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

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If k, m, and t are positive integers and k/6 + m/4 = t/12 , [#permalink] New post 05 Sep 2004, 02:10
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-k-m-and-t-are-positive-integers-and-k-6-m-4-t-127989.html
[Reveal] Spoiler: OA
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Re: Common Factor [#permalink] New post 07 Aug 2010, 14:13
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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.

Answer: A.
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Re: A quant question [#permalink] New post 12 Jul 2006, 03:19
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ishtmeet 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.





(1) k is a multiple of 3, so k=3n for some positive integer n

k/6+m/4= 3n/6+m/4= (6n+3m)/12

Thus t=6n+3m=3(2n+m)

Since n and m are both +ve integers, t is a multiple of 3, and t and 12 have a common factor of 3.
SUFFICIENT


(2) m is a multiple of 3, so k=3p for some positive integer p

k/6+m/4= k/6+3p/4= (2k+9p)/12

If k=p=1, t=11 and the only factor 11 and 12 have in common is 1
If k=1 and p=2, t is a multiple of 2, so t and 12 have a common factor of 2.
NOT SUFFICIENT


Answer: A
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Re: DS [#permalink] New post 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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 [#permalink] New post 05 Sep 2004, 02:51
Answer A.

t is a multiple of 3 so 3 is a grater than 1 common factor between t and 12
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 [#permalink] New post 05 Sep 2004, 08:07
A.

t = 2k + 3m

I. If k is a multiple of 3,
t = 2.A.3 + 3m = 3(2A+m)
3 is a common multiple. Ans: Yes
Suff.

II. If m is a multiple of 3,
t = 2k + 3.3.B = 2k + 9B
Ans is Yes for m = 3, k = 3; Ans is No for m = 3, k = 5
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A quant question [#permalink] New post 12 Jul 2006, 03:05
Can you please give a detailed answer to this question


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.



A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.
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do t and 12 have a common factor greater than 1 ? [#permalink] New post 24 Jun 2007, 14:46
Q5:
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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


OA: sorry, I don't have the OA
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 [#permalink] New post 24 Jun 2007, 14:59
I get A.

You can set this up by adding the fractions with a common denominator. You get

(2k/12) + (3m/12) = t/12

so now we know that

2k + 3m = t

And we want to know if t and 12 share a factor other than 1.

1 says that k is a multiple of 3. There is a rule that when you add together numbers that are all a multiple of the same number, the sum is also a multiple of that number. So with statement 1 we know that 2k is a multiple of 3, and it's given that 3m is also a multiple of 3, so t must be a multiple of 3. Since 12 is also a multiple of 3, both t and 12 share at least 3 as a common factor.

2 says that m is a multiple of 3. Using the same rule, we don't know anything about 2k, so we really don't know what t is a multiple of at all. It could be even or odd, and it doesn't have to have any of the factors that 12 has. So it's not enough information.
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Great Help!! [#permalink] New post 24 Jun 2007, 15:15
Thanks...ian7777..Wonderful explanation. I got it now.
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 [#permalink] New post 24 Jun 2007, 21:10
ian7777 wrote:
I get A.

You can set this up by adding the fractions with a common denominator. You get

(2k/12) + (3m/12) = t/12

so now we know that

2k + 3m = t

And we want to know if t and 12 share a factor other than 1.

1 says that k is a multiple of 3. There is a rule that when you add together numbers that are all a multiple of the same number, the sum is also a multiple of that number. So with statement 1 we know that 2k is a multiple of 3, and it's given that 3m is also a multiple of 3, so t must be a multiple of 3. Since 12 is also a multiple of 3, both t and 12 share at least 3 as a common factor.

2 says that m is a multiple of 3. Using the same rule, we don't know anything about 2k, so we really don't know what t is a multiple of at all. It could be even or odd, and it doesn't have to have any of the factors that 12 has. So it's not enough information.


Given : 2k + 3m = t

statement 1 -- k is a multiple of 3.

let k=3x

so 2*3x + 3m = t

6x + 3m = t

t = 3(m + 2x)

so, 3 is a multiple of t and hence 3 is the common factor of t and 12.

nothing can be inferred from statement 2.

Hence A.
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DS: Common multiple [#permalink] New post 05 Sep 2007, 07:37
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.

Last edited by sidbidus on 06 Sep 2007, 02:46, edited 1 time in total.
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 [#permalink] New post 05 Sep 2007, 08:58
I think it is C.

k+m = t/12.

Statements 1 and 2 imply 6 = t/12. Therefore, t =72. 72 and 12 have several common factors - 3, 4, 6.
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 [#permalink] New post 05 Sep 2007, 09:47
if given equation is simplified than

k+m=t/12 k, m, t positive integers

that means ANY number since it is positive integer will be enough to conclude
that t and 12 has common factor higher than 1.

Ans: D

(Just can you double check the problem, whether it is copied correctly)
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 [#permalink] New post 05 Sep 2007, 09:48
i think its D

cuz finally v get 12(k+m)=t
so wotever value u substitute T will hav a common factors with 12
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 [#permalink] New post 06 Sep 2007, 02:47
Sorry, posted incorrect question.

Correct Version:

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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 [#permalink] New post 06 Sep 2007, 03:01
The equation can be simplified as -
t=2k + 3m

Stmt1: K is multiple of 3. This means T will be a mulitple of 3. So SUFF

Stmt2: M is a multiple of 3. So T may or may not be the multiple of 3. So INSUFF

Answer is 'A'
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 [#permalink] New post 06 Sep 2007, 08:32
St1:
Represent k as 3a where a = any integer.

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

So t = 6a+3m = 3(m+2a). So t is also a multiple of 3 and must have 3 as it's factor. 12 will also have 3 as it's factor, so they have at least one factor in common. Sufficient.

St2:
Represent m as 3b where b = any integer.

k/6 + 3b/4 = t/12
2k + 9b/12 = t/12
2k + 9b = t
If k = 2, b = 3, then t = 31 then t and 12 has no common factor
If k = 1, b = 2, t = 20 and t and 12 have 2 as common factor.
Insufficient.

Ans A
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 [#permalink] New post 06 Sep 2007, 08:45
St1:
Represent k as 3a where a = any integer.

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

So t = 6a+3m = 3(m+2a). So t is also a multiple of 3 and must have 3 as it's factor. 12 will also have 3 as it's factor, so they have at least one factor in common. Sufficient.

St2:
Represent m as 3b where b = any integer.

k/6 + 3b/4 = t/12
2k + 9b/12 = t/12
2k + 9b = t
If k = 2, b = 3, then t = 31 then t and 12 has no common factor
If k = 1, b = 2, t = 20 and t and 12 have 2 as common factor.
Insufficient.

Ans A
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Re: DS: Common multiple [#permalink] New post 06 Sep 2007, 09:11
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.


One more A..

first I simplify the Q.

k/6 + m/4 = t/12
2k + 3m = t ------------1

3m is a multiple of 3.

St-1 : k is multiple of 3. = > 3k is multiple of 3.
if a & b are multiples of 3, then a+b is also multiple of 3. Hence T is multiple of 3 => T and 12 have atleast 1 common factor.
Suff.

St2: M is a multiple of 3. There is no new info given here. We already know this. (Eq1)
Let k = 1 and m = 1 in eq 1,
we have t = 5 which has no common factor with 12. Hence In-suff.

Hence A.
Re: DS: Common multiple   [#permalink] 06 Sep 2007, 09:11
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