SOLUTION: a salesman drives from Ajax to Barrington, a distance of 192 km, at a steady speed. He then increases his speed by 16 km/h to drive the 240 km from Barrington to Collins. If the se

Question 1083311: a salesman drives from Ajax to Barrington, a distance of 192 km, at a steady speed. He then increases his speed by 16 km/h to drive the 240 km from Barrington to Collins. If the second leg of his trip took 6 min more than the first leg, how fast was he driving between Ajax and Barrington?
as the question says, time of sec. trip tool six more mins than first leg, does this mean sec. leg - first leg = 1/10 hour?
btw, to set up equations, i have to find something in the question that is equal to each other in terms of same variable, am i correct???
i have hard time solving this up, please guide me!!!
thank you
adamchen894@gmail.com

Answer by josgarithmetic(39621) (Show Source): You can put this solution on YOUR website!

                   SPEED        TIME          DISTANCE

A to B             r                          192

B to C             r+16                       240

Difference

The B to C part of the trip took MORE time than the A to B part of the trip.

Solve this equation for speed r, the speed for the first part of the trip.
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