47 questions. Try each one, then open the answer to check yourself and read the explanation.
Calculate 33105 − 14425.
Answer: A (13135)
In base ten, 33105 = 455 and 14425 = 247, so the difference is 208. 208 = 1×125 + 3×25 + 1×5 + 3, which is 13135.
Convert 3.1415926 to 5 decimal places.
Answer: B (3.14159)
Keep five digits after the point: 3.14159|26. The next digit is 2 (less than 5), so it stays 3.14159.
The length of a notebook 15cm, was measured as 16.8cm. Calculate the percentage error to 2 significant figures.
Answer: A (12.00%)
Error = 16.8 − 15 = 1.8cm. Percentage error = 1.8/15 × 100 = 12%.
A worker's present salary is N24,000 per annum. His annual increment is 10% of his basic salary. What would be his annual salary at the beginning of the third year?
Answer: A (N28,800)
Each increment is 10% of N24,000 = N2,400. By the start of the third year he has had two increments: 24,000 + 2 × 2,400 = N28,800.
Express the product of 0.0014 and 0.011 in standard form.
Answer: D (1.54 × 10−5)
14 × 11 = 154, and there are 4 + 3 = 7 decimal places, so the product is 0.0000154 = 1.54 × 10−5.
Evaluate (81¾ − 27⅓)/(3 × 23)
Answer: B (1)
81¾ = 33 = 27 and 27⅓ = 3, so the top is 24. The bottom is 3 × 8 = 24. 24/24 = 1.
Find the value of (16)3/2 + log100.0001 + log232
Answer: D (65.000)
163/2 = 43 = 64, log100.0001 = −4 and log232 = 5. Total 64 − 4 + 5 = 65.
Simplify (√12 − √3)/(√12 + √3)
Answer: A (⅓)
√12 = 2√3, so the fraction is (2√3 − √3)/(2√3 + √3) = √3/(3√3) = ⅓.
Four members of a school first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams?
Answer: B (21)
Count both teams and remove the double-counted boys: 11 + 14 − 4 = 21.
If S = {x : x2 = 9, x > 4}, then S equal to
Answer: C (∅)
x² = 9 gives x = 3 or −3, and neither is greater than 4. So S has no elements: it is the empty set ∅.
If x − 1 and x + 1 are both factors of the equation x3 + px2 + qx + 6 = 0, evaluate p and q.
Answer: A (−6, −1)
x = 1 gives 1 + p + q + 6 = 0, so p + q = −7. x = −1 gives −1 + p − q + 6 = 0, so p − q = −5. Adding: 2p = −12, p = −6 and q = −1.
Find a positive value of p if the expression 2x2 − px + p leaves a remainder 6 when divided by x − p
Answer: B (2)
By the remainder theorem, put x = p: 2p² − p² + p = 6, so p² + p − 6 = 0, (p + 3)(p − 2) = 0. The positive value is p = 2.
Find T in terms of K, Q and S if S = 2r√(π(QT + K)).
Answer: B (S2/(4πr2Q) − K/Q)
Square both sides: S² = 4r²π(QT + K). So QT + K = S²/(4πr²), and T = S²/(4πr²Q) − K/Q.
The graph of f(x) = x2 − 5x + 6 crosses the x-axis at the points.
Answer: D ((2, 0), (3, 0))
Solve x² − 5x + 6 = 0: (x − 2)(x − 3) = 0, so x = 2 or 3. The points are (2, 0) and (3, 0).
Factorize completely the expression abx2 + 6y − 3ax − 2byx
Answer: A ((ax − 2y)(bx − 3))
Group: abx² − 2bxy = bx(ax − 2y) and −3ax + 6y = −3(ax − 2y). Together: (ax − 2y)(bx − 3).
Solve the following inequality (x − 3)(x − 4) ≤ 0
Answer: A (3 ≤ x ≤ 4)
The product is zero at 3 and 4 and negative between them, so (x − 3)(x − 4) ≤ 0 for 3 ≤ x ≤ 4 (the ends are included because of '='.)
The 4th term of an A.P is 13 while the 10th term is 31. Find the 21st term.
Answer: C (64)
a + 3d = 13 and a + 9d = 31, so 6d = 18, d = 3 and a = 4. The 21st term is a + 20d = 4 + 60 = 64.
Express (5x − 12)/((x − 2)(x − 3)) in partial fractions.
Answer: B (2/(x − 2) + 3/(x − 3))
Write 5x − 12 = A(x − 3) + B(x − 2). x = 2 gives −2 = −A, so A = 2; x = 3 gives 3 = B. So it is 2/(x − 2) + 3/(x − 3).
