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# A pub owner has three different kinds of beer of quantities

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Senior Manager
Joined: 30 Aug 2003
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A pub owner has three different kinds of beer of quantities [#permalink]

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09 Jan 2004, 14:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A pub owner has three different kinds of beer of quantities 82 litres, 123 litres, and 205 litres respectively. Find the least number of cannisters of equal size required to store all the beer without mixing.

a. 9
b. 10
c. 12
d. 15
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Pls include reasoning along with all answer posts.
****GMAT Loco****
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Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE

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09 Jan 2004, 14:39
your answers are useless to us without reasoning
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Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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09 Jan 2004, 14:56
10

82 = 2*41
123 = 3*41
205 = 5*41
Total is 2+3+5 cannisters of equal size (41liters)
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Paul

Senior Manager
Joined: 30 Aug 2003
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09 Jan 2004, 15:03

_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE

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09 Jan 2004, 15:04

_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Director
Joined: 28 Oct 2003
Posts: 501
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09 Jan 2004, 15:17
Why isn't it three?

The question does not state that the canisters must be filled to the rim.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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09 Jan 2004, 15:18
Stoolfi is doing some serious CR - Weaken the question lately
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Paul

Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE

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09 Jan 2004, 15:26

_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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10 Jan 2004, 10:26
82+123+205 = 410

this can be factorised as 41*10
So the minimum number of cans required to completely fill each can without mixing = 10

Again largest common factor among them is 41. The problem can be solved this way as well
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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10 Jan 2004, 11:25
I see. But I still think it's better by breaking it down to factors. It's easier to see the 41 stand out. But for an accustomed eye, your way is sure easier
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Paul

Intern
Joined: 09 Dec 2003
Posts: 20
Location: Stockholm

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10 Jan 2004, 14:05
I didn't like this problem. We could have a canister of 82 liters, and in that case the answer would be 5
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Salvador aÃ­ vou eu!!!

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10 Jan 2004, 14:09
It cannot be 82 because it is not a common factor to 123 and 205...
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Best Regards,

Paul

10 Jan 2004, 14:09
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# A pub owner has three different kinds of beer of quantities

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