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:

x, y, x + y, x – 4y, xy, 2y For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be

(x + y) 2

y + 3

y

3y/2

x /3 + y

adding up all the elements of set we get (3x + xy)/6 = y+3 => solving this we get x=6 we have y>6 so we get x=6,y>6,x+y >12, x-4y < -18,xy>36 and 2y >12 arranging these we get x-4y, x, y, x+y, 2y, xy

median = (y + x+y) /2 = (2y+6)/2 = y+3 will go with option 2

Find the average and set it equal to the y+3 then find out that x=6 then you find that y and x+y are the middle 2 numbers and sub x for 6 then you find the median number is y+3
_________________

x, y, x + y, x - 4y, xy, 2y For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be A. (x+y)/2 B. y+3 C. y D. 3y/2 E. x/3+y

First of all: the median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

Next, given that the mean of the set equals y+3: x+y+(x+y)+(x-4y)+xy+2y=6*(y+3) --> 3x+xy=6*(y+3) --> x(3+y)=6(y+3), notice that generally you cannot reduce by 3+y here as suggested above --> (3+y)(x-6)=0 --> x=6 or y=-3 (not a valid solution as y>6), so x=6;

Arrange the set in ascending order: {x-4y, x, y, x+y, 2y, xy} --> median=(y+x+y)/2=(2y+6)/2=y+3.

Answer: B.

Or: pick some number greater than 6 for y, let's say y=7. Then our set will be: {6, 7, 13, -22, 42, 14} --> arrange in ascending order: {-22, 6, 7, 13, 14, 42}.

Thus the median of the set is (7+13)/2=10 --> 10=7+3=y+3.

Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]

Show Tags

14 Oct 2014, 04:37

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: For the set of terms shown above, if y > 6 and the mean of [#permalink]

Show Tags

15 Apr 2015, 20:20

Bunuel wrote:

x, y, x + y, x - 4y, xy, 2y For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be A. (x+y)/2 B. y+3 C. y D. 3y/2 E. x/3+y

First of all: the median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

Next, given that the mean of the set equals y+3: x+y+(x+y)+(x-4y)+xy+2y=6*(y+3) --> 3x+xy=6*(y+3) --> x(3+y)=6(y+3), notice that generally you cannot reduce by 3+y here as suggested above --> (3+y)(x-6)=0 --> x=6 or y=-3 (not a valid solution as y>6), so x=6;

Arrange the set in ascending order: {x-4y, x, y, x+y, 2y, xy} --> median=(y+x+y)/2=(2y+6)/2=y+3.

Answer: B.

Or: pick some number greater than 6 for y, let's say y=7. Then our set will be: {6, 7, 13, -22, 42, 14} --> arrange in ascending order: {-22, 6, 7, 13, 14, 42}.

Thus the median of the set is (7+13)/2=10 --> 10=7+3=y+3.

Just a small clarification needed. The highlighted portion: Is it implying that (y+3) on both sides cannot be discarded? if so, then why exactly because as per my understanding when the term in question can be ZERO, only then we must avoid cancelling it on both LHS and RHS.

As you continue to study, you're going to find that TESTing VALUES is a great approach for many types of GMAT questions (and not just Stats questions....DS questions, in particular, can often be beaten in this way). To spot 'TEST VALUES' questions, you should be on the lookout for variables, especially those that involve 'descriptions' (odd, even, >, <, integer, positive, negative, etc.). You'll also come to find that trying this approach during your studies (you can go back and re-attempt past questions) can help you to better define when to TEST VALUES and when to just 'do the math.'

Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]

Show Tags

30 Dec 2015, 18:18

I can't seem to find the easier version of this question (stated below) so I've posted to this forum: --

6, y, y + 6, 6 – 4y, 6y, 2y

For the set of terms shown above, if the mean of the set equals 10 then, in terms of y, the median must be A) (Y+6)/2 B) y + 3 C) y D) 3y/2 E) y + 2

I can't seem to get this answer in under 3 mins, even though it's an "easy" question. Any thoughts on how to this quicker than finding y, plugging in y for all the answers, then finding the median between the two middle numbers?

I can't seem to find the easier version of this question (stated below) so I've posted to this forum: --

6, y, y + 6, 6 – 4y, 6y, 2y

For the set of terms shown above, if the mean of the set equals 10 then, in terms of y, the median must be A) (Y+6)/2 B) y + 3 C) y D) 3y/2 E) y + 2

I can't seem to get this answer in under 3 mins, even though it's an "easy" question. Any thoughts on how to this quicker than finding y, plugging in y for all the answers, then finding the median between the two middle numbers?

I think you should be able to get it in under 3 minutes.

6 + y + y + 6 + 6 – 4y + 6y + 2y = 10*6 18 + 6y = 60 6y = 42 y = 7

I can't seem to find the easier version of this question (stated below) so I've posted to this forum: --

6, y, y + 6, 6 – 4y, 6y, 2y

For the set of terms shown above, if the mean of the set equals 10 then, in terms of y, the median must be A) (Y+6)/2 B) y + 3 C) y D) 3y/2 E) y + 2

I can't seem to get this answer in under 3 mins, even though it's an "easy" question. Any thoughts on how to this quicker than finding y, plugging in y for all the answers, then finding the median between the two middle numbers?

Hi jwamala,

Since you've brought up the issue of 'time', when you dealt with this question, what were you doing for those 3 minutes? If you were to review your 'steps', what took the most time to complete (and WHY?)?

Since the 6 terms are not necessarily in order (from least to greatest), we really DO need to figure out the value of Y first so that we can determine how all of the other pieces of information 'inter-connect.'

Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]

Show Tags

31 Dec 2015, 10:50

Hey Rich, it was the first question of the section for me, so I think I was just adjusting. If I look back at my scrap, I didn't even fully compute all of the numbers in the set. I knocked off the two highs and two lows to find the median between 7 & 13, then switched to (y + y + 6 )/2 to find the answer.

Hey Rich, it was the first question of the section for me, so I think I was just adjusting. If I look back at my scrap, I didn't even fully compute all of the numbers in the set. I knocked off the two highs and two lows to find the median between 7 & 13, then switched to (y + y + 6 )/2 to find the answer.

Hi jwamala,

You wrote something interesting just now - you "knocked off the two highs and two lows...", but HOW did you know which ones were the biggest and which ones were the smallest? Wouldn't the value of Y impact those calculations (for example, what if Y=10? Y=0? Y=something negative?).

When dealing with ANY type of GMAT question (not just Quant questions), it's important to think about what you KNOW and what you DON'T KNOW. The correct answer will be based on the information that you're GIVEN, so it helps to be a little cynical about how the information is presented (in this prompt, you should be thinking "are these 6 numbers in numerical order or not?"

Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]

Show Tags

31 Dec 2015, 11:13

1

This post received KUDOS

That makes sense. I meant after I found y=7 I had written down: 6, 7, 13, 6 - 4(7), 6(7), 2(7) and was able to cross out the 2 highs and 2 lows terms from here w/o simplifying.

That makes sense. I meant after I found y=7 I had written down: 6, 7, 13, 6 - 4(7), 6(7), 2(7) and was able to cross out the 2 highs and 2 lows terms from here w/o simplifying.

Hi jwamala,

What you describe (above) is the PERFECT approach, so you had the right idea. It might just be that you're still honing your skills right now, so the work might be taking a little longer to complete than it will in the future. Most GMAT questions involve work that is actually pretty straight-forward, so you shouldn't shy away from that work (or try to do it in your head) - just put the pen on the pad and get it done.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...