Use the graph of the curve y = f(x) above to solve the inequality f(x) ≥ 0.
Answer: A (−1 ≤ x ≤ 1, x ≥ 2)
f(x) ≥ 0 where the curve is on or above the x-axis. It cuts the axis at −1, 1 and 2, is above it between −1 and 1, dips below between 1 and 2, then rises after 2. So −1 ≤ x ≤ 1 or x ≥ 2.
Which of the following binary operations is commutative in the set of integers?
Answer: B (a*b = a + b − ab)
Commutative means a*b = b*a. Swapping a and b in a + b − ab gives b + a − ba, which is the same. The others change, e.g. 1 + 2(2) ≠ 2 + 2(1).
If a*b = +√(ab), evaluate 2*(12*27).
Answer: C (6)
Work out the bracket first: 12*27 = √324 = 18. Then 2*18 = √36 = 6.
Find the sum to infinity of the following sequence: 1, 9/10, (9/10)2, (9/10)3
Answer: D (10)
This is a GP with a = 1 and r = 9/10. Sum to infinity = a/(1 − r) = 1/(1/10) = 10.
If X = (1 2; 0 3) and Y = (2 1; 4 3) find XY.
Answer: A ((10 7; 12 9))
Multiply rows of X by columns of Y: first row (1×2 + 2×4, 1×1 + 2×3) = (10, 7); second row (0×2 + 3×4, 0×1 + 3×3) = (12, 9).
Determine the value of x in the fig. above.
Answer: A (134°)
In the triangle, the third angle (bottom left) is 180° − 81° − 53° = 46°. The angle x is opposite it in a cyclic quadrilateral, so x = 180° − 46° = 134°.
PT is a tangent to the circle TYZX, YT = YX and ∠PTX = 50°. Calculate ∠TZY
Answer: B (65°)
By the alternate segment theorem, angle XYT = angle PTX = 50°. Triangle XYT is isosceles (YT = YX), so angle TXY = (180° − 50°)/2 = 65°. Angle TZY stands on the same arc TY, so it is also 65°.
In a triangle XYZ, ∠YXZ = 44° and ∠XYZ = 112°. Calculate the acute angle between the internal bisectors of ∠XYZ and ∠XZY.
Answer: C (68°)
∠XZY = 180° − 44° − 112° = 24°. The bisectors make 56° and 12° with YZ, so they meet at 180° − 56° − 12° = 112°. The acute angle between them is 180° − 112° = 68°.
Find the distance between two towns P(45°N, 30°W) and Q(15°S, 30°W) if the radius of the earth is 7 000km. (π = 22/7)
Answer: C (22 000/3)
Both towns are on longitude 30°W, so the angle between them is 45° + 15° = 60°. Distance = 60/360 × 2 × 22/7 × 7000 = 44 000/6 = 22 000/3 km.
Two perpendicular lines PQ and QR intersect at (1, −1). If the equation of PQ is x − 2y + 4 = 0, find the equation of QR.
Answer: D (2x + y − 1 = 0)
PQ has gradient ½, so QR has gradient −2. Through (1, −1): y + 1 = −2(x − 1), i.e. 2x + y − 1 = 0.
P is on the locus of point equidistant from two given points X and Y. UV is a straight line through Y parallel to the locus. If ∠PYU is 40°, find ∠XPY
Answer: B (80°)
The locus is the perpendicular bisector of XY, and it bisects angle XPY. UV is parallel to it, so angle PYU equals the angle between PY and the bisector (alternate angles) = 40°. So angle XPY = 2 × 40° = 80°.
In the diagram above, k, m and n are parallel lines. What is the marked x?
Answer: B (63°)
The 117° angle and x lie on opposite sides of the transversal with the parallel lines, so together they make a straight angle: x = 180° − 117° = 63°.
In the diagram above, the base diameter is 14cm while the height is 12cm. Calculate the total surface area if the cylinder has both a base and a top. [π = 22/7]
Answer: A (836cm2)
r = 7cm. Curved surface = 2πrh = 2 × 22/7 × 7 × 12 = 528cm². Top and base = 2πr² = 308cm². Total 836cm².
In the diagram above, find PQ if the area of triangle PQR is 35cm2.
Answer: C (14cm)
Use area = ½ab sin C with the 30° angle at P: ½ × PQ × 10 × ½ = 35. So 2.5 × PQ = 35 and PQ = 14cm.
