45 questions. Try each one, then open the answer to check yourself and read the explanation.
Suppose x varies inversely as y, y varies directly as the square of t and x = 1 when t = 3. Find x when t = 1/3
Answer: A (81)
x ∝ 1/y and y ∝ t², so x = k/t². x = 1 at t = 3 gives k = 9. At t = ⅓, x = 9 ÷ (1/9) = 81.
If sine x equals cosine x, what is x in radians?
Answer: C (π/4)
sin x = cos x means tan x = 1, so x = 45° = π/4 radians.
The ratio of the price of a loaf of bread to the price of a packet of sugar in 1975 was r:t. In 1980 the price of a loaf went up by 25% and that of a packet of sugar by 10%. Their new ratio is now
Answer: C (50r : 44t)
The new prices are 1.25r and 1.10t. 1.25r : 1.10t = 125r : 110t = 50r : 44t.
Find a two-digit number such that three times the tens digit is 2 less than twice the units digit, and twice the number is 20 greater than the number obtained by reversing the digits.
Answer: D (47)
With tens digit x and units digit y: 3x = 2y − 2 and 2(10x + y) = 10y + x + 20, i.e. 19x − 8y = 20. Solving gives x = 4, y = 7, so the number is 47.
Find the value of x satisfying x/2 − 1/3 < 2x/5 + 1/6
Answer: A (x < 5)
Collect terms: x/2 − 2x/5 < 1/6 + 1/3, so x/10 < 1/2 and x < 5.
If sin x° = a/b, what is sin (90 − x)°?
Answer: C (√(b2 − a2)/b)
sin(90 − x) = cos x. In a right triangle with opposite side a and hypotenuse b, the adjacent side is √(b² − a²), so cos x = √(b² − a²)/b.
7 pupils of average age 12 years leave a class of 25 pupils of average age 14 years. If 6 new pupils of average age 11 years join the class, what is the average age of the pupils now in the class?
Answer: D (13 years 10 months)
Total age = 25 × 14 − 7 × 12 + 6 × 11 = 350 − 84 + 66 = 332 years for 24 pupils. 332 ÷ 24 ≈ 13.83 years = 13 years 10 months.
A sum of money invested at 5% per annum simple interest amounts to $285.20 after 3 years. How long will it take the same sum to amount to $434.00 at 7½% per annum simple interest?
Answer: B (10 years)
After 3 years at 5% the amount is 1.15P = 285.20, so P = $248. The interest needed is 434 − 248 = $186, and 248 × 7.5% = $18.60 a year, so it takes 186 ÷ 18.60 = 10 years.
By selling an article for ₦45.00 a man makes a profit of 8%. For how much should he have sold it in order to make a profit of 32%?
Answer: E (₦55.00.)
Cost price = 45 × 100/108. A 32% profit needs 132/100 of that: 45 × 132/108 = ₦55.00.
An isosceles triangle of sides 13cm, 13cm, 10cm is inscribed in a circle. What is the radius of the circle?
Answer: A (71/24cm)
The height to the 10cm side is √(13² − 5²) = 12cm, so the area is 60cm². The circumradius R = abc/(4 × area) = (13 × 13 × 10)/240 = 169/24 = 7 1/24cm.
A group of 14 children received the following scores in a reading test: 35, 35, 26, 26, 26, 29, 29, 29, 12, 25, 25, 25, 25, 17. What was the median score?
Answer: B (26)
In order: 12, 17, 25, 25, 25, 25, 26, 26, 26, 29, 29, 29, 35, 35. The 7th and 8th scores are both 26, so the median is 26.
A micrometre is defined as one millionth of a millimetre. A length of 12,000 micrometres may be represented as
Answer: C (0.000012m)
Using the question's definition, 12,000 micrometres = 12,000 ÷ 1,000,000 mm = 0.012mm, and 0.012mm ÷ 1000 = 0.000012m.
In one and a half hours, the minute hand of a clock rotates through an angle of
Answer: E (540°)
The minute hand turns 360° every hour, so in 1½ hours it turns 1.5 × 360° = 540°.
In the figure above, the chords XY and ZW are produced to meet at T such that YT = WT, ZYW = 40° and YTW = 30°. What is YXW?
