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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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2,3,8,11 and x,y

we are asked maximal x, so can start backsolving from E. If x=4 it will be 3th term and median=4+8/2=6, so out.
If x=3, median=3+8/2=5.5, so right

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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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All solutions mentioned above are confusion below one seems to be very understandble.

You could also say that since the mean is 7, x + y = 18. Since we have an even number of terms in the set, the median must be the average of the two middle numbers. Our possibilities are

(x + y)/2 = 5.5
(3 + 8)/2 = 5.5
(x + 8)/2 = 5.5

The first one doesn't work, as it contradicts our equation above (that x + y = 18).

If x = 3, the other two both work, so x = 3 is a possible solution. x > 3 is not, as we won't have a median of 5.5, so we're set!
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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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Abdulla wrote:
Set S includes elements {8, 2, 11, x, 3, y} and has a mean of 7 and a median of 5.5. If x < y, then which of the following is the maximum possible value of x?

A. 0
B. 1
C. 2
D. 3
E. 4



{8, 2, 11, x, 3, y} ~ {(x,y),2,3,8,11} : Mean = 7
So sum of values = 7*6 = 42
8+2+11+x+3+y = 42
x+y = 42-24 = 18
(x,y) can be (1,17), (2,16), (3,15), ......................, (9,9)

median = 5.5
Since there are even number of items in the set, so the median will be average of mid numbers.
So, average of 3, 8 = 5.5 . hence mid numbers must be 3,8
Also x<y , so x will be on left side of 3 and y will be on right side of 8.

So, x must be less than or equal to 3 and y must be greater than or equal to 8.
Since we need to find the maximum value of x .. x= 3 and y = 15

Answer D
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Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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IanStewart wrote:
Abdulla wrote:
Set S includes elements {8, 2, 11, x, 3, y} and has a mean of 7 and a median of 5.5. If x < y, then which of the following is the maximum possible value of x?

OA is 3


If any one knows how to solve this problem fast please share it.

Thanks..


Using the fact that the mean is 7, then x+y = 18. Since x and y are different, one must be less than 9, the other greater than 9. Since x is smaller than y, we know that x < 9.

Now we need to use the median, which is the average of the two middle elements. Say x is greater than 3. Then we will have (since y is larger than 9), the elements x and 8 in the middle of our set, and the median will be (8 + x)/2. If x is greater than 3, then this will be larger than 11/2 = 5.5, so x cannot be greater than 3. Can x be exactly equal to 3? Sure. Then y = 15. So the largest possible value for x is 3.


IanStewartVeritasKarishma

If x = 3, the set will include two values which are 3. Can a set have duplicate values ?
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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
AbhiroopGhosh If it is explicitly not mentioned distinct value then yes, it can have duplicate value.
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Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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AbhiroopGhosh wrote:
If x = 3, the set will include two values which are 3. Can a set have duplicate values ?


That's a good point -- no, a mathematical set cannot contain repeated values. That's not something you actually need to know for the GMAT; most similar GMAT questions will talk about "lists" or "data sets", both of which can contain repeated values, and when the GMAT talks about a set without mentioning repeated values, you'll get the same answer whether you permit them or not. Prep company questions on the other hand are often careless about the definition of a set, and this is one example (and because it's a prep company question, I didn't even consider that technicality when I posted the solution you quote).
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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean [#permalink]
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