| 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? |
A) \[6\frac{2}{7}\min \]
B) \[6\frac{3}{7}\min \]
C) \[6\frac{5}{7}\min \]
D) \[6\frac{1}{7}\min \]
Correct Answer: B
Solution :
| Part filled by X In 1st min and Y in the 2nd min |
| \[=\left( \frac{1}{6}+\frac{1}{7} \right)=\frac{13}{42}\] |
| Part filled by (X + Y) working alternatively is 6min |
| \[=\frac{1}{2}\times \frac{13}{42}\times 6=\frac{13}{14}\] |
| \[\therefore \]Remaining part \[=\left( 1-\frac{13}{14} \right)=\frac{1}{14}\] |
| Now, it is the turn of X, one-sixth part is filled in 1 min. |
| One-fourteenth part is filled in \[\left( 6\times \frac{1}{14} \right)\min =\frac{3}{7}\min \] |
| \[\therefore \]Required time \[=\left( 6+\frac{3}{7} \right)=6\frac{3}{7}\min \] |
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