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Two taps A and B can fill a tank in 20 minutes and 30 minutes respectively,both the taps open in an empty tank and when the tank should be filled it comes into the notice that a leakage remains opened by mistake ,The leakage immediately closed and after that in 3 minutes three mintes tank filled by water.In what time the leakage empty the tank fill of water?

I have tried:

Both the pipes can fill a tank in 20 and 30 minutes respectively

The time taken to complete the tank by both pipes is

1/30 + 1/20 = 1/12 for one hour , completely filled the tank is 12 hours

Another statement leakage is not noticed,for first three minutes, then i have calculate the tank filled by water with the leakage

1/30 + 1/20 - 1/x = 3/12

which gives x=6

for leakge time calculating to fill the tank

1/30 + 1/20 -1/6 for one hour it gives 1/2 to completely fill the tank by 2 hours only na

But the Answer is 48 hours, please anyone guide me what i am doing mistake and guide the solution

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    Part way through this you changes units of measurement from minutes to hours and you did not multiply or divide by 60, just changed the wording. This is very onfusing. It would help if you edited your post, – 2017-02-08

2 Answers 2

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You are on the right track but it helps to treat each case as a separate step.

Also be sure to add a word or two to explain the meaning of each equation.

First pipe fills tank in 20 minutes --> 1/20 of tank filled per minute.

Second pipe fills tank in 30 minutes --> 1/30 of tank filled per minute.

Both pipes together fill at a rate of 1/20 + 1/30 = 5/60 = 1/12 of tank filled per minute.

At this rate we expect the tank to fill with both pipes in 12 minutes.

After the leakage is closed, the tank finishes filling (with both pipes) in 3 minutes. Therefore the amount that the tank was below full is 1/12 tank per minute x 3 minutes = 3/12 = 1/4 of the tank.

The leakage is 1/4 of the tank in the 12 minutes when expected to be full so 1/4 divided by 12 = 1/4 x 1/12 = 1/48 tank per minute. So if the leakage was left open starting with a full tank, it would empty in 48 minutes.

(Note - you changed units while typing this. If the original problem was in hours, change all the time units to hours.)

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    yes I have typed worng it is minutes,question is also in Minutes only – 2017-02-08
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    Thank you everyone guiding the answer – 2017-02-08
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There are no hours in the problem. Your first computation shows the two taps together fill the tank in $12$ minutes, not $12$ hours. Where did you get the second equation? I don't understand it. As when the leak is blocked they fill the tank in $3$ minutes, the tank was $3/4$ full when the leak was discovered. Then in $12$ minutes the leak emptied $\frac 14$ of the tank, so it can empty the whole tank in $48$ minutes.

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    the leakage is noticed after 3 minutes 1/20 +1/30 - 1/x = 9/12 @Ross Millikan why this equation is wrong, sorry instead i have to put 9/12, is this equation is correct? – 2017-02-08
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    before three minutes with the leakage the tank should be filled by above equation correct ah guide me @Ross Millikan – 2017-02-08
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    No, it says the leakage is noticed at $12$ minutes, the time when the tank should have been full. Then they plug the leak and the tank fills in another $3$ minutes. When you solved that second equation you dropped the minus sign on the $1/x$ to find $x=6$. – 2017-02-08
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    ok, after the tank got filled , they noticed the leakage when? – 2017-02-08
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    From the question "when the tank should be filled it comes into the notice that a leakage remains opened by mistake ". You computed that it should be filled at $12$ minutes, so that is when it is noticed. – 2017-02-08
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    ok i have understand after that how you solved the equation,after 12 minutes then? – 2017-02-08
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    After $12$ minutes there was no leak, and the tank filled in $3$ minutes. In $3$ minutes you fill $1/4$ tank, so the tank must have been $3/4$ full at that time. Since the tank should have been full, the leak drained $1/4$ tank in $12$ minutes. – 2017-02-08