47 questions. Try each one, then open the answer to check yourself and read the explanation.
Simplify 3⅓ − 1¼ × ⅔ + 1⅖
Answer: B (3 9/10)
Multiply first: 1¼ × ⅔ = 5/6. Then 10/3 − 5/6 + 7/5 = (100 − 25 + 42)/30 = 117/30 = 3 9/10.
If 2257 is the result of subtracting 4577 from 7056 in base n find n
Answer: A (8)
The digit 7 rules out bases below 8. In base 8, 7056 = 3630 and 4577 = 2431 in base ten; 3630 − 2431 = 1199, and 2257 in base 8 is also 1199. So n = 8.
Find correct to 3 decimal places [1/0.05] ÷ [1/5.005] − [0.05 × 20.5]
Answer: A (99.998)
1/0.05 ÷ 1/5.005 = 20 × 5.005 = 100.1. Taking the second bracket as 0.05 × 2.05 = 0.1025 (as in the book's working), 100.1 − 0.1025 = 99.9975 = 99.998.
Express 62 ÷ 3 as a decimal correct to 3 significants figures.
Answer: D (20.7)
62 ÷ 3 = 20.666…, and to 3 significant figures that is 20.7.
Factory P produces 20,000 bags of cement per day while Q factory produces 15,000 bags per day. If P reduces production by 5% and Q increases production by 5%, determine the effective loss in the number of bags produced per day by the two factories.
Answer: A (250)
P loses 5% of 20,000 = 1,000 bags and Q gains 5% of 15,000 = 750 bags. The overall loss is 1,000 − 750 = 250 bags.
Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after four months. How much does he still owes?
Answer: C (N2.80)
Interest = 10 × 2/100 × 4 = N0.80, so he owes N10.80 in all. After repaying N8.00 he still owes N2.80.
If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water?
Answer: B (16 7/8%)
Water = 20% of 3 + 15% of 5 = 0.6 + 0.75 = 1.35 gallons out of 8 gallons. 1.35/8 × 100 = 16⅞%.
What is the product of 27/5, (3)−3 and (1/5)−1?
Answer: C (1)
(3)−3 = 1/27 and (1/5)−1 = 5, so 27/5 × 1/27 × 5 = 1.
Simplify 2log 2/5 − log 72/125 + log 9
Answer: D (1 − 2 log 2)
Combine: log(4/25 × 125/72 × 9) = log(5/2) = log 10 − log 4 = 1 − 2 log 2.
Rationalize (2√3 + 3√2)/(3√2 − 2√3)
Answer: B (5 + 2√6)
Multiply top and bottom by 3√2 + 2√3. The top becomes (2√3 + 3√2)2 = 30 + 12√6 and the bottom 18 − 12 = 6, giving 5 + 2√6.
Simplify 1/(3 + √5) − 1/(3 − √5)
Answer: A (−½√5)
Over the common denominator (3 + √5)(3 − √5) = 4, the top is (3 − √5) − (3 + √5) = −2√5. So the answer is −2√5/4 = −½√5.
Multiply (x2 − 3x + 1) by (x − a)
Answer: A (x3 − (3 + a) x2 + (1 + 3a) x − a)
Expanding gives x3 − ax2 − 3x2 + 3ax + x − a. Grouping like terms: x3 − (3 + a)x2 + (1 + 3a)x − a.
Evaluate (xy2 − x2y)/(x2 − xy) when x = −2 and y = 3
Answer: A (−3)
The expression is xy(y − x)/[x(x − y)] = −y. With y = 3, the value is −3.
A car travels from Calabar to Enugu, a distance of p km with an average speed of u km per hour and continues to Benin, a distance of q km, with an average speed of w km per hour. Find its average speed from Calabar to Benin.
Answer: C (uw(p + q)/(wp + uq))
Total time = p/u + q/w = (pw + qu)/(uw). Average speed = total distance ÷ total time = uw(p + q)/(wp + uq).
If w varies inversely as uv/(u + v) and is equal to 8 when u = 2 and v = 6, find a relationship between u, v, w
Answer: C (urw = 12(u + r))
w = k(u + v)/(uv). With u = 2, v = 6: 8 = k × 8/12, so k = 12 and uvw = 12(u + v). This is option C, where the book prints r in place of v.
