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A car with 15 inch radius tires was driven on a trip of a distance equal to 400 miles. Two months later, with snow tires, the odometer indicated 390 miles for the same trip. Find the radius of the snow tires.

This is how I attempted to solve it.

$2\pi \left( 15 \right)n\; =\; 400\cdot 5280\cdot 12$

Where n is the number of revolutions of the tire, and I converted everything to inches.

You get that n = $\frac{844800}{\pi }$

Then $2\pi \left( r \right)\left( \frac{844800}{\pi } \right)\; =\; 390\cdot 5280\cdot 12$

Solving for r gives $\frac{117}{8}$ inches, but after checking the answer, it seems that it is $\frac{200}{13}$ inches.

Where did I go wrong? I have a feeling it has to do with my approach, as n - the number of revolutions - will not be same for both tires.

1110101001
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2 Answers2

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If the first trip was $400$ miles, then it was also $400 * 5280 * 12$ inches, and so the $15$ inch radius tires, having a circumference of $30\pi$ inches, would have made $\frac{844800}{\pi}$ revolutions, as you calculated.

However, it seems that the odometer measures by number of revolutions, and not by actual distance, and so on the trip with the snow tires, there must have been $\frac{390}{400}$ times as many revolutions, or $\frac{823680}{\pi}$ revolutions.

Now, as this number of revolutions multiplied by $2\pi r$ should end up actually being equal to the actual distance of $400 * 5280 * 12 = 25344000$ inches, we get that $r = \frac{25344000}{1647360}$, which then simplifies to $\frac{200}{13}$, the answer that you were given.

Edit: Just in equations:

$$\frac{n_1}{n_2} = \frac{400}{390} \quad\text{[As this is how the odometer measures.]}$$

$$2\pi(15) * n_1 = 2\pi r * n_2 \quad\text{[Because the distance travelled was the same in both cases.]}$$

$$r = 15 * \frac{n_1}{n_2} \quad\text{[Rearranging the second equation.]}$$

$$r = \frac{200}{13} \quad\text{[By the previous and the first equations.]}$$

qaphla
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Try more symbolics: Let $r_1$ the radius of the tire, $d_1$ the first measured distance, $d_2$ the second measured distance and $r_2$ the radius of the snow tires and $n_1, n_2$ the respective number of revolutions. Then the measurements are $$2\pi r_1 \cdot n_1 = d_1 = 2\pi r_2 \cdot n_2, 2\pi r_1 \cdot n_2 = d_2$$ so we have $$n_2 = \frac{d_2}{2\pi r_1} = \frac{d_2}{\frac{d_1}{n_1}} = \frac{d_2 n_1}{d_1}$$ and then $$r_2 = \frac{d_1}{2\pi n_2} = \frac{d_1}{2\pi \frac{d_2 n_1}{d_1}} = \frac{d_1}{2\pi n_1 \frac{d_2}{d_1}} = \frac{d_1}{\frac{d_1}{r_1}\frac{d_2}{d_1}} = \frac{d_1}{d_2} r_1$$ So finally $$ r_2 = \frac{400}{390} \cdot 15 = \frac{600}{39} = \frac{200}{13} \text{[in]}$$

AlexR
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  • $2\pi r_{1}; \cdot ; n_{1}; =; 2\pi r_{2}\cdot n_{2}; =; d_{1}$ is because the odometer measures in revolutions, not miles - right? Because otherwise wouldn't $2\pi r_{2}\cdot n_{2}; =; d_{2}; =; 390$, as per the text in the problem? – 1110101001 Aug 05 '13 at 21:03
  • Yes, that is correct. The odometer calculates the distance on basis of the constant circumference $2\pi r_1$ and the revolution count $n_1, n_2$ resp. – AlexR Aug 06 '13 at 10:29