SOLUTION: A cargo plane flew to the maintenance facility and back. It took one hour less time to get there than it did to get back. The average speed on the trip there was 220 mph. The av

Algebra ->  Length-and-distance -> SOLUTION: A cargo plane flew to the maintenance facility and back. It took one hour less time to get there than it did to get back. The average speed on the trip there was 220 mph. The av      Log On


   

Question 95190: A cargo plane flew to the maintenance facility and back. It took one hour less time to get there than it did to get back. The average speed on the trip there was 220 mph. The average speed on the way back was 200 mph. How many hours did the trip take?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r
Let t=time coming back
Then t-1=time going
distance going =rate going times time going =220*(t-1)
distance coming back=rate coming back times time coming back=200*t
distance going=distance coming back so our equation to solve is:
220(t-1)=200t get rid of parens
220t-220=200t subtract 220t from both sides
220t-220t-220=200t-220t collect like terms
-20t=-220 divide both sides by -20
t=11 hours ----------------------------time coming back
t-1=11-1=10 hours ------------------time going
Total time of trip =11+10=21 hours
CK
distance going =220*10=2200 mi
distance coming =200*11=2200mi