Problema Solution
a salesman drives from ajax to barrington a distance of 111 miles, at a steady speed. he then increases his speed by 7 mi/h to drive 130 mi from barrington to collins. if the second leg of his trip took 2 min more time than the first leg, how fast was he driving between ajax and barrington
Answer provided by our tutors
let 'x' represent the speed from ajax to barrington, then 'x+7' represents the speed from barrington to collins
let 't' represent the time, in hours, taken to drive from ajax to barrington, then 't+2/60' represents the time taken to drive from barrington to collins
distance = speed * time
111 = tx, so x=111/t
130 = (t+2/60)(x+7)
substituting x=111/t in the second equation:
130 = (t+2/60)(111/t + 7)
solving for 't' we have a positive value of t=2.47
111=2.47x
x= 111/2.47 = 44.9
the salesman was driving approximately 45mph between ajax and barrington