Question 270675: A train traveled the first 1/6 of the distance from station A to station B at an average speed of 90 km/h, 2/3 of the distance at an average speed of 120 km/h, and the remaining 9 km at an average speed of 90 km/h. Find the average speed at which the train was traveling from station A to station B.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
OK.
We know that average speed or rate: r(av)=(total distance)/(total time)
Let d=total distance from station A to Station B
Then total time is the sum of the times listed below:
time to travel 1/6 distance =(d/6)/90=d/540
time to travel 2/3 or 4/6 the distance=(4d/6)/120=d/180
time to travel the remaining 9 km=9/90=1/10
Now lets look at what we have, first:
1/6 of the distance plus 2/3 of the distance = 5/6 of the distance leaving 1/6 of the distance remaining which we are told is 9 km
So, if (1/6)d=9, then the total distance (d)=54 km
Therefore total time would be:
54/540 + 54/180 + 1/10 simplify
1/10 + 27/90 +1/10 multiply each term by 90/90
9/90 +27/90 +9/90=45/90=1/2 hr
So, Average Speed (rav)=54/(1/2)=108 km/hr
Hope this helps---ptaylor
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