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# Help and please explain why! Thanks!

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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!

Sep 19, 2018

#1
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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   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
#2
+1

Thank you so much that makes sense!

HelpPLZ  Sep 20, 2018