A man leaves from P at 6 am and reaches Q at 2 pm on the same day. Another man leaves Q at 8 am and reaches at 3 pm on the same day. At what time do they meet?

If PQ = x km, then

Speed of person starting from P = $$\frac{x}{8}$$ kmph

Speed of person starting from Q = $$\frac{x}{7}$$ kmph

Relative speed = $$\frac{x}{8} + \frac{x}{7}$$

= $$\frac{15x}{56} $$kmph

Distance covered by first person till 8 a.m. = 2 $$\times \frac{x}{8} = \frac{x}{4}$$

Remaining distance = x -$$\frac{x}{4} km = \frac{3x}{4} $$

Time of meeting = $$\frac{Remaining distance}{Relative speed}$$

= $$\frac{\frac{3x}{4}}{\frac{15x}{56}}$$

= 2.08 hours

∴∴ Required time = 10:48 a.m.