Pipes & Cisterns - Solved Problem
Pipes & Cisterns - Solved Problem - 1
Question 1
Pipe A can fill a tank in 60 minutes and Pipe B can empty the tank in 120 minutes.How long will they take to fill the tank if both pipes are opened simultaneously ?
a)120 minutes b)30 minutes c)60 minutes d)45 minutes
Answer: a)120 minutes
Solution:
Pipe A can fill in 1 hour(60 minutes) is 1/1 of the tank.
Pipe B can empty in 1 hour 1/2 of the tank [120 mins= 2hrs]
Both pipes together can fill the tank in 1 hour = 1/1 - 1/2 = 1/2 of the tank.
Since 1/2 part of the tank is filled in 1 hour, the remaining part left is 1/2 of the tank.
The remaining 1/2 part will be filled in other 1 hour.
So both the pipes take 2 hours(120 minutes) to fill the tank.
Question 2
Pipe 1 and pipe 2 can fill a cistern in 2 and 6 hours respectively.Pipe 3 can empty the cistern in 9 hrs.If all the pipes are opened together find the time taken to full the cistern.
a) 1.5 hrs b) 1.4 hrs c) 1.8 hrs d)1.6 hrs
Answer: c)1.8 hrs
Solution :
Pipe1 can fill 1/2 of the cistern in 1 hour
Pipe 2 can fill 1/6 of the cistern in 1 hour
Pipe 3 can empty 1/9 of the cistern in 1 hour
Time taken to full the cistern = 1/2 + 1/6 - 1/9 = 5/9
5/9 of the cistern will be filled in 1 hour.
Full cistern will be filled in 9/5 x 1 = 1.8 hours
Question 3
Two pipes P and Q can fill a cistern in 8 and 12 minutes respectively.Find the time taken to full the cistern if pipe Q is turned off after 4 minutes.
a) 5.3 minutes b)4.5 minutes c)5 minutes d)12 minutes.
Answer : a)5.3 minutes.
Solution:
In 8 mins P can fill the cistern.
Therefore in 1 minute P can fill 1/8 of the cistern.
Similarly Q can fill 1/12 of the cistern in 1 minute
In 4 mins both the pipes can fill 4(1/8 + 1/12)
4 ( 3+2 / 24 ) = 5/6 part of the cistern
The remaining part to be filled = 1 - 5/6 = 1/6
Pipe P Minutes
1/8 1
1/6 ?
8 x 1/6 = 4/3 = 1.3 minutes.
Already P & Q opened together for 4 minutes and P alone for 1.3 minutes.
The total time taken to full the tank is 4 + 1.3 = 5.3 minutes.
Question 4
A pipe P alone can fill a tank in 5 hours and pipe Q alone can fill the same tank in 10 hours. If both P and Q are opened together then find the time taken to fill the tank.
a)2 hours b)5 1/5 hours c)3 1/3 hours d)4 hours
Answer : c)3 1/3 hours
Solution :
Pipe P can fill in one hour = 1/5
Pipe Q can fill in one hour = 1/10
Both the Pipe can fill in one hour = 1/5 + 1/10 = 3/10
Tank Filled Time Taken
3/10 1
1 ?
= 1/(10/3)
Therefore both p and q together fill the tank in 10/3 hours.
i.e., 3 1/3 hours.
Question 5
Two pipes A and B Can fill a tank in 12 hours and 1 day respectively. A pipe C can empty the tank in 10 hours. If Pipe C is opened after 3 hours from the pipe A & B are opened, then find the time taken to full the tank.
a) 24 hours b) 28 hours c) 45 hours d) none of these
Answer : b) 28 hours
Solution :
The part of the tank filled in 1 hour = 1/12 + 1/24
Then, the part filled in 3 hours = 3[1/12 + 1/24] = 3/8
Remaining part to be filled = 1 - 3/8 = 5/8
Net part filled in 1 hour when A, B & C are opened = 1/12 + 1/24 - 1/10 = 1/40
Now, 1/40 part is filled in 1 hour then 5/8 part will be filled in [40 x 5/8] = 25 hours
Therefore, 3/8 part will be filled in 3 hours and 5/8 part filled in 25 hours.
Hence the total time taken to full the tank is 28 hours.
Question 6
Two taps P and Q fill a cistern in 2 days and 3 days respectively and another tap R empties the full cistern in 4 days. If all the taps P,Q and R switched simultaneously then the time taken to fill the cistern is:
a)1day + 120/7 hours b)2days + 12/7hours c)3days + 24/7hours d)1day + 121/7hours
Answer : a)1day + 120/7 hours
Solution :
The tap P can fill a cistern in 1 hour = 1/48 part
The tap Q can fill a cistern in 1 hour = 1/72 part
The tap R can empty a cistern in 1 hour = 1/96 part
Then the net part filled in 1 hour = 1/48 + 1/72 - 1/96
= 7/288
Tank Filled Time Taken
7/288 1
1 ?
