Discussion

A runs 1 3/4 times as fast as B. If A gives B a start of 60 meters. How far must the winning post be in order that A and B reach it at the same time?

Answer: B

A is \( 1\tfrac{3}{4} = \tfrac{7}{4} \) times as fast as B, and B starts 60 m ahead. Let the post be \( d \) metres away.

\[ \frac{d}{7/4} = \frac{d - 60}{1} \]

\[ 4d = 7d - 420 \quad\Longrightarrow\quad 3d = 420 \]

\[ d = 140 \text{ m} \]

Checking: A runs 140 m while B runs 80, and \( \tfrac{140}{80} = \tfrac{7}{4} \).

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