Question 597434: A man walks 31 miles, partly at the rate of 2 miles per hour and the rest at the rate of 5 miles per hour. If he had walked at the rate of 5 miles per hour for the same time as he actually walked at the rate of 2 miles per hour, and vice versa, he would have covered 15 miles more than he did. How long did it take him to walk the 31 miles?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A man walks 31 miles, partly at the rate of 2 miles per hour and the rest at the rate of 5 miles per hour.
If he had walked at the rate of 5 miles per hour for the same time as he actually walked at the rate of 2 miles per hour, and vice versa, he would have covered 15 miles more than he did.
How long did it take him to walk the 31 miles?
:
Let x = time walked at 2 mph originally
and
Let y = time walked at 5 mph originally
:
In the 2nd scenario, he would have walked, 31 + 15 = 46 mi
:
Write a distance equation for each scenario
2x + 5y = 31
and
5x + 2y = 46
:
Multiply the 1st equation by 2, multiply the 2nd equation by 5
25x + 10y = 230
4x + 10y = 62
----------------subtraction eliminates y, find x
21x = 168
x = 168/21
x = 8 hrs at 2 mph originally
;
Find the time at 5 mph
2(8) + 5y = 31
5y = 31 - 16
5y = 15
y = 15/3
y = 3 hrs at 5 mph originally
therefore
8 + 3 = 11 hrs to walk the 31 miles
:
:
Check our solutions in the 2nd equation
2(3) + 5(8) = 46 min
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