Fun Evo Math Brain Teaser!!!
This is the common assumption. However, you cannot average MPH over two different time periods. For example, in your answer, Warrt would drive for 1 hour @ 40mph then for 20 minutes @ 120mph. You cannot average these two numbers for average total speed, because you'd have to keep driving @ 120mph for a full hour (to match the full hour @ 40mph).
Try again.
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Clamps is correct, because nowhere in your problem statement did you put a time limit on how long the trip needs to be. Just because Warr planned to originally complete his "trip" in a time period t (e.g. 1 hour) doesn't make it any less of a "trip" if he takes time 2*t (2 hours). A trip is a trip. Now if you had specified how quickly he needs to get there AND the average speed, then you would have a noodle scratcher. As it stands, this isn't a trick question at all.
Try again.
Try again.
Plug in some numbers and try...
80 miles @ 80mph = 80/80 hours
40 miles @ 80mph = 40/80 hours
160 miles @ 80mph = 160/80 hours
So...
If he's going 80 miles, his limit is 80 / 80 hours... if he's already done 40 miles @ 40mph, he's already used his hour (80 / 80)!
If he's going 160 miles, his limit is 160 / 80 hours... if he's already done 80 miles @ 40mph, he's already used his 2 hours (160 / 80)!
If he's going 40 miles, his limit is 40 / 80 hours... if he's already done 20 miles @ 40mph, he's already used his half hour (40 / 80)!
Last edited by recompile; Dec 27, 2006 at 01:18 PM.
This is the common assumption. However, you cannot average MPH over two different time periods. For example, in your answer, Warrt would drive for 1 hour @ 40mph then for 20 minutes @ 120mph. You cannot average these two numbers for average total speed, because you'd have to keep driving @ 120mph for a full hour (to match the full hour @ 40mph).
Last edited by lords3t; Dec 27, 2006 at 01:14 PM.


