Gilda walks to the train station. If she walks at the rate of 3 mph, she misses her train by 7 minutes. However, if she walks at the rate of 4 mph, she reaches the station 5 minutes before the arrival of the train. Find the distance Gilda walks to the station.
Using d = rt, the distance is the same, but the rate and time are different.
If Gilda misses the train, it means the time t needs 7 more minutes so
d = 3(t + 7). If she gets to the station 5 minutes early means tcan be 5 minutes less so d = 4(t - 5).
3(t + 7) = 4(t - 5)
3t + 21 = 4t - 20
t = 41
d = rt, so d = 3(41) = 123

Question

Gilda walks to the train station. If she walks at the rate of 3 mph, she misses her train by 7 minutes. However, if she walks at the rate of 4 mph, she reaches the station 5 minutes before the arrival of the train. Find the distance Gilda walks to the station. 
Using d = rt, the distance is the same, but the rate and time are different.
If Gilda misses the train, it means the time t needs 7 more minutes so 
d = 3(t + 7). If she gets to the station 5 minutes early means tcan be 5 minutes less so d = 4(t - 5).
3(t + 7) = 4(t - 5)
3t + 21 = 4t - 20
t = 41
d = rt, so d = 3(41) = 123

ellie202000 · Answer

Find David's mistake in his calculations. In two or more complete sentences, explain his mistake. Include the correct calculations and solutions in your answer.

anonymous · Answer

The units for time in this problem is hours.  Minutes have to be converted to hours.

The distance to the train station is 2 miles.

Refer to the attachment from Mathematica.

ellie202000 · Answer

So his mistake was not convert the hours to minutes?

anonymous · Answer

I chose to convert minutes to hours.  Hours to minutes should also work.