SOLUTION: A jet, traveling at an average speed of 420 miles per hour in still air, flew 240 miles and back in 1 hour and 10 minutes. What is the average speed of the wind in mph?

Question 184155: A jet, traveling at an average speed of 420 miles per hour in still air, flew 240 miles and back in 1 hour and 10 minutes. What is the average speed of the wind in mph?
Found 2 solutions by ptaylor, checkley77:
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
Let s=speed of the wind
With the wind, add wind speed; against the wind subtract
Time travelling with the wind= 240/(420+s)
Time travelling against the wind=240/(420-s)
Now we are told that the above two times add up to 1 hr 10min=1 1/6=7/6 hr so:
240/(420+s) +240/(420-s)=7/6 multiply each term by 6(420+s)(420-s)
1440(420-s)+1440(420+s)=7(420+s)(420-s) get rid of parens
1440*420-1440s+1440*420+1440s=7*176400-7s^2
YOU KNOW, THERE MAY BE AN EASIER WAY BUT THE ONLY THING I CAN SEE TO DO RIGHT NOW IS GRIND OUT THE BIG NUMBERS.
604800+604800=1234800-7s^2 this equals
1209600=1234800-7s^2 subtract 1234800 from each side
-7s^2=-25200 divide each side by -7
s^2=3600 take square root of each side
s=+or- 60 mph-------------------------------------------------speed of the wind
disregard the negative value for s; wind speed is positive
CK
240/(420+60) +240/(420-60)=7/6
240/480 + 240/360=7/6
1/2 + 2/3=7/6
3/6+ 4/6=7/6
7/6=7/6
Hope this helps---ptaylor

Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
240/(420-w)+240/(420+w)=70/60
[240(420-w)+240(420+w)]/(420+w)(420-w)=70/60
(100,800-240w+100,800+240w/(176,400-w^2)=70/60
201,600/(176,400-w^2)=70/60
60*201,600=70(176,400-70w^2)
12,096,000=12,348,000-70w^2
70w^2=-12,096,000+12,348,000=0
70w^2=252,000
w^2=252,000/70
w^2=3600
w=sqrt3600
w=60 mph. for the wind.
Proof:
240/(420-60)+240/(420+60)=70/60
240/360+240/480=70/60
2/3+1/2=7/6
7/6=7/6

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