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**+2 **

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

CPhill Jan 9, 2018