Question 319987
A motorist travels a distance of 120 miles.  Had she increased her average speed by 10 mi/hr she would have cut 36 minutes off of her travel time.  Find her average speed.


Let her average speed be S


Then we'll have: {{{120/S - 36/60 = 120/(S + 10)}}}-----> {{{120/S - 3/5 = 120/(S + 10)}}}


Multiply by LCD, 5S(S + 10) to get: 120(5)(S + 10) - 3S(S + 10) = 120(5S)


{{{600S + 6000 - 3S^2 - 30S = 600S}}}


{{{0 = 3S^2 + 30S - 6000}}}


{{{3(0) = 3(S^2 + 10S - 2000)}}}----- Dividing by GCF, 3 


{{{S^2 + 10S - 2000 = 0}}} ----- (S + 50)(S - 40) = 0


Ignoring the negative value (- 50), we get S, or average speed as: {{{highlight_green(40)}}} mph


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Check
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Initial speed: 40 mph


Time taken: 120/40 = 3 hours


Increased speed = 50 (40 + 10)mph


New time = 120/50 = 2.4 hours


Difference in travel times = .6 hour (3 - 2.4)


.6 hour = {{{(6/10) * 60}}} = 36 minutes