|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 08 Apr 2004
Posts: 137
Location: Corea
Followers: 1
Kudos [?]:
3
[0], given: 0
|
If k, m, and t are positive integers and k/6 + m/4 = t/12 , [#permalink]
 05 Sep 2004, 03:10
Question Stats:
49% (02:13) correct
49% (01:15) wrong based on 34 sessions
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
|
|
|
|
|
|
|
|
|
Director
Joined: 31 Aug 2004
Posts: 619
Followers: 1
Kudos [?]:
8
[0], given: 0
|
Answer A.
t is a multiple of 3 so 3 is a grater than 1 common factor between t and 12
|
|
|
|
|
|
Director
Joined: 20 Jul 2004
Posts: 601
Followers: 1
Kudos [?]:
4
[0], given: 0
|
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
|
|
|
|
|
|
Intern
Joined: 05 Mar 2006
Posts: 20
Followers: 0
Kudos [?]:
0
[0], given: 0
|
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.
|
|
|
|
|
|
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1278
Location: Madrid
Followers: 10
Kudos [?]:
72
[0], given: 0
|
Re: A quant question [#permalink]
12 Jul 2006, 04:19
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
|
|
|
|
|
|
Manager
Joined: 28 Aug 2006
Posts: 146
Followers: 1
Kudos [?]:
6
[0], given: 0
|
do t and 12 have a common factor greater than 1 ? [#permalink]
24 Jun 2007, 15:46
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
|
|
|
|
|
|
CIO
Joined: 09 Mar 2003
Posts: 471
Followers: 1
Kudos [?]:
16
[0], given: 0
|
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.
|
|
|
|
|
|
Manager
Joined: 28 Aug 2006
Posts: 146
Followers: 1
Kudos [?]:
6
[0], given: 0
|
Thanks...ian7777..Wonderful explanation. I got it now.
|
|
|
|
|
|
Senior Manager
Joined: 06 Jul 2004
Posts: 480
Location: united states
Followers: 1
Kudos [?]:
6
[0], given: 0
|
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.
_________________
for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..
|
|
|
|
|
|
Senior Manager
Joined: 19 Feb 2007
Posts: 328
Followers: 1
Kudos [?]:
7
[0], given: 0
|
DS: Common multiple [#permalink]
05 Sep 2007, 08:37
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, 03:46, edited 1 time in total.
|
|
|
|
|
|
Intern
Joined: 25 Aug 2007
Posts: 14
Followers: 0
Kudos [?]:
0
[0], given: 0
|
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.
|
|
|
|
|
|
Senior Manager
Joined: 27 Aug 2007
Posts: 260
Followers: 1
Kudos [?]:
7
[0], given: 0
|
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)
|
|
|
|
|
|
Manager
Joined: 09 Jul 2007
Posts: 178
Followers: 0
Kudos [?]:
4
[0], given: 0
|
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
|
|
|
|
|
|
Senior Manager
Joined: 19 Feb 2007
Posts: 328
Followers: 1
Kudos [?]:
7
[0], given: 0
|
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.
|
|
|
|
|
|
Director
Joined: 14 Jan 2007
Posts: 787
Followers: 1
Kudos [?]:
32
[0], given: 0
|
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'
|
|
|
|
|
|
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5134
Location: Singapore
Followers: 9
Kudos [?]:
89
[0], given: 0
|
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
|
|
|
|
|
|
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5134
Location: Singapore
Followers: 9
Kudos [?]:
89
[0], given: 0
|
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
|
|
|
|
|
|
Director
Joined: 09 Aug 2006
Posts: 531
Followers: 2
Kudos [?]:
20
[0], given: 0
|
Re: DS: Common multiple [#permalink]
06 Sep 2007, 10:11
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.
|
|
|
|
|
|
Manager
Joined: 29 Aug 2007
Posts: 112
Followers: 1
Kudos [?]:
0
[0], given: 0
|
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.
|
|
|
|
|
|
CEO
Joined: 29 Mar 2007
Posts: 2618
Followers: 13
Kudos [?]:
144
[0], given: 0
|
Re: DS: Common multiple [#permalink]
06 Sep 2007, 20:51
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.
|
|
|
|
|
|
|
Re: DS: Common multiple
[#permalink]
06 Sep 2007, 20:51
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
If k,m, and t are positive integers and k/6+m/4=t/12 , do t
|
falcor |
2 |
16 Sep 2004, 08:42 |
 |
|
|
If k, m, and t are positive integers and k/6 + m/4 = t/12 ,
|
pb_india |
10 |
23 Feb 2005, 20:07 |
|
|
|
|
If k, m, and t are positive integers and (k/6)+(m/4) =(t/12)
|
apollo168 |
3 |
24 Aug 2006, 10:43 |
|
|
1
|
|
If k, m, and t are positive integers and k/6 + m/4 = t/12 ,
|
leonidas |
5 |
07 Oct 2008, 23:07 |
|
|
2
|
 |
If k, m, and t are positive integers and k/6 + m/4 = t/12
|
Stiv |
3 |
23 Feb 2012, 02:28 |
|
|
|
|
|
|
close
