Two pipelines X and Y fill a tank in 7.5 hours and 10 hours respectively, while a third pipeline Z can empty the full tank in 12.5 hours. All the three pipelines are opened in the beginning. Z is closed after 5 hours. In how much time, will the tank be full?

7.5 hrs

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6 hrs

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2.5 hrs

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10 hrs

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3.5 hrs

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Solution

The correct option is B
6 hrs

$F o r 5 h o u r s = 5 (\frac{2}{15} + \frac{1}{10} - \frac{2}{25}) = \frac{23}{30}$

$R e m a i n i n g p a r t = 1 - \frac{23}{30} = \frac{7}{30}$

$(X + Y) ‘ s 1 h o u r w o r k = \frac{2}{15} + \frac{1}{10} = \frac{7}{30}$

$T h e r e f o r e, X a n d Y w i l l f i l l r e s t o f t h e t a n k i n = \frac{7}{30} \times \frac{30}{7} = 1 h o u r$

$H e n c e, t o t a l t i m e i s = 5 + 1 = 6 h o u r s$

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Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tak be full?

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(b) 7 hrs 30 min

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Two pipelines X and Y fill a tank in 7.5 hours and 10 hours respectively, while a third pipeline Z can empty the full tank in 12.5 hours. All the three pipelines are opened in the beginning. Z is closed after 5 hours. In how much time, will the tank be full?

The correct option is B 6 hrs For 5 hours =5(215+110−225)=2330 Remaining part =1−2330=730 (X+Y)‘s 1 hour work =215+110=730 Therefore, X and Y will fill rest of the tank in =730×307=1 hour Hence, total time is =5+1=6 hours