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I am getting E. What is the OA.
2K+3m=t
S1: k=3 implies 2K=6 Let m=1 therefore t=9 which is not a multiple of 12
k=3 implies 2K=6 Let m=6 therfore t=24 which is a multiple of 12

St2: k=1 m=3 t=11
k=1 m=6 t=20
k=3 m=6 t=24

Together S1 & St2 does not satisfy as well
_________________

I am getting E. What is the OA. 2K+3m=t S1: k=3 implies 2K=6 Let m=1 therefore t=9 which is not a multiple of 12 k=3 implies 2K=6 Let m=6 therfore t=24 which is a multiple of 12

St2: k=1 m=3 t=11 k=1 m=6 t=20 k=3 m=6 t=24

Together S1 & St2 does not satisfy as well

The question is asking not if t divisible by 12, but if 12 and t have common multiple:

'do t and 12 have a common factor greater than 1 ? ' if t is divisible by either 2,3,4,6, the answer is 'yes'

If k, m, and t are positive integers and \(\frac{k}{6} + \frac{m}{4} = \frac{t}{12}\), 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.
_________________

Trying hard to achieve something unachievable now....

If k, m, and t are positive integers and \(\frac{k}{6} + \frac{m}{4} = \frac{t}{12}\), 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.

(1) k is a multiple of 3 --> \(k=3x\), where \(x\) is a positive integer --> \(2k+3m=6x+3m=3(2x+m)=t\) --> \(t\) is multiple of 3, hence \(t\) and 12 have a common factor of 3>1. Sufficient.

(2) m is a multiple of 3 --> \(m=3y\), where \(y\) is a positive integer --> \(2k+3m=2k+9y=t\) --> \(t\) and 12 may or may not have a common factor greater than 1. Not sufficient.

Re: If k, m, and t are positive integer [#permalink]

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26 Oct 2010, 07:03

Given:

k > 0, m > 0, t > 0 and k, m and t are integers && \(\frac{k}{6}\) + \(\frac{m}{4}\) = \(\frac{t}{12}\)

To prove, t and 12 have a common factor greater than 1 ... we have to show that t is multiple of any factor of 12 other than 1. i.e., 2 (or) 3 (or) 4 (or) 6 (or) 12

(1) K is a multiple of 3 ==> k = 3a If we substitute k in (**) then we get ==> 2 * 3a + 3m = t ==> 3 (2a + m) = t ==> clearly t is multiple of 3 and this is enough to answer the question ==> Sufficient

(2) m is a multiple of 3 ==> m = 3a If we substitute m in (**) then we get ==> 2k + 3 * 3a = t ==> 2k + 9a = t ==> Not sufficient

Hence, Ans: A
_________________

Cheers! Ravi

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t and 12 have a common factor greater than 1 [#permalink]

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05 Mar 2011, 01:57

90) 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

Rephrase-- Is t multiple of 2 ---- minimum requirement 2k+3m/12 = t/12

a) k is multiple of 2 nd 3

Sufficient

b) m is multiple of 3 => m is multiple of 9

Pls tell where m i wrong as the question has multiple of two and alone sufficient. Pls rephrase the question.I think i did mistake while rephrasing but when i attempted this question w/o rephrase first time i mark E second time i mark C. Both the times i was wrong.
_________________

The proof of understanding is the ability to explain it.

Re: t and 12 have a common factor greater than 1 [#permalink]

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05 Mar 2011, 02:13

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

12 = 2*2*3. Rephrasing the question - Is t a multiple of 2 or 3?

k/6 + m/4 =t/12 Multiply both sides by 12 => 2k + 3m = t

1) Sufficient. k=3 2*3 + 3m = t t is a multiple of 3. The answer is YES

2) Insufficient m=3 2k+3*3 = t

k=1, t=11. The answer is NO k=3, t=15. The answer is YES

GMATD11 wrote:

90) 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

Rephrase-- Is t multiple of 2 ---- minimum requirement 2k+3m/12 = t/12

a) k is multiple of 2 nd 3

Sufficient

b) m is multiple of 3 => m is multiple of 9

Pls tell where m i wrong as the question has multiple of two and alone sufficient. Pls rephrase the question.I think i did mistake while rephrasing but when i attempted this question w/o rephrase first time i mark E second time i mark C. Both the times i was wrong.

No worries! Think of DS paraphrase as the CR must be true statement. If you infer too much, you are wrong. You you infer too less, you are wrong. I think there is high correlation between verbal and math sections of gmat - high scores in CR will correlate with high DS caliber basically you are reasoning you way to the answer. Do I sound like a philosopher? Pardon me, this is not the psychology forum.

1 says k is multiple of 3, then just by looking at the above equation one can say t is also multiple of 3. Sufficient 2. m is multiple of 3 . that doesnt help us in any way here, also we dont know anything about k . so insufficient.

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