| 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? [SSC (CPO) 2008] |
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
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 \[\frac{40}{11}h.\] |
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