The cistern will be filled in = 288/7
288 / 7 can be expressed as = (24 x 7 + 120) / 7 hours = 24 120/7 hours.
i.e., 1 day and 120 / 7 hours
Question 7
Two taps can fill a cistern in 30 and 40 minutes respectively. If both the taps are opened simultaneously then the approximate time taken to fill the cistern is:
a) 17 minutes b) 12 minutes c) 19 minutes d) 21 minutes
Answer : a) 17 minutes
Solution :
Remember the shortcut method:
"Two pipes A and B can fill (or empty) a tank in X and Y minutes respectively, while working alone. If both the pipes are opened together, then the time taken to fill (or empty) the cistern is given by XY/(X+Y) minutes."
Here, X = 30 minutes and Y = 40 minutes
Therefore, the required time = 30 x 40 /(30 + 40) = 1200/70 = 120/7 = 17 1/7 minutes.
Hence the answer is 17 minutes (approximately)
Question 8
A Pipe P can fill a tank in 16 minutes and the other pipe Q can empty the whole tank in 32 minutes. If both P and Q are opened simultaneouly then the time taken to fill the tank is:
a) 16 minutes b) 32 minutes c) 48 minutes d) 40 minutes
Answer : b) 32 minutes
Solution :
Let X hours be the time taken to fill a tank by P.
Let Y hours be the time taken to empty the tank by Q.
Then the time taken to fill the tank when P and Q are switched together : XY / Y-X hours.
Here, X = 16 minutes And Y = 32 minutes
Therefore required time = 16 x 32/(32 - 16) = 32 x 16 / 16 = 32 minutes.
Question 9
Three taps A,B and C are used to fill a cistern. Tap A alone can fill the cistern in 9 minutes. Tap B can fill in 6 minutes and Tap C can fill in 3 minutes. How many minutes will it take to fill this cistern if all the three taps are used simultaneously?
a) 2 3/7 b) 1 7/11 c) 3 2/11 d) 5 6/7
Answer : b) 1 7/11
Solution :
Let the time taken to fill the cistern by 3 taps A, B and C be X, Y, and Z minutes respectively.
Then the short cut formula for,
Time taken to fill the tank when all the pipes are opened = XYZ/(XY + YZ + ZX) minutes
Here, X = 9 minutes, Y = 6 minutes and Z = 3 minutes.
Now the required time = (9)(6)(3)/(9x6)+(6x3)+(3x9) minutes
= 9x6x3 / 54 + 18 + 27
= 9x6x3 / 9{6+2+3} = 6x3/6+2+3 = 18/11 = 1 + 7/11 minutes
Hence the answer is 1 7/11 minutes.
Question 10
Two taps X and Y filled a tank in 22 and 34 minutes respectively. If both X and Y are switched simultaneously then after how much time will the tank get filled?
a) 11 1/14 minutes b) 9 3/14 minutes c) 13 5/14 minutes d) 8 3/14 minutes
Answer : c) 13 5/14 minutes
Solution :
In these type of questions, we first find part of tank filled in 1 minute by both pipes then we add them to get the result.
Time taken by X to fill the tank = 22 minutes.
Then, part filled by X in 1 minute = 1/22
Time taken by Y to fill the tank = 34 minutes.
Then, part filled by Y in 1 minute = 1/34
Net part filled by (X+Y) in 1 minute = 1/22 + 1/34 = (17+11)/374 = 28/374 = 14/187.
Time Part of tank filled
1 minute 14/187
? 1 (here 1 denotes completely filled tank)
Time taken to fill the tank = 187/4 x 1 = 13 5/14 minutes.
Question 11
A pipe A can fill a tank in 120 minutes. Another pipe B can empty the tank in 270 minutes. Find the time taken to fill the tank, if A and B are opened together.
a) 196 minutes b) 296 minutes c) 270 minutes d) 216 minutes
Answer : d) 216 minutes
Solution :
Time taken by A to fill the tank = 120 minutes.
Part filled by A in 1 minute = 1/120
Time taken by B to empty the tank = 270 minutes.
Part filled by B in 1 minute = 1/270
Net Part filled by (A+B) in 1 minute = 1/120 - 1/270 = 5/1080.
Time Part of tank filled
1 minute 5/1080
? 1 (here 1 denotes completely filled tank)
Time taken to fill the tank = 1080/5 x 1 = 216 minutes.
Question 12
Half of the water tank is filled manually. Tap A can fill the tank in 20 minutes and B can empty the tank in 12 minutes. If A and B are opened together, then the time taken to empty or fill the tank is:
a) 30 minutes b) 15/2 minutes c) 60 minutes d) 45/2 minutes
Answer : b) 15/2 minutes
Solution :
Given that, A takes 20 minutes to fill and B takes 12 minutes to empty.
Clearly, tap B is faster than tap A. And so, the tank will be emptied.
