Question: 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?
Options:
A) $ 6\frac{2}{7}\min $
B) $ 6\frac{3}{7}\min $
C) $ 6\frac{5}{7}\min $
D) $ 6\frac{1}{7}\min $
Show Answer
Answer:
Correct Answer: B
Solution:
- Part filled by X In 1st min and Y in the 2nd min
$ =( \frac{1}{6}+\frac{1}{7} )=\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 $ =( 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 $ ( 6\times \frac{1}{14} )\min =\frac{3}{7}\min $
$ \therefore $ Required time $ =( 6+\frac{3}{7} )=6\frac{3}{7}\min $