LovinLife is right. Your title could have stopped you getting an answer.
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. Find my average speed, in miles per hour, during the time I was not on the highway. Please explain all the math.
To be honest I do not know exactly where I am going as I progress through my answer.
I am following a bouncing ball and with luck it will take me where I need to go.
Why not let the distance to work be D km. and the time taken be T
Highway distance = 5/6 of D
Highway time = 3/4 of T
Speed = distance /time
Highway speed =
\(=\frac{5D}{6}\div \frac{3T}{4}\\ =\frac{5D}{6}\times \frac{4}{3T}\\ =\frac{5D}{3}\times \frac{2}{3T}\\ =\frac{10D}{9T}\)
so
\(\frac{10D}{9T}=60\\ \frac{D}{9T}=6\\ D=54T\\\)
Minor road distance = 1/6 of D = 1/6 of 54T = 9T
Minor road time = 1/4 of T
Minor road speed
= 9T / 1/4 of T
\(=9T \div \frac{T}{4}\\ =9T \times \frac{4}{T}\\ =36 miles /hour\)
I have not checked my answer. You should do that :)
+118725 
