Mean and Median

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Mean and Median

by vinay1983 » Wed Sep 25, 2013 10:02 pm
If x is an integer and x,5,6,11,19 are a set of numbers, is the median of this set greater than the arithmetic mean of this set?

1. X >= 8
2. X is greater than the median of the 5 numbers
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!

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by theCodeToGMAT » Wed Sep 25, 2013 10:17 pm
Is the Answer {E}?
Last edited by theCodeToGMAT on Thu Sep 26, 2013 12:05 am, edited 1 time in total.
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by [email protected] » Thu Sep 26, 2013 12:01 am
Hi vinay1983,

For this type of DS question, TESTing values is a great way to prove what the answer is.

We're given a set of 5 values: 5, 6, 11, 19 and X (although X can be ANY INTEGER). We're asked IS the median > average for the set? This is a YES/NO question and will clearly depend on the value of X.

Fact 1: X >= 8

If X = 8, then we have 5,6,8,11,19; the median = 8, the average = 49/5 = 9.8 and the answer is NO
If X = 11, then we have 5,6,11,11,19; the median = 11, the average = 52/5 = 10.4 and the answer is YES
Fact 1 is INSUFFICIENT.

Fact 2: X is > median.

This means that X CANNOT be the median; 11 MUST BE the median and X > 11

If X = 12, then we have 5,6,11,12,19; the median = 11, the average = 53/5 = 10.6 and the answer is YES
If X = 1,000, then we have 5,6,11,12,1000; the median = 11, the average = BIG and the answer is NO
Fact 2 is INSUFFICIENT.

Combined, we know that X > 8 AND X > 11; thus, X MUST BE > 11
All the info from Fact 2 applies here; we have a YES and a NO answer.
Combined = INSUFFICIENT.

Final Answer: E

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by Brent@GMATPrepNow » Thu Sep 26, 2013 5:10 am
vinay1983 wrote:If x is an integer and x,5,6,11,19 are a set of numbers, is the median of this set greater than the arithmetic mean of this set?

1. X >= 8
2. X is greater than the median of the 5 numbers
This is a very small twist on this question: https://www.beatthegmat.com/mean-median-t269562.html


Target question: Is the median greater than the average?

Given: We have the set {x, 5, 6, 11, 19}
Notice that there are 3 possible scenarios we need to consider:
Scenario #1: x is less than 6, in which case the median is 6
Scenario #2: 6 < x < 11, in which case the median is x
Scenario #3: x is greater than 11, in which case the median is 11

The average of this set will be (41+x)/5.

Okay, now onto the statements

Statement 1: x > 6
This rules our scenario #1, but we must still consider scenarios #2 and #3
Here are two possible values of x that yield conflicting answers to the target question:
Case a: x = 9 (median = 9 and average = 10), in which case, the median is NOT greater than the average
Case b: x = 12 (median = 11 and average = 53/5), in which case, the median IS greater than the average
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x is greater than the median of the 5 numbers
This rules our scenarios #1 and #2, which leaves scenario #3
Here are two possible values of x that yield conflicting answers to the target question:
Case a: x = 12 (median = 11 and average = 53/5), in which case, the median IS greater than the average
Case b: x = 59 (median = 11 and average = 20), in which case, the median is NOT greater than the average
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that the x-values we used to show that statement 2 is not sufficient ALSO satisfy statement 1. So, we know immediately that the combined statements are NOT SUFFICIENT.
To see what I mean, here are the two conflicting cases:
Case a: x = 12 (median = 11 and average = 53/5), in which case, the median IS greater than the average
Case b: x = 59 (median = 11 and average = 20), in which case, the median is NOT greater than the average
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT.

Answer = E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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