46 questions. Try each one, then open the answer to check yourself and read the explanation.
Find the value of 1101112 + 101002
Answer: D (10010112)
1101112 = 55 and 101002 = 20, so the sum is 75. In base two, 75 = 64 + 8 + 2 + 1 = 10010112.
A woman bought a grinder for ₦60,000. She sold it at a loss of 15%. How much did she sell it?
Answer: D (₦51,000)
15% of ₦60,000 is ₦9,000. Selling price = ₦60,000 − ₦9,000 = ₦51,000.
Express the product of 0.00043 and 2000 in standard form.
Answer: D (8.6 × 10-1)
0.00043 × 2000 = 0.86, and 0.86 = 8.6 × 10-1.
A man donates 10% of his monthly net earnings to his church. If it amounts to ₦4,500, what is his net monthly income?
Answer: C (₦45,000)
If 10% is ₦4,500, then 100% is 10 × ₦4,500 = ₦45,000.
If log 7.5 = 0.8751, evaluate 2 log 75 + log 750.
Answer: C (6.6253)
log 75 = 1.8751 and log 750 = 2.8751. So 2(1.8751) + 2.8751 = 3.7502 + 2.8751 = 6.6253.
Solve for x: 8x-2 = 2/25
Answer: A (10)
x-2 = 2/(25 × 8) = 1/100, so x2 = 100 and x = 10.
Evaluate log2 8 + log2 16 − log2 4.
Answer: D (5)
log2 8 = 3, log2 16 = 4 and log2 4 = 2, so the value is 3 + 4 − 2 = 5.
If P = {1,2,3,4,5} and P∪Q = {1,2,3,4,5,6,7}, list the elements in Q.
Answer: D ({6, 7})
6 and 7 are in the union but not in P, so they must come from Q. The only option containing both is {6, 7}.
From the venn diagram above, the shaded parts represent
Answer: B ((P∩Q)∪(P∩R))
The shading covers the region common to P and Q together with the region common to P and R. Joining two regions is a union, so the answer is (P∩Q)∪(P∩R).
If gt2 − k − w = 0, make g the subject of the formula.
Answer: B ((k + w)/t2)
Move k and w across: gt2 = k + w. Divide by t2: g = (k + w)/t2.
Factorise 2y2 − 15xy + 18x2.
Answer: C ((2y − 3x)(y − 6x))
(2y − 3x)(y − 6x) = 2y2 − 12xy − 3xy + 18x2 = 2y2 − 15xy + 18x2.
Find the value of k if y − 1 is a factor of y3 + 4y2 + ky − 6.
Answer: D (1)
If y − 1 is a factor, putting y = 1 gives zero: 1 + 4 + k − 6 = 0, so k = 1.
y varies directly as w2. When y = 8, w = 2, find y when w = 3.
Answer: B (18)
y = kw2; 8 = 4k gives k = 2. When w = 3, y = 2 × 9 = 18.
P varies directly as Q and inversely as R. When Q = 36 and R = 16, P = 27. Find the relation between P, Q and R.
Answer: C (P = 12Q/R)
P = kQ/R, so 27 = 36k/16, giving k = 12. Hence P = 12Q/R.
What is the solution of (x − 5)/(x + 3) < −1?
Answer: B (−3 < x < 1)
Add 1 to both sides: (x − 5 + x + 3)/(x + 3) < 0, i.e. (2x − 2)/(x + 3) < 0. The top and bottom must have opposite signs, which happens only when −3 < x < 1.
Evaluate the inequality x/2 + 3/4 ≤ 5x/6 − 7/12
Answer: B (x ≥ 4)
Multiply through by 12: 6x + 9 ≤ 10x − 7. Then 16 ≤ 4x, so x ≥ 4.
The 4th term of an A.P. is 13 while the 10th term is 31. Find the 24th term.
