A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received ₦5,000 more than X. What is the total profit of the company for the year?
A.₦20,000
B.₦25,000
C.₦30,000
D.₦15,000
E.₦45,000.
Show the answer and explanation
Answer: E (₦45,000.)
Explanation
Y gets 1 part more than X, so 1 part = ₦5,000. The total is 4 + 5 = 9 parts = ₦45,000.
Given that cos z = L, where z is an acute angle, find an expression for (cot z − cosec z)/(sec z + tan z)
A.(1 − L)/(1 + L)
B.(L2 − √(1 − L2))/(L2 + L − 1)
C.−L√(1 − L)/(√(1 + L)(1 + √(1 − L2)))
D.√(L − 1)/(√(1 − L2) + 1 − L2)
E.L(L − 1)/(1 + √(1 − L2) − L2)
Show the answer and explanation
Answer: E (L(L − 1)/(1 + √(1 − L2) − L2))
Explanation
With sin z = √(1 − L²): cot z − cosec z = (L − 1)/sin z and sec z + tan z = (1 + sin z)/L. Dividing gives L(L − 1)/(sin z + sin²z) = L(L − 1)/(√(1 − L²) + 1 − L²).
In a class of 60 pupils, the statistical distribution of the number of pupils offering Biology, History, French, Geography and Additional Mathematics is as shown in the pie chart above. How many pupils offer Additional Mathematics?
A.15
B.10
C.18
D.12
E.28.
Show the answer and explanation
Answer: B (10)
Explanation
The angles add to 360°: (2x − 24) + (3x − 18) + (2x + 12) + (x + 12) + x = 9x − 18 = 360, so x = 42. Additional Mathematics has 60°, i.e. 60/360 × 60 = 10 pupils.
y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.
A.y = 10x2/31 + 52/(31√x)
B.y = x2 + 1/√x
C.y = x2 + 1/x
D.y = x2/31 + 1/(31√x)
E.y = 10/31 (x2 + 1/√x)
Show the answer and explanation
Answer: A (y = 10x2/31 + 52/(31√x))
Explanation
y = ax² + b/√x. x = 1 gives a + b = 2; x = 4 gives 16a + b/2 = 6, i.e. 32a + b = 12. Subtracting: 31a = 10, so a = 10/31 and b = 52/31.
The scores of a set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.
A.47
B.48½
C.50
D.48
E.49.
Show the answer and explanation
Answer: B (48½)
Explanation
In order: 21, 23, 29, 40, 41, 43, 47, 50, 50, 55, 56, 70, 70, 73. With 14 scores the median is the average of the 7th and 8th: (47 + 50)/2 = 48½.
Which of the following equations represents the above graph?
A.y = 1 + 2x + 3x2
B.y = 1 + 2x + 3x2
C.y = 1 + 2x − 3x2
D.y = 1 − 2x − 3x2
E.y = 3x2 + 2x − 1.
Show the answer and explanation
Answer: D (y = 1 − 2x − 3x2)
Explanation
Roots −1 and ⅓ give the factors (x + 1)(3x − 1) = 3x² + 2x − 1. The curve is an upside-down parabola through (0, 1), so y = −(3x² + 2x − 1) = 1 − 2x − 3x².
PQRS is a desk of dimensions 2m × 0.8m which is inclined at 30° to the horizontal. Find the inclination of the diagonal PR to the horizontal.
A.23°35'
B.30°
C.15°36'
D.10°
E.10°42'.
Show the answer and explanation
Answer: E (10°42'.)
Explanation
The 0.8m edge rises at 30°, so R is 0.8 sin 30° = 0.4m higher than P. The diagonal is √(2² + 0.8²) ≈ 2.154m, so sin θ = 0.4/2.154 ≈ 0.186 and θ ≈ 10°42'.
If O is the centre of the circle in the figure above, find the value of x.
A.50
B.260
C.100
D.65
E.130.
Show the answer and explanation
Answer: C (100)
Explanation
The 130° angle at the circumference stands on the major arc, so the reflex angle at the centre is 260°. The angle x on the other side is 360° − 260° = 100°.
The score of 16 students in a Mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65, 70. What is the sum of the median and modal scores?
A.125
B.130
C.140
D.150
E.137.5.
Show the answer and explanation
Answer: B (130)
Explanation
65 occurs most often (6 times), so the mode is 65. In order, the 8th and 9th scores are both 65, so the median is 65. The sum is 130.
The letters of the word 'MATRICULATION' are cut and put into a box. One letter is drawn at random from the box. Find the probability of drawing a vowel.
A.7/13
B.5/13
C.6/13
D.8/13
E.4/13
Show the answer and explanation
Answer: C (6/13)
Explanation
MATRICULATION has 13 letters, of which 6 are vowels (A, I, U, A, I, O). The probability is 6/13.
PQRS is a cyclic quadrilateral in which PQ = PS. PT is a tangent to the circle and PQ makes an angle of 50° with the tangent as shown in the figure below. What is the size of QRS?
A.50°
B.40°
C.110°
D.80°
E.100°.
Show the answer and explanation
Answer: E (100°.)
Explanation
By the alternate segment theorem <PSQ = 50°, and PQ = PS makes <PQS = 50° too, so <QPS = 80°. Opposite angles of a cyclic quadrilateral add to 180°, so <QRS = 100°.
In a sample survey of a university community, the following table shows the percentage distribution of the number of members per household. No. of members per household: 1, 2, 3, 4, 5, 6, 7, 8 (Total) Number of households: 3, 12, 15, 28, 21, 10, 7, 4 (100) What is the median?
A.4
B.3
C.5
D.4.5
E.None.
Show the answer and explanation
Answer: A (4)
Explanation
The running totals are 3, 15, 30, 58, … so the 50th and 51st households both have 4 members. The median is 4.
On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram correct to 3 significant figures.
The farm yields of four crops on a piece of land in Ondo are represented on the pie chart above. What is the angle of the sector occupied by Okro in the chart?
A.91⅓°
B.19⅓°
C.33⅓°
D.11°
E.91°.
Show the answer and explanation
Answer: B (19⅓°)
Explanation
The total yield is 14.5 + 25.6 + 45.4 + 184.5 = 270kg. Okro's angle = 14.5/270 × 360° ≈ 19⅓°.
PQR is the diameter of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = QRT = 60°, find the perimeter of the figure PTRS (π = 22/7)
A.25cm
B.18cm
C.36cm
D.29cm
E.25.5cm.
Show the answer and explanation
Answer: A (25cm)
Explanation
Triangle PTR has two 60° angles, so it is equilateral with side PR = 7cm. The semicircular arc is ½ × 22/7 × 7 = 11cm. Perimeter = 11 + 7 + 7 = 25cm.
A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Altogether he covered 600 kilometres. At what speed did he drive for the last 3 hours?
A.120km/hr
B.60km/hr
C.600/7km/hr
D.50km/hr
E.100km/hr.
Show the answer and explanation
Answer: A (120km/hr)
Explanation
With first speed v: 4v + 3(2v) = 10v = 600, so v = 60km/hr. The last 3 hours were at 2v = 120km/hr.