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### Exercise 3: Car moving at steady speed
The problem states that a car is moving at a constant speed, so the distance \(d\) traveled is proportional to the time \(t\).

#### (a) Ratio of distance to time
The car travels 170 miles in 2.5 hours. We are asked to find the ratio of distance to time and express it as a unit rate.

\[
\text{Unit rate} = \frac{170 \text{ miles}}{2.5 \text{ hours}} = 68 \text{ miles per hour}
\]

This value represents the speed of the car. The car is traveling at 68 miles per hour.

#### (b) Distance traveled in 4 hours
Now, we need to set up a proportion to find how far the car will travel in 4 hours at the same speed.

Let \( d \) represent the distance traveled in 4 hours:
\[
\frac{170 \text{ miles}}{2.5 \text{ hours}} = \frac{d \text{ miles}}{4 \text{ hours}}
\]

We can solve this proportion:
\[
d = \frac{170}{2.5} \times 4 = 68 \times 4 = 272 \text{ miles}
\]

The car will travel 272 miles in 4 hours.

#### (c) Time needed to travel 425 miles
We are given that the car travels 425 miles and need to find the time it would take. Set up a proportion:

\[
\frac{170 \text{ miles}}{2.5 \text{ hours}} = \frac{425 \text{ miles}}{t \text{ hours}}
\]

Solve for \( t \):
\[
t = \frac{425}{170} \times 2.5 = 2.5 \times 2.5 = 6.25 \text{ hours}
\]

The car would take 6.25 hours to travel 425 miles.

---

### Exercise 4: Weight and volume of water
This problem describes the weight of water being proportional to its volume.

#### (a) Find the weight of 55 gallons of water
The relationship is given that 20 gallons of water weigh 166 pounds. We need to set up a proportion to find the weight \( w \) of 55 gallons of water.

\[
\frac{166 \text{ pounds}}{20 \text{ gallons}} = \frac{w \text{ pounds}}{55 \text{ gallons}}
\]

Solving for \( w \):
\[
w = \frac{166}{20} \times 55 = 8.3 \times 55 = 456.5 \text{ pounds}
\]

The weight of 55 gallons of water is 456.5 pounds.

#### (b) Find how many gallons weigh 1,028 pounds
We are given 1,028 pounds of water and need to find the number of gallons \( g \).

\[
\frac{166 \text{ pounds}}{20 \text{ gallons}} = \frac{1028 \text{ pounds}}{g \text{ gallons}}
\]

Solving for \( g \):
\[
g = \frac{1028}{166} \times 20 = 6.2 \times 20 = 124 \text{ gallons}
\]

The tank would hold 124 gallons of water if the weight is 1,028 pounds.

---

Would you like more details on any of these solutions or further clarification? Here are some additional questions to deepen understanding:

1. What does it mean for two variables to be proportional?
2. How do you interpret unit rates in real-life problems?
3. What happens to the unit rate if the time increases but the distance remains the same?
4. How do you determine the relationship between volume and weight in the water tank problem?
5. How would the formulas change if the car's speed varied instead of being constant?

**Tip:** When working with proportions, always ensure that the units on both sides of the equation match for accurate calculations.

A car is moving at a steady speed on a highway such that the distance it travels is proportional to the time it has traveled. The car travels a distance of 170 miles in 2.5 hours. (a) What is the ratio of the distance to time traveled expressed as a fraction and then as a unit rate? What does this value represent? (b) If the car travels at this rate for a total of 4 hours, what distance has it traveled? Set up a proportion to find the distance. (c) If the car traveled 425 miles, set up and solve a proportion that could be used to find the time it had been driving.

Explore how to solve real-world proportion problems involving a car moving at a constant speed. The solution covers unit rates, setting up proportions, and solving for unknowns, with questions related to distance and time. Suitable for students in Grades 7-9.

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