to two decimal places, the total surface area. [Take π = 22/7]
Answer: C (1,106.29 cm2)
Total surface area = 2πr(r + h) = 2 × 22/7 × 8 × 22 = 7744/7 = 1106.2857…, which is 1,106.29 cm² to two decimal places.
In the diagram, O is the centre of the circle NST. |NT| = |ST| and ∠NTS = 36°. Find the measure of the angle marked l.
Answer: A (36°)
The angle l lies between the tangent RV and the chord SN. By the alternate segment theorem it equals the angle NTS in the other segment, 36°.
A sphere has radius 3 cm. Find, in terms of π, its volume.
Answer: C (36π cm3)
V = (4/3)πr³ = (4/3)π × 27 = 36π cm³.
Arrange the following numbers in ascending order of magnitude; 110two, 31eight, 42five
Answer: B (110two, 42five, 31eight)
In base ten: 110two = 6, 42five = 22 and 31eight = 25, so the order is 110two, 42five, 31eight.
A notebook of length 15 cm was measured by a student as 16.8 cm. Calculate, correct to two decimal places, the percentage error in the measurement.
Answer: A (12.00%)
Error = 16.8 − 15 = 1.8 cm. Percentage error = 1.8/15 × 100% = 12.00%.
Find the value of m in the diagram.
Answer: B (130°)
Draw a line through the right-angled corner parallel to the arrowed lines. The 40° angle is split off below, so the upper slanting line meets that parallel at 90° − 40° = 50°. Then m and 50° are co-interior angles, so m = 180° − 50° = 130°.
A line L, passing through the point (6, −13) is parallel to the line which passes through (7, 4) and (−3, 9). Find the equation of the line, L.
Answer: C (y = −½x − 10)
Gradient = (9 − 4)/(−3 − 7) = −½. Then y + 13 = −½(x − 6), so y = −½x + 3 − 13 = −½x − 10.
An empty cyclindrical tank is 140 cm in diameter. If 200 litres of water is poured into the tank, calculate, correct to the nearest centimetre, the height of water in the tank. [Take π = 22/7]
Answer: C (13 cm)
200 litres = 200 000 cm³ and r = 70 cm. h = 200 000 ÷ (22/7 × 70 × 70) = 200 000 ÷ 15 400 ≈ 12.99, so about 13 cm.
Mrs. Kebeh stands at a distance of 110 m away from a building of vertical height 58 m. If Kebeh is 2 m tall, find the angle of elevation of the top of the building from her eye.
Answer: B (27°)
Above her eye level the building rises 58 − 2 = 56 m. tan θ = 56/110 = 0.509, so θ ≈ 27°.
Find the mean deviation of the set of numbers: 14, 15, 16, 17, 18 and 19.
Answer: B (1.5)
The mean is 99/6 = 16.5. The distances from the mean are 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, which add to 9, and 9 ÷ 6 = 1.5.
The interior angle of the regular polygon is 6 times its exterior angle. Find the number of sides of the polygon.
Answer: B (14)
Exterior + interior = 180° and interior = 6 × exterior, so 7 × exterior = 180° and exterior = 180°/7. Sides = 360° ÷ (180°/7) = 14.
The length of the diagonal of a square is 12 cm. Calculate the area of the square.
Answer: C (72 cm2)
If the side is x, then x² + x² = 12² = 144, so the area x² = 72 cm².
Consider the statements:
p: Siah is from Foya.
q: Foya is in Lofa.
Write in symbolic form the statement: "If Siah is from Foya, then Foya is in Lofa".
Answer: C (p ⇒ q)
"If p, then q" is the implication p ⇒ q.
What is the name of a triangle with vertices (1, −3), (6, 2) and (0, 4)?
Answer: A (Isosceles triangle)
The squared side lengths are 5² + 5² = 50, 6² + 2² = 40 and 1² + 7² = 50. Two sides are equal (√50) and 40 + 50 ≠ 50, so it is isosceles, not right-angled.
