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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!

HelpPLZ  Sep 19, 2018

Best Answer 

 #1
avatar+92674 
+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

 

 

cool cool cool

CPhill  Sep 19, 2018
 #1
avatar+92674 
+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

 

 

cool cool cool

CPhill  Sep 19, 2018
 #2
avatar+144 
+1

Thank you so much that makes sense!

HelpPLZ  Sep 20, 2018

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