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word problem

by Mclaughlin » Sun Aug 03, 2008 12:05 pm
A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55

OA is C

I have no idea how to set this up"? any tricks from this awesome group of brainiacs?
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by newera » Sun Aug 03, 2008 12:46 pm
Ok, this isn't the shortest way, but its how I got to the answer.

S=450/t
S+5=450/(t-1) this represents the "1 hour less if the average speed S for the trip had been greater by 5 miles per hour" statement.

Before finding S, we need to find out what the original time was. So, I made the equations equal one another.

So to isolate the "S" in this equation, we just move the 5 to the other side. S+5=450/(t-1) will be S=450/(t-1)-5 --> S=(450-5(t-1))/(t-1) --> S=(455-5t)/(t-1).

Lets make this equal to our original S=450/t.

450/t=(455-5t)/(t-1). Cross multiply to get 455t-5t^2=450t-450 --> 5t-5t^2+450=0 --> 5t^2-5t-450=0. Simply by dividing by 5. t^2-t-90=0 --> (t-10)(t+9)=0. So t=10.

Now plug this into the original S=450/t and you get answer choice C or 45.
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by vishubn » Tue Aug 05, 2008 10:02 am
formual=d=st

intial condition =t=450/s

when the speed is increased => t-1=450/s+5

so equate he same and u get the beow calcualtion i got it withing 2min :)

hope it helps

450/s-1=450/s+5

(450-s)s+5=450s
s^2+5s-2250=0

s^2+50s-45s-2250=0
s(s+50)-45(s+50)=0

(s-45)(s+50)

cant be negative so it has to be 45
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