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 350,000 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: Are all of the numbers in a certain list of 15 numbers [#permalink]
04 Mar 2013, 00:02

1

This post received KUDOS

Hello !

Here is how I solved it. Please correct me if I'm wrong :

(1) : Since you don't have any contraints regarding the numbers : the fifteen numbers can all equal 4 or you can have fourteen 0 and one 60. Not sufficient.

(2) : There are several ways to reach 12 by adding 3 numbers together : 4 + 4 + 4 = 12 3 + 3 + 6 = 12 8 + 2 +2 = 12 etc...

Let's consider the ways where you have at least 2 different numbers. For examples : 3 + 3 + 6. Let's say your fifteen numbers are divided in 5 groups of numbers composed by 3, 3 and 6 :

Statements 2 tells us we can pick any 3 numbers and get 12 by adding them. If you pick one full group : 3+3+6, you get 12. But if you pick 3, 3 from one group and another 3 from another group, you get 3+3+3 = 9. It is therefore impossible to have different numbers, they all have to be the same. Sufficient.

Re: Equal number DS-find the shortcut [#permalink]
18 Jul 2009, 13:34

Expert's post

sondenso wrote:

Are all of the numbers in a certain list of 15 numbers equal?

1. The sum of all the numbers in the list is 60 2. The sum of any 3 numbers in the list is 12

Guys, can you tell me what is the logic disguided in the second stat.? Thanks!

There are a few ways to look at this. One is to reverse the problem: say they aren't all equal. Write the set in increasing order: {a, b, c, ..., m, n, o}, and while some of these might be equal, we must have a < o. Well clearly then the sum of the three smallest numbers is less than the sum of the three largest, (a+b+c < m+n+o), so the sum of any three numbers in the list isn't always the same. So the only way S2 can be true is if all the numbers are equal. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: Equal number DS-find the shortcut [#permalink]
18 Jul 2009, 17:29

IanStewart wrote:

sondenso wrote:

Are all of the numbers in a certain list of 15 numbers equal?

1. The sum of all the numbers in the list is 60 2. The sum of any 3 numbers in the list is 12

Guys, can you tell me what is the logic disguided in the second stat.? Thanks!

There are a few ways to look at this. One is to reverse the problem: say they aren't all equal. Write the set in increasing order: {a, b, c, ..., m, n, o}, and while some of these might be equal, we must have a < o. Well clearly then the sum of the three smallest numbers is less than the sum of the three largest, (a+b+c < m+n+o), so the sum of any three numbers in the list isn't always the same. So the only way S2 can be true is if all the numbers are equal.

Re: Equal number DS-find the shortcut [#permalink]
01 Aug 2011, 19:34

For B, any three number's sum is 12. So what if 2,5,5, the sum is 12 but they are not equal.

Could you explain this ????

IanStewart wrote:

sondenso wrote:

Are all of the numbers in a certain list of 15 numbers equal?

1. The sum of all the numbers in the list is 60 2. The sum of any 3 numbers in the list is 12

Guys, can you tell me what is the logic disguided in the second stat.? Thanks!

There are a few ways to look at this. One is to reverse the problem: say they aren't all equal. Write the set in increasing order: {a, b, c, ..., m, n, o}, and while some of these might be equal, we must have a < o. Well clearly then the sum of the three smallest numbers is less than the sum of the three largest, (a+b+c < m+n+o), so the sum of any three numbers in the list isn't always the same. So the only way S2 can be true is if all the numbers are equal.

Re: Equal number DS-find the shortcut [#permalink]
07 Aug 2011, 00:00

i would say both statments together are correct.

because taking statment 1 and 2 will reveal that 60 divided by 15 is 4 so we know the set is containing the number 4 and statment 2 saying that 3 times 4 is 12

Re: Equal number DS-find the shortcut [#permalink]
07 Aug 2011, 03:06

Quote:

There are a few ways to look at this. One is to reverse the problem: say they aren't all equal. Write the set in increasing order: {a, b, c, ..., m, n, o}, and while some of these might be equal, we must have a < o. Well clearly then the sum of the three smallest numbers is less than the sum of the three largest, (a+b+c < m+n+o), so the sum of any three numbers in the list isn't always the same. So the only way S2 can be true is if all the numbers are equal.

Thanks for this explanation, IanStewart. _________________

Re: Are all of the numbers in a certain list of 15 numbers [#permalink]
04 Mar 2013, 02:10

Expert's post

pancakeFR wrote:

Hello !

Here is how I solved it. Please correct me if I'm wrong :

(1) : Since you don't have any contraints regarding the numbers : the fifteen numbers can all equal 4 or you can have fourteen 0 and one 60. Not sufficient.

(2) : There are several ways to reach 12 by adding 3 numbers together : 4 + 4 + 4 = 12 3 + 3 + 6 = 12 8 + 2 +2 = 12 etc...

Let's consider the ways where you have at least 2 different numbers. For examples : 3 + 3 + 6. Let's say your fifteen numbers are divided in 5 groups of numbers composed by 3, 3 and 6 :

Statements 2 tells us we can pick any 3 numbers and get 12 by adding them. If you pick one full group : 3+3+6, you get 12. But if you pick 3, 3 from one group and another 3 from another group, you get 3+3+3 = 9. It is therefore impossible to have different numbers, they all have to be the same. Sufficient.

Are all of the numbers in a certain list of 15 numbers equal?

(1) The sum of all the numbers in the list is 60. Clearly insufficient.

(2) The sum of any 3 numbers in the list is 12. Since the sum of ANY 3 numbers is 12 then ALL numbers must equal to 12/3=4, because if not all the numbers equal to 4, then we could pick certain set of 3 numbers so that their sum is not 12. Sufficient.

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

I have not posted in more than a month! It has been a super busy period, wrapping things up at Universal Music, completing most of the admin tasks in preparation for Stanford...