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Question 65910: A plane flies 720 miles against a steady 30 mi/hr headwind and then returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane’s speed in still air?
Note: must use algebra for this problem!
If at all possible, whoever helps me solve this problem, if you could explain to me, step by step, what it is you are doing I would greatly appreciate it. I don't know where to go after 720/x-30 + 720/x+30 = 10. Someone please help me understand what to do next.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! A plane flies 720 miles against a steady 30 mi/hr headwind and then returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane’s speed in still air?
Note: must use algebra for this problem!
If at all possible, whoever helps me solve this problem, if you could explain to me, step by step, what it is you are doing I would greatly appreciate it. I don't know where to go after
720/x-30 + 720/x+30 = 10. Someone please help me understand what to do next.
OK
LCM OF DENOMINATORS =(X+30)(X-30)...HENCE MULTIPLY WITH LCM THROUGH OUT TO GET
[720(X+30)(X-30)/(X-30)]+[720(X+30)(X-30)/(X+30)]=10(X+30)(X-30)
[720(X+30)+720(X-30)]=10[X^2-30^2]
720X+720*30+720X-720*30]=10X^2-10*900
1440X=10X^2-9000...................DIVIDE WITH 10 THROUGH OUT
144X=X^2-900
X^2-144X-900=0
X^2-150X+6X-900=0
X(X-150)+6(X-150)=0
(X-150)(X+6)=0
X-150=0.....THAT IS .....X=150
[THE OTHER EQN X+6=0....OR ...X=-6 IS NOT POSSIBLE.]
HENCE PLANES SPEED IN STILL AIR =150 MPH
CHECK
TIME UP WIND = 720/(150-30)=720/120=6 HRS
TIME PRO WIND =720(150+30)=720/180=4 HRS
TOTAL TIME = 6+4=10 HRS................OK
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