Pipes and Cistern
- A tap can fill an empty tank in 12 h and a leakage can empty the tank in 20 h. If tap and leakage both work together, then how long will it take to fill the tank?
-
View Hint View Answer Discuss in Forum
Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20
Net part filled in 1 h when both (tap and leakage) work = 1/12 - 1/20
= (5 - 3)/60 = 2/60 = 1/30Correct Option: C
Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20
Net part filled in 1 h when both (tap and leakage) work = 1/12 - 1/20
= (5 - 3)/60 = 2/60 = 1/30
∴ Required time to fill the tank = 30 h
- Three taps A, B and C together can fill an empty cistern in 10 min . The tap A alone can fill it in 30 min and the tap B alone can fill it in 40 min. How long will the tap C alone take to fill it?
-
View Hint View Answer Discuss in Forum
Part filled by (A + B + C ) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40
Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120Correct Option: B
Part filled by (A + B + C ) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40
Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120
∴ Part filled by C in 1 min = 1/10 - 7/120 = (12 - 7)/120 = 5/120 = 1/24
∴ Tap C will fill the cistern in 24 min.
- Two pipes A and B can fill a cistern in 15 and 20, respectively. Both the pipes are opened together, but after 2 min, pipe A is turned off. What is the total time required to fill the tank?
-
View Hint View Answer Discuss in Forum
Part filled by both in 2 min = 2 x ( 1/15 + 1/20 ) = 2 x ( 4 + 3)/60 = 7/30
Part unfilled = 1 - 7/30 = (30 - 7)/30 = 23/30
Now, B fills 1/20 part in 1 min.Correct Option: B
Part filled by both in 2 min = 2 x ( 1/15 + 1/20 ) = 2 x ( 4 + 3)/60 = 7/30
Part unfilled = 1 - 7/30 = (30 - 7)/30 = 23/30
Now, B fills 1/20 part in 1 min.
∴ 23/30 part will be filled by B in (20 x 23)/30 min or in 46/3 min.
∴ Required time taken to fill the tank = 2 + 46/3 = 52/3 min.
- If two pipes function together, the tank will be filled in 12 h . One pipe fills the tank in 10 h faster than the other. How many hours does the faster pipe take to fill up the tank?
-
View Hint View Answer Discuss in Forum
Let one pipe takes m h to fill the tank.
Then, the other pipe takes (m - 10) h.
According to the question.
∴ 1/m + 1/(m - 10) = 1/12Correct Option: A
Let one pipe takes m h to fill the tank.
Then, the other pipe takes (m - 10) h.
According to the question.
∴ 1/m + 1/(m - 10) = 1/12
⇒ (m - 10 + m)/{m(m - 10)} = 1/12
⇒ 12m( m - 10 + m ) = ( m - 10 )
⇒ m2 - 34m + 120 = 0
⇒ m2 - 30m - 4m + 120 = 0
⇒ ( m - 30 )( m - 4 ) = 0
∴ m = 30 or 4
∴ Faster pipe will take ( 30 - 10 ) h = 20 h to fill the tank.
- A pipe P can fill a tank in 12 min and another pipe R can fill it in 15 min. But, the 3rd pipe M can empty it in 6 min. The 1st two pipes P and R are kept open for double the 2.5 min in beginning and then the 3rd pipe is also opened. In what time is the tank emptied?
-
View Hint View Answer Discuss in Forum
According to the question,
Double the 2.5 min = 5 min
Now, part filled in 5 min = 5 x ( 1/12 + 1/15 ) = [5 x ( 5 + 4)]/60 = (5 x 9)/60 = 3/4
Part emptied in 1 min when P, R and M, all are opened. = 1/6 - ( 1/12 + 1/5 ) = 1/6 - (5 + 4)/60 = 1/60Correct Option: C
According to the question,
Double the 2.5 min = 5 min
Now, part filled in 5 min = 5 x ( 1/12 + 1/15 ) = [5 x ( 5 + 4)]/60 = (5 x 9)/60 = 3/4
Part emptied in 1 min when P, R and M, all are opened. = 1/6 - ( 1/12 + 1/5 ) = 1/6 - (5 + 4)/60 = 1/60
One - sixtieth part is emptied in 1 min.
∴ Three - fourth part will be emptied in 60 x (3/4) = 15 x 3 = 45 min.
