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: 6 3/7

Part filled by X in 1st min and Y in the 2nd min

= (1/6+1/7)=13/42

Part filled by (X+Y) working alternately in 6 min

= 1/2×13/42×6=13/14

Therefore, Remaining part = 1−13/14=1/14

Now, it is the turn of x, one-sixth part is filled in 1 min. One-fourteenth part is filled in (6×1/14) min = 3/7min

Therefore, Required time = (6+3/7)=
6 3/7 min