SOLUTION: two friends live 1.1 miles apart along the same road. If they walk toward each other at rates that differ by 1/2 mph and they meet 12 minutes later, how fast is each person walking

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Question 1049517: two friends live 1.1 miles apart along the same road. If they walk toward each other at rates that differ by 1/2 mph and they meet 12 minutes later, how fast is each person walking?
Found 3 solutions by jorel555, ewatrrr, advanced_Learner:
Answer by jorel555(1290) About Me  (Show Source):
You can put this solution on YOUR website!
Let n be the slower walker. Then:
1/5(n)+1/5(n+1/2)=11/10
2n+2(n+1/2)=11
4n+1=11
4n=10
n=2.5 mph
n+1/2=3 mph. ☺☺☺☺

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
D = rt
D/t = r
1.1/.2 = average speed
5.5mph = combined speed
(r + (r+.5))= 5.5
2r = 5
r = 2.5mph
r + .5 = 3mph
2.5mph and 3mph2.5%2B.5 the two rates

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
first friend
speed r
time 1%2F5
distance r%2F5
second friend
speed r%2B1%2F2
time 1%2F5
distance 1%2F5*%28r%2B1%2F2%29
given also
11%2F10 distance
the equations are
r%2F5+1%2F5*%28r%2B1%2F2%29=11%2F10
r%2F5+%28r%2F5%2B1%2F10%29=11%2F10
multiply by 10 on each side
2r+2r%2B1=11
4r%2B1=11
4r=11-1
4r=10
r=10%2F4
r=5%2F2
the rate of the first friend is r=5%2F2 and the rate of the second firend is r=3