SOLUTION: Two cars enter the Florida Turnpike at Commercial Boulevard at 8AM, each heading for Wildwood. One car's average speed is 10 miles per hour more than the other's. THe faster car
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Question 338967: Two cars enter the Florida Turnpike at Commercial Boulevard at 8AM, each heading for Wildwood. One car's average speed is 10 miles per hour more than the other's. THe faster car arrives at Wildwood at 11AM, 1/2 hour before the other car. What is the average speed of each car? How far did each car travel?
Found 2 solutions by mananth, stanbon:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let one car's speed be x
the other car's speed = x+10
..
time taken by faster car = 3 hours
time taken by slower car = 3.5 hours
..
distance = speed * time
the distance they travel is the same
3(x+10)=3.5x
3x+30=3.5x
3.5x-3x=30
0.5x=30
x=30/0.5
x=60 mph. speed of one car.
the other car's speed = x+10 = 70 mph
distance traveled = speed * time
= 70*3=210 miles
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Two cars enter the Florida Turnpike at Commercial Boulevard at 8AM, each heading for Wildwood.
One car's average speed is 10 miles per hour more than the other's.
THe faster car arrives at Wildwood at 11AM, 1/2 hour before the other car.
------
Faster car DATA:
time = 3 hrs ; rate = x+10 mph ; distance = rt = 3(x+10) miles
-----------------------------
Slower car DATA:
time = 3.5 hrs ; rate = x mph ; distance = rt = 3.5x miles
-------------------
What is the average speed of each car? How far did each car travel?
==================
Equation:
distance = distance
3(x+10) = 3.5x
3x+30 = 3.5x
0.5x = 30
x = 60 mph (rate of the slower car)
x+10 = 70 mph (rate of the faster car)
===========================================
Cheers,
Stan H.
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