In the diagram, NR is a diameter, ∠MNR = x° and ∠SRN = (5x + 20)°. Find the value of 2x.
Answer: B (35°)
∠NMR = 90° (angle in a semicircle), so ∠NRM = 90° − x. S, R and M lie on a straight line, so ∠SRN = 180° − (90° − x) = 90° + x. Then 5x + 20 = 90 + x gives x = 17.5 and 2x = 35°.
Find the value of α in the equation: cos (α + 14)° = sin (4α + 6)°.
Answer: A (14)
cos A = sin B when A + B = 90°: (α + 14) + (4α + 6) = 90, so 5α = 70 and α = 14.
A bag contains 4 white marbles and 3 blue marbles. Another bag contains 5 red and 6 blue marbles. If a marble is picked at random from each bag, find the probability that they are of the same colour.
Answer: B (18/77)
Only blue is in both bags, so P(both blue) = 3/7 × 6/11 = 18/77.
The angle of a sector of a circle of radius 3.4 cm is 115°. Find the area of the sector. [Take π = 22/7]
Answer: B (11.6 cm2)
Area = 115/360 × 22/7 × 3.4² = 115/360 × 22/7 × 11.56 ≈ 11.6 cm².
The diagonals of a rhombus are 16 cm and 12 cm. Find the length of the side.
Answer: B (10 cm)
The diagonals bisect each other at right angles, giving half-lengths 8 cm and 6 cm. Side = √(8² + 6²) = 10 cm.
The angle of elevation of the top of a vertical building from a point Z on the ground is 50°. If the height of the building is 124 m, find the distance from Z to the foot of the building.
Answer: A (104.05 m)
tan 50° = 124/d, so d = 124 ÷ 1.1918 ≈ 104.05 m.
A student measured the height of a pole as 5.98 m which is less than the actual height. If the percentage error is 5%, find, correct to two decimal places, the actual height of the pole.
Answer: C (6.29 m)
The measured value is 95% of the actual height: 0.95h = 5.98, so h = 6.2947… ≈ 6.29 m.
In the diagram, O is the centre of the circle QRS and ∠SQR = 28°. Find ∠ORS.
Answer: B (62°)
The angle at the centre is twice the angle at the circumference: ∠SOR = 56°. Triangle SOR is isosceles (OS = OR), so ∠ORS = (180° − 56°)/2 = 62°.
John was facing S35°E. If he turned 90° in the anticlockwise direction, find his new direction.
Answer: A (N55°E)
S35°E is a bearing of 145°. Turning 90° anticlockwise takes it to 145° − 90° = 055°, which is N55°E.
If 2x − 3y = −11 and 3x + 2y = 3, evaluate (y − x)2.
Answer: C (16)
Multiply to get 4x − 6y = −22 and 9x + 6y = 9; adding gives 13x = −13, so x = −1 and y = 3. (y − x)² = 4² = 16.
An equilateral triangle has side 2 cm. Calculate the height of the triangle.
Answer: A (√3 cm)
The height bisects the base: h² = 2² − 1² = 3, so h = √3 cm.
The bar chart represents the distribution of marks scored by students in an Economics examination. If the failed mark was 4, what is the probability that a student selected at random passed?
Answer: A (0.64)
Students who passed scored 5 to 8 marks: 12 + 7 + 5 + 10 = 34 out of 53 students. 34/53 ≈ 0.64.
The bar chart represents the distribution of marks scored by students in an Economics examination. What percentage of the students scored at most 5 marks?
Answer: B (58.5%)
At most 5 marks: 3 + 6 + 9 + 1 + 12 = 31 students out of 53. 31/53 × 100% ≈ 58.5%.
The bar chart represents the distribution of marks scored by students in an Economics examination. How many students scored at least 3 marks?
Answer: C (44)
Only those scoring 1 or 2 marks are excluded: 53 − (3 + 6) = 44.
