SOLUTION: A tourist can bicycle 33 miles in the same time as he can walk 12 miles. If he can ride 7 mph faster than he can walk, how much time should he allow to walk a 30-mile trail?

Question 618505: A tourist can bicycle 33 miles in the same time as he can walk 12 miles. If he can ride 7 mph faster than he can walk, how much time should he allow to walk a 30-mile trail?
Found 2 solutions by John10, swincher4391:
Answer by John10(297) (Show Source): You can put this solution on YOUR website!
A tourist can bicycle 33 miles in the same time as he can walk 12 miles. If he can ride 7 mph faster than he can walk, how much time should he allow to walk a 30-mile trail?
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Let x be the speed that he walks
since he can bicycle 33 miles AS SAME TIME as he walks 12 miles.
we have the equation for time:
33/(x + 7) = 12/x
33x = 12(x + 7)
33x = 12x + 84
33x - 12x = 84
21x = 84
x = 4 mph
If the trail is 30 miles then the time he must walk is:
t = distance / speed = 30/4 = 7.5 hours or 7 hours and 30 minutes.
Hope it helps you:)
John10
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
This is a proportion.
Let b be how fast a tourist can ride on is bicycle.
Then b-7 is how fast he can walk.
Recall that d = rt.
Then 33 = b*t and 12 = (b-7)*t
We need to solve for b.
Since 12 = bt - 7t but recall that 33=bt so we have that:
12 = 33-7t
-21 = -7t
t = 3 hours.
33miles = b* 3hours
b = 11 miles per hour.
That means b-7 = 4 miles per hour
Then following d =rt
30 = 4*t
30/4 = t
t = 7.5 hours
So he should allow himself 7.5 hours.

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