Question 182939

Distance(d) equals Rate(r) times Time(t) or d=rt;  r=d/t and t=d/r

Average Speed =(Total Distance)/(Total Time)
Let t1=time going
And t2=time returning
So t1+t2=Total Time
Distance going=50* t1
Distance returning=30*t2
Total Distance=50t1+30t2
Average Speed=(50t1+30t2)/(t1+t2)--------------------eq1
but
Distance going=Distance returning, so:
50t1=30t2
t1=(3/5)t2
Substitute into eq1
Average Speed=(50*(3/5)t2+30t2)/((3/5)t2+t2)
Average Speed=60t2/(8/5)t2  the t2's cancel, so we have
Average Speed=60/(8/5)=60*5/8=300/8=37.5 mph-------------------Ans

Hope this helps---ptaylor