Quantitative Aptitude Ques 1418
Question: Three taps are fitted to a cistern. The empty cistern is filled by the first and 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, then the empty cistern will be filled up in
Options:
A) $ 1\frac{14}{23}h $
B) $ 2\frac{14}{23}h $
C) $ 2h40\min $
D) $ 1h56\min $
Show Answer
Answer:
Correct Answer: B
Solution:
- Given, time taken by first tap to fill the cistern $ =3h $
Part of cistern filled by first tap in $ 1h=\frac{1}{3} $ … (i)
(if a pipe fills a tank in x h, then the part of tank filled in $ 1h=\frac{1}{x} $ )
Similarly, part of cistern filled by second tap in $ 1h=\frac{1}{4} $
(ii)
and part of cistern emptied by third tap in $ 1h=\frac{1}{5} $
(iii)
Now, part of cistern filled by all taps In 1 h
$ =\frac{1}{3}+\frac{1}{4}-\frac{1}{5}=\frac{20+15-12}{60}=\frac{23}{60} $
Hence, all three taps can fill the cistern in $ \frac{60}{23} $ or $ 2\frac{14}{23}h. $
(If a pipe fills $ \frac{1}{x} $ part of the tank in 1 h, then the time taken by the pipe to fill the full tank $ =xh $ )
BETA
