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# If k, m, and t are positive integers and k/6 + m/4 = t/12 ,

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Another A,

after simplifying the equation it comes to this form:

2k+3m=t

1) Suff because t will also be divisible by 3
2) Insuff because whether 2k and 3m (*3) do not provide any useful info about common factor

Ans: A
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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
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subhen wrote:
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'

It is another "twister" Gmat trap =)
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DS [#permalink]  03 Oct 2008, 10:50
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.
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Re: DS [#permalink]  03 Oct 2008, 11:00
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vksunder wrote:
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.

k/6 + m/4 = t/12
2k + 3m = t

1: if k is a multipe of 3, k = 3x where x is an integer
2k + 3m = t
2(3x) + 3m = t
3 (2x + m) = t

so 3 is a factor of t. suff

2: m is a multiple of 3 doesnot help us as we do not know about k.

A.
so suff...
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Common Factor [#permalink]  07 Aug 2010, 13:40
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.
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Re: Common Factor [#permalink]  07 Aug 2010, 14:13
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Expert's post
dkverma wrote:
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.

$$\frac{k}{6} + \frac{m}{4} = \frac{t}{12}$$ --> $$2k+3m=t$$.

(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.

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Re: Common Factor [#permalink]  07 Aug 2010, 14:44
Thanks Bunuel for the detailed explanation.
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Re: If k, m, and t are positive integer [#permalink]  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

$$\frac{k}{6}$$ + $$\frac{m}{4}$$ = $$\frac{t}{12}$$
==> 2k + 3m = t ----- (**)

(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
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t and 12 have a common factor greater than 1 [#permalink]  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.
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Re: t and 12 have a common factor greater than 1 [#permalink]  05 Mar 2011, 02:09
Expert's post
Merging similar topics.
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Re: t and 12 have a common factor greater than 1 [#permalink]  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.
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Re: Common Factor [#permalink]  05 Mar 2011, 02:46
Hi

There r n number of ways that u can lead to wrong conclusion.

But few ways to lead to correct conclusion.
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Re: Common Factor [#permalink]  05 Mar 2011, 03:56
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.
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Re: Common Factor [#permalink]  07 Mar 2011, 18:42
equation can be solved to 2k+3m = t

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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Re: DS - Common Factor [#permalink]  10 May 2011, 08:42
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.

Corrected Question Stem.

1. K= 3*a then, 4K + 6M = 2T => 2k + 3M = T

substituting, 2* 3*a + 3M = T => 3( 2a + M) = T

Thus factor 3 is always present. Hence Sufficient.

2. M= 3*b then 2k + 3M = T => 2k + 3*3M = T will depend upon the value of M. Not sufficient.

Thus A.
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Re: DS - Common Factor   [#permalink] 10 May 2011, 08:42

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