Math, asked by sag387, 1 year ago

Two pipes A and B can fill an empty cistern in 32 and 48 hours, respectively. Pipe C can drain the entire cistern in 64 hours when no other pipe is in operation. Initially, when the cistern was empty Pipe A and Pipe C were turned on After a few hours, Pipe A turned off and Pipe B was tuned on instantly. In all it took 112 hour to fill the cistern. for how many hours was Pipe B turned on ?

Answers

Answered by CharlieBrown2
16

Answer:

Pipe B was turned on for 72 hours.

Step-by-step explanation:

In 1 hour pipe A can fill 1/32, pipe B 1/48 and pipe C can drain 1/64 of the cistern.

When A and C are opened :  1/32 - 1/64 = 1/64

When B and C are opened : 1/48 - 1/64 = 4/192 - 3/192 = 1/192

If t is time while A is turned on:

1/64 t + 1/ 192 ( 112 - t ) = 1  /*192  ( we multiply both sides by 192 )

3 t + 112 - t = 192

2 t = 192 - 112

2 t = 80

t = 80 : 2 = 40

So fot pipe B :   112 - 40 = 72 hours.

Answered by Anonymous
0

Pipe B was turned on for 72 hours.....

Answer in the picture given upper.....

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