SOLUTION: A car completed a 1000 mile race in 35 hours. For the 600 miles traveled driving daylight, the car averaged 20 mph more than it did for the 400 miles traveled at night. What was

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A car completed a 1000 mile race in 35 hours. For the 600 miles traveled driving daylight, the car averaged 20 mph more than it did for the 400 miles traveled at night. What was       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 96888: A car completed a 1000 mile race in 35 hours. For the 600 miles traveled driving daylight, the car averaged 20 mph more than it did for the 400 miles traveled at night. What was the average speed of the car during the daytime?

Not from a text book, was a handed out assignment
Found 2 solutions by scott8148, edjones:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
(600/x)+(400/(x-20))=35 ... multiply by x(x-20) ... 600(x-20)+400(x)=35x(x-20)

600x-12000+400x=35x^2-700x ... 35x^2-1700x+12000=0 ... 7x^2-340x+2400=0

factoring gives (7x-60)(x-40) ... so x=40 and x=60/7

60/7 results in a negative night speed, which is not realistic

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
d=s*t
distance=speed*time
Daytime: d=600 s=x time=35-t
nighttime: d=400 s=x-20 t=t
A. x(35-t)=600
B.(x-20)t=400
in equ. A. divide 35-t into both sides x=600/(35-t)
Substitute 600/(35-t) for x in B. t%28%28600%2F%2835-t%29%29-20%29=400
%28%28600t%2F%2835-t%29%29-20t%29=400
multiply 35-t times both sides600t-20t%2835-t%29=400%2835-t%29
600t-700t%2B20t%5E2=14000-400t
20t%5E2%2B300t-14000=0
20(t^2+15t-700)=0
Quadratic formula: t=20hrs driving time at night. 35-t=15hrs driving time in day
[t=-35(thrown out)]
put t=20 into equation A. x(35-20)=600
x=600/15=40mph daytime speed. Answer
20mph=night driving speed
Check:
Day:40*15=600
Night:20*20=400
That one was tough!
Ed