Question 1034414: solve. Round to the nearest tenth of necessary
A bicycle club took a weekend trip to a national forest and rode 60 mi at a certain average rate. On their return trip over the same route, they traveled 4 mi/hr faster and took 4 hrs less. What were their average rates going and returning?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! r*t=d
r = rate
t = time
d = distance
in your problem:
d = 60
when the club was going, the formula became r*t=60
when the club was coming back, the formula became (r+4)*(t-4) = 60
these are 2 equations that need to be solved simultaneously.
since r*t = 60, the second equation can be shown as (r+4)*(t-4) = r*t
simplify the left side of that equation to get:
r*t - 4*r + 4*t - 16 = r*t
the r*t in both sides of the equation cancels out and you are left with -4*r + 4*t - 16 = 0
from the first equation of r*t = 60, solve for t to get t = 60/r.
replace t in the second equation of -4*r + 4*t - 16 = 0 with 60/r to get -4*r + 4*60/r - 16 = 0
multiply both sides of this equation by r to get -4*r^2 + 240 - 16*r = 0
multiply both sides of this equation by -1 and rearrange the terms in descending order of degree to get 4*r^2 + 16*r - 240 = 0
factor out the gcf of 4 to get r^2 + 4*r - 60 = 0
factor this quadratic equation to get (r+10) * (r-6) = 0
solve for r to get r = -10 or r = 6.
since r can't be negative, you get r = 6.
when r = 6, t = 10 because r*t = 60
when they are going, their rate is 6 miles per hours and it takes 10 hours.
when they are coming back, their rate is 10 miles per hour (6+4) and it take 6 hours (10-4).
going r*t = 60 becomes 6*10 = 60
coming back r*t = 60 becomes 10*6 = 10.
you were asked for their average rates going and coming.
they are 6 and 10 miles per hour.
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