Half of the tank or 1/2 part of the tank is already filled.
Therefore, we have to find the time taken to empty that 1/2 part.
Part filled by A in 1 minute = 1/20
Part emptied by B in 1 minute = 1/12.
Part emptied by (A+B) in 1 minute = 1/12 – 1/20 = (5-3)/60 = 2/30.
Therefore, time taken by (A+B) to empty the full tank is 15 minutes.
Time taken to empty 1/2 part of the tank is 30/2 = 15/2 minutes.
Question 13
Tap X can fill a cistern in 10 minutes and tap Y can fill in 12 minutes. There is a leakage in the bottom of the cistern that empties the tank in 24 minutes. If both taps X and Y are opened together, then how long will it take to fill the cistern?
a) 5 5/17 minutes b) 9 12/17 minutes c) 2 16/17 minutes d) 7 1/17 minutes
Answer : d) 7 1/17 minutes.
Solution :
Time taken by X to fill the tank = 10 minutes.
Part filled by X in 1 minute = 1/10
Time taken by Y to fill the tank = 12 minutes.
Part filled by Y in 1 minute = 1/12
Time taken by the leakage to empty the tank = 24 minutes.
Part emptied by the leakage in 1 minute = 1/24
Therefore, net part filled by (X + Y + leakage) in 1 minute = 1/10 + 1/12 – 1/24 = (24+20-10)/240 = 34/240 = 17/120.
Time Part of tank filled
1 minute 17/120
? 1 (here 1 denotes completely filled tank)
Time taken to fill the tank = 120/17 x 1 = 7 1/17 minutes.
Question 14
Two pipes X and Y together can fill a tank in 72 minutes. If the size of the pipe X is thrice as Y then Y alone can fill the tank in:
a) 5 hours and 12 minutes
b) 3 hours and 56 minutes
c) 4 hours and 48 minutes
d) none of these
Answer : c) 4 hours and 48 minutes.
Solution :
Let the time taken by Y alone to fill the tank be A minutes.
Given that, the size of the pipe X is thrice as Y.
Then, X fills the tank in A/3 minutes.
Part filled by X in 1 minute = 1/(A/3) = 3/A
Part filled by Y in 1 minute = 1/A.
Since, X and Y together take 72 minutes.
Part filled by (X+Y) in 1 minute = 1/72
i.e., (1/A + 3/A) = 1/72
4/A = 1/72
A = 288 minutes = 288/60 hours = 4 + 48/60 = 4 + 4/5 hours
= 4 hours and 4/5 x 60 minutes = 4 hours and 48 minutes.
Hence, the pipe Y alone takes 4 hours and 48 minutes to fill the tank.
Question 15
Pipe X can fill a cistern thrice as fast as another pipe Y and the pipe Y is thrice as fast as pipe Z. If X, Y and Z together fill the cistern in 10 minutes then the time taken by X to fill the cistern is:
a) 1 hour and 42 minutes
b) 2 hours and 10 minutes
c) 1 hour and 23 minutes
d) none of these
Answer : b) 2 hours and 10 minutes
Solution :
Let the pipe Z alone takes A minutes to fill the tank.
Given that, Y is thrice as fast as Z.
Then, Y takes A/3 minutes to fill the tank.
And, X is thrice as fast as Y.
X takes (A/3)/3 = A/9 minutes to fill the tank.
Now,
Part filled by X in 1 minute = 9/A
Part filled by Y in 1 minute = 3/A
Part filled by Z in 1 minute = 1/A
Net part filled by (X+Y+Z) in 1 minute = 9/A + 3/A + 1/A = 13/A
(X+Y+Z) take 10 minutes to fill the cistern.
Part filled by (X+Y+Z) in 1 minute = 1/10
Thus, we have, 1/10 = 13/A
A = 130
Therefore, Z alone takes 130 minutes i.e., 2 hours and 10 minutes.
Question 16
Pipe X takes 6 hours to fill a cistern and another pipe Y takes 7 and half hours to fill the same cistern. If the pipes X and Y are switched together at the same time and X is closed after 1 and half hours then the extra time taken by Y to fill the cistern is:
a) 39/8 hours b) 41/8 hours c) 37/8 hours d) none of these.
Answer : a) 39/8 hours
Solution :
X takes 6 hours, part filled by X in 1 hour = 1/6
Y takes 7 and half hours, part filled by Y in 1 hour = 1/7.5 or 2/15.
Therefore, the part filled by X and Y together = 1/6 + 2/15
X and Y together fill the cistern for 1 and half hour i.e., 3/2 hours.
Part filled by (X+Y) in 3/2 hours = (3/2)(1/6 + 1/15) = (3/2)(7/30) = 7/20
Remaining part filled by Y alone = 1 - 7/20 = 13/20.
Part Filled Time Taken by Y
2/15 1 hour
13/20 ?
Time taken to fill 13/20 part by Y alone = 13/20 x 15/2 = 39/8 hours.