49 questions. Try each one, then open the answer to check yourself and read the explanation.
Simplify ((2/3 − 1/5) − 1/3 of 2/5)/(3 − 1/1½)
Answer: A (1/7)
Top: 2/3 − 1/5 = 7/15, and 1/3 of 2/5 = 2/15, so the top is 5/15 = 1/3. Bottom: 3 − 2/3 = 7/3. Then 1/3 ÷ 7/3 = 1/7.
If 263 + 441 = 714, what number base has been used?
Answer: D (9)
Units: 3 + 1 = 4. Next column: 6 + 4 = 10 is written as 1 carry 1, so 10 = base + 1 and the base is 9. Check: 2 + 4 + 1 = 7.
0.0001432/1 940 000 = K × 10n where 1 ≤ K < 10 and n is a whole number. The values of K and n are
Answer: A (7.381 and −11)
1.432 × 10⁻⁴ ÷ 1.94 × 10⁶ = 0.7381 × 10⁻¹⁰ = 7.381 × 10⁻¹¹.
P sold his bicycle to Q at a profit of 10%. Q sold it to R for ₦209 at a loss of 5%. How much did the bicycle cost P?
Answer: A (₦200)
Q paid 209 ÷ 0.95 = ₦220. That was 110% of P's cost, so P paid 220 ÷ 1.1 = ₦200.
If the price of oranges was raised by ½k per orange, the number of oranges a customer can buy for ₦2.40 will be less by 16. What is the present price of an orange?
Answer: A (2½k)
With price x kobo: 240/x − 240/(x + ½) = 16. This simplifies to 2x² + x − 15 = 0, i.e. (2x − 5)(x + 3) = 0, so x = 2½k.
A man invested a total of ₦50,000 in two companies. If these companies pay dividend of 6% and 8% respectively, how much did he invest at 8% if the total yield is ₦3,700?
Answer: E (₦35,000.)
If all ₦50,000 earned 6% the yield would be ₦3,000. The extra ₦700 comes from 2% on the 8% part, so that part is 700 ÷ 0.02 = ₦35,000.
Thirty boys and x girls sat for a test. The mean of the boys' scores and that of the girls were respectively 6 and 8. Find x if the total score was 468.
Answer: C (36)
The boys scored 30 × 6 = 180 in total, so the girls scored 468 − 180 = 288. 288 ÷ 8 = 36 girls.
The cost of production of an article is made up as follows: Labour ₦70, Power ₦15, Materials ₦30, Miscellaneous ₦5. Find the angle of the sector representing labour in a pie chart.
Answer: A (210°)
The total cost is ₦120. Labour's angle = 70/120 × 360° = 210°.
Bola chooses at random a number between 1 and 300. What is the probability that the number is divisible by 4?
Answer: B (1/4)
Every fourth number is divisible by 4 (4, 8, …, 300), which is 75 numbers out of 300, so the probability is ¼.
Find, without using logarithm tables, the value of (log327 − log¼64)/log3(1/81)
Answer: C (−3/2)
log₃27 = 3, log_¼ 64 = −3 (since (¼)⁻³ = 64) and log₃(1/81) = −4. So (3 + 3)/(−4) = −3/2.
A variable point P(x,y) traces a graph in a two dimensional plane. (0,−3) is one position of P. If x increases by 1 unit, y increases by 4 units the equation of the graph is
Answer: D (y + 3 = 4x)
The gradient is 4 and the line passes through (0, −3), so y = 4x − 3, i.e. y + 3 = 4x.
A trader in a country where their currency 'MONT' (M) is in base five bought 103(5) oranges at M14(5) each, what will be his gain? If he sells them at M24 each
Answer: B (M1030(5))
In base ten: 103₅ = 28 oranges, 14₅ = 9 and 24₅ = 14, so the gain is 28 × 5 = 140. In base five 140 = 1030₅.
Rationalize (5√7 − 7√5)/(√7 − √5)
Answer: C (−√35)
Multiply top and bottom by √7 + √5. The bottom becomes 2 and the top 35 + 5√35 − 7√35 − 35 = −2√35, so the result is −√35.
Simplify (3n − 3n−1)/(33 × 3n − 27 × 3n−1)
Answer: C (1/27)
27 × 3ⁿ − 27 × 3ⁿ⁻¹ = 27(3ⁿ − 3ⁿ⁻¹). Cancelling (3ⁿ − 3ⁿ⁻¹) leaves 1/27.
P varies directly as the square of q and inversely as r. If p = 36 when q = 3 and r = 4, find p when q = 5 and r = 2
Answer: D (200)
p = kq²/r. 36 = 9k/4 gives k = 16. Then p = 16 × 25/2 = 200.
Factorize 6x2 − 14x − 12
Answer: C (2(x − 3)(3x + 2))
Take out 2: 2(3x² − 7x − 6) = 2(3x + 2)(x − 3).
