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SSC Sample Paper Mock Test-8 SSC CGL Tear-II Paper-1

  • question_answer
       Two pipes can fill a cistern in 14 h and 16 h, respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom. In how many hours the tank will be empty?  

    A) 90

    B)  85      

    C)  115     

    D)  112

    Correct Answer: D

    Solution :

    Work done by two pipes in 1 h \[=\left( \frac{1}{4}+\frac{1}{16} \right)=\frac{15}{112}\] Total time taken by two pipes and leak in 1 h \[=\left( \frac{112}{15}+\frac{32}{60} \right)\]h = 8 h Work done by the leak in 1 h \[=\left( \frac{15}{112}-\frac{1}{8} \right)=\frac{1}{112}\] Hence, the leak will empty the full cistern in 112 h


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