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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-5

  • question_answer
    Three pipes P, Q and R can separately fill a cistern in 4, 8 and 2 hours respectively. Another pipe \[\text{S}\]can empty the completely filled cistern in 10 hours. Which of the following arrangements will fill the empty cistern in less time than others?

    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 filled in 1 hour 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 \[\frac{20}{3}\] hours. Part of the cistern filled in 1 hour 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 \[\frac{7}{30}\] hours. Part of the cistern filled in I hour when pipes P, Q 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\[\frac{40}{11}\]Hours. Hence, option [d] is correct.


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