45 questions. Try each one, then open the answer to check yourself and read the explanation.
Find x and y respectively in the subtraction below carried out in base 5.
4 2 4 3
− 1 3 x 4
= y 3 4 4
Answer: C (4, 2)
Units: borrow, 8 − 4 = 4. Fives: 3 − x needs a borrow, 8 − x = 4, so x = 4. Twenty-fives: 6 − 3 = 3. Hundred-and-twenty-fives: 4 − 1 − 1 = 2, so y = 2.
Find p, if 4516 − p7 = 3056.
Answer: C (1167)
4516 = 175 and 3056 = 113 in base ten, so p = 62. In base seven, 62 = 1 × 49 + 1 × 7 + 6 = 1167.
A farmer planted 5,000 grains of maize and harvested 5,000 cobs, each bearing 5000 grains. What is the ratio of the number of grains sowed to the number harvested?
Answer: B (1 : 5,000)
Grains harvested = 5,000 × 5,000 = 25,000,000. The ratio 5,000 : 25,000,000 simplifies to 1 : 5,000.
Three teachers shared a packet of chalk. The first teacher got 2/5 of the chalk and the second teacher received 2/15 of the remainder. What fraction did the third teacher receive?
Answer: C (13/25)
The remainder after the first teacher is 3/5; the second gets 2/15 × 3/5 = 2/25. The third gets 1 − 2/5 − 2/25 = 13/25.
Given that ∛(42x) = 16, find the value of x.
Answer: B (3)
∛(42x) = 42x/3 and 16 = 42, so 2x/3 = 2 and x = 3.
Simplify 1/(√3 + 2) in the form a + b√3
Answer: C (2 − √3)
Multiply top and bottom by 2 − √3: (2 − √3)/(4 − 3) = 2 − √3.
If 6logx2 − 3logx3 = 3log50.2, find x.
Answer: B (3/4)
The left side is logx(64/27) and the right side is 3 × (−1) = −3. So x−3 = 64/27, x3 = 27/64 and x = 3/4.
In a class of 40 students, each student offers at least one of Physics and Chemistry. If the number of students that offer Physics is three times the number that offer both subjects and the number that offer Chemistry is twice the number that offer Physics, find the number of students that offer Physics only.
Answer: C (10)
Let x offer both. Physics = 3x, Chemistry = 6x, so 3x + 6x − x = 40 and x = 5. Physics only = 3x − x = 10.
Find the values of x where the curve y = x3 + 2x2 − 5x − 6 crosses the x-axis
Answer: C (2, −1 and −3)
x = 2 makes the cubic zero, and dividing by (x − 2) leaves x2 + 4x + 3 = (x + 1)(x + 3). So x = 2, −1, −3.
Find the remainder when 3x3 + 5x2 − 11x + 4 is divided by x + 3.
Answer: B (1)
By the remainder theorem, put x = −3: −81 + 45 + 33 + 4 = 1.
Factorize completely ac − 2bc − a2 + 4b2.
Answer: B ((a − 2b)(c − a − 2b))
ac − 2bc = c(a − 2b) and 4b2 − a2 = −(a − 2b)(a + 2b). Taking out (a − 2b) leaves c − a − 2b.
y is inversely proportional to x and y = 4 when x = ½. Find x when y = 10.
Answer: B (1/5)
y = k/x, so k = 4 × ½ = 2. When y = 10, x = 2/10 = 1/5.
The length L of a simple pendulum varies directly as the square of its period T. If a pendulum with period 4 sec. is 64cm long, find the length of a pendulum whose period is 9 sec.
Answer: D (324cm)
L = kT2: 64 = 16k, so k = 4. Then L = 4 × 81 = 324cm.
The shaded area in the diagram above is represented by
Answer: A ({(x,y) : y + 3x < 6})
The line through (0, 6) and (2, 0) is y = −3x + 6, i.e. y + 3x = 6. The shaded side contains the origin, where y + 3x = 0 < 6.
