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Question 5 : There are two taps, T1 and T2, at the bottom of a water tank, either or both of which may be opened to empty the water tank, each at a constant rate. If T1 is opened keeping T1 closed, the water tank (initially full) becomes empty in half an hour. If both T1 and T2 are kept open, the water tank (initially full) becomes empty in 20 minutes. Then, the time (in minutes) it takes for the water tank (initially full) to become empty if T2 is opened while T1 is closed is


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Explanatory Answer

The tank has two taps T1 and T2.
It is given that the Tap T1 empties the tank in 30 minutes and also that the taps T1 and T2 together can empty the tank in 20 minutes.
So, understanding this in terms of ‘1 minute’.
Tap T1 will empty \(\frac{1}{30} t h\) of the tank in 1 minute
And T1 and T2 will together empty \(\frac{1}{20} t h\) of the tank in 1 minute
In this 1 minute, Out of the \(\frac{1}{20} t h\) tank empties by the two taps together, T1 empties \(\frac{1}{30} t h\) of the tank and the rest is emptied by T2.
So, to find out the portion of the tank emptied by T2 in 1 minute we can simply subtract T1 from the total.
Which is, \(\frac{1}{20}\) - \(\frac{1}{30}\) 
= \(\frac{1}{60}\)
So T2 empties \(\frac{1}{60}\)th of the tank in 1 minute, and so to empty the entire tank it will require 60 minutes.


The answer is '60'

 

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