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median

by dreamv » Tue Feb 28, 2012 4:54 pm
x, 3, 1, 12, 8

If x is an integer, is the median of the 5 numbers shown greater than the average of the 5 numbers?

1) x>6
2) x is greater than the median of the 5 numbers
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Source: — Data Sufficiency |
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by GMATGuruNY » Tue Feb 28, 2012 7:47 pm
dreamv wrote:x, 3, 1, 12, 8

If x is an integer, is the median of the 5 numbers shown greater than the average of the 5 numbers?

1) x>6
2) x is greater than the median of the 5 numbers
Look for values that satisfy both statements.
Be sure to try extreme values.

If x=9:
The numbers are 1, 3, 8, x=9, 12.
Average = (1+3+8+9+12)/5 = 33/5.
Median = 8.
Average < median.

If x=100:
The numbers are 1, 3, 8, 12, x=100.
Average = (1+3+8+12+100)/5 = 124/5.
Median = 8.
Average > median.

Since in the first case the average is less than the median, and in the second case the average is greater than the median, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by krusta80 » Tue Feb 28, 2012 8:35 pm
dreamv wrote:x, 3, 1, 12, 8

If x is an integer, is the median of the 5 numbers shown greater than the average of the 5 numbers?

1) x>6
2) x is greater than the median of the 5 numbers
Integers can be negative or postive, therefore we do not have sufficient info to solve the problem with neither (1) nor (2). Let's move on to each line item:

(1) x > 6

Let's analyze the minimum possible value and then the max.

(1a) x = 7

So, we have 1,3,7,8,12. Mean is 31/5 = 6.2 <=> Median is 7

(1b) x = 1000000

So, we have 1,3,8,12,1000000. Mean is much bigger than the median, which is 8.

INSUFFICIENT


(2) x is greater than the median

(2a) Minimum valid value of x is 9, which gives is 1,3,8,9,12 <=> Mean is 33/5 = 6.6
(2b) x is 1000000, which gives us 1,3,8,12,1000000 <=> Mean is larger than median, which is 8

INSUFFICIENT


(1) and (2)
Since part 2 is just a subset of part 1 (for x), still INSUFFICIENT

Answer is E
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