Question 1164293: amanda drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Amanda drove home, there was no traffic and the trip only took 5 hours. If her average rate was 18 miles per hour faster on the trip home, how far away does Amanda live from the mountains?
Found 3 solutions by greenestamps, Theo, josgarithmetic:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
The distances there and back are the same. So, because the ratio of times was 7:5, the ratio of speeds was 5:7.
Given that ratio, let the speed going be 5x and the speed returning be 7x. The difference in speeds, 7x-5x=2x, was 18mph; so x=9.
So her speed going was 5x=45mph and her speed returning was 7x=63mph.
The distance for each trip was then 45*7 = 315 miles, or 63*5 = 315 miles.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! r * t = d
r = rate
t = time
d = distance
going there, the formula becomes r * 7 = d
going home, the formula becomes r * 5 = d
her average rate going home was 18 miles per hour faster than her average rate going home.
going there, the formula stays at r * 7 = d
coming home, the formula becomes (r + 18) * 5 = d
simplify both formulas to get:
7r = d
5r + 90 = d
subtract the second equation from the first to get:
2r - 90 = 0
add 90 to both sides of that equation to get:
2r = 90
solve for r to get:
r = 45
7r = d becomes 7 * 45 = d which gets you d = 315
5r + 90 = d gets you 5 * 45 + 90 = d which gets you d = 315
the distance is 315 miles.
it takes her 7 hours at 45 miles per hour to get there.
it takes her 5 hours at 63 miles per hour to get home.
7 * 45 = 315 miles getting there.
5 * 63 = 315 miles going home.
she lives 315 miles from the mountains.
Answer by josgarithmetic(39620)  (Show Source):
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