A straight line y = mx meets the curve y = x2 − 12x + 40 in two distinct points. If one of them is (5,5) find the other.
Answer: B ((8,8))
(5,5) on y = mx gives m = 1. Then x = x² − 12x + 40, i.e. x² − 13x + 40 = 0, so x = 5 or 8. The other point is (8,8).
The table below is drawn for a graph y = x3 − 3x + 1
x: −3, −2, −1, 0, 1, 2, 3
y: 1, −1, 3, 1, −1, 3, 1
From x = −2 to x = 1, the graph crosses the x-axis in the range(s)
Answer: B (−2 < x < −1 and 0 < x < 1)
y changes sign between x = −2 (−1) and x = −1 (3), and between x = 0 (1) and x = 1 (−1). So the graph crosses the x-axis in −2 < x < −1 and 0 < x < 1.
In a racing competition Musa covered a distance of 5x km in the first hour and (x + 10)km in the next hour. He was second to Ngozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?
Answer: E (0 ≤ x < 18.)
Musa covered less than Ngozi: 5x + x + 10 < 118, so 6x < 108 and x < 18. Distances cannot be negative, so 0 ≤ x < 18.
If 2x + 3y = 1 and x − 2y = 11, find (x + y)
Answer: D (2)
From the second equation x = 11 + 2y. Then 22 + 4y + 3y = 1, so y = −3 and x = 5. x + y = 2.
Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?
Answer: C (x2 − 21x − 270 = 0)
In one day they do 1/x + 1/(x + 15) = 1/18 of the work. Multiplying out: 18(2x + 15) = x(x + 15), which gives x² − 21x − 270 = 0.
The quadratic equation whose roots are 1 − √13 and 1 + √13 is
Answer: E (x2 − 2x − 12 = 0.)
The roots add to 2 and multiply to 1 − 13 = −12, so the equation is x² − 2x − 12 = 0.
Find a factor which is common to all three binomial expressions 4a2 − 9b2, 8a3 + 27b3, (4a + 6b)2
Answer: C (2a + 3b)
4a² − 9b² = (2a − 3b)(2a + 3b), 8a³ + 27b³ = (2a + 3b)(4a² − 6ab + 9b²) and (4a + 6b)² = 4(2a + 3b)². All contain 2a + 3b.
What is the volume of the regular three dimensional figure drawn above?
Answer: B (48cm3)
The end is a right-angled triangle with hypotenuse 5cm and side 4cm, so its other side is 3cm and its area is ½ × 3 × 4 = 6cm². Volume = 6 × 8 = 48cm³.
If (x − 2) and (x + 1) are factors of the expression x3 + px2 + qx + 1, what is the sum of p and q?
Answer: B (−3)
x = 2: 8 + 4p + 2q + 1 = 0. x = −1: −1 + p − q + 1 = 0, so p = q. Then 6p = −9, p = q = −3/2 and p + q = −3.
A cone is formed by bending a sector of a circle having an angle of 210°. Find the radius of the base of the cone if the diameter of the circle is 12cm.
Answer: D (3.50cm)
The arc becomes the base circumference: 2πr = 210/360 × 2π × 6, so r = 210/360 × 6 = 3.5cm.
Using ΔXYZ in the figure above find XYZ.
Answer: D (31°18')
By the sine rule, sin Y/3 = sin 120°/5, so sin Y = 3 × 0.866/5 ≈ 0.5196 and Y ≈ 31°18'.
The sides of a triangle are (x + 4)cm, x cm and (x − 4)cm respectively. If the cosine of the largest angle is 1/5, find the value of x.
Answer: A (24cm)
The largest angle faces x + 4. Cosine rule: (x + 4)² = x² + (x − 4)² − 2x(x − 4)/5. Multiplying by 5 and simplifying gives 3x² − 72x = 0, so x = 24.
If a = 2x/(1 − x) and b = (1 + x)/(1 − x) then a2 − b2 in the simplest form is
Answer: A ((3x + 1)/(x − 1))
a² − b² = (4x² − (1 + x)²)/(1 − x)² = (3x² − 2x − 1)/(1 − x)² = (3x + 1)(x − 1)/(x − 1)² = (3x + 1)/(x − 1).
Simplify (1 + (x − 1)/(1/(x + 1)))(x + 2)
Answer: C (x2(x + 2))
Dividing by 1/(x + 1) multiplies by (x + 1): 1 + (x − 1)(x + 1) = 1 + x² − 1 = x². So the expression is x²(x + 2).
In the figure above PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS.
Answer: A (60°)
Each interior angle is 120°. Triangles QRS and RST are isosceles, so <RSQ = <SRT = 30°. Then <RVS = 120° and <TVS = 180° − 120° = 60°.
