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Mr. Blue drove from Allston to Brockton, a distance of 105 miles. On his way back, he increased his speed by 10 mph. If the journey back took him 15 minutes less time, what was his original speed?

2 Answers

3 votes

Answer:

60 mph

Explanation:

User EJTH
by
7.3k points
1 vote

Answer:

His original speed was 60 mph.

Explanation:

The average speed formula is shown below:

speed = distance / time

For the first leg of the trip Mr. Blue drove at a speed of "x" mph, for a distance of 105 miles, therefore the time that took to complete the trip is:

time 1 = distance / speed

time 1 = 105 / x h

On the second leg of the trip Mr. Blue drove faster at a speed of "x + 10" mph, so the time it took him to complete the trip was 15 minutes less, therfore:

time 2 = time1 - (15/60) = (105/x) - 0.25 = (105 - 0.25*x)/x h

Applyint the time and speed from the second leg to the average speed formula we have:

x + 10 = {105/[(105 - 0.25*x)/x]}

x + 10 = 105*x/(105 - 0.25*x)

(x + 10)*(105 - 0.25*x) = 105*x

105*x + 1050 -0.25*x² - 2.5*x = 105*x

-0.25*x² -2.5*x + 1050 = 0 /(-0.25)

x² + 10*x - 4200 = 0

x1 = [-(10) + sqrt((10)² - 4*(1)*(-4200))]/2 = 60

x2 = [-(10) - sqrt((10)² - 4*(1)*(-4200))]/2 = -70

Since his speed couldn't be negative the only possible value was 60 mph.

User Pnklein
by
7.7k points
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