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Mean & Median (Faster Way Needed)

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Mean & Median (Faster Way Needed)

by Abdulla » Tue Apr 28, 2009 6:52 pm
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 some one knows fast way to solve this problem please share it.

Thanks,
Abdulla
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by Feep » Wed Apr 29, 2009 12:55 am
So, we know that the average is 7, which means that (2 + 3 + 8 + 11 + x + y) = 7, or 24 + x + y = 42, or that x + y = 18. We also know that x < y, so x may in fact only be x < 9.

For the median, we need to look at possible ranges of x. If x > 3, then the ordered set becomes {2, 3, x, 8, 11, y} or {2, 3, x, 8, y, 11} (if x = 8 and y = 10). In either of these cases, the median becomes (x + 8) / 2, which we know MUST equal 5.5. In this case, x must be equal to 3. If x < 3, well...I mean, whatever, we're looking for the greatest possible value, and x = 3 is higher than any of those possible values.

OA is 3.
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by honey_babee » Thu May 17, 2012 4:28 pm
How come x=0, y =18 is not a viable solution?
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by Thrills4ever » Thu May 17, 2012 6:11 pm
honey_babee,

x=0 and y=18 is definitely a solution, but not the right one

If we want to "maximize" x, while keeping the constraints of the problem then x must be 3

Let's look at the case if x = 0 and y=18. The set would look like this:

0,2,3,8,11,18 --> the constraints of the problem are met with the median being 5.5 and the average being 7

But can x=1? Well it can! If x=1 and y=17 we get:

1,2,3,8,11,17 --> constraints of the problem are still met.

Using this logic we can conclude that x must be 3 where the set looks like this:

2,3,3,8,11,15

If x were to equal 4, and y=14 then the set would look like this:

2,3,4,8,11,14 --> while we maintain the average of 7, the median now becomes 6..so in order to MAXIMIZE x it must equal 3.
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