Answer: C (35°)
Triangle YTW is isosceles, so <TWY = 75°. The exterior angle of cyclic quadrilateral XYWZ at W equals the opposite interior angle: <YXZ = 75°. <WXZ = <WYZ = 40° (same segment), so <YXW = 75° − 40° = 35°.
In the figure above, find the area of XYZW.
Answer: C (27cm2)
XY ∥ WZ. The height is 6 sin 30° = 3cm, so the area of the trapezium is ½(8 + 10) × 3 = 27cm².
In the figure above, find x in terms of a, b and c.
Answer: A (a + b + c)
Join the top vertex through the inner vertex. Each exterior angle of a triangle equals the sum of the two opposite interior angles, so the two parts of x add up to a + b + c.
Factorize 3x3 + 4x2 − 13x + 6 completely, given that x − 1 is a factor.
Answer: E ((x − 1)(x + 3)(3x − 2).)
Dividing by x − 1 gives 3x² + 7x − 6, which factorises as (x + 3)(3x − 2).
Simplify √2/(√3 − √2) − (√3 − √2)/(√3 + √2)
Answer: B (3(√6 − 1))
√2/(√3 − √2) = √2(√3 + √2) = √6 + 2, and (√3 − √2)/(√3 + √2) = (√3 − √2)² = 5 − 2√6. The difference is 3√6 − 3 = 3(√6 − 1).
Solve 1/(x + 1) − 1/(x + 3) = 1/4
Answer: C (x = 1 or −5)
The left side is 2/((x + 1)(x + 3)). Setting it equal to ¼ gives x² + 4x + 3 = 8, so x² + 4x − 5 = 0 and x = 1 or −5.
In ΔXYZ, XY = 3cm, XZ = 5cm and YZ = 7cm. If the bisector of XYZ meets XZ at W, what is the length of XW?
Answer: A (1.5cm)
The angle bisector divides XZ in the ratio of the adjacent sides: XW : WZ = 3 : 7. So XW = 3/10 × 5 = 1.5cm.
A cuboid has a diagonal of length 9cm and a square base of side 4cm. What is its height?
Answer: D (7cm)
The diagonal satisfies 9² = 4² + 4² + h², so h² = 81 − 32 = 49 and h = 7cm.
Given that 10x = 0.2 and log10 2 = 0.3010, find x.
Answer: B (−0.6990)
x = log₁₀ 0.2 = log₁₀ 2 − log₁₀ 10 = 0.3010 − 1 = −0.6990.
Two cars X and Y start at the same point and travel towards a point P which is 150km away. If the average speed of Y is 60km per hour and X arrives at P 25 minutes earlier than Y. What is the average speed of X?
Answer: B (72km per hour)
Y takes 150 ÷ 60 = 2½ hours, so X takes 2 hours 5 minutes = 125/60 hours. X's speed = 150 ÷ 125/60 = 72km per hour.
Simplify (62n + 1 × 9n × 42n)/(18n × 2n × 122n)
Answer: D (6)
Count the powers. Of 2: top 2n + 1 + 4n = 6n + 1, bottom n + n + 4n = 6n. Of 3: top 2n + 1 + 2n = 4n + 1, bottom 2n + 2n = 4n. What is left is 2 × 3 = 6.
Evaluate correct to 4 decimal places 827.51 × 0.015.
Answer: B (12.4127)
827.51 × 0.015 = 12.41265, which is 12.4127 to 4 decimal places.
The number 25 when converted from the tens and units base to the binary base (base two) is one of the following
Answer: D (11001)
25 = 16 + 8 + 1, so in base two it is 11001.
Evaluate (6.3 × 105)/(8.1 × 103) to 3 significant figures
Answer: A (77.80)
6.3 ÷ 8.1 = 0.7778 and 10⁵ ÷ 10³ = 100, giving 77.78, or 77.8 to three significant figures.
Marks scored by some children in an arithmetic test are 5,3,6,9,4,7,8,6,2,7,8,4,5,2,1,0,6,9,0,8. The arithmetic mean of the marks is
Answer: B (5)
The 20 marks add up to 100, so the mean is 100 ÷ 20 = 5.
