Question 521731: amy drove 200 miles at an average speed 10 miles per hour faster than her usual speed. if she completed the trip in 1 hour less time than usual, what is her usual driving speed in miles per hour?
Found 2 solutions by stanbon, mamiya:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! amy drove 200 miles at an average speed 10 miles per hour faster than her usual speed. if she completed the trip in 1 hour less time than usual, what is her usual driving speed in miles per hour?
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Early trip DATA:
distance = 200 miles; rate = r+10 mph ; time = 200/(r+10) hrs
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Later trip DATA:
distance = 200 miles : rate = r mph ; time = 200/r hrs
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Equation:
early time - later time = 1 hr
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200/r - 200/(r+10) = 1 hr.
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200(r+10) - 200r = r(r+10)
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200*10 = r^2+10r
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r^2 + 10r - 2000 = 0
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(r-40)(r+50) = 0
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Positive Solution:
rate = 40 mph
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Cheers,
Stan H.
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Answer by mamiya(56) (Show Source):
You can put this solution on YOUR website! let x be the usual average speed and t the among of times she usually took to complete the trip. let s be the average speed , t' the among of time she takes now and d the distance.
s= x + 10 since her average speed now is 10 miles per hour more than her usual speed
d= 200 miles
t'= t - 1 since she completed the trip in 1 hour less time usual
we know that distance= speed x time
so d= st' and d= xt
so xt= 200
(x+10)(t-1)=200
xt=200
xt - x + 10t -10= 200
xt=200
200 - x + 10t - 10 =200
xt=200
10t - x= 10
x= 10(t - 1)
xt=200
x= 10(t-1)
10(t-1)*t = 200
10t^2 - 10t - 200 =0
t^2 - t - 20 =0
let's call p(x)= t^2 - t - 20 and let's factorize this polynomial.
the determinant D= (-1)^2 - 4(1)(-20)= 81= (9)^2
t1= (1+9)/2 = 5
t2= (1-9)/2 = -4
so the answers of the equation ( t^2-T-20=0) are 5 and -4, but since t is an among of time , t has to be a positive number, so answer can't be -4 , therefore the answer is 5.
so t=5
x=10(t-1)
= 10(5-1)
= 40
so, her usual driving speed is 40 miles per hour.
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