48 questions. Try each one, then open the answer to check yourself and read the explanation.
Change 7110 to base 8
Answer: A (1078)
71 ÷ 8 = 8 remainder 7, and 8 ÷ 8 = 1 remainder 0, and 1 ÷ 8 = 0 remainder 1. Reading the remainders upwards gives 1078.
Evaluate 3524/0.05 correct to 3 significant figures.
Answer: D (70 500)
3524 ÷ 0.05 = 3524 × 20 = 70 480. To 3 significant figures this is 70 500.
If 9(x − ½) = 3x2, find the value of x.
Answer: B (1)
Write 9 as 32: 32x − 1 = 3x², so x² − 2x + 1 = 0, i.e. (x − 1)² = 0 and x = 1.
Solve for y in the equation
10y × 5(2y − 2) × 4(y − 1) = 1
Answer: B (⅔)
Write everything in powers of 2 and 5: 2y·5y × 52y−2 × 22y−2 = 103y−2. This equals 1 = 100, so 3y − 2 = 0 and y = ⅔.
If 2 log3y + log3x2 = 4, then y is
Answer: D (9/x)
2 log3y + log3x² = log3(x²y²) = 4, so x²y² = 34 = 81. Then xy = 9 and y = 9/x.
Solve without using tables log5(62.5) − log5(½)
Answer: A (3)
Subtracting logs means dividing: 62.5 ÷ ½ = 125 = 53, so the value is log5125 = 3.
If N225.00 yields N27.00 in x year simple interest at the rate of 4% per annum, find x.
Answer: A (3)
Simple interest = PRT/100, so 27 = 225 × 4 × x/100 = 9x, giving x = 3 years.
The shaded portion in the Venn diagram above is
Answer: C (X ∩ Y′ ∩ Z)
The shaded part lies inside both X and Z but outside Y, which is X ∩ Y′ ∩ Z.
If √(x2 + 9) = x + 1, solve for x.
Answer: B (4)
Square both sides: x² + 9 = x² + 2x + 1, so 2x = 8 and x = 4 (check: √25 = 5 = 4 + 1).
Make x the subject of the relation
(1 + ax)/(1 − ax) = p/q
Answer: B ((p − q)/(a(p + q)))
Cross-multiply: q + aqx = p − apx. Collect x terms: ax(p + q) = p − q, so x = (p − q)/(a(p + q)).
Which of the following is a factor of 15 + 7x − 2x2?
Answer: C (x − 5)
15 + 7x − 2x² = −(2x² − 7x − 15) = −(2x + 3)(x − 5), so x − 5 is a factor (x = 5 gives 15 + 35 − 50 = 0).
Evaluate (x + 1/x + 1)2 − (x − 1/x − 1)2
Answer: D (4(1 + x))
Use a² − b² = (a + b)(a − b). Here a + b = 2x and a − b = 2/x + 2, so the product is 2x(2/x + 2) = 4 + 4x = 4(1 + x).
Solve the following simultaneous equations for x
x2 + y − 5 = 0
y − 7x + 3 = 0
Answer: D (1, −8)
From the second equation y = 7x − 3. Substituting: x² + 7x − 8 = 0, i.e. (x + 8)(x − 1) = 0, so x = 1 or −8.
Solve the following equation (3x − 2)(5x − 4) = (3x − 2)2
Answer: C (⅔, 1)
Bring everything to one side: (3x − 2)[(5x − 4) − (3x − 2)] = (3x − 2)(2x − 2) = 0. So x = ⅔ or x = 1. Don't divide by (3x − 2), or you lose a root.
The figure above represents the graphs of y = x(2 − x) and y = (x − 1)(x − 3). What are the x coordinates of P, Q and R respectively?
Answer: D (1, 2, 3)
P, Q and R are where the curves cut the x-axis. y = (x − 1)(x − 3) cuts it at 1 and 3; y = x(2 − x) cuts it at 0 and 2. From the figure P = 1, Q = 2 and R = 3.
If the function f is defined by
f(x + 2) = 2x2 + 7x − 5, find f(−1).
Answer: B (−8)
x + 2 = −1 when x = −3. Then f(−1) = 2(9) + 7(−3) − 5 = 18 − 21 − 5 = −8.
Divide the expression x3 + 7x2 + x + 7 by 1 + x2
Answer: D (x + 7)
Group the terms: x³ + 7x² + x + 7 = x²(x + 7) + (x + 7) = (x² + 1)(x + 7). Dividing by 1 + x² leaves x + 7.
