Two persons A and B start simultaneously from P and Q respectively. A meets B at a distance of 60m from P. After A reaches Q and B reaches P, they turn around and start walking in opposite direction, now B meets A at a distance of 40m from Q. Find distance between P and Q?

P_____60m___R__Xm____40m____Q
Total distance = 100 + X
Initially, A traveled 60m and B traveled (40 + X)m. Since time is same. So, Speed ∝ Distance.
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}}$$
Second time,
So, now total distance traveled by A = 100 + X + 40
Total distance traveled by B = 100 + X + 60 + X
These will be in the ratio of Speed of A and B
Thus,
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}} = \frac{{100 + {\text{X}} + 40}}{{100 + {\text{X}} + 60 + {\text{X}}}}$$
X = 40
Thus Total distance = 100 + X = 100 + 40 = 140 m