Bill drove to town hall and back. the trip there took 1.5 hours and 2hrs on trip back. if bill's speed there was 12 mph faster than on the way back , find his average speed on the way back.

 36 mph

Step-by-step explanation:

Let s represent the speed on the way back. Time is inversely proportional to speed, so the speed ratios are ...

  (speed going)/(speed coming) = (s+12)/s = 2/1.5 = (time coming back)/(time going there)

  1 +(12/s) = 4/3 . . . . . divide, simplify the ratio

  12/s = 1/3 . . . . . . .  subtract 1

  s = 36 . . . . . . . . .  multiply by 3s

Bill's average speed on the way back was 36 mph.

I sometimes like to think of these problems in terms of "ratio units." We have the speed ratio:

  (s+12)/s = 2/1.5 = 4/3

The difference between s+12 and s is 12 mph, and that corresponds to the difference 4-3 = 1 ratio unit. So 3 ratio units will be 36 mph, corresponding to the value of s. Bill's speed on the way back was 36 mph.

  (48 mph)/(36 mph) = 4/3 . . . . check

c > -10.55

step-by-step explanation: