SOLUTION: solve. Round to the nearest tenth of necessary
Me. Truman drove to his cottage in the mountains at a rate of 45 mi/hr. On the way back he drove at a rate of 50 mi/hr but took a
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Me. Truman drove to his cottage in the mountains at a rate of 45 mi/hr. On the way back he drove at a rate of 50 mi/hr but took a
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Question 1034135: solve. Round to the nearest tenth of necessary
Me. Truman drove to his cottage in the mountains at a rate of 45 mi/hr. On the way back he drove at a rate of 50 mi/hr but took a route 15 mo shorter than the route by which he traveled to the cottage. How many total miles did he travel if it took him 1/2 hr longer to go than to return? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Truman drove to his cottage in the mountains at a rate of 45 mi/hr.
On the way back he drove at a rate of 50 mi/hr but took a route 15 mi shorter than the route by which he traveled to the cottage.
How many total miles did he travel if it took him 1/2 hr longer to go than to return?
:
let m = the distance driven to the cottage
The return was 15 mi shorter, therefore;
(m-15) = the return distance
:
Write a time equation; time = dist/speed
To time - return time = .5 hrs - = .5
get rid of the denominator, multiply eq by 450, cancel the denominators and you have.
10m - 9(m-15) = 450(.5)
10m - 9m + 135 = 225
m = 225 - 135
m = 90 mi to the cottage
then
90 - 15 = 75 mi return
:
Total dist driven: 90 + 75 = 165 mi
:
:
Check this by finding the actual time each way
90/45 = 2.0 hr to the cottage
75/50 = 1.5 hr return
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Time dif: .5 hrs