If g(x) = x2 + 3x + 4 find g(x + 1) − g(x)
Answer: B (2(x + 2))
g(x + 1) = (x + 1)2 + 3(x + 1) + 4 = x2 + 5x + 8. Subtracting g(x) leaves 2x + 4 = 2(x + 2).
Factorize m3 − 2m2 − m + 2
Answer: C ((m − 2)(m + 1)(m − 1))
Group: m2(m − 2) − 1(m − 2) = (m2 − 1)(m − 2) = (m − 2)(m + 1)(m − 1).
Factorize 1 − (a − b)2
Answer: B ((1 − a + b)(1 + a − b))
Difference of two squares: [1 − (a − b)][1 + (a − b)] = (1 − a + b)(1 + a − b).
Which of the following is a factor of rs + tr − pt − ps?
Answer: C ((r − p))
Group: r(s + t) − p(t + s) = (s + t)(r − p). So r − p is a factor.
Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x2 + xy = 6
Answer: A (−1 and 5)
From the first equation y = 8 − 3x. Substituting: x2 + 8x − 3x2 = 6, so x2 − 4x + 3 = 0 and x = 1 or 3. Then y = 5 or −1.
Find the range of value of x which satisfy the inequality (x/2 + x/3 + x/4) < 1
Answer: A (x < 12/13)
x/2 + x/3 + x/4 = 13x/12. So 13x/12 < 1 gives x < 12/13.
Find the positive number n, such that thrice its square is equal to twelve times the number.
Answer: D (4)
3n2 = 12n gives n(n − 4) = 0. The positive answer is n = 4.
Solve the equation (x − 2)(x − 3) = 12
Answer: C (−1, 6)
Expand: x2 − 5x + 6 = 12, so x2 − 5x − 6 = 0, i.e. (x − 6)(x + 1) = 0. So x = −1 or 6.
Evaluate x2(x2 − 1)−½ − (x2 − 1)½
Answer: D ((x2 − 1)−½)
Write both terms over (x2 − 1)½: [x2 − (x2 − 1)]/(x2 − 1)½ = 1/(x2 − 1)½ = (x2 − 1)−½.
Find the gradient of the line passing through the points (−2, 0) and (0, −4)
Answer: C (−2)
Gradient = (−4 − 0)/(0 − (−2)) = −4/2 = −2.
What is the nth term of the progression 27, 9, 3,...?
Answer: A (27(1/3)n−1)
It is a G.P. with first term 27 and common ratio ⅓, so the nth term is 27(⅓)n−1.
Find the sum of the 20 terms in an arithmetic progression whose first term is 7 and last term is 117
Answer: B (1240)
Sum = n/2 × (first + last) = 20/2 × (7 + 117) = 10 × 124 = 1240.
The angles of a quadrilateral are 5x − 30, 4x + 60, 60 − x and 3x + 61. Find the smallest of these angles.
Answer: C (60 − x)
The angles add up to 360°: 11x + 151 = 360, so x = 19. The angles are 65°, 136°, 41° and 118°, so the smallest is 60 − x.
The area of a square is 144sq cm. Find the length of its diagonal.
Answer: C (12√2cm)
The side is √144 = 12cm, so the diagonal is 12√2cm.
One angle of a rhombus is 60°. The shorter of the two diagonals is 8cm long. Find the length of the longer one.
Answer: A (8√3)
The shorter diagonal cuts the rhombus into two equilateral triangles, so each side is 8cm. The longer diagonal is twice the height of such a triangle: 2 × 4√3 = 8√3cm.
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x.
Answer: C (62°)
Exterior angles add up to 360°: 5x + 50 = 360, so x = 62.
Use the figure below to answer questions 35 and 36. PMN and PQR are two secants of the circle MQTRN and PT is a tangent.
If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimetres
Answer: C (12.5, 7.7)
For secants from the same point, PM × PN = PQ × PR, so 60 = 4.8 × PR and PR = 12.5. For the tangent, PT2 = PM × PN = 60, so PT ≈ 7.7.
Use the figure below to answer questions 35 and 36. PMN and PQR are two secants of the circle MQTRN and PT is a tangent.
If PN̂R = 110° and PM̂Q = 55°, find MP̂Q
Answer: D (15°)
MQRN is cyclic, so the exterior angle PMQ equals the interior opposite angle QRN = 55°. In triangle PNR, angle P = 180° − 110° − 55° = 15°.
