during a trip on a expressway, Don drove a total of x miles. HI speed on a certain 5 mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been, if he had traveled at a constant rate of 60 miles per hour for the entire trip?
please go step by step figuring this out, if you had this appear on your test, on test day.
answer is 500/x
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RAGS
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x mile he drove in total
5 miles at 30/hr
x-5 at 60/hr
at 60 for whole trip he would take x/60 hrs
he actually takes 5/30 + x-5)/60
it greater by 1/6 -1/12= 1/12
as a percentage (1/12 )*100 /(x/60) =500/x
5 miles at 30/hr
x-5 at 60/hr
at 60 for whole trip he would take x/60 hrs
he actually takes 5/30 + x-5)/60
it greater by 1/6 -1/12= 1/12
as a percentage (1/12 )*100 /(x/60) =500/x
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Cybermusings
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Time = D/S
Time for 5 miles = 5/30 = 1/6 m/hr
Time for x-5 miles = x-5/60 m/hr
Time if the entire trip was for x miles @ 60m/hr = x/60
Greater = [(x-5/60)+5/30] - x/60 = 5/60
% Greater = [(5/60) / x/60]*100 = 500/x
Time for 5 miles = 5/30 = 1/6 m/hr
Time for x-5 miles = x-5/60 m/hr
Time if the entire trip was for x miles @ 60m/hr = x/60
Greater = [(x-5/60)+5/30] - x/60 = 5/60
% Greater = [(5/60) / x/60]*100 = 500/x