Find all integral values of x which satisfy the inequalities −3 < 2 − 5x < 12
Answer: C (−1, 0)
Subtract 2: −5 < −5x < 10. Divide by −5 and reverse the signs: −2 < x < 1. The integers are −1 and 0.
Find the area of the shaded portion of the semi-circular figure above.
Answer: B (r2/4 (2π − 3√3))
The semicircle is three 60° sectors, each minus an equilateral triangle. Each segment is πr²/6 − (√3/4)r², so three of them give r²/4(2π − 3√3).
In the figure above, QRS is a line, PSQ = 35°, SPR = 30° and O is the centre of the circle. Find OQP.
Answer: D (25°)
<PRQ is an exterior angle of triangle PRS: 35° + 30° = 65°. The angle at the centre <POQ = 130°, and triangle OPQ is isosceles, so <OQP = (180° − 130°)/2 = 25°.
If pq + 1 = q2 and t = 1/p − 1/pq express t in terms of q.
Answer: C (1/(q + 1))
t = (q − 1)/pq, and pq = q² − 1. So t = (q − 1)/((q − 1)(q + 1)) = 1/(q + 1).
The cumulative frequency function of the data below is given by the equation y = cf(x). What is cf(5)?
Scores (n): 3, 4, 5, 6, 8
Frequency (f): 30, 32, 30, 35, 20
Answer: E (92.)
cf(5) counts all scores up to 5: 30 + 32 + 30 = 92.
In the figure below, determine the angle marked y
Answer: A (66°)
QTSR is cyclic, so the exterior angle <PQT = 70° equals the opposite interior angle <TSR. In triangle PRS, y = 180° − 44° − 70° = 66°.
A right circular cone has a base radius r cm and a vertical angle 2y°. The height of the cone is
Answer: C (r cot y° cm)
Half the vertical angle is y, and tan y = r/h. So h = r/tan y = r cot y.
Two fair dice are rolled. What is the probability that both show up the same number of points?
Answer: E (1/6)
Of the 36 equally likely outcomes, 6 are doubles (1-1, …, 6-6). 6/36 = 1/6.
The larger value of y for which (y − 1)2 = 4y − 7 is
Answer: B (4)
y² − 2y + 1 = 4y − 7 gives y² − 6y + 8 = 0, so (y − 2)(y − 4) = 0. The larger value is 4.
Find the x co-ordinates of the points of intersection of the two equations in the graph above.
Answer: E (0, 4.)
Set 2x + 1 = x² − 2x + 1: x² − 4x = 0, so x = 0 or 4.
If sin θ = x/y and 0° < θ < 90° then find 1/tan θ
Answer: C (√(y2 − x2)/x)
With opposite side x and hypotenuse y, the adjacent side is √(y² − x²). 1/tan θ = adjacent/opposite = √(y² − x²)/x.
In the figure above, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP.
Answer: C (9cm)
Triangles PRQ and PST share angle P and have <PRQ = <PST, so they are similar: PR/PS = QR/TS. 6/PS = 8/12 gives PS = 9cm.
The bar chart above shows the mark distribution in a class test. Find the number of students in the class.
Answer: E (34)
The number of students is the sum of the bar heights (frequencies) for all the mark groups.
In the figure above, O is the centre of circle PQRS and PS∥RT. If PRT = 135°, then PSO is
Answer: B (45°)
PS ∥ RT, so <SPR = 180° − 135° = 45° (co-interior angles). OP = OS, so <PSO = <SPO = 45°.
XYZ is a triangle and XW is perpendicular to YZ at W. If XZ = 5cm and WZ = 4cm, calculate XY.
Answer: B (3√5cm)
XW = √(5² − 4²) = 3cm and YW = 10 − 4 = 6cm. XY = √(6² + 3²) = √45 = 3√5cm.
Measurements of the diameters, in centimeters, of 20 copper spheres are distributed as shown below:
Class boundary in cm: 3.35 − 3.45, 3.45 − 3.55, 3.55 − 3.65, 3.65 − 3.75
Frequency: 3, 6, 7, 4
What is the mean diameter of the copper sphere?
Answer: C (3.56cm)
Use the class mid-points 3.4, 3.5, 3.6, 3.7: Σfx = 10.2 + 21.0 + 25.2 + 14.8 = 71.2, and 71.2 ÷ 20 = 3.56cm.
Use the construction (Fig. 1) to answer this question. What is the obtuse angle formed when the point U is joined to Q?
Answer: D (105°)
QT is the perpendicular to PQ at Q (90°). The arcs through T and S bisect a further angle, putting QU 15° beyond QT, so the obtuse angle PQU is 90° + 15° = 105°.
Use the construction (Fig. 1) to answer this question. What is the acute angle formed when the point V is joined to Q?
Answer: B (30°)
The arcs at V mark a bisected angle on the P side of the perpendicular, so the line QV makes 30° with QP.