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how to solve this problem?

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Intern
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how to solve this problem? [#permalink]

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New post 20 Jun 2006, 02:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can anyone tell me how to solve this problem?

Jack and joe are 150 miles apart in a straight line. Jack leaves from A at a speed of 30 miles per hour. Joe leaves from B at a speed of 50 miles per hour. When they meet each other, how far is joe from the starting point of Jack i.e, A.?
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New post 20 Jun 2006, 03:35
As it will take the same time for jack and joe to meet at some point

Therefore,

x/30 = (150-x)/50

x = 56.25 miles
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Thanks srikanth [#permalink]

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New post 20 Jun 2006, 03:54
klsrikanth wrote:
As it will take the same time for jack and joe to meet at some point

Therefore,

x/30 = (150-x)/50

x = 56.25 miles



Wow! such a simple problem and simple concept! Thanks srikanth.
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New post 20 Jun 2006, 12:41
Jack and Joe are moving towards each other, hence relative speed = 50 + 30 = 80 mph

Time taken to meet = 150/80 = 15/8 hr

Distance of meeting point from A = Distance travelled by Jack = 30*15/8
= 56.25 miles.
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New post 20 Jun 2006, 22:41
56.25 is the answer. Nice explanations provided by srikanth and paddyboy.
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Re: how to solve this problem? [#permalink]

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New post 22 Jun 2006, 18:24
beshtesh wrote:
Can anyone tell me how to solve this problem?

Jack and joe are 150 miles apart in a straight line. Jack leaves from A at a speed of 30 miles per hour. Joe leaves from B at a speed of 50 miles per hour. When they meet each other, how far is joe from the starting point of Jack i.e, A.?


Using d = st,

for Jack, d = 30t
for Joe , (150-d) = 50t

d/(150-d) = 3/5 => 5d = 450-3d => d = 450/8 => 56.25 miles.
Re: how to solve this problem?   [#permalink] 22 Jun 2006, 18:24
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