<PRE><B>Question 862927</B>
A student drove a distance of 125 miles at an average speed of 55 mph. How much faster would she have to drive on the return trip to save 30 minutes of driving time?
***
let x=speed on the return trip to save 30 minutes of driving time
travel time=distance/speed
{{{125/55-125/x=1/2}}}
LCD:55x
125x-55*125=55x/2
250x-13750=55x
195x=13750
x=70.51
speed on the return trip to save 30 minutes of driving time=70.51 mph
</PRE>