Expert: Sherman D. - 10/28/2008

Question
Gene spent 10 min riding his bicycle to a friend's house. He left his bike there and, with his friend, walked for 15 min to the gym. Gene rides his bike 10 km/h faster than he walks. If the entire trip covered a distance of 2.75 km, how far is it from his friend's house to the gym?

The problem asks for a table and I used the following:
d= distance r= rate t= time    here's my numbers:
                            WALK:rate=r time=1/6 dist=1/6r
                            BIKE:rate=r+10 time=1/4 d=1/4(r+10)

Answer
s = d/t
d = st

2.75 = (r + 10)(10/60) + r(15/60)

2.75 = (r + 10)(1/6) + r(1/4)
11/4 = (r + 10)/6 + (r/4)
multiply everything by 12
33 = 2r + 20 + 3r
33 = 5r + 20
13 = 5r
x = 13/5

((13/5) + 10)(1/6)
((13/5) + (50/5))(1/6)
(63/5)(1/6)
(63/30) or 2.1 km

(13/5)(1/4) = 13/20 = .65km

He bike rode 2.1km
He walked .65km

It's 650m from his friend's house to the gym.