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Joined: 03 Apr 2009
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Set S includes elements {8, 2, 11, x, 3, y} and has a mean
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Updated on: 13 Jul 2014, 03:26
Question Stats:
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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
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Originally posted by Abdulla on 28 Apr 2009, 18:47.
Last edited by Bunuel on 13 Jul 2014, 03:26, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean
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29 Apr 2009, 07:59
First, reach to the point of getting the equation : x + y = 18 It should take about 20 seconds.
Then Substitute the answer choices into the equation. I don't know what the answer choices in this case are. But I'm sure, you would be able to eliminate at least 2 or 3 answer choices. (about 10 seconds).
Say you are left with 2 answer choices. (If you are short on time, guess One of the two and you'll have a 50% probability of getting it right.)
The Median (of 6 numbers) = 5.5. See if the AVERAGE of any two numbers among (2,3,8,11) results in the median. In this case, it does for 3 and 8. (15 seconds). Once you know that the numbers that contribute towards Median are 3 and 8, and not x or y, then given x < y, x ≤ 3. (about 10 seconds)
In less than a minute, you have the Correct answer.



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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean
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01 May 2009, 12:33
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.
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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean
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11 Sep 2015, 00:55
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
D



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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean
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10 Dec 2016, 17:13
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
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09 Apr 2018, 20:33
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 = 4224 = 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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Re: Set S includes elements {8, 2, 11, x, 3, y} and has a mean
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09 Apr 2018, 20:33






