Two pipes A and B can fill a tank in 10 hours and 30 hours respectively. Due to a leak in the tank, it takes 2.5 hours more to fill the tank. How much time would the leak alone will take to empty the tank ?
Answer: C
Let the capacity of the tank be LCM (10, 30) = 30 units
=> Efficiency of pipe A = 30 / 10 = 3 units / hour
=> Efficiency of pipe B = 30 / 30 = 1 units / hour
=> Combined efficiency of both pipes = 4 units / hour
Now, total time taken by A and B working together to fill the tank if there was no leak = 30 / 4 = 7.5 hours
=> Actual time taken = 7.5 + 2.5 = 10 hours
The tank filled by A and B in these 2.5 hours is the extra work done to compensate the wastage by the leak in 10 hours.
=> 2.5 hours work of A and B together = 10 hours work of the leak
=> 2.5 x 4 = 10 x E, where ‘E’ is the efficiency of the leak.
=> E = 1 unit / hour
Therefore, time taken by the leak alone to empty the full tank = 30 / 1 = 30 hours
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A tap can fill a tub in 24 hours. Due to a leak at the bottom of the tub, the tap fills the tub in 36 hours. If the tub is full, how much time will the leak take to empty it?
Answer: A
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Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
Answer: C
Work done by the waste pipe in 1 minute = | 1 | - | ![]() |
1 | + | 1 | ![]() |
15 | 20 | 24 |
= | ![]() |
1 | - | 11 | ![]() |
15 | 120 |
= - | 1 | . [-ve sign means emptying] |
40 |
![]() |
1 | part = 3 gallons. |
40 |
Volume of whole = (3 x 40) gallons = 120 gallons.
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One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in .. ?
Answer: C
No answer description available for this question.
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A cistern has two taps attached to it. Tap B can empty the cistern in 45 minutes. But Tap A can fill the cistern in just 30 minutes. Rohit started both taps unknowingly but realized his mistake after 30 minutes. He immediately closed Tap B. Now after this, in how much time will the cistern be filled?
Answer: D
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Pipe P can fill a tank in 4 hours, pipe Q in 8 hours and pipe R in 24 hours. If all the pipes are open, in how many hours will the tank be filled?
Answer: B
Part filled by (P + Q + R) in 1 hour = (1/4 + 1/8 + 1/24) = 10/24
All the three pipes together will fill the tank = 24/10 = 2.4 hours
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A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.
Answer: D
Half tank will be filled in 3 hours
Lets calculate remaining half,
Part filled by the four taps in 1 hour = 4 × (1/6) = 2/3
Remaining part after 1/2 filled = 1-1/2 = 1/2
Total time = 3 hours + 45 mins = 3 hours 45 mins
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Working alone, two pipes A and B require 9 hours and 6.25 hours more respectively to fill a pool than if they were working together. Find the total time taken to fill the pool if both were working together.
Answer: D
Let the time taken if both were working together be ‘n’ hours.
=> Time taken by A = n + 9
=> Time taken by B = n + 6.25
In such kind of problems, we apply the formula :
n2 = a x b, where ‘a’ and ‘b’ are the extra time taken if both work individually than if both work together.
Therefore, n2 = 9 x 6.25
=> n = 3 x 2.5 = 7.5
Thus, working together, pipes A and B require 7.5 hours.
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To fill a cistern, pipes A, B and C take 20 minutes, 15 minute and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern is:
Answer: A
A can do a piece of work in x minutes
B can do a piece of work in y minutes
C can do a piece of work in z minutes
Minute’s work of each of the three is 1/x +1/y + 1/z
1 minutes work of each of the three pipes = 1/20 + 1/15 + 1/12
= 1/5 or = 5 minutes
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Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.
Answer: B
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