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For the set of terms shown above, if y > 6 and the mean of

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

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Feb 2012, 16:59, edited 1 time in total.
Edited the question and added the OA

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Re: Sequence [#permalink]

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adarsh12345 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

(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

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Re: Sequence [#permalink]

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New post 02 Nov 2009, 12:22
adarsh12345 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

(x + y) 2

y + 3

y

3y/2

x /3 + y


I get:

(x + y + x + y + x - 4y + xy + 2y)/6 = y + 3

Combine like terms and you get:

3x + xy = 6y + 18
x(3+y)=6(y+3)
3+y cancel
x = 6

Substitute 6 for x in the above equation
Also, since y>6 pick any number for y and solve. Let's say I pick a 7

the following six numbers are:
6, 7, 13, -22, 42, 14

median = (7+13)/2 = 10

my answer is B = y + 3

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Re: Sequence [#permalink]

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New post 11 Feb 2012, 16:33
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
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Re: Sequence [#permalink]

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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.

Answer: B.

Hope it helps.
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For the set of terms shown above, if y > 6 and the mean of the s [#permalink]

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New post 13 Dec 2012, 00:05
x y, x + y, x – 4y, xy, 2y

\(\frac{{x + y + (x+y) + (x-4y) + xy + 2y}}{{6}} = y+3\)
\({x + y + (x+y) + (x-4y) + xy + 2y}=6(y+3)\)
\(3x + xy = 6(x+y)\)
\(x(y+3)=6(3+y)\)
\(x=6\)

Let y=7 and x=6: Set={-18, 6, 7, 13, 14, 42}
Thus, median is \frac{(7+13)}{2} = 10


A. (x + y)/2 = 13/2
B. y + 3 = 10
C. y = 7
D. (3y)/2 = 21/2
E. x/3 + y = 9

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

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

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New post 15 Apr 2015, 21:20
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.

Answer: B.

Hope it helps.



Hi Bunuel,

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.

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

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New post 16 Apr 2015, 08:02
adarsh12345 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



assume y = 7

x would come out to be 6 (as mean is given)

median = 10

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

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New post 06 Jun 2015, 12:31
plugging in seems to be the easier way out here. can anyone suggest ways to identify questions in stats that could be solved by plugging in numbers

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

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New post 08 Jun 2015, 21:55
Hi Shreks1190,

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.'

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

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New post 30 Dec 2015, 19:18
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?

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

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jwamala wrote:
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

Numbers are 6, 7, 13, 6 – 22, 42, 14
Rearranging, -22, 6, 7, 13, 14, 42

Median = 10
Option B
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Re: For the set of terms shown above, if y > 6 and the mean of [#permalink]

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New post 31 Dec 2015, 11:37
jwamala wrote:
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.'

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

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New post 31 Dec 2015, 11:50
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.

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

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New post 31 Dec 2015, 11:54
jwamala wrote:
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?"

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

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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.

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

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jwamala wrote:
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.

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

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New post 10 Dec 2016, 17:02
I hope below solution will be of some help.

All we know is that y>6, so let's pick y=7. We now know that the mean is 7+3=10.

So:

Avg = (sum of terms)/(# of terms)
10 = (x + 7 + x+7 + x-28 + 7x + 14)/6
60 = 10x + 28 - 28
60 = 10x
6 = x

We want the median; since we have an even number of terms, the median is the average of the two middle terms.

Plugging in x=6 and y=7, our terms are:

{6, 7, 13, -22, 42, 14}

arranging them:

{-22, 6, 7, 13, 14, 42}

and our median is (7+13)/2 = 20/2 = 10

Now we sub into the choices:

a) 13/2... nope
b) 7+3... yup!
c) 7... nope
d) 21/2... nope
e) 2 + 7 = 9... nope

Only (B) is a match - choose (B)!
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Re: For the set of terms shown above, if y > 6 and the mean of   [#permalink] 10 Dec 2016, 17:02
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