45 questions. Try each one, then open the answer to check yourself and read the explanation.
Let P = {1, 2, u, v, w, x}, Q = {2, 3, u, v, w, 5, 6, y} and R = {2, 3, 4, v, x, y}. Determine (P − Q) ∩ R.
Answer: C ({x})
P − Q keeps the elements of P that are not in Q: {1, x}. Of these, only x is in R, so (P − Q) ∩ R = {x}.
If the population of a town was 240 000 in January 1998 and it increased by 2% each year, what would be the population of the town in January 2000?
Answer: C (249 600)
After one year: 240 000 × 1.02 = 244 800. After two years: 244 800 × 1.02 = 249 696, which is closest to option C (249 600).
If (2√3 − √2)/(√3 + 2√2) = m + n√6, find the values of m and n respectively.
Answer: B (−2, 1)
Multiply top and bottom by (√3 − 2√2): the top becomes 6 − 4√6 − √6 + 4 = 10 − 5√6 and the bottom 3 − 8 = −5. So the value is −2 + √6, giving m = −2 and n = 1.
In a youth club of 94 members, 60 like modern music and 50 like traditional music. The number of members who like both traditional and modern music is three times those who do not like any type of music. How many members like only one type of music?
Answer: C (62)
Let x like neither, so 3x like both. Then 60 + 50 − 3x + x = 94, so x = 8 and 24 like both. Only one type: (60 − 24) + (50 − 24) = 36 + 26 = 62.
Evaluate (2.813 × 10−3 × 1.063)/(5.637 × 10−2) reducing each number to two sig. fig. and leaving your answer in two significant figures.
Answer: B (0.055)
To 2 s.f. the numbers are 2.8 × 10−3, 1.1 and 5.6 × 10−2. (0.0028 × 1.1)/0.056 = 0.00308/0.056 = 0.055.
A man wishes to keep some money in savings deposit at 25% compound interest so that after 3 years he can buy a car for N150000. How much does he need to deposit now?
Answer: D (N76 800.00)
The deposit P must satisfy P × 1.253 = 150 000. Since 1.253 = 1.953125, P = 150 000 ÷ 1.953125 = N76 800.
If 31410 − 2567 = 340x, find x
Answer: A (7)
2567 = 2×49 + 5×7 + 6 = 139, so the left side is 314 − 139 = 175. Then 3x2 + 4x = 175, i.e. (x − 7)(3x + 25) = 0, so x = 7.
Audu bought an article for N50 000 and sold it to Femi at a loss of x%. Femi later sold the article to Oche at a profit of 40%. If Femi made a profit of N10000, find the value of x.
Answer: B (50)
Femi's 40% profit was N10 000, so he paid 10 000 ÷ 0.4 = N25 000. Audu lost 50 000 − 25 000 = N25 000, which is 50% of N50 000.
Simplify (3(2n + 1) − 4(2n − 1))/(2n + 1 − 2n)
Answer: C (4)
Write everything in terms of 2n: the top is 6(2n) − 2(2n) = 4(2n) and the bottom is 2(2n) − 2n = 2n. The quotient is 4.
If P3446 − 23P26 = 2PP26, find the value of digit P.
Answer: D (5)
In base ten: (216P + 136) − (542 + 6P) = 434 + 42P. This gives 168P = 840, so P = 5.
Evaluate 5−3log52 × 22log23
Answer: B (1⅛)
5−3log52 = 5log5(2−3) = 1/8 and 22log23 = 32 = 9. The product is 9/8 = 1⅛.
A binary operation * is defined by a*b = ab. If a*2 = 2 − a, find the possible value of a.
Answer: D (1, −2)
a*2 = a2, so a2 = 2 − a, i.e. a2 + a − 2 = 0. Factorising, (a − 1)(a + 2) = 0, so a = 1 or −2.
The 3rd term of an A.P. is 4x − 2y and the 9th term is 10x − 8y. Find the common difference.
Answer: C (x − y)
The 9th term minus the 3rd term is 6d. So 6d = (10x − 8y) − (4x − 2y) = 6x − 6y, giving d = x − y.
Find the inverse of p under the binary operation * defined by p*q = p + q − pq, where p and q are real numbers and zero is the identity.
Answer: C (p/(p − 1))
The inverse q satisfies p*q = 0, so p + q − pq = 0. Then q(1 − p) = −p, so q = p/(p − 1).
