48 questions. Try each one, then open the answer to check yourself and read the explanation.
Arrange the following numbers in ascending order of magnitude 6/7, 13/15, 0.865
Answer: A (6/7 < 0.865 < 13/15)
6/7 ≈ 0.8571 and 13/15 ≈ 0.8667, so the order is 6/7 < 0.865 < 13/15.
A sum of money was invested at 8% per annum simple interest. If after 4 years the money amounts to ₦330.00, find the amount originally invested.
Answer: E (₦250.00)
In 4 years at 8% the interest is 32% of the principal, so 1.32P = 330 and P = ₦250.00.
In the equation below, solve for x if all the numbers are in base 2.
11/x = 1000/(x + 101)
Answer: B (11)
In base ten the equation is 3/x = 8/(x + 5), so 3x + 15 = 8x and x = 3, which is 11 in base 2.
List all integers satisfying the inequality −2 ≤ 2x − 6 < 4
Answer: B (2, 3, 4)
Add 6: 4 ≤ 2x < 10, so 2 ≤ x < 5. The integers are 2, 3 and 4.
Find correct to two decimal places 100 + 1/100 + 3/1000 + 27/10000
Answer: A (100.02)
100 + 0.01 + 0.003 + 0.0027 = 100.0157, which is 100.02 to two decimal places.
Simplify ½ + 1/(2 + 1/(2 − 1/(4 + 1/5)))
Answer: C (169/190)
Work from the bottom: 4 + 1/5 = 21/5, 2 − 5/21 = 37/21, 2 + 21/37 = 95/37, and ½ + 37/95 = (95 + 74)/190 = 169/190.
If three numbers p, q, r are in the ratio 6:4:5, find the value of (3p − q)/(4q + r)
Answer: B (2/3)
Put p = 6, q = 4, r = 5: (18 − 4)/(16 + 5) = 14/21 = 2/3.
Without using tables, evaluate log2 4 + log42 − log255
Answer: E (2.)
log₂4 = 2, log₄2 = ½ (since √4 = 2) and log₂₅5 = ½. So 2 + ½ − ½ = 2.
John gives one-third of his money to Janet who had ₦105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money John had at first.
Answer: A (₦45.00)
John keeps 2x/3 and Janet has 105 + x/3. 2x/3 = ¼(105 + x/3) gives 8x = 315 + x, so x = 45: ₦45.00.
Find x if log9 x = 1.5
Answer: B (27.0)
x = 9^1.5 = (√9)³ = 3³ = 27.
Write h in terms of a, b, c, d if a = b(1 − ch)/(1 − dh)
Answer: A (h = (a − b)/(ad − bc))
a − adh = b − bch, so a − b = adh − bch = h(ad − bc), giving h = (a − b)/(ad − bc).
22½% of the Nigerian naira is equal to 171/10% of a foreign currency M. What is the conversion rate of the M to the naira?
Answer: C (1M = 118/57₦)
0.225 naira = 0.171 M, so 1 M = 0.225/0.171 = 225/171 = 1 54/171 = 1 18/57 naira.
Find the values of p for which the equation x2 − (p − 2)x + 2p + 1 = 0 has equal roots.
Answer: A ((0, 12))
Equal roots need b² = 4ac: (p − 2)² = 4(2p + 1), i.e. p² − 12p = 0, so p = 0 or 12.
If ex = 1 + x + x2/1.2 + x3/1.2.3 + … Find 1/e½
Answer: A (1 − x/2 + x2/1.23 − x3/24.3 + …)
Replace x by −x/2 in the series: 1 − x/2 + (x/2)²/2 − (x/2)³/6 + … = 1 − x/2 + x²/8 − x³/48 + …, and 1.2³ = 8, 2⁴.3 = 48.
(∜3 + ∜2)(∜3 − ∜2)(√3 + √2) is equal to
Answer: E (1.)
The first two brackets give (∜3)² − (∜2)² = √3 − √2. Then (√3 − √2)(√3 + √2) = 3 − 2 = 1.
In a restaurant, the cost of providing a particular type of food is partly constant and partly inversely proportional to the number of people. If the cost per head for 100 people is 30k and the cost for 40 people is 60k, find the cost for 50 people.
Answer: D (50k)
Cost per head C = a + k/N. 30 = a + k/100 and 60 = a + k/40 give k = 2000 and a = 10. For 50 people: 10 + 2000/50 = 50k.
