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Hi all, 

 

I recently discovered that I wasn't as good as Algebra than Pre-Algebra (tear), and can't find how to do algebraically do this simple word problem.

 

Paul walked to school at 6 kph and rode home from school at a speed 6 times as fast. If his total traveling time was 56 minutes, how many minutes did he spend walking?

 

Thanks so much,


 

MrPatel  Jan 9, 2018
 #1
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Paul walked to school at 6 kph and rode home from school at a speed 6 times as fast. If his total traveling time was 56 minutes, how many minutes did he spend walking?

 

Speed riding  =  36 kph

 

And

 

Distance  / Rate  =  Time  in hours

 

And  56  minutes =    14/15  hrs 

 

So

 

D / 6   +   D / 36   =    14/15

 

[ 6D + D ]   /  36   =  14/15     multiply both sides by 36

 

7D  =  (36)(14/15)     divide both sides by 7

 

D  =  (14/7) (36/15)   =   2 (12/5)  =  24/5  =  4.8  km

 

So......the time spent walking  =

 

4.8  km  /  6 kph  =   .8 h  =  . 8  * 60 min  =   48 minutes walking

 

And  8 minutes  riding

 

This makes sense......he rides 6 times as fast as he walks....so.....the time spent riding should be  1/6  of the time he spent walking  =   8 / 48   =   1 / 6

 

 

cool cool cool

CPhill  Jan 9, 2018

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