SOLUTION: This was originally a test question that I tried to figure out. The answer is 160 miles per hour. I just need to know how they got it because I can't. Thank you. A plane flew 48

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Question 112878: This was originally a test question that I tried to figure out. The answer is 160 miles per hour. I just need to know how they got it because I can't. Thank you.
A plane flew 480 miles at a certain speed; and then increased its speed by 20 miles per hour, and continued on the same course. After having flown a total distance of 840 miles in a total of 5 hours, it landed. What was the original speed of the plain?
Found 2 solutions by scott8148, ptaylor:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
let x="original speed", so x+20="increased speed" ___ time=distance/rate

the plane flies 480 mi @ x; and (840-480) mi @ x+20

(480/x)+(360/(x+20))=5 ___ 480(x+20)+360(x)=5(x)(x+20) ___ 96(x+20)+72(x)=x(x+20)

0=x^2-148x-1920 ___ 0=(x+12)(x-160)

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=original rate (speed) of the plane
Then r+20=the increased speed of the plane
Time that elapsed while travelling at the original speed=480/r
Time that elapsed while travelling at the increased speed=(840-480)/(r+20)
Now we are told that the time that elapsed travelling at the original speed plus the time that elapsed travelling at the increased speed equals 5 hours.
So our equation to solve is:
(480/r)+360/(r+20)=5 multiply each term by r(r+20)
480(r+20)+360r=5r(r+20)
480r+9600+360r=5r^2+100r subtract 5r^2 and also 100r from both sides
480r+9600+360r-5r^2-100r=0 collect like terms
-5r^2+740r+9600=0 divide each term by -5
r^2-148r-1920=0------------------quadratic in standard form and it can be factored
(r-160)(r+12)=0
r=160 mph----original rate (speed) of the plane
and
r=-12-----------------------------neglect negative speed
CK
(480/160)+(360/180)=5
3+2=5
5=5

Hope this helps ----ptaylor