If loga3 = m and loga5 = p, find loga 75.
Answer: C (m + 2p)
75 = 3 × 5², so loga75 = loga3 + 2 loga5 = m + 2p.
In the diagram, M, N, R are points on the circle centre O. ∠ORN = 48° and ∠RNM = 124°. Find ∠OMN.
Answer: A (76°)
OR = ON, so ∠ONR = 48° and ∠MNO = 124° − 48° = 76°. OM = ON, so ∠OMN = ∠ONM = 76°.
The truth set of 8 + 2x − x2 = 0 is {p, q}. Evaluate p + q.
Answer: C (2)
x² − 2x − 8 = 0 gives (x − 4)(x + 2) = 0, so the roots are 4 and −2 and p + q = 2.
For what values of x is (x2 + 2)/(10x2 − 13x − 3) undefined?
Answer: C (3/2, −1/5)
The fraction is undefined when the denominator is zero: 10x² − 13x − 3 = (5x + 1)(2x − 3) = 0, so x = −1/5 or 3/2.
(a) A car travels a distance of 112 km at an average speed of 70 km/h. It then travels further for 60 km at an average speed of 50 km/h. Calculate, for the entire journey, the total time taken.
(b) If x/y = 2 and y/z = 3, find the value of (x + y)/(y + z).
(a) Time for first part = 112 ÷ 70 = 1.6 h
Time for second part = 60 ÷ 50 = 1.2 h
Total time = 2.8 h = 2 hours 48 minutes
(b) x = 2y and z = y/3
(x + y)/(y + z) = 3y ÷ (4y/3) = 9/4 = 2¼
Time = distance ÷ speed for each part; then write x and z in terms of y.
In a football match, the tickets for children and adults were sold at D 3.00 and D 5.00 respectively. If 400 people attended the football match and D1,700.00 was collected in ticket sales:
(a) how many tickets were sold to adults?
(b) Mr. Sonko sold 250 tickets. If 175 of the tickets were for adults, how much sales did he make altogether?
(a) Let c = children's tickets, a = adults' tickets: c + a = 400 and 3c + 5a = 1700
3(400 − a) + 5a = 1700 ⇒ 2a = 500 ⇒ a = 250 adults' tickets
(b) Children's tickets = 250 − 175 = 75
Sales = 175 × D5 + 75 × D3 = D875 + D225 = D1,100.00
Form two simultaneous equations: one for people, one for money.
In the diagram, PQR is an equilateral triangle of side 18cm. M is the midpoint of QR. An arc of a circle with centre P touches QR at M and meets PQ at A and PR at B. Calculate, correct to two decimal places, the area of the shaded region. [Take π = 22/7]
Radius PM = height of triangle = 18 sin 60° = 9√3 ≈ 15.588 cm
Area of triangle PQR = ½ × 18 × 18 × sin 60° = 81√3 ≈ 140.30 cm²
Area of sector PAB (angle 60°) = 60/360 × 22/7 × 243 ≈ 127.29 cm²
Shaded area = 140.30 − 127.29 ≈ 13.01 cm²
The radius of the arc is the height PM of the triangle.
In the diagram, P, Q, R and S are points on the circle centre K. KR is a bisector of ∠SRQ, ∠KSP = 41° and ∠SKR = 80°. Find: (a) ∠RQP; (b) ∠SPQ
KS = KR (radii), so ∠KSR = ∠KRS = (180° − 80°)/2 = 50°
(a) ∠RSP = 50° + 41° = 91°; PQRS is cyclic, so ∠RQP = 180° − 91° = 89°
(b) KR bisects ∠SRQ, so ∠SRQ = 2 × 50° = 100°; opposite angles of the cyclic quadrilateral give ∠SPQ = 180° − 100° = 80°
Use the isosceles triangle KSR first, then opposite angles of a cyclic quadrilateral.
