SOLUTION: Gauss and Euler run a 12-mile race. Euler starts 20 minutes after Gauss and finishes 40 minutes before Gauss. If Gauss's rate is 1 mph more than 50% of Euler's rate, how many hours

Question 926219: Gauss and Euler run a 12-mile race. Euler starts 20 minutes after Gauss and finishes 40 minutes before Gauss. If Gauss's rate is 1 mph more than 50% of Euler's rate, how many hours does it take Euler to finish the race? List all possible answers.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=Euler's rate
50% of Euler's rate is 0.5r and 1 mph more is (0.5r+1)
(I prefer to deal in fractions rather than decimals. It eliminates round off errors and generally is not as messy). Sooo
(0.5r+1)=((r/2)+1)=(r+2)/2)
So, Gauss's rate=(r+2)/2 mph
(20 min=20/60=1/3 hr)
When Euler starts, Gauss has already ran a distance of (1/3)(r+2)/2)=(r+2)/6 miles leaving (12-(r+2)/6) =(72-r-2)/6=(70-r)/6 miles to go
Time it takes Euler to run 12 miles=12/r hours
Time it takes Gauss to finish the 12 miles after Euler starts:
((70-r)/6)/((r+2)/2)=2(70-r)/6(r+2)=(70-r)/3(r+2)
Now if we add 40 min=(2/3) hr to Euler�s time, we get Gauss�s time, sooo:
(12/r)+2/3=(70-r)/3(r+2) multiply each term by 3r(r+2)
36(r+2)+2r(r+2)=r(70-r)
36r+72+2r^2+4r=70r-r^2
2r^2+r^2+36r+4r-70r+72=0
3r^2+-30r+72=0
r^2-10r+24=0 ---quadratic in standard form and it can be factored
(r-6)(r-4)=0
So Euler�s rate is either
r=6 mph or
r=4 mph
If Eulers rate is 6 mph, then Gauss�s rate is (r+2)/2=8/2= 4mph
If Euler�s rate is 4 mph, then Gauss�s rate is ( 4+2)/2=3 mph
If r=6, then time it takes Gauss to finish the race is
(70-r)/(3(r+2)=64/24=8/3 hr= 2 and 2/3 hr=160 min; Euler would have taken 12/6=2 hr=120 minutes---40 min faster
If r=4, then time it takes Gauss to finish the race is 66/18 hr=
3 and 2/3 hr=220 min; Euler would have taken 12/4=3 hr=180 min----40 min faster
Hope this helps---ptaylor

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