48 questions. Try each one, then open the answer to check yourself and read the explanation.
Evaluate
1/3 ÷ [5/7(9/10 − 1 + 3/4)]
Answer: A (28/39)
Inside the bracket: 9/10 − 1 + 3/4 = 13/20, and 5/7 × 13/20 = 13/28. Then 1/3 ÷ 13/28 = 1/3 × 28/13 = 28/39.
Evaluate (0.36 × 5.4 × 0.63)/(4.2 × 9.0 × 2.4) correct to 2 significant figures.
Answer: B (0.014)
The top is 1.22472 and the bottom is 90.72, giving 0.0135. Rounded to 2 significant figures this is 0.014.
Evaluate log5(0.04)/(log318 − log32)
Answer: B (−1)
0.04 = 1/25 = 5−2, so the top is −2. The bottom is log3(18 ÷ 2) = log39 = 2. So the value is −2/2 = −1.
Without using table, solve the equation
8x−2 = 2/25
Answer: D (10)
x−2 = 2/25 ÷ 8 = 1/100, so x² = 100 and x = 10.
Simplify √48 − 9/√3 + √75
Answer: B (6√3)
√48 = 4√3, 9/√3 = 3√3 and √75 = 5√3. So 4√3 − 3√3 + 5√3 = 6√3.
In a science class of 42 students, each student offers at least one of Mathematics and Physics. If 22 students offer Physics and 28 students offer Mathematics, find how many students offer Physics only?
Answer: D (14)
Students taking both = 22 + 28 − 42 = 8. Physics only = 22 − 8 = 14.
Given that for sets A and B, in a universal set E, A ⊆ B then A ∩ (A ∩ B)′ is
Answer: B (∅)
If A ⊆ B then A ∩ B = A. So the expression is A ∩ A′, which has no elements: the empty set ∅.
Solve for x if 25x + 3(5x) = 4
Answer: B (0)
Let p = 5x, so p² + 3p − 4 = 0, i.e. (p + 4)(p − 1) = 0. 5x can't be −4, so 5x = 1 and x = 0.
Simplify ((2m − u)2 − (m − 2u)2)/(5m2 − 5u2)
Answer: A (3/5)
The top is a difference of two squares: (3m − 3u)(m + u) = 3(m − u)(m + u). The bottom is 5(m − u)(m + u). Cancelling leaves 3/5.
Factorize a2x − b2y − b2x + a2y
Answer: D ((x + y)(a − b)(a + b))
Group as a²(x + y) − b²(x + y) = (x + y)(a² − b²) = (x + y)(a − b)(a + b).
Find the value of p and q such that (x − 1) and (x − 3) are factors of px3 + qx2 + 11x − 6.
Answer: B (1, −6)
Put x = 1: p + q + 5 = 0. Put x = 3: 27p + 9q + 27 = 0, i.e. 3p + q + 3 = 0. Subtracting gives 2p = 2, so p = 1 and q = −6.
The equation of the graph above is
Answer: A (y = (x − 3)3)
The graph passes through (0, −27) and (3, 0), and its flat point (where it levels off) is at (3, 0). y = (x − 3)³ has exactly this: it is y = x³ shifted 3 units right.
If a = 1, b = 3, solve for x in the equation a/(a − x) = b/(x − b)
Answer: C (3/2)
Substitute: 1/(1 − x) = 3/(x − 3). Cross-multiply: x − 3 = 3 − 3x, so 4x = 6 and x = 3/2.
Solve for r in the following equation
1/(r − 1) + 2/(r + 1) = 3/r
Answer: A (3)
Multiply through by r(r − 1)(r + 1): r(r + 1) + 2r(r − 1) = 3(r² − 1). This gives 3r² − r = 3r² − 3, so r = 3.
Find P if (x − 3)/((1 − x)(x + 2)) = P/(1 − x) + Q/(x + 2)
Answer: A (−2/3)
Multiply out: x − 3 = P(x + 2) + Q(1 − x). Put x = 1 to remove Q: −2 = 3P, so P = −2/3.
Find the range of values of x for which 1/x > 2 is true.
Answer: C (0 < x < ½)
1/x can only be greater than 2 if x is positive. Then multiplying by x gives 1 > 2x, so x < ½. Together: 0 < x < ½.
Find the inequality which represents the shaded portion in the diagram above.
Answer: B (2x − y − 2 ≥ 0)
Test a point in the shaded part, such as (3, 0): 2(3) − 0 − 2 = 4, which is positive. The boundary line is solid, so it is included: 2x − y − 2 ≥ 0.
