1. A clock is set at 5 a.m. The clock loses 16 minutes in 24 hours. What will be the approximate near time when the clock indicates 10 p.m. on 4th day?
(a) 9 P.M (b) 10 P.M. (c) 11 P.M. (d) 12 P.M.
Upto 5 A.M. on the 4th day the clock would have lost 48 minutes. From 5 A.M. to 5 P.M.on the 4th day it loses further 8 minutes. From 5.00 P.M. to 10 P.M. on the 4th day (for about 5 hours) the clock will lose further 3.33 minutes. Thus the total minutes lost by the clock will be 48 + 8 + 3.3 = 59.3 minutes or nearly 1 hour. Thus the actual time will be 9.00 P.M.
2. Find the compound interest on Rs.16,000 at 20% per annum for 9 months, compounded quarterly.
20% rate of interest per annum would amount to 5% per quarter. Now applying the formula C.I. = P (1 + R/100)n – P and substituting the values of P as 16000, R as 5% and n = 3 we get the value Rs 2522.
3. The value of (4.7×13.26 + 4.7×9.43+4.7×77.31) is
(a) 0.47 (b) 4 (c) 470 (d) 4700
4.7 is common in all the three multiplication. The sum of the three other multipliers aggregates to 100. Hence the answer is 4.7 x 100 = 470.
4. A ladder learning against a wall makes an angle of 60° with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall.
(a) 9 m (b) 9.5 m (c) 10.5 m (d) 12 m
Let the distance from the wall be ‘x’. The angle to the ground is 600. The length of the ladder (hypotenuse) is 19 m. Thus we have cos 600 = ½ = x/19. Hence x -> 9.5m
5. The value of log343 (7) is
(a) 1/3 (b) – 3 (c) – 1/3 (d) 3
6. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?
(a) 46 (b) 81 (c) 126 (d) 252
Since we are talking about a rectangle, we have the following information.
The length of the unpainted side is 9 feet and hence the opposite of this side also will be 9 ft.
Since the total length of the three painted sides is 37 feet the lengths of the other two painted sides
are -> 37 – 9 = 28 / 2 = 14. Thus the sides of the rectangle are 14 feet and 9 feet and the
area -> 14 x 9 = 126 sq.ft.
7. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
(a) 40 (b) 60 (c) 200 (d) 120
The total expense for fencing at the rate of Rs 26.50 per metre is Rs 5300.00
Hence the perimeter of the rectangle is 5300 / 26.50 = 200 metre.
Let the breadth of the rectangle be ‘x’ metres. Then the length of the rectangle is (x + 20) metres
So the perimeter is -> x + (x+20) + x + (x+20) -> 4x + 40 = 200.
Solving we get ‘x’ = 40 and the length is (x+20) = 40 + 20 = 60 metres.
8. There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of sections A is 40 kg and that of section b is 35 kg. Find the average weight of the whole class?
(a) 36.25 (b) 37.25 (c) 38.35 (d) 39.25
Total weight of Class A -> 36 x 40 = 1440 kg
Total weight of Class B -> 44 x 35 = 1540
Total weight of both Classes A & B = 2980
Total number of students in both classes = 36 + 44 -> 80 students
Average weight of both the classes -> 2980 / 80 = 37.25 kg
9. A batsman makes a score of 87 runs in the 17th innings and thus increases his averages by 3. Find his average after the 17th innings?
(a) 19 (b) 29 (c) 39 (d) 49
Let the average after 16 innings be ‘x’. Then Total after 16 innings -> 16x.
Total after 17 innings -> 16x + 87 ……………… (i)
The average after 17 innings is (x+3)
Total after 17 innings -> 17 (x+3) ……………… (ii)
Both (i) and (ii) are the same and equal. So we have an equation
16x + 87 = 17(x+3) solving we get the value of ‘x’ as 31
Hence the average after 17 innings is (x+3) -> 36 + 3 = 39.
10. The banker’s discount on Rs. 1800 at 12% per annum is equal to the true discount on Rs.1872 for the same time at the same rate. Find the time?