A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in how much time?

20 minutes

Let the capacity of tank = 60 ltrs. (the LCM of 20 & 60, to solve easily)

We have,

Efficiency of 1st tab (A) = 3 ltr/min

Efficiency of 2nd tab (B) = 1 ltr/min

Efficiency of both (A + B) = 3 + 1 = 4 ltr/min

Now, ATQ,

Both taps are open for 10 minutes

So, tank filled by them in 10 minutes = 10 × ( 3 + 1 ) = 10 × 4 = 40 ltr

Remaining capacity of tank to be filled = 60 - 40 = 20 ltr

Now, Only 2nd tap (B) is in use.

So, time taken by pipe B to fill the remaining quantity = \(\frac{20}{1}\) = 20 minutes