The factors of 9 − (x2 − 3x − 1)2 are
Answer: A (−(x − 4)(x + 1)(x − 1)(x − 2))
Difference of two squares: (3 − x² + 3x + 1)(3 + x² − 3x − 1) = −(x² − 3x − 4)(x² − 3x + 2) = −(x − 4)(x + 1)(x − 1)(x − 2).
If 32y − 6(3y) = 27, find y
Answer: C (2)
Put u = 3ʸ: u² − 6u − 27 = 0, so (u − 9)(u + 3) = 0. 3ʸ cannot be negative, so 3ʸ = 9 and y = 2.
Factorize abx2 + 8y − 4bx − 2axy
Answer: A ((ax − 4)(bx − 2y))
Group the terms: bx(ax − 4) − 2y(ax − 4) = (ax − 4)(bx − 2y).
At what real value of x do the curves whose equations are y = x3 + x and y = x2 + 1 intersect?
Answer: E (1.)
x³ + x = x² + 1 gives x(x² + 1) − (x² + 1) = (x − 1)(x² + 1) = 0. Since x² + 1 is never zero, x = 1.
If the quadratic function 3x2 − 7x + R is a perfect square, find R
Answer: D (49/12)
A perfect square has b² = 4ac: 49 = 4 × 3 × R, so R = 49/12.
Solve the following equation 2/(2r − 1) − 5/3 = 1/(r + 2)
Answer: B ((1, −5/2))
2/(2r − 1) − 1/(r + 2) = 5/((2r − 1)(r + 2)) = 5/3, so (2r − 1)(r + 2) = 3, i.e. 2r² + 3r − 5 = 0. Then (r − 1)(2r + 5) = 0 and r = 1 or −5/2.
Solve for (x,y) in the equations 2x + y = 4; x2 + xy = −12
Answer: A ((6, −8); (−2, 8))
y = 4 − 2x, so x² + x(4 − 2x) = −12, i.e. x² − 4x − 12 = 0 and x = 6 or −2. Then y = −8 or 8.
If f(x − 2) = 4x2 + x + 7 find f(1)
Answer: D (46)
f(1) needs x − 2 = 1, so x = 3: 4(9) + 3 + 7 = 46.
In ΔXYZ, XY = 13cm, YZ = 9cm, XZ = 11cm and XYZ = θ°. Find cos θ°
Answer: E (43/78.)
By the cosine rule, cos θ = (13² + 9² − 11²)/(2 × 13 × 9) = 129/234 = 43/78.
Find the missing value in the table below
x: −4, −3, −2, −1, 0, 1, 2, 3
y = 4 − 3x − x3: 80, ___, 18, 8, 4, 0, 10, 32
Answer: C (40)
At x = −3: y = 4 − 3(−3) − (−3)³ = 4 + 9 + 27 = 40.
The number of goals scored by a football team in 20 matches is shown below
No. of goals: 0, 1, 2, 3, 4, 5
No. of matches: 3, 5, 7, 4, 1, 0
What are the values of the mean and the mode respectively?
Answer: B ((1.75, 2))
Total goals = 0 + 5 + 14 + 12 + 4 = 35 in 20 matches, so the mean is 1.75. 2 goals happened most often (7 times), so the mode is 2.
If the hypotenuse of a right angled isosceles triangle is 2, what is the length of each of the other sides?
Answer: A (√2)
With equal sides x: x² + x² = 2², so x² = 2 and x = √2.
If two fair coins are tossed, what is the probability of getting at least one head?
Answer: E (3/4)
The only outcome with no head is TT, with probability ¼. So P(at least one head) = 1 − ¼ = ¾.
The ratio of the length of two similar rectangular blocks is 2:3. If the volume of the larger block is 351cm3, then the volume of the other block is
Answer: E (104.00cm3.)
Lengths 2:3 means volumes 2³ : 3³ = 8 : 27. So the smaller volume is 351 × 8/27 = 104cm³.
The bearing of a bird on a tree from a hunter on the ground is N72°E. What is the bearing of the hunter from the bird?
Answer: B (S72°W)
A back bearing points the opposite way: north becomes south and east becomes west, with the same angle. So N72°E becomes S72°W.
In ΔXYZ above, XKZ = 90°, XK = 15cm, XZ = 25cm and YK = 8cm. Find the area of ΔXYZ.
Answer: B (210sq.cm)
KZ = √(25² − 15²) = 20cm, so YZ = 8 + 20 = 28cm. Area = ½ × 28 × 15 = 210sq.cm.
Without using tables, calculate the value of 1 + sec2 30°.
