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

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If k, m, and t are positive integers and 6k + 4m = 12t, do t [#permalink]

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If k, m, and t are positive integers and 6k + 4m = 12t, 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- Question Set 7 [#permalink]

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New post 13 Oct 2008, 09:53
rnemani wrote:
If k, m, and t are positive integers and 6k + 4m = 12t, 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.


I am confused between C and E but will go with E

Statement 1: only K is defined , there is no information of M. Not suff.
Statement 2: only m is defined, there is no information of K. not suff.
now taking both together for some multiples of 3, 12 and T have common factors but for some they don't have a common factor.
for instance 6*6+4*6= 12*5 . here 12 and 5 do not have any multiples common other than 1

let's see what others have to say.
what is the OA

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Re: DS- Question Set 7 [#permalink]

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New post 13 Oct 2008, 10:36
rnemani wrote:
If k, m, and t are positive integers and 6k + 4m = 12t, 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.


2(3k+2M)=12t

3k+2M=6t <-- stem

1) K=3N

9N+2M=6t, if N=1 then 9+2M=6t if M and t are positive integers then, 2M=6t-9 or 2M=3(t-3)

then t-3 has to be even, so suppose t=5, then M=3, if t=5 and 12 have no factor greater than 1..

if 2M=3(t-3), suppose 2M=18, the t-3=6 t=9, then 12 and t have factors, 1 and 3 in common..insuff

2)I just picked example above of M=multiple of 3..
6K+4*3N=12t

3(2K+4N)=12t -> 2k+4N=4t -> 2(K+2n)=4t ; k+2n=2t ; then we know K has to be even,

suppose K=2, n=2, then 2+4=2t or 6=2t t=3 then 12 and 3 have factors 1 and 3 in common

suppose K=4, N=2 4+4=2t, 8, t=4 then 12 and t have 1 and 2 and 4 in common..

B is sufficient

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Re: DS- Question Set 7 [#permalink]

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New post 13 Oct 2008, 12:27
E for me.

The original equation can be simplified to 3k + 2m = 2*3*k

From stmt1: k is a multiple of 3. But, k has to be an even integer, hence k will be an even multiple of 3. That means, k = 2*3*x for x = 1,2,3,.....

Thus, from the original equation, 3*2*3*x + 2m = 2*3*k

In order for the above equation to be true, m must be a multiple of 3. That means, m = 3y for y = 1,2,3....

Thus, in the original equation, 3*2*3*x + 2*3*y = 2*3*k
or 3x + y = k or, k = 4, 5, 7, 8, etc. and hence insufficient.

Stmt2 is the same as stmt1 and hence insufficient.

Combining the two also is insufficient.

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Re: DS- Question Set 7 [#permalink]

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New post 13 Oct 2008, 19:20
imo B.
3(2K+4N)=12t -> 2k+4N=4t -> 2(K+2n)=4t ; k+2n=2t ; then we know K has to be even,

this makes- sense..

oa?

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Re: DS- Question Set 7 [#permalink]

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New post 14 Oct 2008, 06:05
rnemani wrote:
If k, m, and t are positive integers and 6k + 4m = 12t, 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.


t=k/2 + m/3

We know that for t to be an integer, k must be an even integer and m must be a multiple of 3

(1) If k is a multiple of 3, so is k/2. However m/3 could be any positive integer. NOT SUFF
(2) No new imformation. NOT SUFF
(T) If k=6 and m=3 YES
If k=6 and m=6 NO

NOT SUFF

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Re: DS- Question Set 7 [#permalink]

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New post 14 Oct 2008, 10:28
OA is A..
I think OA of A makes sense as the solved equation is :

2k + 3m = t

if k is multiple of 3; then 2k+3m will be multiple of 3 as a result t and 12 will have common multiple

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Re: DS- Question Set 7 [#permalink]

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New post 14 Oct 2008, 12:57
rnemani wrote:
If k, m, and t are positive integers and 6k + 4m = 12t, 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.


given

2(3k+2m) =12t ie 3k+2m = 6t

from one

3k = 9f , f can be even or odd

2m is even and so is 6t , as such f is even , rewriting 3k = 9f = 18j

18j+2m = 6t , 9j+m = 3t
thus surely m is a multiple of 3

9j+3u =3t

3j+u = t ( t = 4,5...) ...........insuff

from 2

m = 3p , 2m = 6p

3k+6p = 6t

k+2p = 2t , k is even

2s+2p =2t , s+p = t ..........insuff

both together

18j+6p = 6t

3j+p = t .......insuff..........E

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Re: DS- Question Set 7 [#permalink]

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New post 15 Oct 2008, 02:24
rnemani wrote:
OA is A..
I think OA of A makes sense as the solved equation is :

2k + 3m = t

if k is multiple of 3; then 2k+3m will be multiple of 3 as a result t and 12 will have common multiple



Are you sure there is no typo in the question? In the current form, OA cannot be A.
Also, I do not know how you got the above equation.

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Re: DS- Question Set 7 [#permalink]

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New post 16 Oct 2008, 12:01
Yes the OA is A.
I see an issue with my equation. But the OA says A.

Now I am confused

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Re: DS- Question Set 7 [#permalink]

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New post 17 Oct 2008, 01:35
rnemani wrote:
Yes the OA is A.
I see an issue with my equation. But the OA says A.

Now I am confused


One more time......

6k + 4m = 12t or, 3k + 2m = 6t.

From stmt1: k = 3x, for x = 1,2,3,4,....

Hence, 3*3*x + 2*m = 3*2*t

In order for the above equation to be true for x, m and t as integers, x has to be even and m has to be a multiple of 3.

Thus, with x = 2, m = 3, t will be 4 and hence, sufficient.
With, any other values of x and m also, this will be true.

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Re: DS- Question Set 7 [#permalink]

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New post 21 Oct 2008, 09:50
If k, m, and t are positive integers and 6k + 4m = 12t, 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) Since LHS is a multiple of 12 we can write the LHS as

we have to express it as a multiple of 12

So it takes the form 36a + 12b......

Which takes the form 12(3a +1)

Now, 3a + 1 can never be a multiple of 12...........There fore they are prime to each other.

Taking 2) 6m + 12b =12t

this takes the form 12a +12b = 12t

which is 12(a+b) = 12 t

Now here (a+b) can take any value. It can be prime to 12 as well as have common factor with 12.

Therefore insufff....

Hence answer is A
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Re: DS- Question Set 7 [#permalink]

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New post 21 Oct 2008, 10:34

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Re: DS- Question Set 7 [#permalink]

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New post 22 Oct 2008, 11:50
bigfernhead wrote:


Aren't the two problems different? I mean some one changed the equation or typed it wrong

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Re: DS- Question Set 7 [#permalink]

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New post 22 Oct 2008, 12:09
Sorry, I should have been more clear.

I think the question in this post is typed incorrectly, and should have been the same question in the post I linked. That would explain why A) is the correct answer.

icandy wrote:
bigfernhead wrote:


Aren't the two problems different? I mean some one changed the equation or typed it wrong

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Re: DS- Question Set 7 [#permalink]

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New post 23 Oct 2008, 19:20
Ok .... the above posted question may have a few typos .... But shouldn't E be the answer for the above posted question ??
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Re: DS- Question Set 7   [#permalink] 23 Oct 2008, 19:20
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