Answer: D (73)
a + 3d = 13 and a + 9d = 31, so 6d = 18, d = 3 and a = 4. The 24th term is a + 23d = 4 + 69 = 73.
What is the common ratio of the G.P. (√10 + √5) + (√10 + 2√5) + …?
Answer: B (√2)
r = (√10 + 2√5)/(√10 + √5). Multiply top and bottom by (√10 − √5): the top is 10 − √50 + 2√50 − 10 = √50 = 5√2 and the bottom is 5, so r = √2.
A binary operation * is defined by x * y = xy. If x * 2 = 12 − x, find the possible values of x.
Answer: C (3, −4)
x * 2 = x2, so x2 = 12 − x, i.e. x2 + x − 12 = 0 = (x − 3)(x + 4). So x = 3 or −4.
Find y, if (5 −6 / 2 −7)(x / y) = (7 / −11)
(the first matrix has rows 5, −6 and 2, −7; the column vectors are (x, y) and (7, −11))
Answer: D (3)
The equations are 5x − 6y = 7 and 2x − 7y = −11. Doubling the first and multiplying the second by 5 gives 10x − 12y = 14 and 10x − 35y = −55; subtracting, 23y = 69, so y = 3.
If the determinant |−x 12 / −1 4| = −12, find x.
(rows: −x, 12 and −1, 4)
Answer: A (6)
The determinant is (−x)(4) − (12)(−1) = −4x + 12. Setting −4x + 12 = −12 gives x = 6.
Find the value of the determinant with rows (0 3 2), (1 7 8), (0 5 4).
Answer: A (−2)
Expanding along the first row: 0(28 − 40) − 3(4 − 0) + 2(5 − 0) = 0 − 12 + 10 = −2.
How many sides has a regular polygon whose interior angle is 135° each?
Answer: A (8)
Exterior angle = 180° − 135° = 45°. Number of sides = 360° ÷ 45° = 8.
In the figure above, KL∥NM, LN bisects ∠KNM. If angle KLN is 54° and angle MKN is 35°, calculate the size of angle KMN.
Answer: D (37°)
Since KL∥NM, ∠LNM = ∠KLN = 54° (alternate angles). LN bisects ∠KNM, so ∠KNM = 2 × 54° = 108°. In triangle KMN, ∠KMN = 180° − 108° − 35° = 37°.
From the figure above, what is the value of p?
Answer: C (90°)
q° and 30° are vertically opposite, so q = 30. (p + 2q)° and 30° lie on a straight line, so p + 60 + 30 = 180, giving p = 90°.
Find the value of x in the figure above.
Answer: C (10√3cm)
The third angle is 90°. By the sine rule, x/sin 60° = 10/sin 30°, so x = 10 × (√3/2) ÷ (1/2) = 10√3 cm.
If the angle of a sector of a circle with radius 10.5 cm is 120°, find the perimeter of the sector. [π = 22/7]
Answer: C (43cm)
Arc length = (120/360) × 2 × (22/7) × 10.5 = 22 cm. Perimeter = 22 + 2 × 10.5 = 43 cm.
A cylindrical tank has a capacity of 6160 m3. What is the depth of the tank if the radius of its base is 28m? [π = 22/7]
Answer: A (2.5 m)
V = πr2h, so h = 6160 ÷ ((22/7) × 784) = 6160 ÷ 2464 = 2.5 m.
The locus of a dog tethered to a pole with a rope of 4 m is a
Answer: C (circle with radius 4 m)
With the rope pulled tight, the dog is always 4 m from the pole. All points 4 m from a fixed point form a circle of radius 4 m.
Find the mid point of S(−5,4) and T(−3,−2)
Answer: D (−4, 1)
Midpoint = ((−5 + −3)/2, (4 + −2)/2) = (−4, 1).
The gradient of the line joining (x,4) and (1,2) is ½. Find the value of x.
Answer: B (5)
Gradient = (4 − 2)/(x − 1) = ½, so x − 1 = 4 and x = 5.
