SOLUTION: If you increase your rate by 25 mph and you are going 25 miles you get there 10 min earlier, what was your initial rate?

Question 214655: If you increase your rate by 25 mph and you are going 25 miles you get there 10 min earlier, what was your initial rate?
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If you increase your rate by 25 mph and you are going 25 miles you get there 10 min earlier, what was your initial rate?
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d = rt
25 = rt
25 = (r+25)*(t- 1/6) [10 mins = 1/6 hour]
rt = (r+25)*(t- 1/6) = rt + 25t - r/6 - 25/6
6rt = 6rt + 150t - r - 25
150t = r+25
r = 25/t
150t = (25/t) + 25
150t^2 = 25 + 25t
6t^2 - t - 1 = 0
(3t + 1)*(2t - 1) = 0
t = 1/2 hour (Ignore negative solution)
r = 50 mph

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
If you increase your rate by 25 mph and you are going 25 miles you get there 10 min earlier, what was your initial rate?
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Initial DATA:
distance = 25 miles ; rate = x mph ; time = d/r = 25/x hrs.
------------------------
Modified DATA:
distance = 25 miles ; rate = (x+25) mph ; time = d/r = 25/(x+25) hrs.
------------------------
Equation:
old time - new time = 1/6 hr
25/x - 25/(x+25) = 1/6
---
6*25(x+25) - 6*25x = x(x+25)
150x + 6*625 - 150x = x^2 + 25x
x^2 + 25x - 6*625 = 0
(x-50)(x+75) = 0
Positive solution:
x = 50 mph (initial rate)
=============================
Cheers,
Stan H.
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