1. A person got an article after getting two successive discounts of 20% and 10% at Rs 7200. What is the marked price of this article ?
A. Rs 12000
B. Rs 13900
C. Rs 14000
D. Rs 10000
E. None of these
Sol. Single equivalent discount = (20 + 10) – (20*10/100) = 28%
So MP is 100/(100-28) * 7200 = Rs 10,000.
2. A cube of side 8 feet is to be painted from outside. The cost of paint is Rs 25 per kg. If 48 square feet are covered by 1 kg, then what is the cost of painting the cube ?
A. Rs 200
B. Rs 250
C. Rs 280
D. RS 230
E. None of these.
Sol. Surface area of cube = 6a^2 = 6*8^2 = 384 sq. feet
So quantity of paint required = (384/48) = 8 kg
So cost of painting = 25*8 = Rs 200
3. The students of section X and section Y have average weights of 50 kg and 45 kg respectively. Find the approximate average weight of both the sections together such that the number of students in the sections X and Y is in the ratio 3 : 2.
A. 45
B. 48
C. 47
D. 46
E. None of these
Sol. Students in A section = 3x, students in B section = 2x
So total weight of students of X section = 50*3x = 150x
And total weight of students of Y section = 45*2x = 90x
So average weight of both sections = [(150x+90x)/(3x+2x)] = 48 kg.
4. There are two trains of lengths 300 m and 200 m respectively. Both start at same time. If they travel in same direction they cross each other in 20 hours and if they travel in opposite direction towards each other, they cross each other in 2 hours. What is the speed of the fastest train ?
A. 135
B. 137
C. 137.5
D. 140
E. None of these
Sol. Let speed of 1st train = x km/hr, of 2nd train = y km/hr
Then when they travel in same direction, relative speed = x-y
So (300+200) = (x-y)*20. This gives x – y = 25
When they travel in opposite direction, relative speed = x+y
So (300+200) = (x+y)*2. This gives x + y = 250
Solve both equations, x = 137.5
5. If the price of wheat is reduced by 5%. How many kilograms of wheat a person can buy with the same money which was earlier sufficient to buy 38 kg of wheat ?
A. 60
B. 50
C. 40
D. 30
E. None of these
Sol. Let the original price = 100 Rs per kg
Then money required to buy 38 kg = 38*100 = Rs 3800
New price per kg is 95% of Rs 100 = 95
So quantity of wheat bought in 3800 Rs is 3800/95 = 40 kg.
6. 20 litres of the water is drawn out of a jar and filled with milk. This operation is performed 1 more times. If the ratio of the quantity of water left in jar to that of milk in jar is 25 : 11, what was the initial quantity of water present in the jar ?
A. 116
B. 115
C. 118
D. 120
E. None of these
Sol. Let initial quantity of milk = x litres
After two times, quantity of milk left in jar = x [1 – 20/x]^2
So x [1 – 20/x]^2 / x = 25/25+11
[1 – 20/x]2 = 25/36
Square root both sides, so [1 – 20/x] = 5/6
Solve, x = 120.
7. X and Y can complete a work in 20 days and 30 days respectively. They start the work and X left after some days. Now Z who can alone complete the same work in 20 days joined with Y and they completed the remaining work in 5 days. For how many days did Y work ?
A. 17
B. 18
C. 12
D. 15
E. None of these
Sol. Y worked for 5 days + the days after which X left.
Let X left after x days. so
(1/20)*x + (1/30)*(5+x) + (1/20)*5 = 1
Solve, x = 7
So Y worked for 5+7 days = 12 days.
8. 45 people start a work working 10 hours each day. After 8 days it was found that only 3/5th of the work was done. How many additional people need to be employed in order to complete the remaining work in 4 days working 10 hours each day ?
A. 12
B. 20
C. 15
D. 18
E. None of these
Sol. Let x people need to be employed to complete remaining work (1 – 3/5) = 2/5
M1*D1*H1*W2 = M2*D2*H2*W1
45*8*10*(2/5) = (45+x)*4*10*(3/5)
Solve x = 15.
9. X and Y invested Rs 16000 and Rs 12000 respectively in a business. After 3 months A withdrew Rs 5000 while Y invested Rs 5000 more. After 3 more months Z joined them with Rs 21000. After 1 year, out of a total profit of Rs 26,400, the share of Y exceeded that of Z by
A. 2400
B. 1800
C. 3600
D. 4800
E. None of these
Sol. X : Y : Z
16000*3 + 11000*9 : 12000*3 + 17000*9 : 21000*6
7 : 9 : 6
So, difference in shares of B and C = (9-6)/(7+9+6) * 26400 = 3600
10. A pipe ‘X’ can empty the tank in 30 minutes and pipe ‘Y’ can fill the same tank in 15 minutes. The tank is filled with water and then pipe ‘X’ is opened. After 10 minutes, pipe Y is also opened. Find in how much total time the tank will be full again ?
A. 10
B. 15
C. 20
D. 25
E. None of these
Sol. Emptying pipe X is opened first for 10 minutes, so in 10 minutes the part of tank it has emptied is (1/30)*10 = 1/3
Now filling pipe is also opened, now since only 1/3 of the tank is empty so 1/3 is only to be filled by both pipes, let it take now x minutes, so
(1/15 – 1/30)*x = 1/3
Solve, x= 10
So total = 10+10 = 20 minutes [total time is asked – 10 minutes when emptying pipe was only opened, 10 minutes when both were operating.]
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