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

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) (Show Source): You can put this solution on YOUR website!
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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 - = 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 = hours. The slower car spends = hours. The difference is = = = 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) (Show Source): You can put this solution on YOUR website!
r, speed of slow car
r+10, speed of fast car

Time difference,

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

Slow Car, 30 mph
Fast Car, 40 mph
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