A matrix P = [a, b; c, d] is such that PT = −P, where PT is the transpose of P. If b = 1, then P is
Answer: B ([0, 1; −1, 0])
PT = [a, c; b, d] must equal [−a, −b; −c, −d]. So a = −a and d = −d give a = d = 0, and c = −b = −1. Hence P = [0, 1; −1, 0].
The solution of the simultaneous inequalities 2x − 2 ≤ y and 2y − 2 ≤ x is represented by
Answer: B (Graph B)
The boundary lines y = 2x − 2 and y = ½x + 1 both slope upwards and meet at (2, 2). The region lies above the steeper line and below the gentler one, the wedge opening to the left, which is graph B.
Find the values of t, for which the determinant of the matrix [t − 4, 0, 0; −1, t + 1, 1; 3, 4, t − 2] is zero
Answer: D (4, −2, 3)
Expanding along the first row: (t − 4)[(t + 1)(t − 2) − 4] = (t − 4)(t2 − t − 6) = (t − 4)(t − 3)(t + 2). It is zero when t = 4, −2 or 3.
If (x − 1), (x + 1) and (x − 2) are factors of the polynomial ax3 + bx2 + cx − 1, find a, b, c respectively.
Answer: A (−½, 1, ½)
The polynomial equals a(x − 1)(x + 1)(x − 2) = a(x3 − 2x2 − x + 2). Matching the constant term, 2a = −1, so a = −½, b = −2a = 1 and c = −a = ½.
A trader realises 10x − x2 naira profit from the sale of x bags of corn. How many bags will give him the maximum profit?
Answer: B (5)
The profit is greatest where its derivative 10 − 2x is zero, i.e. at x = 5 bags.
Solve the inequality 2 − x > x2.
Answer: D (−2 < x < 1)
Rearrange to x2 + x − 2 < 0, i.e. (x + 2)(x − 1) < 0. The product is negative between the roots, so −2 < x < 1.
If α and β are the roots of the equation 3x2 + 5x − 2 = 0, find the value of 1/α + 1/β.
Answer: D (5/2)
1/α + 1/β = (α + β)/(αβ). Here α + β = −5/3 and αβ = −2/3, so the value is (−5/3) ÷ (−2/3) = 5/2.
Find the minimum value of the function f(θ) = 2/(3 − cosθ) for 0 ≤ θ ≤ 2π
Answer: A (½)
cosθ lies between −1 and 1, so 3 − cosθ lies between 2 and 4. The fraction 2/(3 − cosθ) is smallest when the denominator is 4, giving 2/4 = ½.
A frustum of pyramid with square base has its upper and lower sections as square of sizes 2cm and 5cm respectively and the distance between them 6cm. Find the height of the pyramid from which the frustum was obtained.
Answer: D (10.0m)
By similar shapes, height/5 = (height − 6)/2. So 2H = 5H − 30, giving H = 10.
A ship sails a distance of 50km in the direction S50°E and then sails the distance of 50km in the direction N40°E. Find the bearing of the ship from its original position.
Answer: D (N85°E)
The two legs (bearings 130° and 040°) are at right angles and equal, so the direct line makes 45° with the first leg. Its bearing is 130° − 45° = 085°, i.e. N85°E.
3y = 4x − 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K.
Answer: A (−4/3)
The gradients are 4/3 and 1/K. For perpendicular lines their product is −1, so 4/(3K) = −1 and K = −4/3.
In the diagram above, if ∠RPS = 50°, ∠RPQ = 30° and PQ = QR, find the value of ∠PRS.
Answer: B (70°)
Since PQ = QR, ∠QRP = 30° and ∠PQR = 120°. Opposite angles of a cyclic quadrilateral add to 180°, so ∠PSR = 60°. In triangle PRS, ∠PRS = 180° − 50° − 60° = 70°.
In the diagram above EFGH is a circle centre of O. FH is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH.
Answer: C (26cm)
The diameter bisects the chord, so EN = NG = 12 cm. By intersecting chords, FN × NH = EN × NG, so FN × 8 = 144 and FN = 18. FH = 18 + 8 = 26 cm.
If P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is
Answer: D (the perpendicular bisector of PQ)
Points that are the same distance from two fixed points lie on the perpendicular bisector of the line joining them.
In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon.
Answer: B (6)
Interior + exterior = 180° and interior = 2 × exterior, so the exterior angle is 60°. Number of sides = 360° ÷ 60° = 6.
A predator moves in a circle of radius √2 centre (0, 0), while a prey moves along the line y = x. If 0 ≤ x ≤ 2, at which point(s) will they meet?
Answer: A ((1, 1) only)
The circle is x2 + y2 = 2. Putting y = x gives 2x2 = 2, so x = ±1. Only x = 1 is in the range, so they meet at (1, 1) only.