A boy stands at a point M on the same horizontal level as the foot, T, of a vertical building. He observes an object on the top, P of the building at an angle of elevation of 66°. He moves directly backwards to a new point C and observes the same object at an angle of elevation of 53°. If |MT| = 50 m:
(a) illustrate the information in a diagram;
(b) calculate, correct to one decimal place: (i) the height of the building; (ii) |MC|
(a) Vertical PT with right angle at T; C, M, T in a straight line with MT = 50 m; angle of elevation 66° at M and 53° at C
(b)(i) PT = 50 tan 66° ≈ 112.3 m
(ii) CT = 112.30 ÷ tan 53° ≈ 84.63 m, so |MC| = 84.63 − 50 ≈ 34.6 m
C is further from the building than M, so the smaller angle is at C.
(a) M = {n: 2n − 3 ≤ 37}, where n is a counting number.
(i) Write down all the elements in M.
(ii) If a number is selected at random from M, what is the probability that it is a: (α) multiple of 3; (β) factor of 10.
(b) A shopowner gave an end-of-year bonus to two of his attendants, Kontor and Gapson in the ratio of their ages. Kontor's age is one and half times that of Kontor who is 20 years old. If Kontor received Le 200,000.00: find: (i) the total amount shared; (ii) Gapson's share.
(a)(i) 2n ≤ 40 ⇒ n ≤ 20, so M = {1, 2, 3, …, 20}
(ii)(α) Multiples of 3: 3, 6, 9, 12, 15, 18 ⇒ P = 6/20 = 3/10
(β) Factors of 10: 1, 2, 5, 10 ⇒ P = 4/20 = 1/5
(b) Gapson is 20, so Kontor is 1½ × 20 = 30; ratio Kontor : Gapson = 3 : 2
(i) 3/5 of total = Le 200,000 ⇒ total = Le 333,333.33 (= 1,000,000/3)
(ii) Gapson's share = 2/5 of total = Le 133,333.33 (= 400,000/3)
Read (b) as 'one and half times that of Gapson, who is 20 years old'.
(a) The sum of three numbers is 81. The second number is twice the first. Given that the third number is 6 more than the second, find the numbers.
(b) Given the points P(3, 5) and Q(−5, 7) on the cartesian plane such that R(x, y) is the midpoint of PQ, find the equation of the line that passes through R and perpendicular to PQ.
(a) Let the numbers be x, 2x, 2x + 6: 5x + 6 = 81 ⇒ x = 15. The numbers are 15, 30 and 36
(b) R = ((3 − 5)/2, (5 + 7)/2) = (−1, 6)
Gradient of PQ = (7 − 5)/(−5 − 3) = −¼, so the perpendicular gradient = 4
y − 6 = 4(x + 1) ⇒ y = 4x + 10
Perpendicular gradients multiply to −1.
(a) Copy and complete the table of values for y = 2x2 − x − 4 for −3 ≤ x ≤ 3.
x: −3, −2, −1, 0, 1, 2, 3
y: 17, __, __, −4, __, __, __
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graph of y = 2x2 − x − 4 for −3 ≤ x ≤ 3.
(c) Use the graph to find the: (i) roots of the equation 2x2 − x − 4 = 0; (ii) values of x for which y increases as x increases. (iii) minimum point of y.
(a) y = 17, 6, −1, −4, −3, 2, 11
(b) Plot the points and join with a smooth U-shaped curve
(c)(i) Where the curve crosses the x-axis: x ≈ −1.2 and x ≈ 1.7
(ii) y increases for x > 0.25 (about x > 0.2 from the graph)
(iii) Minimum point ≈ (0.25, −4.1)
The roots are where the curve cuts the x-axis; the lowest point is the minimum.
The table shows the height of teak trees harvested by a farmer.
Height (m): 3, 4, 5, 6, 7, 8
Number of trees: 4, 6, 4, 5, 6, 2
(a) Find the median height.
