SOLUTION: I am needing help with this problem. I am totally confused on how to solve this problem. Two airplanes are leaving O'Hare at the same time, one bound for New York and the other

Question 288160: I am needing help with this problem. I am totally confused on how to solve this problem.
Two airplanes are leaving O'Hare at the same time, one bound for New York and the other for California. The one heading to California has a tail wind and is thus going 25 MPH faster than the one headed for New York. If after one hour they are 575 miles apart, what are their speeds?

Thanks,
Jen

Found 3 solutions by checkley77, Alan3354, ptaylor:
Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
Assuming one is going due West & the other is going due East then their distances are additive.
d=rt
575=(x+25)*1+x*1
575=x+25+x
2x=575-25
2x=550
x=550/2
x=275 mph for the plane headed for New York.
175+25=300 mph for the plane headed for California.
Proof:
275+300=575
575=575

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Assuming that one is due west and the other due east is unreasonable.
California is 2000 miles from north to south, so the direction is not known.
-------------
The best answer is "insufficient data."

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 r=speed of the NY bound plane
Then r+25=speed of the CA bound plane
Distance travelled by NY bound plane in 1 hour=r*1
Distance travelled by CA bound plane in 1 hour=(r+25)*1
Now we are told that these distances add up to 575 mi, so:
r+r+25=575
2r+25=575 subtract 25 from each side of the qe
2r+25-25=575-25
2r=550
r=275 mph------------speed of NY bound plane
r+25=275+25=300 mph--speed of CA bound plane
CK
275*1+300*1=575
575=575
Does this help?---ptaylor
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