A tank can be filled by one tap in 20 minutes | vedclass

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(C) Let the capacity of the tank be $1$ unit.Rate of the first tap $= \frac{1}{20}$ units per minute.Rate of the second tap $= \frac{1}{25}$ units per

Vedclass · Accepted Answer

$11$

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(C) Let the capacity of the tank be $1$ unit.Rate of the first tap $= \frac{1}{20}$ units per minute.Rate of the second tap $= \frac{1}{25}$ units per minute.Combined rate of both taps $= \frac{1}{20} + \frac{1}{25} = \frac{5+4}{100} = \frac{9}{100}$ units per minute.In $5$ minutes,the part of the tank filled by both taps $= 5 	imes \frac{9}{100} = \frac{45}{100} = \frac{9}{20}$ units.Remaining part of the tank to be filled $= 1 - \frac{9}{20} = \frac{11}{20}$ units.Since the second tap is turned off,the first tap fills the remaining part.Time taken by the first tap to fill $\frac{11}{20}$ units $= \frac{11/20}{1/20} = 11$ minutes.

$A$ tank can be filled by one tap in $20$ $minutes$ and by another in $25$ $minutes$. Both the taps are kept open for $5$ $minutes$ and then the second is turned off. In how many $minutes$ more will the tank be completely filled?

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