Tap Time

This is my solution:

If both taps are on, in 1 minute, the tank will be filled $\frac { 1 }{ 40 }$ of its volume.

If only tap X is on, in 1 minute, the tank will be filled $\frac { 1 }{ 60 }$ of its volume.

So, in 1 minute, if only tap X is on, the tank will be filled $\frac { 1 }{ 40 } -\frac { 1 }{ 60 } =\frac { 1 }{ 120 }$ of its volume.

So the time will it take for only tap Y to fill up the tank is: $1\div \frac { 1 }{ 120 } =120$ (minutes).

So the answer of this problem is 120 minutes

Let t be time in which only Y tap can fill whole tank of volume V.
Now as per given, filling speed of X tap is V/60 and as per our assumption that of Y tap is V/t. now both if tap are on for 40 min. then they can fill volume V. So we have
(V/60)x40+(V/t)x40=V solving this we have t=120 min.

We calculate the filling rate/minute .for both it is 1/40,for X it is 1/60 difference is 1/40-1/60 =1/120 ,therefore the other tap Y will fill this tank in 120 mts Ans

K.K.GARG,India

1/60 +1/n=1/40 so n=120

x 40+y 40 = x 60; x = 2 y => x 60 => y 120 x, y in L/s;

60 = 2^2 . 3 .5 40 = 2^3 . 3

KPK = 2^3 . 3 . 5 = 8.3.5 = 24.5 = 120