Henry Ernest Dudeney/Puzzles and Curious Problems/71 - The Three Motor-Cars/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $71$

The Three Motor-Cars
Three motor-cars travelling along a road in the same direction are, at a certain moment, in the following positions in relation to one another.
Andrews is a certain distance behind Brooks,
and Carter is twice that distance in front of Brooks.
Each car travels at its own uniform rate of speed,
with the result that Andrews passes Brooks in seven minutes,
and passes Carter five minutes later.
Now, in how many minutes after Andrews would Brooks pass Carter?


Solution

$6 \tfrac 2 3$ minutes.


Proof

The trick of this puzzle is to consider the relative speeds instead of absolute speeds.

Let $A$, $B$ and $C$ denote the current positions of Andrews, Brooks and Carter respectively.

Let $d$ be the distance between where $A$ and $B$ are now.

Let $v_A$, $v_B$ and $v_C$ units per minute be the speeds of $A$, $B$ and $C$ respectively.

Let $t$ be the number of minutes since the start for Brooks to pass Carter.

We have:

\(\text {(1)}: \quad\) \(\ds \dfrac d {v_A - v_B}\) \(=\) \(\ds 7\) Andrews passes Brooks in seven minutes,
\(\text {(2)}: \quad\) \(\ds \dfrac {3 d} {v_A - v_C}\) \(=\) \(\ds 12\) and passes Carter five minutes later.
\(\text {(3)}: \quad\) \(\ds \dfrac {2 d} {v_B - v_C}\) \(=\) \(\ds t\)

Write $x = v_A - v_B$ and $y = v_A - v_C$.

Then $v_B - v_C = y - x$.

So we have:

\(\text {(4)}: \quad\) \(\ds d\) \(=\) \(\ds 7 x\) from $(1)$
\(\text {(5)}: \quad\) \(\ds \) \(=\) \(\ds 4 y\) from $(2)$
\(\ds \) \(=\) \(\ds \frac t 2 \paren {y - x}\) from $(3)$
\(\ds \) \(=\) \(\ds \frac t 2 \paren {\frac d 4 - \frac d 7}\) substituting $x$ and $y$ from $(4)$ and $(5)$
\(\ds \) \(=\) \(\ds \frac {3 d t} {56}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \frac {56} 3\)
\(\ds \) \(=\) \(\ds 18 \tfrac 2 3\)

therefore Brooks passes Carter $6 \tfrac 2 3$ minutes after Andrew does.

$\blacksquare$


Sources

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