A schoolboy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank is 60°. Calculate the height of the water tank.
Answer: B (30√3m)
tan 60° = height/30, so height = 30 × √3 = 30√3 m.
QRS is a triangle with QS = 12m, ∠RQS = 30° and ∠QRS = 45°. Calculate the length of RS.
Answer: C (6√2m)
Sine rule: RS/sin 30° = QS/sin 45°. So RS = 12 × ½ ÷ (√2/2) = 12/√2 = 6√2 m.
Which of the following is a sketch of y = 3 sin x?
Answer: B (Graph B)
y = 3 sin x passes through the origin, rises to 3 at x = π/2, returns to 0 at π and is negative between −π and 0. Only graph B does this.
The derivative of cosec x is
Answer: B (−cot x cosec x)
cosec x = (sin x)−1, so its derivative is −(sin x)−2 × cos x = −(cos x/sin x)(1/sin x) = −cot x cosec x.
For what value of x is the tangent to the curve y = x2 − 4x + 3 parallel to the x-axis?
Answer: B (2)
A tangent parallel to the x-axis has gradient 0. dy/dx = 2x − 4 = 0 gives x = 2.
Two variables x and y are such that dy/dx = 4x − 3 and y = 5 when x = 2. Find y in terms of x.
Answer: B (2x2 − 3x + 3)
Integrate: y = 2x² − 3x + c. When x = 2, y = 5: 8 − 6 + c = 5, so c = 3 and y = 2x² − 3x + 3.
Find the area bounded by the curve y = 3x2 − 2x + 1, the ordinates x = 1 and x = 3 and the x-axis
Answer: D (20)
Area = ∫ from 1 to 3 of (3x² − 2x + 1) dx = [x³ − x² + x] from 1 to 3 = 21 − 1 = 20 square units.
Ages in years: 13, 14, 15, 16, 17
No. of students: 3, 10, 30, 42, 15
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years-old is.
Answer: D (108°)
There are 100 students, and 30 are 15 years old. Angle = 30/100 × 360° = 108°.
The pie chart above shows the distribution of students in a secondary school class. If 30 students offered French, how many offered CRK?
Answer: C (10)
French fills a right angle, 90°. CRK's angle is x = 360° − 90° − 150° − 90° = 30°. 90° stands for 30 students, so 30° stands for 10 students.
The mean and the range of the set of numbers 1.20, 1.00, 0.90, 1.40, 0.80, 0.80, 1.20 and 1.10 are m and r respectively. Find m + r.
Answer: B (1.65)
The eight numbers add up to 8.40, so m = 1.05. The range is 1.40 − 0.80 = 0.60. m + r = 1.65.
Class: 1-3, 4-6, 7-9
Frequency: 5, 8, 5
Find the standard deviation of the data using the table above.
Answer: D (√5)
Use the class mid-points 2, 5, 8. The mean is 5. Σf(x − 5)² = 5(9) + 8(0) + 5(9) = 90, and 90/18 = 5 is the variance, so the standard deviation is √5.
The variance of the scores 1, 2, 3, 4, 5, is
Answer: C (2.0)
The mean is 3. The squared deviations are 4, 1, 0, 1, 4, total 10. Variance = 10/5 = 2.0.
Class interval (frequency): 1.5-1.9 (2), 2.0-2.4 (1), 2.5-2.9 (4), 3.0-3.4 (15), 3.5-3.9 (10), 4.0-4.4 (5), 4.5-4.9 (3). The class boundaries are 1.45-1.95, 1.95-2.45, … , 4.45-4.95 and the class mid-points 1.7, 2.2, 2.7, 3.2, 3.7, 4.2, 4.7.
Find the mode of the distribution.
Answer: A (3.2)
The modal class is 3.0–3.4 (boundaries 2.95–3.45). Mode = 2.95 + 11/(11 + 5) × 0.5 ≈ 3.29, which is closest to 3.2.
Let P be a probability function on set S, where S = {a1, a2, a3, a4}. Find P{a1} if P{a2} = ⅓, P(a3) = ⅙ and P(a4) = ⅕.
Answer: D (3/10)
The four probabilities must add up to 1. P(a₁) = 1 − (⅓ + ⅙ + ⅕) = 1 − 21/30 = 9/30 = 3/10.
A die has four of its faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive times, the top face will be white in both cases?
Answer: C (4/9)
P(white) = 4/6 = ⅔ on each throw, and the throws are independent: ⅔ × ⅔ = 4/9.