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Banking Sample Paper SBI Junior Associates (Main) Sample Paper Test-2

  • question_answer
    Two pipes P and Q can fill a cistern in 40 minutes and 60 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all three pipes are opened simultaneously, then the cistern is full in 30 minutes. In how much time can the third pipe empty the cistern?

    A)  100 minutes               

    B)  120 minutes

    C)  115 minutes

    D)                   1.10 minutes

    E)  None of these

    Correct Answer: B

    Solution :

    Part of cistern emptied by third pipe in 1 min \[=\frac{1}{30}-\left[ \left( \frac{1}{40}+\frac{1}{60} \right) \right]\] \[=\frac{1}{30}-\left[ \frac{3+2}{120} \right]=\frac{1}{30}-\frac{1}{24}\] \[=\frac{4-5}{120}=-\frac{1}{120}\] Hence, the third pipe alone empties the cistern in 120 minutes. Another Method: LCM of 40, 60 and 30 is 120 units. Let the capacity of the tank be 120 units. Then, A can fill \[\frac{120}{60}=2\]units/m Now, (A+B-C) can fill the tank \[=\left( \frac{120}{30}=4 \right)\]units/m \[\therefore \] C can empty (5 - 4 =) 1 unit per minute Hence C can empty the full tank in \[\frac{120}{1}\] = 120 minutes


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