Pipes and Cistern
- Two pipes A and B can fill a tank in 1 h and 75 min, respectively. There is also an outlet C . If all the three pipes are opened together. The tank is full 50 min. How much time will be taken by C to empty the full tank?
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Work done by C in 1 min = (1/60 + 1/75 - 1/50 )
= (5 + 4 - 6)/300 = 3/300 = 1/100Correct Option: A
Work done by C in 1 min = (1/60 + 1/75 - 1/50 )
= (5 + 4 - 6)/300 = 3/300 = 1/100
Hence, C can empty the full tank in 100 min.
- Through an inlet, a tank takes 8 h to get filled up. Due to a leak in the bottom, it takes 2 h more to get it filled completely. If the tank is full, how much time will the leak take to empty it?
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Let the leak takes x h to empty the tank.
Now, part filled by inlet in 1 h = 1/8
part filled in 1 h when both tap and leak works together = 1/(8+2) = 1/10
According to the question.
1/x = 1/8 - 1/10 = (5 - 4) / 40 = 1/40Correct Option: D
Let the leak takes x h to empty the tank.
Now, part filled by inlet in 1 h = 1/8
part filled in 1 h when both tap and leak works together = 1/(8+2) = 1/10
According to the question.
1/x = 1/8 - 1/10 = (5 - 4) / 40 = 1/40
∴ x = 40 h
- Two pipes can fill a cistern in 14 and 16 h, respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, it took 92 min more to fill the cistern. When the cistern is full, in what time will the leak empty it?
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Part filled by 1st pipe in 1 h = 1/14
Part filled by 2nd pipe in 1 h = 1/16
Part filled by the two pipes in 1 h
= ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
∴ Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
&thee4; Work done by ( two pipes + leak ) in 1 h = 1/9Correct Option: A
Part filled by 1st pipe in 1 h = 1/14
Part filled by 2nd pipe in 1 h = 1/16
Part filled by the two pipes in 1 h
= ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
∴ Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
&thee4; Work done by ( two pipes + leak ) in 1 h = 1/9
Work done by the leak in 1 h = 1/9 - 15/112 = (112 - 135)/1008
= - 23/1008
∴ Time taken by leak to empty the full cistern
= 1008/23 = 4319/23 h
- Three taps are fitted in a cistern. The empty cistern is filled by the first and the second taps in 3 and 4 h, respectively. The full cistern is emptied by the third tap in 5 h. If all three taps are opened simultaneously, the empty cistern will be filled up in ?
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Part of tank filled by first tap in 1 h = 1/3
Part of tank filled by second tap in 1 h = 1/4
Part of tank emptied by third tap in 1 h = 1/5
Part of the tank filled by all pipes opened simultaneously in 1 h
= 1/3 + 1/4 - 1/5 = (20 + 15 - 12)/60 = 23/60Correct Option: B
Part of tank filled by first tap in 1 h = 1/3
Part of tank filled by second tap in 1 h = 1/4
Part of tank emptied by third tap in 1 h = 1/5
Part of the tank filled by all pipes opened simultaneously in 1 h
= 1/3 + 1/4 - 1/5 = (20 + 15 - 12)/60 = 23/60
Time taken by all the taps to fill the tank when it is empty = 23/60 h = 214/23 h
- There are two tank A and B to fill up a water tank. The tank can be filled in 40 min, if both taps are on. The same tank can be filled in 60 min, if tap A alone is on. How much time will tap B alone take, to fill up the same tank ?
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Part filled by tap A in 1 min = 1/60
Let tap B fills the tank in x min
Then, Part filled by tap, B in 1 min = 1/x
According to the question,
1/60 + 1/x = 1/40Correct Option: D
Part filled by tap A in 1 min = 1/60
Let tap B fills the tank in x min
Then, Part filled by tap, B in 1 min = 1/x
According to the question,
1/60 + 1/x = 1/40
⇒ 1/x = 1/40 - 1/60
&rArr ; 1/x = (3 - 2)/120
&rArr ; 1/x = 1/120
∴ Tap B can fill the tank in 120 min.
