Let's break down the problem step by step:
1. Define the variables and relationships:
- Let [tex]\( d \)[/tex] be the rate at which Diana can paint fence posts per hour.
- Let [tex]\( c \)[/tex] be the rate at which Chuck can paint fence posts per hour.
- According to the problem, Diana can paint 10 fence posts more per hour than Chuck. This gives us:
[tex]\[
d = c + 10
\][/tex]
2. Given relationship:
- Diana can paint 150 fence posts in the same time Chuck can paint 130 fence posts. This relationship can be expressed as:
[tex]\[
\frac{150}{d} = \frac{130}{c}
\][/tex]
- We know the relationship between [tex]\( d \)[/tex] and [tex]\( c \)[/tex] from above, we substitute [tex]\( d \)[/tex] from [tex]\( d = c + 10 \)[/tex] into the equation:
[tex]\[
\frac{150}{c + 10} = \frac{130}{c}
\][/tex]
3. Solve the equation:
- Cross-multiplying to get rid of the fractions, we get:
[tex]\[
150c = 130(c + 10)
\][/tex]
- Expanding the equation:
[tex]\[
150c = 130c + 1300
\][/tex]
- Isolate [tex]\(c\)[/tex] by subtracting [tex]\(130c\)[/tex] from both sides:
[tex]\[
20c = 1300
\][/tex]
- Solve for [tex]\(c\)[/tex]:
[tex]\[
c = \frac{1300}{20} = 65
\][/tex]
4. Verification:
- We found that [tex]\( c = 65 \)[/tex]. To verify, let's calculate Diana’s rate:
[tex]\[
d = c + 10 = 65 + 10 = 75
\][/tex]
- Rechecking the original equation:
[tex]\[
\frac{150}{75} = 2 \quad \text{and} \quad \frac{130}{65} = 2
\][/tex]
- Both sides equal 2, confirming the solution is indeed correct.
So, the number of fence posts Chuck can paint per hour is [tex]\( \boxed{65} \)[/tex].
