SOLUTION: A plane travels from city A to city B and it takes two hours...on the second trip the plane travels from city A to City B and it flys 10mph slower and it takes 2 1/2 hours...what i

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Question 83831: A plane travels from city A to city B and it takes two hours...on the second trip the plane travels from city A to City B and it flys 10mph slower and it takes 2 1/2 hours...what is the distance from city A to city B?
Answer by ptaylor(2198)   (Show Source): You can put this solution on YOUR website!

Let d=distance from city A to city B
r=rate on first trip
(r-10)=rate on second trip
distance(d) equals rate(r) times time(t) or d=rt; r=d/t and t=d/r
First trip:
d=rt=r(2) or
d=2r
Second trip:
d=(r-10)(2.5)
Since the distance travelled on the first and second trip is the same, we have:
2r=(r-10)(2.5) get rid of parens
2r=2.5r-25 subtract 2.5r from both sides
2r-2.5r=2.5r-2.5r-25 collect like terms
-0.5r=-25 divide both sides by -0.5
r=50 mph but we know that:
d=2r or d=2*50=100 mi------------distance between city A and City B
CK
First trip:
2*50=100
100=100
and
(2.5)(40)=100
100=100

Hope this helps-----ptaylor
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