Two pipes X and Y can fill a cistern in 6 and 7 min, respectively. Starting with pipe X, both the pipes are opened alternately, each for 1 min. In what time will they fill the cistern?
Correct Answer: |
B) \( \Large 6\frac{3}{7} \) |
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Description for Correct answer:
Part filled by X in 1st min and Y in the 2nd min
= \( \Large \left(\frac{1}{6}+\frac{1}{7}\right) = \frac{13}{42} \)
Part filled by (X+Y) working alternately in 6 min
= \( \Large \frac{1}{2} \times \frac{13}{42} \times 6 = \frac{13}{14} \)
Therefore, Remaining part = \( \Large 1 - \frac{13}{14}=\frac{1}{14} \)
Now, it is the turn of x, one-sixth part is filled in 1 min. One-fourteenth part is filled in \( \Large \left(6 \times \frac{1}{14}\right) \) min = \( \Large \frac{3}{7} \)min
Therefore, Required time = \( \Large \left(6 + \frac{3}{7}\right) = 6\frac{3}{7} \) min
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