Question 947868: How do i solve a problem with time, where a person drives 100miles in rain but when the rain stopped they drove 15miles faster for another 250 miles. If his total drive time was 10 hours how fast did he drive while it was raining? If i can get the equation then I can solve it but i have nothing like this on my notes and nothing like this in my book
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! RT=D for rate, time, distance;
Your description would have rate as miles per hour, and time as hours, and distance as miles. The example trip described is in two portions of distance.
________________speed___________time______________distance
RAINING__________r______________(___)_____________100
CLEAR AIR_______r+15____________(___)_____________250
Total____________________________10
Maybe more than one way to continue this, but you can fill the missing time information from the basic idea  , equivalent to  .
________________speed___________time______________distance
RAINING__________r______________100/r________________100
CLEAR AIR_______r+15____________250/(r+15)_____________250
Total____________________________10
You should be able to determine what to do from there.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! How do i solve a problem with time, where a person drives 100miles in rain but when the rain stopped they drove 15miles faster for another 250 miles. If his total drive time was 10 hours how fast did he drive while it was raining?
***
let x=speed while driving 100 miles in the rain
x+15=speed while driving 250 miles after rain stopped
travel time=distance/speed
..
lcd:x(x+15)
250x+100x+1500=10x(x+15)
10x^2+150x=350x+1500
10x^2-200x-1500=0
x^2-20x-150=0
solve for x by quadratic formula:
a=1, b=-20, c=-150
ans:
x≈25.8
how fast did he drive while it was raining? 25.8 mph
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