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P and Q together can do a job in 6 days. Q and R can finish the same job in $$\frac{{60}}{7}$$ days. P started the work and worked for 3 days. Q and R continued for 6 days to finish the work. Then the difference of days in which R and P can complete the alone is P can complete the job alone is ?

Correct Answer: 10 days
L.C.M. of Total Work = 60
One day work of P + Q = $$\frac{{60}}{{6}}$$ = 10 unit/day efficiency
One day work of Q + R = $$\frac{{60}}{{\frac{{60}}{7}}}$$ = 7 unit/day efficiency
$$\eqalign{ & \left( {{\text{Q}} + {\text{R}}} \right){\text{ 6 days work}} \cr & = 7 \times 6 \cr & = 42{\text{ units}} \cr} $$
Then in 3 days = total work = 18
$$\eqalign{ & {\text{P completes}} \cr & = 60 - 42 \cr & = 18{\text{ units}} \cr & {\text{P's efficiency}} \cr & = \frac{{18}}{3} \cr & = 6{\text{ units/day}} \cr & {\text{Q's efficiency}} \cr & = 10 - 6 \cr & = 4{\text{ units/day}} \cr & {\text{R's efficiency}} \cr & = 7 - 4 \cr & = 3{\text{ units/day}} \cr & {\text{P completes whole work in}} \cr & = \frac{{60}}{6} \cr & = {\text{10 days}} \cr & {\text{R completes whole work in}} \cr & = \frac{{60}}{3} \cr & = {\text{20 days}} \cr & {\text{Difference is}} \cr & = \left( {20 - 10} \right) \cr & = 10{\text{ days }} \cr} $$

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