The nth terms of two sequences are Qn = 3.2n−2 and Um = 3.22m−3. Find the product of Q2 and U2.
Answer: D (18)
Q2 = 3 × 20 = 3 and U2 = 3 × 21 = 6, so the product is 18.
Given that the first and fourth terms of a G.P. are 6 and 162 respectively, find the sum of the first three terms of the progression.
Answer: D (78)
ar3 = 162 and a = 6, so r3 = 27 and r = 3. The first three terms are 6, 18, 54, which add to 78.
Find the sum to infinity of the series ½, 1/6, 1/18, ...
Answer: B (¾)
a = ½ and r = 1/3, so S = a/(1 − r) = ½ ÷ 2/3 = ¾.
If the operation * on the set of integers is defined by p * q = √(pq), find the value of 4 * (8 * 32)
Answer: B (8)
8 * 32 = √256 = 16, then 4 * 16 = √64 = 8.
The inverse of the matrix (2 1; 1 1) is (matrices written row by row)
Answer: D ((1 −1; −1 2))
The determinant is 2 − 1 = 1. Swap the diagonal entries and change the signs of the others: (1 −1; −1 2).
The sum of the interior angles of a pentagon is 6x + 6y. Find y in terms of x.
Answer: B (y = 90 − x)
A pentagon's angles add to 540°, so 6x + 6y = 540, x + y = 90 and y = 90 − x.
PQRSTV is a regular polygon of side 7 cm inscribed in a circle. Find the circumference of the circle PQRSTV. [π = 22/7]
Answer: C (44cm)
A regular hexagon's side equals the radius of its circle, so r = 7cm and the circumference = 2 × 22/7 × 7 = 44cm.
P, R and S lie on a circle centre O as shown above, while Q lies outside the circle. Find ∠PSO.
Answer: A (35°)
In triangle PQR, ∠QRP = 180° − 20° − 35° = 125°, so ∠PRS = 55°. Then ∠POS = 2 × 55° = 110°, and triangle POS is isosceles, so ∠PSO = (180° − 110°)/2 = 35°.
In the diagram above, PQ = 4cm and TS = 6cm. If the area of parallelogram PQTU is 32 cm2, find the area of the trapezium PQRU.
Answer: D (72cm2)
The height is 32 ÷ 4 = 8cm. UR = UT + TS + SR = 4 + 6 + 4 = 14cm, so the area = ½ × (4 + 14) × 8 = 72cm2.
An arc of a circle of length 22cm subtends an angle of 3x° at the centre of the circle. Find the value of x if the diameter of the circle is 14cm. [π = 22/7]
Answer: B (60°)
The circumference is 44cm, so the arc is half the circle: 3x = 180 and x = 60.
Determine the locus of a point inside a square PQRS which is equidistant from PQ and QR.
Answer: B (The diagonal QS)
Points equidistant from two lines meeting at Q lie on the bisector of angle Q, which in a square is the diagonal QS.
The locus of a point which is 5cm from the line LM is a
Answer: A (pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM)
Points 5cm from a line can be on either side of it, giving two parallel lines, one on each side.
Find the value of α2 + β2 if α + β = 2 and the distance between the points (1, α) and (β, 1) is 3 units
Answer: C (11)
(β − 1)2 + (1 − α)2 = 9 gives α2 + β2 − 2(α + β) + 2 = 9, so α2 + β2 = 9 + 4 − 2 = 11.
Find the midpoint of the line joining P(−3, 5) and Q(5, −3).
Answer: D ((1, 1))
Midpoint = ((−3 + 5)/2, (5 − 3)/2) = (1, 1).
Find the value of x in the figure above.
Answer: C (5√6)
By the sine rule, x/sin 45° = 15/sin 60°, so x = 15 × (√2/2)/(√3/2) = 15√2/√3 = 5√6.