Answer: A (2⅓)
sec 30° = 2/√3, so sec² 30° = 4/3 and 1 + 4/3 = 2⅓.
What is the probability that a number chosen at random from the integers between 1 and 10 inclusive is either a prime or a multiple of 3?
Answer: B (3/5)
Primes: 2, 3, 5, 7. Multiples of 3: 3, 6, 9. Together (counting 3 once) that is 6 numbers out of 10, so the probability is 3/5.
Find the area of a regular hexagon inscribed in a circle of radius 8cm.
Answer: B (96√3cm2)
The hexagon is six equilateral triangles of side 8cm: 6 × (√3/4) × 64 = 96√3cm².
In the figure above, MNQP is a cyclic quadrilateral. MN and PQ are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86° and NQP = 122°, find (x°, y°).
Answer: A ((28°, 36°))
Opposite angles of a cyclic quadrilateral add to 180°, so <M = 58° and <MPQ = 94°. In triangle MXP, x = 180° − 58° − 94° = 28°. In triangle MNY, y = 180° − 58° − 86° = 36°.
If cos θ = √3/2 and θ is less than 90°, calculate cot (90 − θ)/sin2 θ
Answer: A (4√3/3)
θ = 30°. cot 60° = 1/√3 and sin² 30° = ¼, so the value is 4/√3 = 4√3/3.
A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?
Answer: C (19kg)
Mass ∝ r³. The shell's volume is proportional to 3³ − 2³ = 19 and the solid sphere's to 4³ = 64, so the mass is 64 × 19/64 = 19kg.
In the figure above POQ is the diameter of the circle PQRS. If PSR = 145°, find x°
Answer: D (55°)
PQRS is cyclic, so <PQR = 180° − 145° = 35°. <PRQ = 90° (angle in a semicircle), so x = <QPR = 180° − 90° − 35° = 55°.
In the figure above GHIJKLMN is a cube of side a. Find the length of HN.
Answer: A (√3a)
HN goes from a bottom front corner to the opposite top back corner. HN² = a² + a² + a² = 3a², so HN = √3a.
PQRS is a trapezium of area 14cm2 in which PQ∥RS. If PQ = 4cm and SR = 3cm, find the area of ΔSQR in cm2.
Answer: B (6.0)
½(4 + 3)h = 14 gives h = 4cm. Triangle SQR has base SR = 3cm and the same height: ½ × 3 × 4 = 6.0cm².
A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?
Answer: A (1/3)
P(first red) = 6/10 and P(second red) = 5/9, so P(both red) = 6/10 × 5/9 = 1/3.
How many 2√2cm diameter discs can be cut out of a sheet of cardboard 218√2π¾ cm long and √2π¼cm wide?
Answer: A (49)
Divide the areas: the sheet is 2¹⁸√2π^¾ × √2π^¼ = 2¹⁹π cm² and each disc is π(√2)² = 2π cm², giving 2¹⁸ = 4⁹ discs.
Two points X and Y both on latitude 60°S have longitudes 147°E and 153°W respectively. Find to the nearest kilometre the distance between X and Y measured along the parallel of latitude. (Take 2πR = 4 × 104km, where R is the radius of the earth).
Answer: E (3,333km.)
Going across the 180° line, the angle between 147°E and 153°W is 33° + 27° = 60°. The parallel at 60° has length 2πR cos 60° = 20 000km, so the distance is 60/360 × 20 000 ≈ 3,333km.
In the figure above the area of the shaded segment is
Answer: C (3(π − 3√3/4))
Sector = 120/360 × π × 3² = 3π. Triangle = ½ × 3 × 3 × sin 120° = 9√3/4. Segment = 3π − 9√3/4 = 3(π − 3√3/4).
In a class of 120 students, 18 of them scored an A grade in Mathematics. If the section representing the A grade students on a pie chart has angle Z° at the centre of the circle, what is Z?
Answer: E (54.)
Z = 18/120 × 360° = 54°.
In the figure above find angle x
Answer: E (140°.)
In this arrowhead shape the angle x at the inner corner equals the sum of the other three angles: 20° + 80° + 40° = 140°.
If a{(x + 1)/(x − 2) − (x − 1)/(x + 2)} = 6x, find a in its simplest form.
Answer: E (x2 − 4.)
The bracket is ((x + 1)(x + 2) − (x − 1)(x − 2))/(x² − 4) = 6x/(x² − 4). So a × 6x/(x² − 4) = 6x and a = x² − 4.