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Buses leave town B at 3 pm and every 10 hours after that.

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Buses leave town B at 3 pm and every 10 hours after that. [#permalink] New post 08 Jul 2004, 19:49
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A
B
C
D
E

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Question Stats:

68% (03:29) correct 32% (01:47) wrong based on 167 sessions
Buses leave town B at 3 pm and every 10 hours after that. Buses leave town C at 4pm and every 15 hours after that. If the buses follow this schedule beginning on a Monday, what is the earliest day on which the buses leave at the same time.

A. Tuesday
B. Wednesday
C. Thursday
D. Sunday
E. The busses will never leave at the same time
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2013, 22:03, edited 1 time in total.
OA added.
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 [#permalink] New post 08 Jul 2004, 22:52
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The answer is E.

I think the best way to do it is to look at the times on a 24 hour clock. Town B busses start at 15:00, and Town C start at 16:00. If you think about it that way, then for Town B you'd add 10 hours each time, and the number will always end in a 5. Town C you'd add 15 hours each time, and the numbers would always end in a 1 or 6. So you can see they'd never coincide.

Alternatively, you could see that if they left at the same time, they'd coincide every 30 hours, but since C is one hour ahead of B, every 30 hours C will still be one hour ahead of B.
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Re: PS Busses leaving station [#permalink] New post 12 Jul 2004, 16:53
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This was a tough one!
Took me 4 mins still not sure if the approach is correct :cry:

Here's my method
Let Bus B make x trips and Bus C make y trips b4 they start at the same time.
The time when they will meet is
Remainder(15+10x)/24 .... B
Remainder(16+15y)/24 .... C

These two must be equal
i.e.
Remainder(15+10x)/24=Remainder(16+15y)/24

Hence I assume we should have integer values of x,y such that
15+10x=16+15y or 10x=15y+1
no integral (x,y) combo exist for this equation

Hence I guess Ans is E

Anyone has a better approach to this problem?
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 [#permalink] New post 12 Jul 2004, 17:54
i think the giveaway is the 3pm vs 4pm start .... if they left at the same time they would eventually meet up (10hrs vs 15hrs common divisor or somesuch) but the offset means they will never meet up anytime soon, if at all.
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 [#permalink] New post 16 Jul 2004, 19:54
tough one for me and E.
Take a & b as the numbers of buses which leave town B & town C after the first ones. a & b must be positive integers.
We got: 3 + 10a = 4 + 15b
---> 10a = 1 + 15b
We see that: (1 + 15 x an positive integer) will never evenly divide to 10
SO E is the ans
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Re: [#permalink] New post 17 Jun 2013, 17:34
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ian7777 wrote:
The answer is E.

I think the best way to do it is to look at the times on a 24 hour clock. Town B busses start at 15:00, and Town C start at 16:00. If you think about it that way, then for Town B you'd add 10 hours each time, and the number will always end in a 5. Town C you'd add 15 hours each time, and the numbers would always end in a 1 or 6. So you can see they'd never coincide.

Alternatively, you could see that if they left at the same time, they'd coincide every 30 hours, but since C is one hour ahead of B, every 30 hours C will still be one hour ahead of B.


Hey, I don't understand this explanation:

Are you saying that the times will end in 5s? I mean, yes, the number of hours elapsed will always end in a 5 or 0, but that doesn't say much about the time, other than demonstrating that the first bus must leave, on a 24-hour clock, at times of 3, 13, 23, 9, 19, 5, 15, 1, 11, 21... and that the second bus must leave at times of 4, 19, 10, 1, 16, 7, 22..

Yes, there is a pattern that is created, but in my opinion, this is not trivial and does not follow easily from the 'number of hours elapsed ending in 5 or 0'.

Any clarification would be appreciated. As of now, I still don't know how to solve this question in a proper way.
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Re: Buses leave town B at 3 pm and every 10 hours after that. [#permalink] New post 17 Jun 2013, 18:56
I think this one should be E.
It took me 4.44mins to manually calculate the whole thing. It turns out they never meet on the same time!
This one was a tough one...
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Re: Buses leave town B at 3 pm and every 10 hours after that. [#permalink] New post 13 Nov 2013, 06:02
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lastochka wrote:
Buses leave town B at 3 pm and every 10 hours after that. Buses leave town C at 4pm and every 15 hours after that. If the buses follow this schedule beginning on a Monday, what is the earliest day on which the buses leave at the same time.

A. Tuesday
B. Wednesday
C. Thursday
D. Sunday
E. The busses will never leave at the same time


Buses B 3,13,23,33 etc....(pattern ending in 3 always)
Buses C 4,19,34,49,54...(pattern ends only in 4 and 9).

Thus E is the correct answer
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Re: Buses leave town B at 3 pm and every 10 hours after that. [#permalink] New post 27 Jul 2014, 19:37
lastochka wrote:
Buses leave town B at 3 pm and every 10 hours after that. Buses leave town C at 4pm and every 15 hours after that. If the buses follow this schedule beginning on a Monday, what is the earliest day on which the buses leave at the same time.

A. Tuesday
B. Wednesday
C. Thursday
D. Sunday
E. The busses will never leave at the same time


10 and 15 are both multiple of 5 . The minimum difference between any multiple of 10 and 15 is always 0 and the next difference is 5. For example , 10 and 15 , 40 and 45.
The offset of their starting time is 1 hour. We can never have account for this 1 hour difference since the difference that we can accommodate is 0 or 5.
Had the departure of Bus C be (3pm + multiple of 5) then there was a possibility of buses leaving at the same time
Re: Buses leave town B at 3 pm and every 10 hours after that.   [#permalink] 27 Jul 2014, 19:37
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