Question

When you travel to work going 60 miles per hour, you arrive there early.
When you travel to work going 30 miles per hour, you arrive there late.
The amount of time you are early in the first case is the same as the amount of time you are late in the second.
How fast should you go to be there on time?

Answer 1

I tried that answer and was told it was wrong

Answer 2

\[time=\frac{distance}{speed}\]
Time to travel at 60 mph is
\[t _{1}=\frac{d}{60}\]
Time to travel at 30 mph is
\[t _{2}=\frac{d}{30}\]
Where d is the distance.
Time to arrive on time is
\[\frac{t _{1}+t _{2}}{2}=\frac{\frac{d}{60}+\frac{d}{30}}{2}=\frac{d}{40}\]
Therefore 40 mph is the optimum speed.

Answer 3

Yeah... that's because I spent 30 seconds looking at it instead of actually thinking about it. Here's the right answer:

Suppose work is M miles from your house, and you want to get there in T hours.

When you go 60 mph, you get there in M/60 hours.
When you go 30 mph, you get there in M/30 hours.

We know that T - M/60 = M/30 - T
Then 2T = M/30 + M/60 = 3M/60 = M/20
So T = M/40, so T is the amount of time it takes to get there going 40 mph.

Answer 4

Let t = time needed to be on time
Going 60 mph you get there in t - x time.
Going 30 mph you get there in t + x time.
s = d/t, so t = d/s
t - x = d/60 ---> t - x = d/60
t + x = d/30 ---> t + x = 2d/60
2t = 3d/60
40t = d
t = d/40 = d/s, so s = speed = 40