SOLUTION: A, B and C started simultaneously from X, towards Y at 9:00 am. A being the fastest of them, reached Y and headed towards X again. He met C half way through at 4:30pm, a

Question 1180446: A, B and C started simultaneously from X, towards Y at 9:00 am. A being the fastest
of them, reached Y and headed towards X again. He met C half way through at
4:30pm, after travelling for 24km more, met B at 5:00pm. Find the ratio of the
speeds of B and C
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A, B and C started simultaneously from X, towards Y at 9:00 am.
let d = dist from x to y
A being the fastest of them, reached Y and headed towards X again.
He met C half way through at 4:30pm,
9:00 Am - 4:30 Pm = 7.5 hrs
A's speed = = .2d km/hr
:
A after travelling for 24km more, met B at 5:00pm.
that means A traveled 24km in 1/2 hr which is 48 km/hr, A's speed
then find the distance from x to y
.2d = 48
d - 48/.2
d = 240 km is the dist from x to y
Now we know B traveled 240-120-24 = 96km, while A traveled 240+120+24 = 384km km
B's speed is 1/4 of A's or 12 km/hr is B's speed
:
Find the ratio of the speeds of B and C,
Find C's speed, C traveled half way (120 km) in 7.5 hrs
Then 120/7.5 = 16 km/hr is C's speed
Their ratio is 12:16 which is 3:4
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