If k, m, and t are positive integers and k/6 + m/4 = t/12 , : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 Feb 2017, 07:26

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 08 Apr 2004
Posts: 133
Location: Corea
Followers: 2

Kudos [?]: 670 [0], given: 0

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

### Show Tags

05 Sep 2004, 02:10
00:00

Difficulty:

65% (hard)

Question Stats:

59% (02:12) correct 41% (01:18) wrong based on 86 sessions

### HideShow timer Statistics

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
Math Expert
Joined: 02 Sep 2009
Posts: 37144
Followers: 7266

Kudos [?]: 96774 [2] , given: 10784

### Show Tags

07 Aug 2010, 14:13
2
KUDOS
Expert's post
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.

_________________
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1264
Followers: 29

Kudos [?]: 300 [1] , given: 0

### Show Tags

12 Jul 2006, 03:19
1
KUDOS
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

SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 69

Kudos [?]: 748 [1] , given: 19

### Show Tags

03 Oct 2008, 11:00
1
KUDOS
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...
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Director
Joined: 31 Aug 2004
Posts: 610
Followers: 3

Kudos [?]: 124 [0], given: 0

### Show Tags

05 Sep 2004, 02:51

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: 593
Followers: 2

Kudos [?]: 126 [0], given: 0

### Show Tags

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
Intern
Joined: 05 Mar 2006
Posts: 20
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

12 Jul 2006, 03:05

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.
Manager
Joined: 28 Aug 2006
Posts: 146
Followers: 2

Kudos [?]: 122 [0], given: 0

do t and 12 have a common factor greater than 1 ? [#permalink]

### Show Tags

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
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2

Kudos [?]: 59 [0], given: 0

### Show Tags

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.
Manager
Joined: 28 Aug 2006
Posts: 146
Followers: 2

Kudos [?]: 122 [0], given: 0

### Show Tags

24 Jun 2007, 15:15
Thanks...ian7777..Wonderful explanation. I got it now.
Senior Manager
Joined: 06 Jul 2004
Posts: 473
Location: united states
Followers: 1

Kudos [?]: 107 [0], given: 0

### Show Tags

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.
_________________

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: 325
Followers: 1

Kudos [?]: 47 [0], given: 0

### Show Tags

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.
Intern
Joined: 25 Aug 2007
Posts: 14
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

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.
Senior Manager
Joined: 27 Aug 2007
Posts: 253
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

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)
Manager
Joined: 09 Jul 2007
Posts: 178
Followers: 1

Kudos [?]: 90 [0], given: 0

### Show Tags

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
Senior Manager
Joined: 19 Feb 2007
Posts: 325
Followers: 1

Kudos [?]: 47 [0], given: 0

### Show Tags

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.
Director
Joined: 14 Jan 2007
Posts: 777
Followers: 2

Kudos [?]: 136 [0], given: 0

### Show Tags

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

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 31

Kudos [?]: 368 [0], given: 0

### Show Tags

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
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 31

Kudos [?]: 368 [0], given: 0

### Show Tags

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
Director
Joined: 09 Aug 2006
Posts: 525
Followers: 2

Kudos [?]: 94 [0], given: 0

### Show Tags

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

Go to page    1   2    Next  [ 36 posts ]

Similar topics Replies Last post
Similar
Topics:
If m is an integer, is m^4 divisible by 5? 4 02 Nov 2016, 02:00
What is the value of [3(k + m) + 4k]/[2k + m]? 1 08 Apr 2015, 13:16
5 If m and k are positive integers, is m!+8k a multiple of k? 3 23 Jun 2014, 06:06
32 If k, m, and t are positive integers and k/6 + m/4 = t/12 7 23 Feb 2012, 01:28
10 If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 5 20 Jun 2011, 06:55
Display posts from previous: Sort by