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leumas: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Aug 21, 2011 1:47 am; Thanked: 3 times

by leumas » Sat Sep 24, 2011 9:17 am

If k, m, and t are positive integers and (k/6)+ (m/4) = (t/12),
do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

If option 2 is m is a multiple of 4, then answer becomes D?
If option 2 is m is a multiple of 6, then answer becomes D?

Correct me if I am wrong, pls.

Last edited by leumas on Sat Sep 24, 2011 9:22 am, edited 1 time in total.

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Source: Beat The GMAT — Data Sufficiency | Sat Sep 24, 2011 9:17 am

akhilsuhag: Master | Next Rank: 500 Posts; Posts: 351; Joined: Mon Jul 04, 2011 10:25 pmFollowed by: 4 members; Thanked: 57 times

by akhilsuhag » Sat Sep 24, 2011 9:20 am

I don't get the prompt. What is the question can u repost it clearly.

Please press "thanks" if you think my post has helped you.. Cheers!!

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leumas: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Aug 21, 2011 1:47 am; Thanked: 3 times

by leumas » Sat Sep 24, 2011 9:24 am

akhilsuhag wrote:I don't get the prompt. What is the question can u repost it clearly.

Question amended now!

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krishnasty: Master | Next Rank: 500 Posts; Posts: 142; Joined: Mon Jan 10, 2011 8:03 am; Thanked: 19 times

by krishnasty » Sun Sep 25, 2011 1:40 am

Wats the OA???

---------------------------------------
Appreciation in thanks please!!

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leumas: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Aug 21, 2011 1:47 am; Thanked: 3 times

by leumas » Sun Sep 25, 2011 2:38 am

krishnasty wrote:Wats the OA???

OA is A

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leumas: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Aug 21, 2011 1:47 am; Thanked: 3 times

by leumas » Sun Sep 25, 2011 2:40 am

If k, m, and t are positive integers and (k/6)+ (m/4) = (t/12),
do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 4.

Will ans be D?

If k, m, and t are positive integers and (k/6)+ (m/4) = (t/12),
do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 6.

Will answer be D?

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Sun Sep 25, 2011 3:12 am

Deleted

Last edited by knight247 on Sun Sep 25, 2011 4:00 am, edited 1 time in total.

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leumas: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Aug 21, 2011 1:47 am; Thanked: 3 times

by leumas » Sun Sep 25, 2011 3:21 am

knight247 wrote:Firstly, make it easier on the eyes by multiplying throughout by 12. So we have,
2k+3m=t

(1)k is a multiple of 3 so 2k has to be a multiple of 3. Remember one ground rule,
If two multiples of a certain number are added/subtracted then the resultant number is also a multiple of the same number. So we have Multiple of 3+Multiple of 3=Multiple of 3. Hence, t is a multiple of 3. And, 12 is also a multiple of 3. So, t and 12 do have a common factor greater than 1 i.e. 3. Sufficient

(2)m is a multiple of 4 so 3m is a multiple of 4. Remember another ground rule, When one multiple of a certain number is added/subtracted to a non multiple the resultant number is NEVER a multiple of the first number.

So we have, 2k+Some multiple of 4=t . Now, 2k might have another 2 in it or it might not. We can't say for certain. All we know is that 2k is multiple of 2 and NOTHING MORE. So we can assume that t would be a multiple of 4 if 2k was a multiple of 4.But t would definitely not be a multiple of 4 if 2k is not a multiple of 4. Since, we get conflicting possibilities from b this statement is insufficient.

Hence A

P.S. The ground rule I mentioned in statement 2 has an exception to it. The number 2. When two odd number are added the resultant number is even which is obviously divisible by 2.

thanks for the post..

I have one doubt.

If an even number (2k) and multiple of 4 added , the result will be an even number which will definitely be a multiple of 2, how can we say option 2 is insufficient?

Am I understanding that exception to the rule in a proper sense?

________
Samuel

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Sun Sep 25, 2011 3:45 am

Hey Samuel,
I'm extremely sorry man. I was kinda shuffling between the 3 different questions u've posted on this thread and kinda jumbled my own thoughts. Can you if possible post them as three separate questions on the forum. And i'll answer then promptly. Cheers

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