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s2008m [1.1K]
3 years ago
9

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 t

han on the way back , find his average speed on the way back.
Mathematics
2 answers:
kogti [31]3 years ago
4 0

Answer:

 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 . . . . <em>check</em>

kipiarov [429]3 years ago
3 0

Answer:

36 mph.

Step-by-step explanation:

Let x be the distance to the town hall.

Trip there:

speed (s) = distance / time

s = x / 1.5 so x = 1-5s

Trip back:

s - 12 = x / 2 so x = 2s - 24

AS they are both = to x:

1.5s = 2s - 24

0.5s = 24

s = 48.

Average speed on the way back = 48-12 = 36 mph.

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