During my trip to work, I spent \(3/4\) of the time driving on the highway, during which I covered \(5/6\) of the distance. My average speed on the highway was 60 miles per hour. What are the possible values of my average speed, in miles per hour, during the time I was not on the highway?

Help and please explain the questions because I'm very confused! Thankyou!

HelpPLZ
Sep 19, 2018

#1**+1 **

Let D be the total distance and T be the total time

So...when on the highway, we have

(5/6)D / 60 = (3/4)T simplify

5D / (6 * 60) = (3/4)T

5D / 360 = (3/4)T

D / 72 = (3/4)T multiply both sides by 4/3

(4/3)D / 72 = T

4D / 216 = T divide both sides by 4

D / 216 = (1/4) T

And when not on the highway....we have (1/6)D left to travel and (1/4) of the time left to travel

Let R be the rate over this distance...and we have

(1/6)D / R = (1/4)T sub D/216 for (1/4)T

(1/6)D / R = D/216 rearrange as

(1/6)D / (D / 216) = R

(1/6)D * (216 / D) = R

36 mph = R = rate while not on the highway

CPhill
Sep 19, 2018

#1**+1 **

Best Answer

Let D be the total distance and T be the total time

So...when on the highway, we have

(5/6)D / 60 = (3/4)T simplify

5D / (6 * 60) = (3/4)T

5D / 360 = (3/4)T

D / 72 = (3/4)T multiply both sides by 4/3

(4/3)D / 72 = T

4D / 216 = T divide both sides by 4

D / 216 = (1/4) T

And when not on the highway....we have (1/6)D left to travel and (1/4) of the time left to travel

Let R be the rate over this distance...and we have

(1/6)D / R = (1/4)T sub D/216 for (1/4)T

(1/6)D / R = D/216 rearrange as

(1/6)D / (D / 216) = R

(1/6)D * (216 / D) = R

36 mph = R = rate while not on the highway

CPhill
Sep 19, 2018