Solve the inequality y2 − 3y > 18
Answer: B (y < −3 or y > 6)
y² − 3y − 18 = (y − 6)(y + 3) > 0. A product of two factors is positive when both have the same sign, so y < −3 or y > 6.
If x is negative, what is the range of value of x within which (x + 1)/3 > 1/(x + 3)
Answer: B (−4 < x < −3)
(x + 1)/3 − 1/(x + 3) = x(x + 4)/(3(x + 3)) > 0. Checking signs between the critical values −4, −3 and 0, the only negative interval that works is −4 < x < −3.
A man's initial salary is N540.00 a month and increases after each period of six months by N36.00 a month. Find his salary in the eighth month of the third year.
Answer: C (N720.00)
The eighth month of the third year is month 32. Rises come at months 7, 13, 19, 25 and 31, which is 5 rises, so the salary is 540 + 5 × 36 = N720.00.
If k + 1, 2k − 1, 3k + 1 are three consecutive terms of geometric progression, find the possible values of the common ratio.
Answer: B (1, 5/3)
For a GP, (2k − 1)² = (k + 1)(3k + 1), which gives k² − 8k = 0, so k = 0 or 8. k = 0 gives 1, −1, 1 (ratio −1) and k = 8 gives 9, 15, 25 (ratio 5/3).
A binary operation * is defined on a set of real numbers by x * y = xy for all real values of x and y. If x * 2 = x, find the possible values of x
Answer: A (0, 1)
x * 2 = x² so x² = x, i.e. x(x − 1) = 0, giving x = 0 or 1.
PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram above. Calculate TUP.
Answer: B (54°)
Each angle of a regular pentagon is 108°. At P, angle TPU = 108° − 90° = 18°. In triangle TUP, angle TUP = 180° − 108° − 18° = 54°.
A regular polygon has 150° as the size of each interior angle. How many sides has the polygon?
Answer: A (12)
Each exterior angle is 180° − 150° = 30°, and the exterior angles add up to 360°, so there are 360 ÷ 30 = 12 sides.
Calculate the length in cm of the arc of a circle of diameter 8cm which subtends an angle of 22° at the centre of the circle.
Answer: D (π/2)
Arc length = (θ/360) × 2πr with r = 4cm. With 22½° this is (22.5/360) × 8π = π/2 cm; with 22° it is about 0.49π, which is still nearest to π/2.
In the diagram above, PQRS is circle with O as centre and PQ‖RT. If RTS = 32°, find PSQ
Answer: C (58°)
PQ is parallel to RT, so angle PQS = angle RTS = 32° (alternate angles). SQ is a diameter, so angle SPQ = 90°. Then angle PSQ = 180° − 90° − 32° = 58°.
In the diagram above, O is the centre of the circle and POQ a diameter. If PÔR = 96°, find the value of OR̂Q
Answer: B (48°)
Angle QOR = 180° − 96° = 84°. Triangle OQR is isosceles (OQ = OR are radii), so angle ORQ = (180° − 84°) ÷ 2 = 48°.
In the diagram above, PQ‖ST, PQR = 34°, QRS = 73° and RS = RT. Find SR̂T.
Answer: B (102°)
Draw a line through R parallel to PQ. It splits the 73° at R into 34° (alternate to PQR) and 39°, so angle RST = 39°. RS = RT makes angle RTS = 39° too, so SRT = 180° − 78° = 102°.
In the figure above, PT is a tangent to the circle at U and QU ‖ RS. If ∠TUR = 35° and ∠SRU = 50°, find ∠x.
Answer: A (95°)
By the alternate segment theorem, angle UQR = angle TUR = 35°. Since QU ‖ RS, angle QUR = angle URS = 50° (alternate angles). In triangle QUR, x = 180° − 35° − 50° = 95°.
In the diagram above, ∠QPS = ∠SPR, PR = 9cm, PQ = 4cm and QS = 3cm. Find SR.
Answer: A (6¾cm)
PS bisects angle P, so it divides QR in the ratio of the other two sides: QS/SR = PQ/PR. So 3/SR = 4/9 and SR = 27/4 = 6¾cm.
The three sides of an isosceles triangle are of length x + 3, 2x + 3, 2x − 3 respectively. Calculate x.
Answer: D (6)
Two sides must be equal. x + 3 = 2x − 3 gives x = 6 (sides 9, 15, 9, a real triangle). x + 3 = 2x + 3 gives x = 0, which makes a side negative, so x = 6.
In the figure above, the line segment ST is tangent to the two circle at S and T, O and Q are the centres of the circles with OS = 5cm, QT = 2cm and OQ = 14cm. Find ST.
