two taps can fill a cistern in 3 and 4 hours respectively.while another tap can emptied the cistern in 2 hours.if the three taps are opened at 7 o clock,8 o clock,9 o clock respectively ,when the cistern will fill?

Dear Student,

Please find below the solution to the asked query:

Two taps can fill a cistern in 3 and 4 hours respectively. Let first tap  =  A and second tap = B  , So

Part filled by pipe A in 1 hour = $\frac{1}{3}$ ,

Part filled by pipe B in 1 hour = $\frac{1}{4}$

Also given : Another tap can emptied the cistern in 2 hours. Let this tap  =  C

Part emptied by pipe C in 1 hour = $\frac{1}{2}$

Also given : If the three taps are opened at 7 o clock,8 o clock,9 o clock respectively ,

So, From 7 to 8 o clock , cistern will be filled by = $\frac{1}{3}$ 

And

And Next hour ( 8 to 9 o clock  ), cistern will filled by A and B =$\frac{1}{3}$ + $\frac{1}{4}$

And From now from next hours( 9 to 10 o clock  ), the work rate of A, B and C = $\frac{1}{3}$ + $\frac{1}{4}$ - $\frac{1}{2}$

At 10 o clock  total cistern filled = $\frac{1}{3}$ + ( $\frac{1}{3}$ + $\frac{1}{4}$ ) + ( $\frac{1}{3}$ + $\frac{1}{4}$ - $\frac{1}{2}$ ) = $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{4}$  +  $\frac{1}{3}$ + $\frac{1}{4}$ - $\frac{1}{2}$ = $\frac{3}{3} + \frac{2}{4}$- $\frac{1}{2}$ = 1 +  $\frac{1}{2}$ - $\frac{1}{2}$ = 1 .

Therefore,


Cistern will be filled at 10 o clock .                                                      ( Ans )


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