# Pipes and Cisterns Questions Answers

Pipes and Cisterns Questions Answers (MCQ) listings with explanations includes questions of calculating filling of tanks, getting tank empty by pipes etc which are important for many competitive exams.

1) A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other ha

• 1) 15 mins
• 2) 20 mins
• 3) 25 mins
• 4) 30 mins
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Ans.   D
Explanation :
Let the total time be x mins. Part filled in first half means in x/2 = 1/40 Part filled in second half means in x/2 = \begin{aligned} \frac{1}{60}+\frac{1}{40} \ = \frac{1}{24} \\text{ Total = } \\frac{x}{2}*\frac{1}{40} + \frac{x}{2}*\frac{1}{24} = 1 \ => \frac{x}{2} \left(\frac{1}{40}+\frac{1}{24} \right) = 1 \=> \frac{x}{2}*\frac{1}{15} = 1 \=> x = 30 mins \end{aligned}

2) Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be ful

• 1) 3 hours
• 2) 5 hours
• 3) 7 hours
• 4) 10 hours
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Ans.   B
Explanation :
(A+B)'s 2 hour's work when opened = \begin{aligned} \frac{1}{6}+\frac{1}{4} = \frac{5}{12} \ (A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \= \frac{5}{6} \text{Remaining work = } 1-\frac{5}{6} \= \frac{1}{6} \ \text{Now, its A turn in 5 th hour} \\frac{1}{6} \text{ work will be done by A in 1 hour}\\text{Total time = }4+1 = 5 hours \end{aligned}

3) A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time

• 1) 10 mins
• 2) 15 mins
• 3) 20 mins
• 4) 25 mins
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Ans.   C
Explanation :
How we can solve this question ? First we will calculate the work done for 10 mins, then we will get the remaining work, then we will find answer with one tap work, As Part filled by Tap A in 1 min = 1/20 Part filled by Tap B in 1 min = 1/60 (A+B)'s 10 mins work = \begin{aligned} 10*\left(\frac{1}{20}+\frac{1}{60}\right) \= 10*\frac{4}{60} = \frac{2}{3} \\text{Remaining work = } 1-\frac{2}{3} \= \frac{1}{3} \ \text{METHOD 1} \=> \frac{1}{60}:\frac{1}{3}=1:X \=> X = 20 \ \text{METHOD 2} \ 1/60 \text{ part filled by B in} = 1 min \1/3 \text{ part will be filled in} \ = \frac{\frac{1}{3}}{\frac{1}{60}} \= \frac{60}{3} = 20 \end{aligned}

4) A cistern can be filled in 9 hours but due to a leak at its bottom it takes 10 hours. If the cistern is full, then the time that the leak will take to make it empty will b

• 1) 20 hours
• 2) 19 hours
• 3) 90 hours
• 4) 80 hours
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Ans.   C
Explanation :
Part filled without leak in 1 hour = 1/9 Part filled with leak in 1 hour = 1/10 Work done by leak in 1 hour \begin{aligned} = \frac{1}{9} - \frac{1}{10} \= \frac{1}{90} \end{aligned} We used subtraction as it is getting empty. So total time to empty the cistern is 90 hours

5) Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the buck

• 1) 8 min 15 sec
• 2) 7 min 15 sec
• 3) 6 min 15 sec
• 4) 5 min 15 sec
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Ans.   A
Explanation :
Part filled in 3 minutes = \begin{aligned} 3*\left(\frac{1}{12} + \frac{1}{15}\right) \= 3*\frac{9}{60} = \frac{9}{20}\\text{Remaining part }= 1-\frac{9}{20} \ = \frac{11}{20} \\ => \frac{1}{15}:\frac{11}{20}=1:X \\ => X = \frac{11}{20}*\frac{15}{1} \=> X = 8.25 mins \end{aligned} So it will take further 8 mins 15 seconds to fill the bucket.

6) A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tan

• 1) 10 hours
• 2) 15 hours
• 3) 17 hours
• 4) 18 hours
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Ans.   B
Explanation :
Suppose, first pipe alone takes x hours to fill the tank . Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank. As per question, we get \begin{aligned} \frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\=> (2x - 5)(x - 9) = x(x - 5)\=> x^2 - 18x + 45 = 0 \end{aligned} After solving this euation, we get (x-15)(x+3) = 0, As value can not be negative, so x = 15

7) One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank

• 1) 144 mins
• 2) 140 mins
• 3) 136 mins
• 4) 132 minw
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Ans.   A
Explanation :
Let the slower pipe alone fill the tank in x minutes then faster will fill in x/3 minutes. Part filled by slower pipe in 1 minute = 1/x Part filled by faster pipe in 1 minute = 3/x Part filled by both in 1 minute = \begin{aligned} \frac{1}{x} + \frac{3}{x}= \frac{1}{36} \=> \frac{4}{x} = \frac{1}{36} \x = 36*4 = 144 mins \end{aligned}

8) 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litr

• 1) 15 bukets
• 2) 17 bukets
• 3) 18 bukets
• 4) 19 bukets
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Ans.   C
Explanation :
Capacity of the tank = (12*13.5) litres = 162 litres Capacity of each bucket = 9 litres. So we can get answer by dividing total capacity of tank by total capacity of bucket. Number of buckets needed = (162/9) = 18 buckets

9) An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty

• 1) 10 hours
• 2) 13 hours
• 3) 17 hours
• 4) 21 hours
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Ans.   D
Explanation :
We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as Work done for 1 hour without leak = 1/3 Work done with leak = \begin{aligned} 3\frac{1}{2} = \frac{7}{2} \\text{Work done with leak in 1 hr= }\frac{2}{7} \\text{Work done by leak in 1 hr }\= \frac{1}{3} - \frac{2}{7} = \frac{1}{21} \end{aligned} So tank will be empty by the leak in 21 hours.

10) Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank

• 1) 10 mins
• 2) 12 mins
• 3) 15 mins
• 4) 20 mins
View Answer View Explanation
Ans.   B
Explanation :
In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as Part filled by A in 1 min = 1/20 Part filled by B in 1 min = 1/30 Part filled by (A+B) in 1 min = 1/20 + 1/30 = 1/12 So both pipes can fill the tank in 12 mins.

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