The shadow of a pole 5√3m high is 5m. Find the angle of elevation of the sun.
Answer: C (60°)
tan θ = height/shadow = 5√3/5 = √3, so θ = 60°.
Find the derivative of (2 + 3x)(1 − x) with respect to x.
Answer: B (1 − 6x)
Expanding gives 2 + x − 3x2, whose derivative is 1 − 6x.
Find the derivative of the function y = 2x2(2x − 1) at the point x = −1.
Answer: C (16)
y = 4x3 − 2x2, so dy/dx = 12x2 − 4x = 12 + 4 = 16 at x = −1.
If y = 3 cos(x/3), find dy/dx when x = 3π/2
Answer: C (−1)
dy/dx = −sin(x/3). At x = 3π/2, x/3 = π/2, so dy/dx = −1.
What is the rate of change of the volume v of a hemisphere with respect to its radius r when r = 2?
Answer: C (8π)
v = (2/3)πr3, so dv/dr = 2πr2 = 8π when r = 2.
Evaluate ∫13 (x2 − 1)dx
Answer: A (6⅔)
[x3/3 − x] from 1 to 3 = (9 − 3) − (1/3 − 1) = 6 + 2/3 = 6⅔.
The pie chart above shows the distribution of the crops harvested from a farmland in a year. If 3000 tonnes of millet is harvested, what amount of beans is harvested?
Answer: D (1200 tonnes)
Maize is a right angle, so beans = 360° − (60° + 90° + 150°) = 60°. Beans = 60/150 × 3000 = 1200 tonnes.
I. Rectangular bars of equal width
II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval.
III. Rectangular bars have common sides with no gaps in-between.
A histogram is described completely by
Answer: C (I, II and III)
In the usual school histogram the bars have equal widths, their heights show the frequencies and they touch with no gaps, so all three statements are needed.
The mean age of a group of students is 15 years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages becomes 18 years. Find the number of students in the group
Answer: B (9)
With n students, (15n + 45)/(n + 1) = 18, so 15n + 45 = 18n + 18 and n = 9.
The weights of 10 pupils in class are 15kg, 16kg, 17kg, 18kg, 16kg, 17kg, 17kg, 17kg, 18kg and 16kg. What is the range of this distribution?
Answer: C (3)
Range = 18 − 15 = 3.
Find the mean deviation of 1, 2, 3 and 4.
Answer: A (1.0)
The mean is 2.5; the absolute deviations are 1.5, 0.5, 0.5, 1.5, which total 4. Mean deviation = 4/4 = 1.0.
In how many ways can 2 students be selected from a group of 5 students in a debating competition?
Answer: A (10 ways)
5C2 = (5 × 4)/2 = 10.
A committee of six is to be formed by a state governor from nine state commissioners and three members of the state House of Assembly. In how many ways can the members of the committee be chosen so as to include one member of the House of Assembly?
Answer: D (378 ways)
Choose 1 of the 3 Assembly members and 5 of the 9 commissioners: 3 × 9C5 = 3 × 126 = 378.
Some white balls were put in a basket containing twelve red balls and sixteen black balls. If the probability of picking a white ball from the basket is 3/7, how many white balls were introduced?
Answer: C (21)
x/(28 + x) = 3/7 gives 7x = 84 + 3x, so x = 21.
An unbiased die is rolled 100 times and the outcome is tabulated as follows:
Number: 1, 2, 3, 4, 5, 6
Frequency: 13, 15, 22, 14, 20, 16
What is the probability of obtaining 5?
Answer: B (1/5)
5 came up 20 times out of 100, so the probability is 20/100 = 1/5.
A container has 30 gold medals, 22 silver medals and 18 bronze medals. If one medal is selected at random from the container, what is the probability that it is not a gold medal?
Answer: A (4/7)
There are 70 medals and 40 are not gold, so the probability is 40/70 = 4/7.