SOLUTION: Solve. Round to the nearest tenth of necessary Mr. Albert 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

Question 1034999: Solve. Round to the nearest tenth of necessary
Mr. Albert 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 miles 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!
Mr. Albert 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 miles 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 d = the distance drove to the cottage
The return trip was 15 mi shorter, therefore:
(d-15) = return distance
:
write a time equation; time = dist/speed
To time - return time = .5 hrs
- = .5
multiply equation by 450, cancel the denominators and you have
10d - 9(d-15) = 450(.5)
10d - 9d + 135 = 225
d = 225 - 135
d = 90 mi to the cottage
then obviously,
90 - 15 = 75 mi return home
:
"How many total miles did he travel "
90 + 75 = 165 mi
:
:
Check this out, find the actual time each way
90/45 = 2.0 hrs
75/50 = 1.5 hrs
-------------------
time dif: .5 hrs
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