The weights of 30 new-born babies are given as follows: 6,9,5,7,6,7,5,8,9,5,7,5,8,7,8,7,5,6,5,7, 6,9,9,7,8,8,7,8,9,8. The mode is
Answer: D (7)
Counting: 5 appears 6 times, 6 four times, 7 eight times, 8 seven times and 9 five times. 7 is the most frequent.
What is the area between two concentric circles of diameters 26cm and 20cm?
Answer: C (69π)
The radii are 13cm and 10cm, so the ring has area π(13² − 10²) = 69π cm².
The positive root of t in the following equation, 4t2 + 7t − 1 = 0, correct to 4 places of decimal is
Answer: C (0.1328)
t = (−7 + √(49 + 16))/8 = (−7 + 8.0623)/8 ≈ 0.1328.
The difference between the length and width of a rectangle is 6cm and the area is 135cm2. What is the length?
Answer: C (15cm)
With width B, B(B + 6) = 135, so B² + 6B − 135 = 0 and B = 9. The length is 9 + 6 = 15cm.
The area of a circular plate is one-sixteenth the surface area of a ball. If the area of the plate is given as Pcm2, then the radius of the ball is
Answer: D (2√(P/π))
P = (1/16) × 4πr² = πr²/4, so r² = 4P/π and r = 2√(P/π).
In the above diagram XZ is the diameter of a circle radius 5/2cm. If XY is 4cm then the area of the triangle XYZ is
Answer: E (6cm2)
XZ = 5cm and the angle at Y is 90° (angle in a semicircle), so YZ = √(25 − 16) = 3cm. Area = ½ × 4 × 3 = 6cm².
Show that sin2x/(1 + cos x) + sin2x/(1 − cos x) is
Answer: E (2.)
Over the common denominator 1 − cos²x = sin²x, the top is sin²x(1 − cos x + 1 + cos x) = 2sin²x. So the sum is 2.
The first term of an Arithmetic Progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum of the terms is 81.
Answer: C (9)
3 + 4d = 9 gives d = 3/2. n/2 × (6 + (n − 1) × 3/2) = 81 simplifies to n² + 3n − 108 = 0, so n = 9.
Multiply x2 + x + 1 by x2 − x + 1.
Answer: B (x4 + x2 + 1)
Write it as ((x² + 1) + x)((x² + 1) − x) = (x² + 1)² − x² = x⁴ + x² + 1.
A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used?
Answer: C (8.75kg)
Sugar is 2.5 out of every 7kg of mixture, so sugar used = 2.5/7 × 24.5 = 8.75kg.
The difference between 45/7 and 21/4 is greater than the sum of 1/14 and 11/2 by
Answer: C (50/56)
4 5/7 − 2 1/4 = 69/28 and 1/14 + 1½ = 44/28. The difference is 25/28 = 50/56.
What is the size of an exterior angle of a regular pentagon?
Answer: C (72°)
The exterior angles of any polygon add to 360°, so each of the 5 equal ones is 360° ÷ 5 = 72°.
In the figure above WU∥YZ, WY∥VU: UZ = 12cm, VZ = 6cm, XU = 8cm. Determine the length of WU.
Answer: E (4cm.)
Triangles XWU and UVZ have parallel sides, so they are similar: WU/VZ = XU/UZ, i.e. WU/6 = 8/12, giving WU = 4cm.
Simplify 1 + 2/(3 − 4/(5 + 6/7))
Answer: A (177/95)
Work from the bottom: 5 + 6/7 = 41/7, 4 ÷ 41/7 = 28/41, 3 − 28/41 = 95/41, 2 ÷ 95/41 = 82/95, and 1 + 82/95 = 177/95.
The sum of the root of a quadratic equation is 5/2 and the product of its roots is 4. The quadratic equation is
Answer: B (2x2 − 5x + 8 = 0)
x² − (sum)x + (product) = 0 gives x² − 5x/2 + 4 = 0, i.e. 2x² − 5x + 8 = 0.
Solve the given equation (log3x)2 − 6log3x + 9 = 0.
Answer: A (27)
Put y = log₃x: y² − 6y + 9 = (y − 3)² = 0, so log₃x = 3 and x = 27.
What is the length of an arc of a circle that subtends 2½ radians at the centre when the radius of the circle is 8cm?
Answer: D (20cm)
With the angle in radians, arc length = radius × angle = 8 × 2½ = 20cm.