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# I need Quicker Approach to solve this problem(Mean & Median)

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Intern
Joined: 03 Apr 2009
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I need Quicker Approach to solve this problem(Mean & Median) [#permalink]  28 Apr 2009, 18:47
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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?

OA is 3

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

Thanks..
Manager
Joined: 08 Feb 2009
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Re: I need Quicker Approach to solve this problem(Mean & Median) [#permalink]  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: I need Quicker Approach to solve this problem(Mean & Median) [#permalink]  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: I need Quicker Approach to solve this problem(Mean & Median)   [#permalink] 01 May 2009, 12:33
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