Quantitative Aptitude Ques 874
Question: Three pipes P, Q and R can separately fill a cistern in 4, 8 and 12 h, respectively. Another pipe S can empty the completely filled cistern in 10 h. Which of the following arrangements will fill the empty cistern in less time than others?
Options:
A) Q alone is open
B) P and S are open
C) P, R and S are open
D) P, Q and S are open
Show Answer
Answer:
Correct Answer: D
Solution:
- Part of the cistern tilled in 1 h when pipes P and S are open
$ =\frac{1}{4}-\frac{1}{10}=\frac{5-2}{20}=\frac{3}{20} $
Hence, the cistern will be filled in $ \frac{20}{3}h. $
Part of the cistern filled in 1 h when pipes P, R and S are open
$ =\frac{1}{4}+\frac{1}{12}-\frac{1}{10} $
$ =\frac{15+5-6}{60}=\frac{14}{60}=\frac{7}{30} $
Hence, the cistern will be filled in $ \frac{30}{7}h. $
Part of the cistern filled in 1 h when pipes P, O and S are open
$ =\frac{1}{4}+\frac{1}{8}-\frac{1}{10} $
$ =\frac{10+5-4}{40}=\frac{11}{40} $
Hence, the cistern will be filled in $ \frac{40}{11}h. $
BETA