Answer: A (7√3cm)
The tangent crosses the line of centres, so make a right triangle with hypotenuse OQ = 14 and one side 5 + 2 = 7. Then ST = √(196 − 49) = √147 = 7√3cm.
In the figure above, the area of the square PQRS is 100cm2. If the ratio of the area of the square TUYS to the area of the square XQVU is 1 : 16, find YR.
Answer: C (8cm)
The big square has side 10cm. Areas 1 : 16 mean sides 1 : 4, so the small squares have sides a and 4a with a + 4a = 10, giving a = 2. YR = 10 − 2 = 8cm.
Find the radius of a sphere whose surface area is 154cm2, (π = 22/7)
Answer: B (3.50cm)
4πr² = 154, so r² = 154 × 7 ÷ 88 = 12.25 and r = 3.5cm.
Find the area of the sector of a circle with radius 3m, if the angle of the sector is 60°
Answer: C (4.7m2)
Sector area = (60/360) × π × 3² = 1.5π ≈ 4.7m².
The angle between latitudes 30°S and 13°N is
Answer: C (43°)
The latitudes are on opposite sides of the equator, so add them: 30° + 13° = 43°.
If sin θ = cos θ, find θ between 0° and 360°
Answer: A (45°, 225°)
Dividing by cos θ gives tan θ = 1. Tan is positive in the first and third quadrants, so θ = 45° or 180° + 45° = 225°.
From the figure above, calculate TH in centimetres
Answer: B (5/(√3 − 1))
Let TH = h. The 45° angle at Q makes QH = h, so PH = 5 + h. Then tan 30° = h/(5 + h), i.e. 5 + h = h√3, so h = 5/(√3 − 1).
If two angles of a triangle are 30° each and the longest side is 10cm, calculate the length of each of the other sides.
Answer: D (10√3/3 cm)
The third angle is 120°, opposite the 10cm side. By the sine rule, each other side = 10 sin 30°/sin 120° = 5 ÷ (√3/2) = 10/√3 = 10√3/3 cm.
Quantities in the proportions 1, 4, 6, 7 are to be represented in a pie chart. Calculate the angle of the sector with proportion 7.
Answer: D (140°)
The proportions add up to 18, so the angle is 7/18 × 360° = 140°.
The bar chart above shows the distribution of marks in a class test. How many students took the test?
Answer: B (20)
Add the heights of the bars: 2 + 5 + 3 + 4 + 3 + 1 + 2 = 20 students.
The following marks were obtained by twenty students in an examination.
53 30 70 84 59 43 90 20 78 48
44 60 81 73 50 37 67 68 64 52
Find the number of students who scored at least 50 marks.
Answer: D (14)
"At least 50" includes 50. Counting 53, 70, 84, 59, 90, 78, 60, 81, 73, 50, 67, 68, 64 and 52 gives 14 students.
Weight (g): 0-10, 10-20, 20-30, 30-40, 40-50
No. of coconuts: 10, 27, 19, 6, 2
Estimate the mode of the frequency distribution above.
Answer: C (16.8g)
The modal class is 10–20 (27 coconuts). Mode = L + d₁/(d₁ + d₂) × c = 10 + (27 − 10)/((27 − 10) + (27 − 19)) × 10 = 10 + 17/25 × 10 = 16.8g.
The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it, the mean becomes 19. Find the age of the teacher.
Answer: D (49)
The ten pupils total 10 × 16 = 160 years. With the teacher, 11 people total 11 × 19 = 209 years, so the teacher is 209 − 160 = 49.
Find the median of the observation in the table above
Answer: D (14.5)
There are 20 observations, so the median is the 10th. Cumulative frequencies are 2, 6, 11, … so it is in class 11–15. Median = 10.5 + (10 − 6)/5 × 5 = 14.5.
A number is selected at random between 20 and 30 both numbers inclusive. Find the probability that number is a prime.
Answer: A (2/11)
There are 11 numbers from 20 to 30, and only 23 and 29 are prime, so the probability is 2/11.
Calculate the standard deviation of the following data 7, 8, 9, 10, 11, 12, 13
Answer: A (2)
The mean is 10. The squared deviations are 9, 4, 1, 0, 1, 4, 9, total 28. Variance = 28/7 = 4, so the standard deviation is 2.
The chances of three independent events X, Y, Z occurring are ½, ⅔, ¼ respectively. What are the chances of Y and Z only occurring?
Answer: C (1/12)
Y and Z happen but X does not: (1 − ½) × ⅔ × ¼ = 1/12.