If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference.
Answer: D (2)
The 6th term is a + 5d, so 1 + 5d = 11 and d = 2.
Find the value of r if log10r2 + log10r4 + log10r8 + log10r16 + log10r32 = 63
Answer: C (10)
Adding logs multiplies the powers: the left side is (2 + 4 + 8 + 16 + 32) log r = 62 log r. So log r ≈ 1 and r ≈ 10 (exactly 10 if the sum also has a log10r term, making 63 log r = 63).
Find the nth term of the sequence 3, 6, 10, 15, 21, ...
Answer: C ((n + 1)(n + 2)/2)
Test n = 1 and n = 2: (n + 1)(n + 2)/2 gives 2×3/2 = 3 and 3×4/2 = 6, matching the sequence. The other formulas fail.
A binary operation * is defined on the set of all positive integers by a*b = ab for all positive integers a, b. Which of the following properties does NOT hold.
Answer: D (Inverse)
Products of positive integers are positive integers (closure), multiplication is associative, and 1 is the identity. But the inverse of 2 would be ½, which is not a positive integer.
The multiplication table above has modulo 10 on the set S = {2, 4, 6, 8}, find the inverse of 2.
Answer: D (8)
First find the identity: 6 leaves every element unchanged (e.g. 6 × 4 = 24 ≡ 4). The inverse of 2 is the element that gives 6 with it: 2 × 8 = 16 ≡ 6, so it is 8.
Solve for x and y in the matrix equation shown above: (1 1; 3 y)(x; 1) = (4; 1)
Answer: C (x = 3, y = −8)
Multiplying out: x + 1 = 4 and 3x + y = 1. So x = 3, and then 9 + y = 1 gives y = −8.
The determinant of matrix (1 2 3; 4 5 6; 2 0 −1) is
Answer: C (−3)
Expand along the first row: 1(5×(−1) − 6×0) − 2(4×(−1) − 6×2) + 3(4×0 − 5×2) = −5 + 32 − 30 = −3.
In the diagram above, PQRS is a parallelogram. Find the value of ∠SQR.
Answer: D (100°)
Angle PQS = 50° (vertically opposite the 50°), and angle SPQ = 30° (alternate to the 30° at S, as PQ ‖ TR). So angle PSQ = 180° − 80° = 100°, and SQR = PSQ = 100° (alternate angles, PS ‖ QR).
In the diagram above O is the centre of the circle. If SOQ is a diameter and ∠PRS is 38°, what is the value of ∠PSQ?
Answer: D (52°)
Angle PQS = angle PRS = 38° (same segment). Angle SPQ = 90° (angle in a semicircle). So angle PSQ = 180° − 90° − 38° = 52°.
If three angles of a quadrilateral are (3y − x − z)°, 3x°, (2z − 2y − x)°, find the fourth angle in terms of x, y and z.
Answer: A ((360 − x − y − z)°)
The three angles add up to (3y − x − z) + 3x + (2z − 2y − x) = x + y + z. The angles of a quadrilateral total 360°, so the fourth is (360 − x − y − z)°.
An open rectangular box is made of wood 2cm thick. If the internal dimensions of the box are 50cm long, 36cm wide and 20cm deep, the volume of wood in the box is
Answer: A (11 520cm3)
Outside measurements: 54 × 40 × 22 (the open top adds no thickness) = 47 520cm³. Inside: 50 × 36 × 20 = 36 000cm³. Wood = 47 520 − 36 000 = 11 520cm³.
Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45°.
Answer: C (16 + 2π)
Arc length = 45/360 × 2π × 8 = 2π. The perimeter adds two radii: 2π + 16.
In the diagram above, PTS is a tangent to the circle TQR at T, calculate ∠RTS.
Answer: B (70°)
In triangle QRT, angle RQT = 180° − 60° − 50° = 70°. By the alternate segment theorem, the angle between tangent TS and chord TR equals angle RQT, so RTS = 70°.
In the diagram above, find h
Answer: B (12/7 √6cm)
Heron's formula: s = (6 + 5 + 7)/2 = 9 and area = √(9 × 3 × 4 × 2) = 6√6. Area = ½ × 7 × h, so h = 12√6/7 cm.
In the frustum of a cone shown above, the top diameter is twice the bottom diameter. If the height of the frustum is h centimeters, find the height of the cone.
Answer: A (2h)
The removed small cone has half the radius of the full cone, so by similar triangles it also has half the height. The frustum is the other half, so the full cone is 2h high.
