SOLUTION: 19. Two hikers leave town A at the same time and walk by two different routes to town B. The average speed of one hiker is 1 mile per hour more than the average speed of the other
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Question 670747: 19. Two hikers leave town A at the same time and walk by two different routes to town B. The average speed of one hiker is 1 mile per hour more than the average speed of the other hiker. The slower hiker reaches town B 1/2 hour before the faster hiker, because the route taken by the faster hiker is 15 miles long, while the route taken by the slower hiker is only 10 miles long. What is the average speed of each hiker?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
19. Two hikers leave town A at the same time and walk by two different routes to town B.
The average speed of one hiker is 1 mile per hour more than the average speed
of the other hiker. The slower hiker reaches town B 1/2 hour before the faster hiker, because the route taken by the faster hiker is 15 miles long, while the route taken by the slower hiker is only 10 miles long.
What is the average speed of each hiker?
:
Let s = be that average speed of the slow hiker
then
(s+1) = average speed of the other hiker
:
Write time equation; time = dist/speed
:
Fast hiker time = slow hiker time +.5 hrs
= + .5
multiply by s(s+1) to clear the denominators
15s = 10(s+1) + .5s(s+1)
:
15s = 10s + 10 + .5s^2 + .5s
:
0 = 10s - 15s + 10 + .5s^2 + .5s
A quadratic equation
.5s^2 + .5s - 5s + 10 = 0
.5s^2 - 4.5s + 10 = 0
Get rid of the decimals mult by 2
s^2 - 9s + 20 = 0
Factors to
(s-4)(s-5) = 0
Two solutions
s = 4 mph, then the faster hiker is 5 mph
and
s = 5 mph, then the faster hiker is 6 mph
;
:
see both of those solutions work; find the time of each:
10/4 = 2.5
15/5 = 3.0
--------------
dif: .5 hrs
and
10/5 = 2.0 hrs
15/6 = 2.5 hrs
---------------
dif: .5 hrs, so either solution is can be used
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