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Let's break down the problem:

### Performance Task: 
Benito and Tyler are painting opposite sides of the same 150-foot fence. 
- **Tyler** has already painted **19.5 feet** (19½ feet) when **Benito** starts.
- **Benito's painting rate**: 15 feet per minute.
- **Tyler's painting rate**: 11 feet per minute.

---

### **Part A:**
How long will it take for both sides to have an equal number of feet painted? How many feet will be painted on Benito's side of the fence when the two sides have an equal number of feet painted?

Let the time it takes for them to paint equal lengths be \( t \) minutes after Benito starts.

- Benito paints at 15 ft/min, so in \( t \) minutes, Benito paints \( 15t \) feet.
- Tyler paints at 11 ft/min, but he already painted 19.5 feet, so in \( t \) minutes, he paints \( 19.5 + 11t \) feet.

We need to solve for \( t \) when they have painted the same number of feet:

\[
15t = 19.5 + 11t
\]

Solve for \( t \):

\[
15t - 11t = 19.5
\]
\[
4t = 19.5
\]
\[
t = \frac{19.5}{4} = 4.875 \text{ minutes}
\]

Now, find how many feet Benito has painted in 4.875 minutes:

\[
15 \times 4.875 = 73.125 \text{ feet}
\]

So, after 4.875 minutes, both Benito and Tyler will have painted 73.125 feet each.

---

### **Part B:**
Tyler claims that because he started painting first, he will finish painting his side of the fence before Benito finishes. Is this true?

Tyler has 150 feet total to paint, and he already painted 19.5 feet. He has:

\[
150 - 19.5 = 130.5 \text{ feet remaining}
\]

At 11 feet per minute, Tyler will take:

\[
\frac{130.5}{11} \approx 11.86 \text{ minutes to finish}
\]

Benito has 150 feet to paint at 15 feet per minute:

\[
\frac{150}{15} = 10 \text{ minutes to finish}
\]

Since Benito will finish in 10 minutes and Tyler will take about 11.86 minutes, Tyler's claim is **false**.

---

### **Part C:**
The painter who finishes first gets to rest while the other finishes. How long will the painter who finishes first get to rest?

Benito finishes in 10 minutes, and Tyler finishes in about 11.86 minutes. The difference is:

\[
11.86 - 10 = 1.86 \text{ minutes}
\]

So, Benito will get to rest for **1.86 minutes**.

---

Let me know if you'd like more details or have questions. Here are 5 follow-up questions for better understanding:

1. How do you set up equations when dealing with two rates and an initial condition?
2. What does it mean when the rates of work or progress differ between two workers?
3. How can the concept of simultaneous work be applied in other real-life situations?
4. What steps should you follow to check the accuracy of your solution?
5. Why does solving for time first help in problems involving different rates?

**Tip**: When comparing rates of work, always pay attention to any head starts or initial amounts—these can greatly impact the overall solution!

Benito and Tyler are painting opposite sides of the same 150-foot fence. Tyler has already painted 19½ feet of his side of the fence when Benito starts painting. Benito paints at 15 feet per minute, and Tyler paints at 11 feet per minute. The task involves determining how long it will take for both sides to have an equal number of feet painted, checking Tyler's claim about finishing first, and calculating rest time for the painter who finishes first.

Learn how to solve a real-world rate problem involving two painters working on opposite sides of a 150-foot fence. This problem explores algebraic equations, rates of work, and time calculations suitable for students in Grades 6-8.

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