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Re: On Sunday John left the place A for B at 3:30 am and on the same day [#permalink]
Assume John and Carsten meet at point M and they take 't' hrs to reach B and A respectively. So the times taken by John and Carsten to cover AB are (t+9) and (t+4) hrs respectively. The ratio of their speeds is the inverse of the times taken by each to cover AB:
John's speed/Carsten's speed = (t+4)/(t+9)
The times taken by John and Carsten to cover MB is 't' and 4 hrs respectively. T/4:
John's speed/Carsten's speed = 4/t. T/4:
(t+4)/(t+9) = 4/t....> t^2 = 36....> t = 6
They reach their respective destinations at 6:30. ANS: B
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Re: On Sunday John left the place A for B at 3:30 am and on the same day [#permalink]
­John's speed: a
Carsten's speed: b

3:30 am: John starts
8:30 am (5 hours later): Carsten starts
12:30 pm (4 hours later): They meet
After that, it take them the same time (t hours) to reach their destination

=> 9a + 4b = (9+t) * a = (4+t) * b
=> 9a + 4b = 9a + ta = 4b + tb
=> \(4b = ta => t = \frac{4b}{a}\)
      \(9a = tb => t = \frac{9a}{b}\)
=> \(t^2 = \frac{4b}{a} * \frac{9a}{b} = 36\)
=> t = 6 (hours)

=> They reach their destination at 6:30 pm
      ­
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Re: On Sunday John left the place A for B at 3:30 am and on the same day [#permalink]
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Great solutions, all, but the question does need a slight correction. It ought to state that both drivers are moving at a constant rate. Otherwise, we have no idea when they get anywhere! :) I know that's picky, but when you deal with DS and other DI section issues of "can I even solve this?" it's important to hold out for those details. As written, this question has no answer!
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Re: On Sunday John left the place A for B at 3:30 am and on the same day [#permalink]
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