Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
[A].
5 min.
[B].
9 min.
[C].
10 min.
[D].
15 min.
Answer: Option B
Explanation:
Let B be turned off after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.
How Part filled by (A + B) in x min + Part filled by A in (30 -x) min = 1?
Gopal said:
(Fri, Oct 29, 2010 12:50:46 PM)
Please explain clearly.
Ashwin said:
(Sun, Nov 7, 2010 05:37:28 AM)
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
Priya said:
(Sun, Jun 19, 2011 08:45:52 AM)
Sorry I didn't understood. Please explain me in detail.
Ricky said:
(Mon, Aug 8, 2011 12:55:52 AM)
Excellent.
Vishnupriya said:
(Wed, Aug 17, 2011 06:35:42 AM)
I can't understand still. Can any one say in detail.
Mahesh Patil said:
(Tue, Aug 23, 2011 08:47:11 AM)
Kasi Srinivas said:
(Wed, Aug 31, 2011 08:28:44 PM)
@ 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.