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Question Detail
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 be ?
- 20 hours
- 19 hours
- 90 hours
- 80 hours
Answer: Option 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
1. 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 bucket?
- 8 min 15 sec
- 7 min 15 sec
- 6 min 15 sec
- 5 min 15 sec
Answer: Option 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.
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 full ?
- 3 hours
- 5 hours
- 7 hours
- 10 hours
Answer: Option 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 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 be ?
- 20 hours
- 19 hours
- 90 hours
- 80 hours
Answer: Option 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
4. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is
- 5780 litres
- 5770 litres
- 5760 litres
- 5750 litres
Answer: Option C
Explanation:
Work done by the inlet in 1 hour =
\begin{aligned}
\frac{1}{6}-\frac{1}{8} = \frac{1}{24} \\
\text{Work done by inlet in 1 min} \\
= \frac{1}{24}*\frac{1}{60}\\
= \frac{1}{1440} \\
=> \text{Volume of 1/1440 part = 4 liters} \\
\end{aligned}
Volume of whole = (1440 * 4) litres = 5760 litres.
5. 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 ?
- 10 hours
- 13 hours
- 17 hours
- 21 hours
Answer: Option 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.
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