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August 1, 2014

August 1, 2014

Posted by **Dani** on Sunday, November 30, 2008 at 12:02pm.

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

- Math (Word Problems) -
**Reiny**, Sunday, November 30, 2008 at 12:42pmlet his first steady speed by x mph

so the time for the first leg of the trip was 120/x hours

on the second leg of the trip his speed was x+10 mph

so the time taken for that was 150/(x+10)

but this took 6 min or 1/10 hour longer

so 150/(x+10) - 120/x = 1/10

multiply each term by x(x+10) to get a quadratic equation.

one of the roots should be extraneous, let me know what you got

- Math (Word Problems) -
**drwls**, Sunday, November 30, 2008 at 12:46pmLet the two speeds be V1 and V2. The times required to travel the two legs of the journey are

120/V1 and 150/V2 = 150/(V1+10)

The second leg took 0.1 hours (6 minutes) more, so

150/(V1+10) = (120/V1) + 0.1

That equation can be solved for V1.

150/(V1 + 10) = (120 + 0.1V1)/V1

150 V1 = 120 V1 + 1200 + V1 + 0.1V1^2

0.1V1^2 -29 V1 + 1200 = 0

V1 = [29 +/-19]/0.2 = 50 or 240 mph

The higher of the two roots is unrealistic. Go with 50 mph

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