In the figure above, find the value of y.
Answer: B (122°)
The 152° bend makes the slanting top line rise at 28° to the parallels. The next line turns a further 30°, so it meets the bottom parallel at 28° + 30° = 58°. y is on the other side of that line, so y = 180° − 58° = 122°.
In the figure above, PQ = PR = PS and SR̂T = 68°, Find QP̂S.
Answer: A (136°)
Angle QRS = 180° − 68° = 112°. Triangles PQR and PRS are isosceles, so angle QPR + angle RPS = (180° − 2 × PRQ) + (180° − 2 × PRS) = 360° − 2 × 112° = 136°.
A flagstaff stands on the top of a vertical tower. A man standing 60m away from the tower observes that the angles of elevation of the top and bottom of the flagstaff are 64° and 62° respectively. Find the length of the flagstaff
Answer: D (60(tan 64° − tan 62°))
The top of the flagstaff is 60 tan 64° high and its bottom (the top of the tower) is 60 tan 62° high. The flagstaff is the difference: 60(tan 64° − tan 62°).
Simplify cos2 x(sec2x + sec2 x tan2x).
Answer: C (sec2 x)
cos2x × sec2x = 1, so the expression is 1 + tan2x = sec2x.
If cos x = √(a/b), find cosec x.
Answer: C (√(b/(b − a)))
sin2x = 1 − a/b = (b − a)/b, so sin x = √((b − a)/b) and cosec x = √(b/(b − a)).
From a point Z, 60m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X.
Answer: C (060°)
From X, Y is 60m north and 60√3m east. tan θ = 60√3/60 = √3 measured from north, so θ = 60° and the bearing is 060°.
A surveyor walks 500m up a hill which slopes at an angle of 30°. Calculate the vertical height through which he rises.
Answer: A (250m)
Vertical height = 500 sin 30° = 500 × ½ = 250m.
In the figure above, PQRS is a square of side 8cm. What is the area of ΔUVW?
Answer: D (10sq. cm)
Take S as (0, 0): then U = (4, 8), V = (8, 2) and W = (6, 0). By the shoelace formula, area = ½|4(2 − 0) + 8(0 − 8) + 6(8 − 2)| = ½ × 20 = 10 sq. cm.
Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm
Answer: B (72πcm2)
Total surface = 2πr2 + 2πrh = 2π(16) + 2π(4)(5) = 32π + 40π = 72πcm2.
Find the volume of the figure above.
Answer: A ((πa2/3)(x + 3y))
The solid is a cylinder (radius a, height y) with a cone (radius a, height x) on top. Volume = πa2y + ⅓πa2x = (πa2/3)(x + 3y).
3% of a family's income is spent on electricity, 58% on food, 20% on transport, 11% on education and 7% on extended family. The angles subtended at the centre of the pie chart under education and food are respectively
Answer: D (39.6° and 212.4°)
Each 1% is 3.6°, so education gets 11 × 3.6° = 39.6°. Only option D has this angle; its 212.4° for food is 59% of 360°.
Use the following information to answer questions 48 and 49.
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result
No. of defective per box: 4, 5, 6, 7, 8, 9
No. of boxes: 2, 7, 17, 10, 8, 6
The mean and the median of the distribution are respectively
Answer: A (6.7, 6)
Σfx = 8 + 35 + 102 + 70 + 64 + 54 = 333 over 50 boxes, so the mean is 6.66 ≈ 6.7. The 25th and 26th boxes both have 6 defective (cumulative frequencies 2, 9, 26), so the median is 6.
Use the following information to answer questions 48 and 49.
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result
No. of defective per box: 4, 5, 6, 7, 8, 9
No. of boxes: 2, 7, 17, 10, 8, 6
Find the percentage of boxes containing at least 5 defective each.
Answer: A (96)
Only 2 boxes have fewer than 5 defective, so 48 of the 50 boxes have at least 5. 48/50 × 100 = 96%.
A crate of soft drinks contains 10 bottles of Coca-cola, 8 of Fanta and 6 of Sprite, if one bottle is selected at random, what is the probability that it is NOT a Coca-cola bottle?
Answer: D (7/12)
There are 24 bottles and 14 are not Coca-cola, so the probability is 14/24 = 7/12.