# 2 pipes A & B can fill a cistern in 12 min and 16 min resp. If both pipes are opend together then after hw mch time B shld be closed so tht the tank is full in 9 min

*print*Print*list*Cite

Pipe A fills the tank in 12 minutes. In a minute 1/12 of the tank is filled. Similarly for tank B, in a minute 1/16 of the tank is filled.

When both the pipes are opened to fill the tank, let the time for which the pipe B is closed to fill the tank in 9 minutes be T.

This gives 9*(1/12) + (9 - T)*(1/16) = 1

=> 9/12 + 9/16 - T/16 = 1

=> T/16 = 9/12 + 9/16 - 1

=> T/16 = 5/16

=> T = 5

**Therefore the pipe B should be kept close for 5 minutes.**

A alone can fill 1/12 of the cistern in 1 minute.

B alone can fill (1/16)th of the cistern.

Let the pipes work together for x minutes to fill the empty cistern

So in x minutes (1/12+1/16)x = 7x/48 of the cistern is filled up.

If B is closed at the xth minute, then the rest of the tank to be filled up by A alone = 1- 7x/48 = (48-7x)/48.

(49-7x)/48 of the cistern can be filled by A in {(48-7x)/48}/(1/12) minutes

Therefore x+ {(48-7x)/48}/(1/12) = 9 minutes

=> x+(48-7x)/4 = 9

=>4x +(48-7x) = 9*4

=> 4x+48-7x = 36

=> 48-36 = 7x-4x

=> 12 = 3x

=> x = 12/3 hrs = 4 mins.

**Therefore, B should be closed after 4 minutes so that the pipe A can fill the cistern is full in the 9 minutes.**