SOLUTION: Bob drives 200 miles to attend a football game. His speed on the return trip was an average of 10mph less than his speed to the game. If the return trip took 1 hour longer, how f
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Question 231976: Bob drives 200 miles to attend a football game. His speed on the return trip was an average of 10mph less than his speed to the game. If the return trip took 1 hour longer, how fast did he drive in both directions?
I come up with 200/(x-10) - 200/x = 1... I don't even know if this is right with the D=RT formula ???
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Bob drives 200 miles to attend a football game.
His speed on the return trip was an average of 10mph less than his speed to the game.
If the return trip took 1 hour longer, how fast did he drive in both directions?
:
But that is right! Just continue: Time = dist/speed:
:
slow time - fast time = 1 hr
- = 1.
Multiply by x{x-10)and you have
200x - 200(x-10) = x(x-10)
:
I made a big mistake here, (-200 * -10 = +2000)
200x - 200x + 2000 = x^2 - 10x
Then
A quadratic equation
x^2 - 10x - 2000 = 0
:
Factors to:
(x-50)(x+40) = 0
:
Positive solution
x = 50 mph is the fast car speed
and obviously
40 mph is the slow car speed
;
:
Check this by finding the times
200/40 = 5 hr
200/50 = 4 hr
:
:
I did this on paper then typed it out wrong, forgive me, I hope I didn't mess you up too much. Carl
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