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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

07 Oct 2008, 23:07

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

50% (00:00) correct
50% (00:04) wrong based on 1 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.
_________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

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

(1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible

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

(1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible

Hence (A)

If there is no common factor visible, shouldn't (B) also be sufficient? as this is a YES or NO type question.
_________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

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

(1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible

Hence (A)

If there is no common factor visible, shouldn't (B) also be sufficient? as this is a YES or NO type question.

(B) does not guarantee of a common factor (A, on the other hand, does). It's possible (in B) for t to have something in common with 12 even when no common factor is obvious.

For instance, if k=3, m=4, then common factor = 2 However, k=5, m=3 does not give any common factor
_________________

This is a good question. Especially when it comes to attacking the second statement

2K + 3M / 12 = T/12

Statement 1-----------SUFFICIENT

K=3N Try N =1, (M CAN TAKE ANY VALUES AND STILL BE A MUTILPLE OF 3 AS IT HAS 3 AS ITS COEFFECIENT) 6 + 3M /12 =T/12 3(2+M)/12 =T/12 Therefore 3 is a factor of T and 12 other than 1 Hence A

Statement 2------------INSUFFICIENT

M=3N Try N=1 2K + 9/ 12 = T/12 No common factor other than 1 But if we put K =6, AND N=1 We will get 12 + 9 / 12 =T/12 =3 (4+3)/12 = T/12 Common factor 3

Two different values. Hence Insufficient

Hence A

gmatclubot

Re: Highest Common Factor
[#permalink]
17 Dec 2009, 02:19