What is the locus of a point P which moves on one side of a straight line XY, so that the angle XPY is always equal to 90°?
Answer: D (A semi-circle)
The angle in a semicircle is 90°, so every point P with angle XPY = 90° lies on the circle with diameter XY. Staying on one side of XY gives a semicircle.
If M(4, q) is the mid-point of the line joining L(p, −2) and N(q, p), find the values of p and q.
Answer: D (p = 6, q = 2)
Midpoint: (p + q)/2 = 4 and (−2 + p)/2 = q. So p + q = 8 and p = 2q + 2, giving 3q + 2 = 8, q = 2 and p = 6.
The angle of depression of a boat from the top of a cliff 10m high is 30°. How far is the boat from the foot of the cliff?
Answer: C (10√3)
The angle of elevation of the cliff top from the boat is also 30°. So tan 30° = 10/d and d = 10/tan 30° = 10√3 m.
What is the value of sin (−690°)?
Answer: D (½)
Adding 720° (two full turns) doesn't change sine: sin(−690°) = sin 30° = ½.
If y = 3t3 + 2t2 − 7t + 3, find dy/dt at t = −1
Answer: C (−2)
dy/dt = 9t² + 4t − 7. At t = −1: 9 − 4 − 7 = −2.
Find the point (x, y) on the Euclidean plane where the curve y = 2x2 − 2x + 3 has 2 as gradient
Answer: A ((1, 3))
dy/dx = 4x − 2 = 2 gives x = 1. Then y = 2 − 2 + 3 = 3, so the point is (1, 3).
Integrate (1 − x)/x3 with respect to x.
Answer: C (1/x − 1/(2x2) + k)
Split it: x−3 − x−2. Integrating gives x−2/(−2) − x−1/(−1) = −1/(2x²) + 1/x, plus a constant.
Evaluate ∫−11 (2x + 1)2 dx
Answer: D (4⅔)
Expand: 4x² + 4x + 1, which integrates to 4x³/3 + 2x² + x. At 1 this is 13/3; at −1 it is −1/3. The difference is 14/3 = 4⅔.
The grades A1, A2, A3, C4 and F earned by students in a particular course are shown in the pie chart above. What percentage of the students obtained a C4 grade?
Answer: D (12.0)
The C4 sector is 43.2°. As a percentage of the whole circle: 43.2/360 × 100 = 12%.
x: 1, 2, 3, 4, 5
f: y + 2, y − 1, 2y − 3, y + 4, 3y − 4
The table above shows the frequency distribution of data. If the mean is 43/14, find y?
Answer: B (2)
Σf = 8y − 2 and Σfx = 28y − 13. Setting (28y − 13)/(8y − 2) = 43/14 and cross-multiplying gives 392y − 182 = 344y − 86, so 48y = 96 and y = 2.
The mean of twelve positive numbers is 3. When another number is added, the mean becomes 5. Find the thirteenth number.
Answer: A (29)
The twelve numbers total 36. Thirteen numbers with mean 5 total 65, so the new number is 65 − 36 = 29.
Find the mean deviation of the set of numbers 4, 5, 9
Answer: B (2)
The mean is 6. The distances from the mean are 2, 1 and 3, whose average is 6/3 = 2.
Class Interval: 1-5, 6-10, 11-15, 16-20, 21-25
Frequency: 6, 15, 20, 7, 2
Estimate the median of the frequency distribution above.
Answer: B (11½)
There are 50 values, so the median is the 25th. Cumulative frequencies are 6, 21, 41, … so it lies in 11–15. Median = 10.5 + (25 − 21)/20 × 5 = 11.5.
x: 1, 2, 3, 4, 5
f: 2, 1, 2, 1, 2
Find the variance of the frequency distribution above.
Answer: B (9/4)
Σf = 8 and Σfx = 24, so the mean is 3. Σf(x − 3)² = 2(4) + 1(1) + 0 + 1(1) + 2(4) = 18. Variance = 18/8 = 9/4.
Age in years: 10, 11, 12
Number of pupils: 6, 27, 7
The table above shows the number of pupils in each group in a class. What is the probability that a pupil chosen at random is at least 11 years old?
Answer: B (17/20)
There are 40 pupils, and 27 + 7 = 34 are 11 or older. Probability = 34/40 = 17/20.
In a survey, it was observed that 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?
Answer: C (3/8)
Students reading both = 20 + 35 − 40 = 15. As a fraction of the 40 students: 15/40 = 3/8.