Travel Suppose a family drives at an average rate of 60 mi/h on the way to visit relatives and then at an average rate of 40 mi/h on the way back. The return trip takes 1 h longer than the trip there. a. Let d be the distance in miles the family traveled to visit their relatives. How many hours did it take to drive there? b. In terms of d, how many hours did it take to make the return trip? c. Write and solve an equation to determine the distance the family drove to see their relatives. What was the average rate for the entire trip?

### a. How many hours did it take to drive to the relatives?

Let \( d \) represent the distance in miles the family traveled to visit their relatives. The family drove at an average rate of 60 miles per hour on the way there. Using the formula:

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

The time taken to drive to their relatives, \( t_1 \), is:

\[
t_1 = \frac{d}{60}
\]

### b. In terms of \( d \), how many hours did it take to make the return trip?

On the way back, the family drove at an average rate of 40 miles per hour, and the return trip took 1 hour longer than the trip there. So, the time for the return trip, \( t_2 \), can be expressed as:

\[
t_2 = \frac{d}{40}
\]

Given that the return trip took 1 hour longer than the trip there, we have:

\[
t_2 = t_1 + 1
\]

Substitute the expressions for \( t_1 \) and \( t_2 \) into this equation:

\[
\frac{d}{40} = \frac{d}{60} + 1
\]

### c. Solve the equation to determine the distance \( d \)

To solve for \( d \), first eliminate the denominators by multiplying the entire equation by the least common multiple of 40 and 60, which is 120:

\[
120 \times \left( \frac{d}{40} \right) = 120 \times \left( \frac{d}{60} + 1 \right)
\]

This simplifies to:

\[
3d = 2d + 120
\]

Now, subtract \( 2d \) from both sides:

\[
d = 120
\]

Thus, the distance the family traveled to visit their relatives is \( d = 120 \) miles.

### What was the average rate for the entire trip?

The total distance traveled for the entire trip is \( 2d = 240 \) miles.

The total time for the trip is:

\[
\text{Total time} = t_1 + t_2 = \frac{120}{60} + \frac{120}{40} = 2 + 3 = 5 \, \text{hours}
\]

Now, the average speed for the entire trip is given by:

\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{240}{5} = 48 \, \text{mi/h}
\]

### Summary:
- The distance to the relatives is 120 miles.
- The total trip took 5 hours.
- The average speed for the entire trip is 48 mi/h.

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Would you like any further clarification or details on this?

Here are 5 related questions for further practice:
1. If the average speed on the way there was 50 mi/h, how would the travel time change?
2. What would the average speed be if the return trip took 2 hours longer instead of 1 hour longer?
3. How would the total travel time change if the distance was 200 miles instead of 120 miles?
4. What would the total time be if the return speed was 45 mi/h?
5. If the family took a break halfway, how would that affect the total trip time?

**Tip:** When solving word problems, it's often helpful to translate words into algebraic expressions and solve step by step.

Suppose a family drives at an average rate of 60 mi/h on the way to visit relatives and then at an average rate of 40 mi/h on the way back. The return trip takes 1 h longer than the trip there. a. Let d be the distance in miles the family traveled to visit their relatives. How many hours did it take to drive there? b. In terms of d, how many hours did it take to make the return trip? c. Write and solve an equation to determine the distance the family drove to see their relatives. What was the average rate for the entire trip?

Learn how to solve a real-world algebra problem involving different travel speeds and times. The family travels to visit relatives at 60 mi/h and returns at 40 mi/h. Find the distance traveled and the average speed for the entire trip. This step-by-step solution is ideal for students in Grades 8-10.

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