OG 90 DS

[leumas]

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.

[akhilsuhag]

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

[leumas]

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

Question amended now!

[krishnasty]

Wats the OA???

[leumas]

Wats the OA???

OA is A

[leumas]

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?

[knight247]

Deleted

[leumas]

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

[knight247]

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