Question

Nastya walked from her apartment to the library at an average speed of 3 mph. Her husband Martin biked the same route at an average speed of 15 mph. Find the distance between their home and the library if Martin biked a third of an hour less than Nastya walked.

Answer 1

To find the distance between Nastya's apartment and the library, we can set up the problem using the formula for distance:

\[
\text{Distance} = \text{Speed} \times \text{Time}
\]

Let \( d \) be the distance to the library. We know the following:

- Nastya's speed: \( 3 \) miles per hour (mph)
- Martin's speed: \( 15 \) mph
- Martin took \( \frac{1}{3} \) hour less than Nastya.

First, let's express the time it took Nastya and Martin to reach the library.

Let \( t \) be the time (in hours) it took Nastya to walk to the library. Then, the distance can be expressed in terms of Nastya's walking time as:

\[
d = 3t
\]

Since Martin took \( \frac{1}{3} \) hour less than Nastya, his biking time would be:

\[
t - \frac{1}{3}
\]

Using the same distance formula for Martin, we can express the distance as:

\[
d = 15\left(t - \frac{1}{3}\right)
\]

Now, since both expressions represent the same distance \( d \), we can set them equal to each other:

\[
3t = 15\left(t - \frac{1}{3}\right)
\]

Now, expand the right side:

\[
3t = 15t - 5
\]

Next, we will isolate \( t \). Subtract \( 15t \) from both sides:

\[
3t - 15t = -5
\]

This gives us:

\[
-12t = -5
\]

Now, divide both sides by \(-12\):

\[
t = \frac{5}{12} \text{ hours}
\]

Next, we can find the distance \( d \) using the distance formula for Nastya:

\[
d = 3t
\]

Substituting \( t \):

\[
d = 3 \times \frac{5}{12} = \frac{15}{12} = \frac{5}{4} \text{ miles}
\]

Thus, the distance between their home and the library is:

\[
\boxed{1.25} \text{ miles}
\]