Let's outline and solve the given problem:
Given:
1. Last month, Miranda rode the train 90 miles.
2. This month, she has ridden 60 miles.
3. Last month, she spent one more hour traveling than this month.
We need to determine how many hours Miranda spent on the train this month.
Assumptions:
- The speed of the train is constant for both last month and this month.
Let's denote:
- [tex]\( t \)[/tex] as the time spent on the train this month (in hours).
- [tex]\( t + 1 \)[/tex] as the time spent on the train last month (in hours), since she traveled one more hour last month.
Steps to solve:
1. Calculate the speed of the train using the time and distance from last month.
[tex]\[ \text{Speed} = \frac{\text{Distance last month}}{\text{Time last month}} = \frac{90 \text{ miles}}{t + 1 \text{ hours}} \][/tex]
2. Since the train’s speed is constant, the speed using this month's distance and time can be given as:
[tex]\[ \text{Speed} = \frac{\text{Distance this month}}{\text{Time this month}} = \frac{60 \text{ miles}}{t \text{ hours}} \][/tex]
3. Equate the speeds:
[tex]\[ \frac{90}{t + 1} = \frac{60}{t} \][/tex]
4. Solve for [tex]\( t \)[/tex]:
First, cross-multiply to get rid of the fractions:
[tex]\[ 90t = 60(t + 1) \][/tex]
Expand and simplify:
[tex]\[ 90t = 60t + 60 \][/tex]
Subtract [tex]\( 60t \)[/tex] from both sides:
[tex]\[ 30t = 60 \][/tex]
Divide both sides by 30:
[tex]\[ t = 2 \][/tex]
Therefore, Miranda spent 2 hours on the train this month.
Let's summarize the findings in the provided table format:
[tex]\[
\begin{tabular}{|l|c|c|c|}
\hline & \text{Distance (mi)} & \text{Rate (mi/h)} & \text{Time (hr)} \\
\hline Last month & 90 & \frac{90}{3} = 30 & 3 \\
\hline This month & 60 & \frac{60}{2} = 30 & 2 \\
\hline
\end{tabular}
\][/tex]
Thus, the correct answer is 2 hours.
