Question

A cyclist rides into the country at an average speed of 10 miles per hour. When his bicycle gets a flat tire he walks back at an average speed of 3 miles per hour if he returns home 6 and a half hours after he starts how far into the country does he go

Answer 1

Let \( d \) be the distance the cyclist rides into the country in miles.

The time taken to ride into the country at an average speed of 10 miles per hour is given by:

\[
\text{Time riding} = \frac{d}{10}
\]

The time taken to walk back at an average speed of 3 miles per hour is given by:

\[
\text{Time walking} = \frac{d}{3}
\]

The total time for the trip (riding out and walking back) is given as 6.5 hours:

\[
\frac{d}{10} + \frac{d}{3} = 6.5
\]

To solve this equation, we first find a common denominator for the fractions. The least common multiple of 10 and 3 is 30. We can rewrite the equation as:

\[
\frac{3d}{30} + \frac{10d}{30} = 6.5
\]

Combining the terms on the left side:

\[
\frac{3d + 10d}{30} = 6.5
\]

\[
\frac{13d}{30} = 6.5
\]

Now, multiply both sides by 30 to eliminate the fraction:

\[
13d = 6.5 \times 30
\]

Calculating \( 6.5 \times 30 \):

\[
6.5 \times 30 = 195
\]

So we have:

\[
13d = 195
\]

Now, divide both sides by 13 to solve for \( d \):

\[
d = \frac{195}{13} = 15
\]

Thus, the distance the cyclist goes into the country is:

\[
\boxed{15} \text{ miles}
\]