+389 Report an Error Problem: Mr. Pillot always rides his bicycle to work, and he begins his ride at the same time every day. If he averages 10 miles per hour, he arrives at work 2 minutes late, but, if he averages 15 miles per hour, he arrives 1 minute early. How many miles does Mr. Pillot ride to work? Express your answer as a decimal to the nearest tenth.
+584 10 miles per hours =1/6 miles per minute
15 miles per hours=1/4 miles per minute
Now, assume he need x minute to rides his bicycle to work.
if he averages 10 miles per hour ,then he need (x+2) minute to arrives at work
if he averages 15 miles per hour,then he need (x-1)minute to arrives at work
distance equal velocity multiply by time. d=v*t
distance did not change,so we can set up an equation like....
1/6*(x+2)=1/4*(x-1)
x+2=1.5(x-1)
x+2=1.5x-1.5
0.5x=3.5
x=7
If if he average speed is 15 miles per hour (1/4 miles per minute), he need 7-1=6 minute to arrive at work
6*(1/4)=3/2 miles=1.5 miles
Same,If he average speed is 10 miles per hour (1/6 miles per minute),he need 7+2=9 minute to arrive at work.
9*(1/6)=3/2 miles=1.5 miles (I just checking) 
+584 10 miles per hours =1/6 miles per minute
15 miles per hours=1/4 miles per minute
Now, assume he need x minute to rides his bicycle to work.
if he averages 10 miles per hour ,then he need (x+2) minute to arrives at work
if he averages 15 miles per hour,then he need (x-1)minute to arrives at work
distance equal velocity multiply by time. d=v*t
distance did not change,so we can set up an equation like....
1/6*(x+2)=1/4*(x-1)
x+2=1.5(x-1)
x+2=1.5x-1.5
0.5x=3.5
x=7
If if he average speed is 15 miles per hour (1/4 miles per minute), he need 7-1=6 minute to arrive at work
6*(1/4)=3/2 miles=1.5 miles
Same,If he average speed is 10 miles per hour (1/6 miles per minute),he need 7+2=9 minute to arrive at work.
9*(1/6)=3/2 miles=1.5 miles (I just checking)
