If k, m, and t are positive integers and k/6 + m/4 = t/12 , : GMAT Data Sufficiency (DS) - Page 2
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# If k, m, and t are positive integers and k/6 + m/4 = t/12 ,

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06 Sep 2007, 11:05
The anwer is A

t = 2K + 3M

Since K is postivie and multiple of 3

K min has to be 3

t = Even > = 6 + 3M

t= Even >= 6 + 3 * m

since M is also positive integer , smallest value m = 1

checking for values, you can safely say that t is a number thats is greater than >= 9 , and for any number to be written as 2K+ 3m.. formfactor.. there is a value.
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06 Sep 2007, 19:51
sidbidus 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.

S1: get the equation 2k+3m=t. So if k is a multiple of 3. 2k is a multiple of 3. 3m is also a multiple of 3. so S1 is suff.

S2: nothing new from this and we don't know if k is a multiple or not.

Ans A.
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06 Sep 2007, 21:40
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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07 Sep 2007, 05:04
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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07 Sep 2007, 05:07
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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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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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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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]

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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

$$\frac{k}{6}$$ + $$\frac{m}{4}$$ = $$\frac{t}{12}$$
==> 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]

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

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05 Mar 2011, 02:09
Merging similar topics.
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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.
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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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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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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]

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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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