Henry Ernest Dudeney/Puzzles and Curious Problems/354 - The Wheels of the Car/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $354$

The Wheels of the Car
"You see, sir," said the motor-car salesman, "at present the fore-wheel of the car I am selling you makes four revolutions more than the hind-wheel in going $120$ yards;
but if you have the circumference of each wheel reduced by $3$ feet, it would make as many as six revolutions more than the hind-wheel in the same distance."
Why the buyer wished that the difference in the number of revolutions between the two wheels should not be increased does not concern us.
The puzzle is to discover the circumference of each wheel in the first case.


Solution

The fore-wheel was $15$ feet in circumference, while the hind-wheel was $18$ feet in circumference.


Proof

We have that $120$ yards is $360$ feet.

We need to find $a$ and $b$ which are divisors of $360$ such that $a n = b \paren {n + 4}$.

We have that:

\(\ds 360\) \(=\) \(\ds 1 \times 360\)
\(\ds \) \(=\) \(\ds 2 \times 180\)
\(\ds \) \(=\) \(\ds 3 \times 120\)
\(\ds \) \(=\) \(\ds 4 \times 90\)
\(\ds \) \(=\) \(\ds 5 \times 72\)
\(\ds \) \(=\) \(\ds 6 \times 60\)
\(\ds \) \(=\) \(\ds 8 \times 45\)
\(\ds \) \(=\) \(\ds 9 \times 40\)
\(\ds \) \(=\) \(\ds 10 \times 36\)
\(\ds \) \(=\) \(\ds 12 \times 30\)
\(\ds \) \(=\) \(\ds 15 \times 24\)
\(\ds \) \(=\) \(\ds 18 \times 20\)

Thus we see that:

$15 \times \paren {20 + 4} = 18 \times 20$

Reducing $15$ and $18$ by $3$ each gives $12$ and $15$, where:

$12 \times \paren {24 + 6} = 15 \times 24$

confirming and corroborating our inkling that the circumferences were $15$ and $18$ feet respectively.

$\blacksquare$


Sources

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