Select the correct answer.

Walter takes 710 of an hour to mow 25 of an acre of lawn. At this rate, how many acres will he mow in an hour?

 **A.** 

725 acre

 **B.** 

47 acre 

 **C.** 

115 acres

 **D.** 

134 acres

We are given that Walter takes \( \frac{7}{10} \) of an hour to mow \( \frac{2}{5} \) of an acre. We need to determine how many acres Walter can mow in 1 hour at this rate.

To solve this, we can set up a proportion. Let \( x \) represent the number of acres Walter can mow in 1 hour. The proportion is:

\[
\frac{\text{Acres}}{\text{Time}} = \frac{2/5}{7/10} = \frac{x}{1}
\]

To solve for \( x \), we first divide \( \frac{2}{5} \) by \( \frac{7}{10} \):

\[
\frac{2/5}{7/10} = \frac{2}{5} \times \frac{10}{7} = \frac{2 \times 10}{5 \times 7} = \frac{20}{35} = \frac{4}{7}
\]

So, Walter can mow \( \frac{4}{7} \) of an acre in 1 hour.

Now, let's compare this to the answer choices:

**A.** \( \frac{7}{25} \) acre  
**B.** \( \frac{4}{7} \) acre  
**C.** \( \frac{1}{15} \) acre  
**D.** \( \frac{1}{34} \) acre

The correct answer is **B** \( \frac{4}{7} \) acre.

Would you like any further details or explanations on the steps? Here are five related questions you might consider:

1. How do you set up proportions to solve time and rate problems?
2. How do you divide fractions step-by-step?
3. Can this method be applied to similar distance-rate-time problems?
4. What are common mistakes to avoid when solving proportion problems?
5. How does simplifying fractions affect the solution?

**Tip:** Always simplify fractions when possible to make the calculations easier.

Walter takes 7/10 of an hour to mow 2/5 of an acre of lawn. At this rate, how many acres will he mow in an hour?
A. 7/25 acre
B. 4/7 acre
C. 1/15 acres
D. 1/34 acres

Solve how many acres Walter can mow in 1 hour if it takes him 7/10 of an hour to mow 2/5 of an acre. This proportion problem involves dividing fractions and working with rates. Suitable for students in Grades 6-8.

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