# 1997 PMWC Problems/Problem T4

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## Problem

In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey?

## Solution

Time $t$, velocity $v$, distance $a$, formula $v=\frac{a}{t}$ or $a=v \cdot t$. Time of departure $T$. First, ferry 1: $a_{1}=v_{1}(8.20-T)$ and ferry 2: $a_{2}=v_{2}(8.20-T)$; sum $a=a_{1}+a_{2}=(v_{1}+v_{2})(8.20-T)$. Second, difference in time: $9.11-8.20-0.15=0.36$. Ferry 1: $a_{2}+a_{4}=v_{1}\cdot (0.36)$ and ferry 2: $a_{1}+a_{3}=v_{2}\cdot (0.36)$; sum $2a=(v_{1}+v_{2})(0.36)$. Dividing: $\frac{1}{2}=\frac{8.20-T}{0.36}$, so $T=8.02$.