In the figure above, what is the equation of the line that passes the y-axis at (0,5) and passes the x-axis at (5,0)?
Answer: C (y = −x + 5)
Gradient = (0 − 5)/(5 − 0) = −1 and the line cuts the y-axis at 5. So y = −x + 5.
Calculate the mid point of the line segment y − 4x + 3 = 0 which lies between the x-axis and y-axis.
Answer: B ((3/8, −3/2))
The line meets the y-axis where x = 0, at (0, −3), and the x-axis where y = 0, at (3/4, 0). The midpoint is (3/8, −3/2).
Find the equation of the straight line through (−2,3) and perpendicular to 4x + 3y − 5 = 0
Answer: B (3x − 4y + 18 = 0)
The given line has gradient −4/3, so the perpendicular gradient is 3/4. y − 3 = ¾(x + 2) gives 4y − 12 = 3x + 6, i.e. 3x − 4y + 18 = 0.
If sin θ = 12/13, find the value of 1 + cos θ.
Answer: C (18/13)
With opposite 12 and hypotenuse 13, the adjacent side is 5, so cos θ = 5/13. Then 1 + 5/13 = 18/13.
If y = 4x3 − 2x2 + x, find dy/dx.
Answer: A (12x2 − 4x + 1)
Differentiate term by term: 4 × 3x2 − 2 × 2x + 1 = 12x2 − 4x + 1.
If y = cos 3x, find dy/dx.
Answer: A (−3sin 3x)
The derivative of cos u is −sin u × du/dx. Here u = 3x, so dy/dx = −3 sin 3x.
Find the minimum value of y = x2 − 2x − 3
Answer: A (−4)
dy/dx = 2x − 2 = 0 at x = 1. The minimum value is y = 1 − 2 − 3 = −4.
Evaluate ∫sin 2x dx.
Answer: D (−½cos 2x + k)
The integral of sin ax is −(1/a) cos ax. With a = 2 the answer is −½cos 2x + k.
Evaluate ∫(2x + 3)½ dx.
Answer: D (⅓(2x + 3)3/2 + k)
Raise the power by one to 3/2, divide by 3/2 and by 2 (the coefficient of x): (2x + 3)3/2 ÷ 3 = ⅓(2x + 3)3/2 + k.
The pie chart above shows the monthly distribution of a man's salary on food items. If he spent ₦8,000 on rice, how much did he spend on yam?
Answer: D (₦16,000)
The yam sector is 360° − (70° + 80° + 50°) = 160°. Rice's 80° is ₦8,000, so 160° is twice that: ₦16,000.
The mean of 2−t, 4+t, 3−2t, 2+t and t−1 is
Answer: D (2)
The t terms cancel (−t + t − 2t + t + t = 0) and the numbers add to 10. Mean = 10 ÷ 5 = 2.
Calculate the standard deviation of 5,4,3, 2 and 1.
Answer: B (√2)
The mean is 3. The squared deviations are 4, 1, 0, 1, 4, which add to 10. Variance = 10 ÷ 5 = 2, so the standard deviation is √2.
In how many ways can a team of 3 girls be selected from 7 girls?
Answer: D (7!/(3!4!))
Order does not matter, so use combinations: 7C3 = 7!/(3!4!).
Numbers: 1, 2, 3, 4, 5, 6
Frequency: 18, 22, 20, 16, 10, 14
The table above represents the outcome of throwing a die 100 times. What is the probability of obtaining at least a 4?
Answer: D (2/5)
'At least 4' means 4, 5 or 6: 16 + 10 + 14 = 40 times out of 100. Probability = 40/100 = 2/5.
A number is chosen at random from 10 to 30 both inclusive, what is the probability that the number is divisible by 3?
Answer: D (1/3)
There are 21 numbers from 10 to 30. The multiples of 3 are 12, 15, 18, 21, 24, 27, 30: 7 of them. Probability = 7/21 = 1/3.