(b) Calculate, correct to one decimal place, the: (i) mean; (ii) standard deviation.
(a) Σf = 27, so the median is the 14th value. Cumulative frequencies 4, 10, 14, … ⇒ median height = 5 m
(b)(i) Σfx = 12 + 24 + 20 + 30 + 42 + 16 = 144; mean = 144/27 ≈ 5.3 m
(ii) Σfx² = 834; variance = 834/27 − (144/27)² ≈ 2.444; standard deviation ≈ 1.6 m
Find the position (n + 1)/2 for the median; use the exact mean for the standard deviation.
In a town, Chief X resides 60 m away on a bearing of 057° from the palace P. While Chief Y resides on a bearing of 150° from the same palace P. The residence of X and Y are 180 m apart.
(a) Illustrate the information in a diagram.
(b) Find, correct to three significant figures, the: (i) bearing of X from Y; (ii) distance between P and Y.
(a) From P, PX = 60 m on 057° and PY on 150°; ∠XPY = 150° − 57° = 93°; XY = 180 m
(b)(i) Sine rule: sin ∠PYX = 60 sin 93° ÷ 180 ≈ 0.3329 ⇒ ∠PYX ≈ 19.4°. Bearing of P from Y = 330°, so bearing of X from Y = 330° + 19.4° ≈ 349°
(ii) ∠PXY = 180° − 93° − 19.4° = 67.6°; PY = 180 sin 67.6° ÷ sin 93° ≈ 167 m (166.6 m)
Find the angle at P first: 150° − 57° = 93°.
(a) Two regular polygons P and Q are such that the number of sides of P is twice the number of sides of Q. The difference between the exterior angle of Q and P is 45°. Find the number of sides of P.
(b) The area of a semi-circle is 32π cm2. Find, in terms of π, the circumference of the semi-circle.
(a) Let Q have n sides, P 2n sides: 360/n − 360/(2n) = 45 ⇒ 180/n = 45 ⇒ n = 4, so P has 8 sides
(b) ½πr² = 32π ⇒ r = 8 cm
Curved part = πr = 8π cm; with the diameter, the perimeter = (8π + 16) cm
A polygon with more sides has a smaller exterior angle.
(a) In the diagram, P, Q, R and S are points on the circle with centre O. QR//OS, ∠QOR = 2m, ∠QPR = n and ∠SOR = 54°. Find the values of m and n.
(b) The length of a rectangle is 4 cm more than the width. If the perimeter is 40 cm, find the area.
(a) ∠ORQ = ∠SOR = 54° (alternate angles, QR // OS)
OQ = OR, so ∠OQR = 54° and ∠QOR = 180° − 108° = 72° ⇒ 2m = 72°, m = 36°
n = ∠QPR = ½∠QOR = 36° (angle at centre is twice angle at circumference)
(b) 2(w + w + 4) = 40 ⇒ w = 8 cm, length = 12 cm; area = 96 cm²
Use alternate angles with the parallel lines, then the isosceles triangle OQR.
(a) In the daigram, the radius of the sector of circle centre O, is 7 cm and ∠MON is 60°. Find, correct to one decimal place, the area of the shaded portion. [Take π = 22/7]
(b) The x and y intercepts of a straight line are −3/4 and 2/7 respectively. Find the equation of the line.
(a) Area of sector OMN = 60/360 × 22/7 × 7 × 7 = 77/3 ≈ 25.67 cm²
OT = 7 cos 60° = 3.5 cm, NT = 7 sin 60° ≈ 6.062 cm; area of triangle ONT = ½ × 3.5 × 6.062 ≈ 10.61 cm²
Shaded area = 25.67 − 10.61 ≈ 15.1 cm²
(b) The line passes through (−3/4, 0) and (0, 2/7); gradient = (2/7)/(3/4) = 8/21
y = (8/21)x + 2/7, i.e. 21y = 8x + 6
The shaded part is the sector minus the right-angled triangle ONT.