Aptitude - Pipes and Cistern - Discussion

Ya I got it now. Thank you Kasi Srinivas.

Excellent explanation Kasi Srinivas

@ Gopal , Priya , Vishnupriya :

Both pipes A and B are opened initially.
Now these both pipes (A and B) fill the cistern together for some time...let us say they are filling the cistern together for x mins.
Which implies pipe B is turned off after x min (according to the question).

If u understood this u can follow the problem..

Pipes A & B are filling the tank for x mins => x*(A+B) --- (1)

As after x mins pipe B closed, pipe A alone filling the tank.

He said cistern will be filled in half an hour ie 30 mins.

=> pipe A alone is filling the remaining tank in (30-x) mins
=> A*(30-x)---(2)

So, part filled by (A + B) in x minutes + Part filled by A in (30 -x) minutes = 1.

here 1 indicates the tank is full.

by substituting we get,


x*(2/75 + 1/45) + (30-x)* 2/75 = 1

on solving we get x = 9.

- K @ $ ! (kasi srinivas)

Shortcut

(75/2)-30
----------- * 45 = 9 ....after solving
(75/2)

I can't understand still. Can any one say in detail.

Excellent.

Sorry I didn't understood. Please explain me in detail.

If pipe B is turned off after x min, which means it is opened for x mins along with pipe A.
therefore for x mins both pipes A & B is filling the tank ie : x*(A+B)----(1)
As after x mins pipe B closed, pipe A alone filling the tank.
In question it asked in total 30 min tank should be filled.
therefore pipe A alone fills the remaining tank in (30-x) mins ie : A*(30-x)-----(2)
Hence the total work is obtained as equ (1) + (2) = 1

Please explain clearly.

How Part filled by (A + B) in x min + Part filled by A in (30 -x) min = 1?