# Math

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A salesman drives from Ajax to Barrington, a distance of 150 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 183 mi from Barrington to Collins. If the second leg of his trip took 3 min more time than the first leg, how fast was he driving between Ajax and Barrington?

• Math -

speed for first leg --- x mph
speed for 2nd leg ---- x+10 mph

time for 1st leg = 150/x
time for 2nd leg = 183/(x+10)

183/(x+10) - 150/x = 3/60 = 1/20

times 20x(x+10)

3660x - 3000(x+10) = x(x+10)
3660x - 3000x - 30000 = x^2 + 10x
x^2 - 650x + 30000 = 0
(x-600)(x - 50) = 0
x = 600 or x = 50

reasonable answer: x = 50 mph for 1st leg

check:
time for 1st leg = 150/50 = 3 hrs
time for 2nd leg = 183/60 = 3.05 hrs
difference = .05 hrs or .05(60) = 3 minutes

time for 1st = 150/600 = .25 hrs
time for 2nd = 186/610 = .3 hrs
difference = .3-.25 = .05 hrs = 3 minutes
Unless he was "driving " a jet, let's go with the 50 mph

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