Author 
Message 
Intern
Joined: 22 Jul 2008
Posts: 42

If k, m, and t are positive integers and 6k + 4m = 12t, do t [#permalink]
Show Tags
13 Oct 2008, 08:34
1
This post was BOOKMARKED
Question Stats:
0% (00:00) correct 100% (00:04) wrong based on 10 sessions
HideShow timer Statistics
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. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Manager
Joined: 15 Apr 2008
Posts: 159

Re: DS Question Set 7 [#permalink]
Show Tags
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



Current Student
Joined: 28 Dec 2004
Posts: 3311
Location: New York City
Schools: Wharton'11 HBS'12

Re: DS Question Set 7 [#permalink]
Show Tags
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=6t9 or 2M=3(t3) then t3 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(t3), suppose 2M=18, the t3=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



SVP
Joined: 17 Jun 2008
Posts: 1504

Re: DS Question Set 7 [#permalink]
Show Tags
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.



Manager
Joined: 09 Oct 2008
Posts: 93

Re: DS Question Set 7 [#permalink]
Show Tags
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?



GMAT Instructor
Joined: 04 Jul 2006
Posts: 1254
Location: Madrid

Re: DS Question Set 7 [#permalink]
Show Tags
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



Intern
Joined: 22 Jul 2008
Posts: 42

Re: DS Question Set 7 [#permalink]
Show Tags
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



Retired Moderator
Joined: 05 Jul 2006
Posts: 1741

Re: DS Question Set 7 [#permalink]
Show Tags
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



SVP
Joined: 17 Jun 2008
Posts: 1504

Re: DS Question Set 7 [#permalink]
Show Tags
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.



Intern
Joined: 22 Jul 2008
Posts: 42

Re: DS Question Set 7 [#permalink]
Show Tags
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



SVP
Joined: 17 Jun 2008
Posts: 1504

Re: DS Question Set 7 [#permalink]
Show Tags
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.



Manager
Joined: 04 Sep 2008
Posts: 238
Location: Kolkata
Schools: La Martiniere for Boys

Re: DS Question Set 7 [#permalink]
Show Tags
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
_________________
Thanks rampuria



Retired Moderator
Joined: 18 Jul 2008
Posts: 920

Re: DS Question Set 7 [#permalink]
Show Tags
21 Oct 2008, 10:34



VP
Joined: 05 Jul 2008
Posts: 1369

Re: DS Question Set 7 [#permalink]
Show Tags
22 Oct 2008, 11:50
bigfernhead wrote: Aren't the two problems different? I mean some one changed the equation or typed it wrong



Retired Moderator
Joined: 18 Jul 2008
Posts: 920

Re: DS Question Set 7 [#permalink]
Show Tags
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



VP
Joined: 30 Jun 2008
Posts: 1019

Re: DS Question Set 7 [#permalink]
Show Tags
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 ?? == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
_________________
"You have to find it. No one else can find it for you."  Bjorn Borg




Re: DS Question Set 7
[#permalink]
23 Oct 2008, 19:20






