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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x and y are unknown positive integers, is the mean of the list {6,

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Math Expert V
Joined: 02 Sep 2009
Posts: 61283
If x and y are unknown positive integers, is the mean of the list {6,  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 55% (02:47) correct 45% (02:43) wrong based on 58 sessions

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If x and y are unknown positive integers, is the mean of the list {6, 7, 1, 5, x, y} greater than the median of the set?

(1) x + y = 7
(2) x - y = 3

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Posts: 547
Re: If x and y are unknown positive integers, is the mean of the list {6,  [#permalink]

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x , y — unknown positive integers
—> set {6,7,1,5,x,y}
Is the mean of the set greater than the median of the set?

(Statement1): x+ y = 7
There are 3 cases for the values of x and y:

(Case1): 1+6 =7
—> the mean = 26/6 = 4.(3)
--> the median=( 5+6)/2= 5.5
4.(3) < 5.5 (No)

(Case2): 2+5= 7
—> the mean =4.(3).
--> The median = (5+5)/2= 5
4.(3) < 5 (No)

(Case3): 3+4 =7
—> the mean = 4.(3)
--> The median = (4+5)/2= 4.5
4.(3) < 4.5 ( No)
Sufficient

(Statement2): x—y =3
If x =5, y= 2, then
The mean= 4.(3),
the median = (5+5)/2 = 5
4.(3) < 5 (No)

If x= 20, y = 17, then
The mean = 56/6= 9.(3)
the median=( 6+7)/2 = 6.5
—> 9.(3) > 6.5 (Yes)
Insufficient

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GMAT Tutor G
Joined: 24 Jun 2008
Posts: 2011
Re: If x and y are unknown positive integers, is the mean of the list {6,  [#permalink]

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There's a problem with the wording here; in math a "set" does not have repeated elements, but presumably they intend for repetition to be possible (if it's not, Statement 1 is immediately sufficient because x and y need to be 3 and 4).

From Statement 1, we know the exact value of the sum of the six elements, so we can find the mean -- it is 26/6 = 4 + 1/3. Since x and y average to 3.5, one of x or y is less than 3.5, and that's the second-smallest value in the set, and one of them is greater than 3.5, so is at least 4. So to find the median we'll be averaging two numbers, one of which is at least 4 and one of which is at least 5, and the median will be 4.5 or greater, and the median is certainly greater than the mean. So Statement 1 is sufficient.

Using Statement 2, we know x = y + 3. If y is a huge number, the mean will clearly exceed the median. But if y = 1 (assuming repetition of values is allowed), then the mean turns out to be less than the median, so Statement 2 is not sufficient. Even if we disallow repeated values, if y = 8 the mean still turns out to be very slightly less than the median. So the answer is A.
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Math Expert V
Joined: 02 Sep 2009
Posts: 61283
Re: If x and y are unknown positive integers, is the mean of the list {6,  [#permalink]

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IanStewart wrote:
There's a problem with the wording here; in math a "set" does not have repeated elements, but presumably they intend for repetition to be possible (if it's not, Statement 1 is immediately sufficient because x and y need to be 3 and 4).

From Statement 1, we know the exact value of the sum of the six elements, so we can find the mean -- it is 26/6 = 4 + 1/3. Since x and y average to 3.5, one of x or y is less than 3.5, and that's the second-smallest value in the set, and one of them is greater than 3.5, so is at least 4. So to find the median we'll be averaging two numbers, one of which is at least 4 and one of which is at least 5, and the median will be 4.5 or greater, and the median is certainly greater than the mean. So Statement 1 is sufficient.

Using Statement 2, we know x = y + 3. If y is a huge number, the mean will clearly exceed the median. But if y = 1 (assuming repetition of values is allowed), then the mean turns out to be less than the median, so Statement 2 is not sufficient. Even if we disallow repeated values, if y = 8 the mean still turns out to be very slightly less than the median. So the answer is A.

_________________________________
Thank you Ian. Replaced "set" with "list".
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Re: If x and y are unknown positive integers, is the mean of the list {6,  [#permalink]

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given ; 19+x+y/ 6 > middle value avg
#1
x + y = 7 we get pairs ( 7,0 ) ( 6,1) ( 5,2) ( 4,3) ( 5,2)
for all cases we get no as answer
sufficient
#2
x - y = 3
its + difference the value of the mean can be >or < mean of the list
insufficient
IMO A

Bunuel wrote:
If x and y are unknown positive integers, is the mean of the list {6, 7, 1, 5, x, y} greater than the median of the set?

(1) x + y = 7
(2) x - y = 3

Are You Up For the Challenge: 700 Level Questions
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Joined: 24 Nov 2016
Posts: 1204
Location: United States
Re: If x and y are unknown positive integers, is the mean of the list {6,  [#permalink]

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Bunuel wrote:
If x and y are unknown positive integers, is the mean of the list {6, 7, 1, 5, x, y} greater than the median of the set?

(1) x + y = 7
(2) x - y = 3

{1,5,6,7,x,y}
median is the average of middle terms
mean is (19+x+y)/6
x,y = positive integers ≥ 1

(1) x + y = 7 sufic

(19+x+y)/6=(19+7)/6=26/6=4.333

(x,y)=(1,6): median {1,5,6,7,x,y}={1,1,5,6,6,7}=5.5 > avg
(x,y)=(2,5): median {1,5,6,7,x,y}={1,2,5,5,6,7}=5 > avg
(x,y)=(3,4): median {1,5,6,7,x,y}={1,3,4,5,6,7}=4.5 > avg

(2) x - y = 3 insufic

Ans (A) Re: If x and y are unknown positive integers, is the mean of the list {6,   [#permalink] 22 Jan 2020, 06:00
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