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:

MySolution Using (1) \((a+b+c+4)/4 > (a +b+c)/3\) or \(a+b+c<12\) No clue about a, b, c. Insufficient

Using (2) four possibilities: (a, b, c, 4), median (b+c)/2 < b or c < b, not possible (4, a, b, c), median (b+a)/2 < b or b > a, already known (a, 4, b, c), median (b+4)/2 < b or b > 4 (a, b, 4, c), median (b+c)/2 < b or b > 4 Considering all the above cases: \(b > 4\) But still no clue about a & c. Insufficient.

Using (1) & (2), \(b_m_i_n = 5\), (a, b, c are integers) assuming all are in A.P. which means, a = b - k c = b + k from (1), \(a+b+c < 12\) or \((b-k)+b+(b+k) < 12\) or\(3b < 12\) or \(b < 4\) but \(b > 4\) (from (2)), thus the assumption is wrong. and a, b, c cannot be in AP

Re: a, b and c are integers such that a < b < c. Do they have a [#permalink]

Show Tags

10 Jun 2015, 11:52

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: a, b and c are integers such that a < b < c. Do they have a [#permalink]

Show Tags

18 Jun 2015, 05:08

Hi, I see that no one responded to this one. I cannot figure it out muself, but will share my thoughts.

[1] a+b+c+4 / 4 > a+b+c/3 I tested the means for these numbers: 2-4-6-4 (if a,b,c did have a common difference). It turns out that the means are the same. I tested the means for these numbers: 2-5-6-4 (if a,b,c did not have a common difference). It turns out that the first means is greater than the second.

However, I have no idea what this means, or if, in the condition when they d have a common difference, they should also have a common difference with 4 (the extra number).

[2] Let's say that these are the numbers in increasing order: a-b-c-4 or 4-a-b-c (and I am not sure here is 4 is supposed to be greatersmaller than the rest or could even be in between).

Then bc (or ab) > b. If the numbers are: 1,2,3,4, then the median is 2.5. For 1,2,3 the median is 3. So, the second median is greater, which should not happen. If the numbers are: 1,3,4,4, then the median is 3.5. For 1,3,4 the median is 3. The first median is greater, which is what should happen.

Re: a, b and c are integers such that a < b < c. Do they have a [#permalink]

Show Tags

22 Jun 2015, 14:06

pacifist85 wrote:

Hi, I see that no one responded to this one. I cannot figure it out muself, but will share my thoughts.

[1] a+b+c+4 / 4 > a+b+c/3 I tested the means for these numbers: 2-4-6-4 (if a,b,c did have a common difference). It turns out that the means are the same. I tested the means for these numbers: 2-5-6-4 (if a,b,c did not have a common difference). It turns out that the first means is greater than the second.

However, I have no idea what this means, or if, in the condition when they d have a common difference, they should also have a common difference with 4 (the extra number).

[2] Let's say that these are the numbers in increasing order: a-b-c-4 or 4-a-b-c (and I am not sure here is 4 is supposed to be greatersmaller than the rest or could even be in between).

Then bc (or ab) > b. If the numbers are: 1,2,3,4, then the median is 2.5. For 1,2,3 the median is 3. So, the second median is greater, which should not happen. If the numbers are: 1,3,4,4, then the median is 3.5. For 1,3,4 the median is 3. The first median is greater, which is what should happen.

But again, I have no idea what this means...

We need to find out whether is a,b,c - progression

1.s (a+b+c+4)/4>(a+b+c)/3 , thus a+b+c<12 or in other words a+b+c<=11 as a,b,c - integers. Let's test it 2,3,4=9 ok 2,3,5=10 also satisfy. Thus s1 is unsuf. 2. median of a,b,c is b as a<b<c. Median in a,b,c,4 could be 4 versions (4,a,b,c) , (a,4,b,c), (a,b,4,c), (a,b,c,4) (4,a,b,c) means b>(a+b)/2 if a>4 . b>a>4 (a,4,b,c) means b>(4+b)2 if a<4<b . b>4 - (a,b,4,c) means b>(b+4)/2 if b<4<c. b>4 - doesn't work as 4>b condition (a,b,c,4) means b>(b+c)/2 if b<c<4. b>c - doesn't work as c>b condition Conclusion: b>a>4 for sure Clearly statement 2 is not suf, as a,b,c could be progression or could be not. Simply input numbers

Let's combine 1 and 2 statements. We know that a+b+c<12, and b>a>4. Let's input numbers a=5, b=6,c=7 total>12. the answer is no, thus sufficient. Answer C.

gmatclubot

Re: a, b and c are integers such that a < b < c. Do they have a
[#permalink]
22 Jun 2015, 14:06

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...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...