average speed Q: help pls!

nishnosh wrote:Bob drove x miles. the first 5 miles he travelled at 30mph and the remainder of the trip he travelled at 60mph.

His travel time for the x-mile trip was what % greater than it would have been had he travelled at a constant rate of 60mph for the entire trip?

8.5%
50%
(x/12)%
(60/x)%
(500/x)%

Total time taken in first case: 5/30 + (x-5)/60 = (x+5)/60.
Time taken if traveled at 60 mph = x/60.

% more time taken = [(x+5/60 - x/60 ]/ x/60 = 5/x%=500/x.
Is E the OA?

Cheers
Ami/-

That's correct. Just wanted to add that this is a great question for plugging in.

Say x = 5. That means he went zero additional miles at 60mph.
Time of trip = 5/30 = 1/6.

If he had gone 60mph, time = 5/60 = 1/12.

Since 1/6 is twice 1/12, the trip would have been 100% longer.
Plugging in x = 5, [spoiler]only E matches 100%.[/spoiler]