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?

His original speed was 60 mph.

Step-by-step 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.

answer: 5100 just subtract

step-by-step explanation: