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If x and y are unknown positive integers, is the mean of the list {6,

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Bunuel
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If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   17 Jan 2020, 02:45
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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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IanStewart
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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   17 Jan 2020, 06:49
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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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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   17 Jan 2020, 05:48
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

The answer is A

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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   17 Jan 2020, 06:54
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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] New post   18 Jan 2020, 01:29
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


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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   22 Jan 2020, 07:00
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)
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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   04 Aug 2020, 01:41
lacktutor wrote:
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

The answer is A

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do we really need to check all three options in case-1. Is there any other shortest way to resolve this question ?
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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   24 Aug 2021, 05:16
lacktutor wrote:
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?

([b]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

The answer is A

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[/b]

Hey, shouldn't the answer be 'D'?

The second statement seems sufficient to me since we can't repeat numbers in a set. So the first case that you've provided for the second statement would be wrong since the number 5 can't be used again.

I could be wrong, please pitch in.
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Re: If x and y are unknown positive integers, is the mean of the list {6, [#permalink] New post   30 Nov 2023, 22:35
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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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