A salesman drives from Ajax to Barrington a distance of 120

A salesman drives from Ajax to Barrington, a distance of 120 mi. at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 min. more time then the first leg, how fast was he driving between Ajax and Barrington?

Solution

A to B DATA:
distance = 120 miles ; rate = r mph ; time = d/r = 120/r hrs.
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B to C DATA:
distance = 150 miles ; rate = (r+10) mph ; time = 150/(r+10) hrs.
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Equation:
B to C time - A to B time = 1/10 hr
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150/(r+10) - 120/r = 1/10
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Multiply thru by 10r(r+10) to get:
1500r - 1200(r+10) = r(r+10)
300r - 12000 = r^2 + 10r
r^2 - 290r + 12000 = 0
(r-50)(r-240) = 0
Realistic solution:
r = 50 mph (His rate from A to B)