Question 63628
<b>
Mindy left at 8am and arrived at 1pm. Kelly left at 8am and arrived at 2pm. How many miles was the trip if kelly traveled 10 miles per hour SLOWER than mindy?</b>

Let Mindy's speed, in miles per hour, = {{{S[m]}}}.
Let Kelly's speed, in miles per hour, = {{{S[k]}}}.
Let Mindy's time = {{{T[m] = 5}}}.
Let Kelly's time = {{{T[k] = 4}}}.

Speed x Time = Distance.
The distance covered by both was the same. Call it D.

So, {{{D=S[m]*T[m] = S[m]*5}}}.
And {{{D=S[k]*T[k] = S[k]*6}}}.

So, {{{S[m]*5 = S[k]*6}}}.
We are told that Kelly traveled 10mph slower than Mindy, so {{{S[m]=S[k]+10}}}.
So, {{{(S[k]+10)*5 = S[k]*6}}}.
Or, {{{5*S[k]+50 = S[k]*6}}}.
Subtract {{{5S[k]}}} from both sides: {{{50=S[k]}}}.
So, Kelly traveled at 50 mph and Mindy traveled 10 mph faster at 60 mph.
The distance traveled is 50 mph * 6 hours (Kelly's speed and time) = {{{highlight(300 miles)}}}.

Nore that Mindy's speed * Mindy's time = 60 mph * 5 hours = 300 miles also.