If the diagram above is the graph y = x2, the shaded area is
Answer: C (64/3 square units)
The shaded region lies under y = x2 from x = 0 to x = 4. Its area is ∫x2 dx from 0 to 4 = 43/3 = 64/3 square units.
Find the value of ∫0π (cos2θ − 1)/sin2θ dθ
Answer: D (−π)
cos2θ − 1 = −sin2θ, so the integrand is −1. Integrating −1 from 0 to π gives −π.
If y = 2x cos 2x − sin 2x, find dy/dx when x = π/4
Answer: B (−π)
dy/dx = 2cos 2x − 4x sin 2x − 2cos 2x = −4x sin 2x. At x = π/4, sin(π/2) = 1, so dy/dx = −π.
A bowl is designed by revolving completely the area enclosed by y = x2 − 1, y = 0, y = 3 and x ≥ 0 around the y-axis. What is the volume of this bowl?
Answer: B (15π/2 cubic units)
Around the y-axis, V = π∫x2 dy with x2 = y + 1. From y = 0 to 3 this is π[y2/2 + y] = π(4.5 + 3) = 15π/2.
If the volume of a hemisphere is increasing at a steady rate of 18πm3s−1, at what rate is its radius changing when it is 6m?
Answer: C (0.25ms−1)
V = (2/3)πr3, so dV/dt = 2πr2 × dr/dt. With r = 6: 18π = 72π × dr/dt, so dr/dt = 0.25 m/s.
A function f(x) passes through the origin and its first derivative is 3x + 2. What is f(x)?
Answer: A (y = (3/2)x2 + 2x)
Integrating 3x + 2 gives (3/2)x2 + 2x + c. Passing through the origin makes c = 0.
The expression ax2 + bx + c equals 5 at x = 1, if its derivative is 2x + 1, what are the values of a, b, c, respectively?
Answer: D (1, 1, 3)
The derivative 2ax + b equals 2x + 1, so a = 1 and b = 1. At x = 1, a + b + c = 5, so c = 3.
X and Y are two events. The probability of X or Y is 0.7 and the probability of X is 0.4. If X and Y are independent, find the probability of Y.
Answer: B (0.50)
For independent events, P(X or Y) = P(X) + P(Y) − P(X)P(Y). So 0.7 = 0.4 + y − 0.4y, giving 0.6y = 0.3 and y = 0.5.
If the mean of the numbers 0, x + 2, 3x + 6 and 4x + 8 is 4, find their mean deviation.
Answer: C (3)
The mean is (8x + 16)/4 = 2x + 4 = 4, so x = 0 and the numbers are 0, 2, 6, 8. Their distances from 4 are 4, 2, 2, 4, whose average is 3.
In how many ways can the word MATHEMATICS be arranged?
Answer: C (11!/(2!2!2!))
MATHEMATICS has 11 letters with M, A and T each appearing twice. The number of arrangements is 11!/(2!2!2!).
No.: 1, 2, 3, 4, 5, 6
Frequency: 30, 43, 54, 40, 41, 32
A dice is rolled 240 times and the result depicted in the table above. If a pie chart is constructed to represent the data, the angle corresponding to 4 is
Answer: D (60°)
The score 4 came up 40 times out of 240. Its angle is 40/240 × 360° = 60°.
If U = {x : x is an integer and 1 ≤ x ≤ 20}, E1 = {x : x is a multiple of 3}, E2 = {x : x is a multiple of 4} and an integer is picked at random from U, find the probability that it is not in E2.
Answer: A (3/4)
The multiples of 4 from 1 to 20 are 4, 8, 12, 16, 20: five numbers. So 15 of the 20 are not in E2, a probability of 15/20 = 3/4.
The variance of x, 2x, 3x, 4x and 5x is
Answer: B (2x2)
The mean is 3x. The squared deviations are 4x2, x2, 0, x2, 4x2, which add to 10x2. Dividing by 5 gives a variance of 2x2.
Find the sum of the range and the mode of the set of numbers 10, 5, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 10, 9, 7, 10, 6, 5.
Answer: A (16)
The range is 10 − 4 = 6 and the mode is 10 (it appears 7 times). Their sum is 16.
In how many ways can a delegation of 3 be chosen from among 5 men and 3 women, if at least one man and at least one woman must be included?
Answer: D (45)
All choices of 3 from 8: 8C3 = 56. Remove the all-men choices (5C3 = 10) and the all-women choice (1): 56 − 10 − 1 = 45.