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:

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

Show Tags

05 Oct 2012, 08:14

5

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

kingb wrote:

If x, y, and z are positive integers and 3x = 4y = 7z, then the least possible value of x + y + z is

A. 33 B. 40 C. 49 D. 61 E. 84

Notice that since 3, 4, and 7 are co-prime (they don't share any common factor but 1), then:

x must be a multiple of 4 and 7, so the least value of x is 28 (LCM of 4 and 7); y must be a multiple of 3 and 7, so the least value of y is 21; z must be a multiple of 3 and 4, so the least value of z is 12;

So, the least possible value of x + y + z is 28+21+12=61.

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]

Show Tags

02 May 2011, 18:19

2

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

MSDHONI wrote:

When I see more than one '=', my instinct is to make them equal to k to get to a 'common ground'.

\(3x = 4y = 7z = k\)

\(x = \frac{k}{3}; y = \frac{k}{4}; z = \frac{k}{7}\) Now, all x, y and z are positive integers so k should be divisible by 3, 4 and 7. The minimum value of k will be 3*4*7.

\(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\) _________________

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

Show Tags

15 Jun 2013, 13:47

1

This post received KUDOS

kingb wrote:

If x, y, and z are positive integers and 3x = 4y = 7z, then the least possible value of x + y + z is

A. 33 B. 40 C. 49 D. 61 E. 84

We know x,y and z are postive integers so based on given information, we can say x= 4/3y and z= 4/7y

then x+y+z= 4/3y+y+4/7y-------> (4/3y+4/7y)+ y ------> 40y/21 +y -----> 61 y/21

61y/21= Sum of positive integers x,y and z. This sum also is a postive integer. Let us say it as Integer "a"

therefoee 61y/21= a now this implies y= a*21/61 ----For y to be an integer, we see that "a" has to be multiple of 61 and the smallest multiple of 61 is 1

Hence ans should D i.e 61 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]

Show Tags

02 May 2011, 14:15

I believe it's D. The solution is the lowest common multiplier. Both 3 and 7 are prime numbers and odd, while 4 is a multiple of 2, which doesn't help with the other two. So if you multiply 3, 4, and 7, you get 84. If you deduce further, x = 28, y = 21, z = 12, just the product of the 2 numbers left after the reference. So the sum is 61.

If there's any hole in my reasoning, I welcome comments. As I can benefit from it too (Thanks in advance.) _________________

= = = = = = = = = =

If you liked my post, please consider a Kudos for me. Thanks!

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

Show Tags

29 Jul 2014, 12:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

Show Tags

08 Nov 2015, 23:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]

Show Tags

01 Feb 2016, 07:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

Show Tags

16 Mar 2016, 09:34

VeritasPrepKarishma wrote:

MSDHONI wrote:

When I see more than one '=', my instinct is to make them equal to k to get to a 'common ground'.

\(3x = 4y = 7z = k\)

\(x = \frac{k}{3}; y = \frac{k}{4}; z = \frac{k}{7}\) Now, all x, y and z are positive integers so k should be divisible by 3, 4 and 7. The minimum value of k will be 3*4*7.

\(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\)

Excellent Approach i actually did it the other way Big takeaway from your Solution => IF YOU SEE ANY THREE OR MORE TERMS EQUATED => MAKE THEM EQUAL TO K . Thanks

gmatclubot

If x, y, and z are positive integers and 3x = 4y = 7z, then
[#permalink]
16 Mar 2016, 09:34

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...