Two persons $P$ and $Q$ start at the same time from city $A$ to city $B$, $60$ km away $P$ travels $40$ km/hr slower than $Q$, $Q$ reaches city $B$ and at once turns back meeting $P$, $12$ km from city $B $.What is the speed of $P$?

When Q meets P, Q has traveled 60 km to city B plus 12 km back, which totals 72 km, while P has traveled 60 minus 12, or 48 km. Since they started at the same time and travel for the same duration, the ratio of their speeds equals the ratio of their distances: speed of Q / speed of P = 72 / 48, which simplifies to 3/2. We are given that P travels 40 km/hr slower than Q, so Q = P + 40; substituting this into the ratio gives (P + 40) / P = 3/2. Cross-multiplying yields 2P + 80 = 3P, which means P = 8 km/hr.

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