SOLUTION: Two cars traveled from Elmhurst to Oakville, a distance of 100 miles. One car traveled 10 mph faster than the other and arrived 5/6 of an hour sooner. Find the speed of the faster

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Question 1138064: Two cars traveled from Elmhurst to Oakville, a distance of 100 miles. One car traveled 10 mph faster than the other and arrived 5/6 of an hour sooner. Find the speed of the faster car.
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(53765) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = the speed of the faster car, in mph

Then the speed of the slower car is (x-10) miles per hour.


From the condition, you have this "time" equation


    100%2F%28x-10%29 - 100%2Fx = 5%2F6   of an hour


To solve it, multiply both sides by  6x*(x-1).  You will get


    600*x - 600*(x-10) = 5x*(x-10).


    6000 = 5x^2 - 50x


    5x^2 - 50x - 6000 = 0


    x^2 - 10x - 1200 = 0


    (x-40)*(x+30) = 0.


The roots are 40 and -30.  Only positive root is meaningful.


ANSWER.  The faster car speed is 40 mph.


CHECK.   The faster car spends  100%2F40 = 10%2F4  hours.

         The slower car spends  100%2F30 = 10%2F3  hours.

         The difference is  10%2F3+-+10%2F4 = 40%2F12+-+30%2F12 = 10%2F12 = 5%2F6 of an hour.   ! Correct !

Solved and completed.

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From my post learn on how to write, to solve and to use the "time" equation.



Answer by josgarithmetic(39800) About Me  (Show Source):
You can put this solution on YOUR website!
r, speed of slow car
r+10, speed of fast car

Time difference, %285%2F6%29%2Ahour
100%2Fr-100%2F%28r%2B10%29=5%2F6
.
.
1200=r%5E2%2B10r
r%28r%2B10%29=1200

Note that 30 and 40 are factors of 1200 and they differ by 10.

Slow Car, 30 mph
Fast Car, 40 mph