To solve this problem, we start with the principle of inverse variation. According to the problem, the time \( t \) it takes to travel to City B varies inversely with the speed \( s \) of the train. This can be expressed mathematically as:
\[ t \times s = k \]
where \( k \) is a constant.
From Carole's journey:
- Speed \( s_C = 70 \) mph
- Time \( t_C = 4 \) hours
We can calculate the constant \( k \):
\[ k = t_C \times s_C = 4 \times 70 = 280 \]
Now, for Steve's journey, we have:
- Speed \( s_S = 50 \) mph
We want to find the time \( t_S \):
\[ t_S \times s_S = k \]
Plugging in the values we have:
\[ t_S \times 50 = 280 \]
Now we can solve for \( t_S \):
\[ t_S = \frac{280}{50} = 5.6 \text{ hours} \]
Thus, the time it will take Steve to arrive at City B is 5.6 hours.
The correct answer is: 5.6 hours.
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