SOLUTION: 2 cars leave at the same time in the same direction. One is going 20 mph faster than the other. After 10 hours, they are 50 miles apart. How fast is the faster one going?
If
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: 2 cars leave at the same time in the same direction. One is going 20 mph faster than the other. After 10 hours, they are 50 miles apart. How fast is the faster one going?
If
Log On
Question 694645: 2 cars leave at the same time in the same direction. One is going 20 mph faster than the other. After 10 hours, they are 50 miles apart. How fast is the faster one going?
If x is the speed of the slower vehicle, then:
distance slower car went: 10x
distance faster car went: 10(x+20)
so 10x + 50 = 10(x+20)
but then, 10x + 50 = 10x + 200
and 50 = 200 (obviously not!)
what am I doing wrong? Found 2 solutions by jim_thompson5910, DrBeeee:Answer by jim_thompson5910(35256) (Show Source):
If car one goes x mph for t hours, then it will travel tx miles.
If car two goes x+20 mph for t hours, then it will travel (x+20)t = tx+20t miles
The difference between the two is
(Car 2) - (Car 1)
(tx+20t) - (tx)
tx + 20t - tx
(tx - tx) + 20t
0tx + 20t
20t
So after t hours, they will be 20t miles apart. So in 10 hours, they are NOT 50 miles apart because when t = 10, 20t is 200. So they *should* be 200 miles apart in 10 hours.
(
