A salesman drives from Ajax to Barrington a distance of 136

A salesman drives from Ajax to Barrington, a distance of 136 mi, at a steady speed. He then increases his speed by 12 mi/h to drive the 180 mi from Barrington to Collins. If the second leg of his trip took 10 min more time than the first leg, how fast was he driving between Ajax and Barrington?

Solution

Let the speed of the salesman from Ajax to Barrington be x mph. Then his speed from Barrington to Collins is x + 12 mph. Since time = distance /speed, the time taken by the salesman for the journey from Ajax to Barrington is 136/x hours or, 136*60/x minutes. Also, the time taken by the salesman for the journey from Barrington to Collins is 180/ ( x + 12) hours or, 180 *60/(x + 12) minutes. Since the second leg of his trip took 10 min more time than the first leg, we have 180*60/(x + 12) = 136*60/x + 10 or, 10800/(x + 12) = 8160/x + 10 = (8160 + 10x)/x or, 10800x = (8160 + 10x)(x + 12) or, 10800x = 8160x+10x²+97920+120x or,10x²+8160x+120x - 10800x +97920 = 0 or, 10x² - 2520x + 97920 = 0 or, x² -252x + 9792 = 0. On using the quadratic formula, we have x = [-(-252)±{ (-252)²- 4*1*9792} ]/2*1 or, x = [ 252 ± (63504- 39168)]/2 or, x = (252± 24336)/2 or, x =(252± 156) /2. Thus x = (252- 156) = 96/2 = 48 or, x = (252+ 156)/2 = 408/2 = 204. At the speed of 48mph, the salesman will complete a journey of 136 miles in (136/48)860 = 170 minutes. Also, at the speed of 48 + 12 = 60 mph, the salesman will complete a journey of 180 miles in 180/60 = 3 hours = 180 minutes which is more than 170 minutes by 10 minutes. Further, at the speed of 204 mph, the salesman will complete a journey of 136 miles in (136/204)60 = 40 minutes.Also, at the speed of 204 + 12 = 216 mph, the salesman will complete a journey of 180 miles in (180/216)60 = 50 minutes which is more than 40 minutes by 10 minutes. Thus, both 48 mph and 204 mph are the correct answers for the salesman\'s speed between Ajax and Barrington.