Expert: Ahmed Salami - 9/29/2004

Question
Hi Ahmed,

Would you mind checking if my approach to solve the following distance is correct? I'd also appreciate if you could suggect any more logical or faster way to slove it. Thank you very much.

Regards,

Ang-sui


**A train makes a 50 mile trip at an average speed of 45 miles per hour. The train travels at an average speed of 40 miles per hour for the first 30 miles of the trip.**

Which is greater in value?

A
The train's average speed for the last 20 miles of the trip

B
45 miles per hour

My answer:
First find out the time that the entire trip
takes. So, 50 = 45T, and T is 1.1...

Then find out T of the first 30 mile trip. So, 30 = 40T, and T is .75 or 45 minutes. To find out the time that took the last 20 miles of trip is 1.1 or 70 miniuts - 45 minutes = 25.

Finally, the train's average speed for the last 20 miles of the trip is 20 = 25/60S. S =48

So the average speed for the last 20 miles of the trip is 48 miles per hour.  

Answer
Hi Ang-sui,
'A' is greater but you made some mistakes. Your approach is, however, good enough.
Now, to point out your mistakes.
For T = 1.111111... hrs
and this is (60)1.111111... = 66.67mins and not 70mins. The remaining time after removing the 45mins for the first 30 mile is then 66.67 - 45 = 21.67mins
As for the average speed for the last 20 miles, this is
20 = S(21.67/60)
S = 20 x 60/21.67
 = 1200/21.67
 = 55.39 miles/hr
which is obviously greater than 45 miles/hr
I hope you get it.
You could, however have gotten the correct option mentally. The 40 miles/hr for the first 30 miles is less than 45 miles/hr, so to still get an overall average speed of 45 miles/hr at the end of the trip, the speed of completion must then be greater than 45 miles/hr. Anyway, this is just mental stuff.
Good luck.
Regards.