SOLUTION: A cyclist traveled to her destination at an average rate of 15mph. By traveling 3 mph faster, she took 30 minutes less to return. What distance did she travel each way?

Algebra ->  Linear-equations -> SOLUTION: A cyclist traveled to her destination at an average rate of 15mph. By traveling 3 mph faster, she took 30 minutes less to return. What distance did she travel each way?      Log On


   

Question 1121435: A cyclist traveled to her destination at an average rate of 15mph. By traveling 3 mph faster, she took 30 minutes less to return. What distance did she travel each way?
Found 2 solutions by josgarithmetic, addingup:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
                SPEED       TIME       DISTANCE

GOING           15          d/15         d

RETURN          18          d/18         d

DIFFERENCE                 1/2

highlight%28d%2F15-d%2F18=1%2F2%29---------solve this.

Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
Let the amount of time she traveled be x
~
15x = 18(x - 0.5)
15x = 18x - 9
9 + 15x = 18x
9 = 3x
x = 3
3*15 = 45
So she traveled for 3 hours at 15mph for a total of 45 miles.
Now, the problem says that on the return she went at 15+3 = 18 mph and that she took 30 minutes less:
45/18 = 2.5 hours. Correct. She traveled 45 miles each way.