Question 524020
Stanley drove 240 miles at a certain rate of speed.
 If he had traveled 15 mph faster, he would have been able to travel 30 miles further in 3/4 of the time that he spent on his trip.
 What was Stanley's rate on the trip?
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This is not hard, we can sum the problem up in an easy to understand statement:
"The fast car goes 270 mi in 3/4 of the time required by slow car to go 240 mi"
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Let s = his speed on the trip
then
(s+15) = his faster speed
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Write a time equation, Time = dist/speed
:
Fast car time = 3/4 slow car time
{{{270/((s+15))}}} = {{{3/4}}}*{{{240/s}}}
Find 3/4 of 240, and you have:
{{{270/((s+15))}}} = {{{180/s}}}
Cross multiply
270s = 180(s+15)
270s = 180s + 2700
270s - 180s = 2700
90s = 2700
s = {{{2700/90}}} 
s = 30 mph is the original speed
then
30 + 15 = 45 mph is the faster speed
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We can check this by finding the time of each
270/45 = 6 hrs
240/30 = 8 hrs; 6 hrs is 3/4 of 8 hrs
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How about that? Make perfect sense to you?