Hi rsarashi,

This is a layered question, but it can be solved with a mix of TESTing VALUES and Algebra.

We're told that the variables A, B, C, D, E, and F are ASCENDING POSITIVE INTEGERS (meaning that they are unique, A is smallest, F is largest, etc.). We're asked if the MEDIAN is greater than the MEAN. This is a YES/NO question.

1) A+E = (3/4)(C+D)

Without information about B and F, your instinct is probably to call this insufficient. We can prove it rather easily though. From this equation, since all of the variables are INTEGERS, we know that A+E must sum to an integer. By extension, (C+D) will have to be a multiple of 4.

IF....
C+D = 12, we could have...
A=1
B=2
C=5
D=7
E=8
F=9 or higher
MEDIAN = (5+7)/2 = 6
MEAN = 32/6 = 5 1/3 OR HIGHER (depending on the value of F)
Thus, the median may or may not be greater than the mean (again, depending on the value of F)
Fact 1 is INSUFFICIENT

2) B+F = (4/3)(C+D)

Here, we have a similar situation to what we had in Fact 1, but here we do not know much about the values of A and E. Here, I'm going to TEST much larger values so that we can see the potential 'swings' in the MEAN...

IF....
C+D = 120, we could have...
A=1-48
B=49
C=50
D=70
E=71 - 110
F=111
MEDIAN = (50+70)/2 = 60
MEAN = 352/6 - 438/6 = A little less than 60 to a little more than 70.
Thus, the median may or may not be greater than the mean (again, depending on the values of A and E)
Fact 2 is INSUFFICIENT

Combined, there's actually an interesting algebra pattern that we can take advantage of. The sum of ALL of the variables can be put in terms of (C+D)....

A+E = (3/4)(C+D)
C+D = (C+D)
B+F = (4/3)(C+D)

Thus, A+B+C+D+E+F = (3/4 + 1 + 4/3)(C+D)
Sum = (9/12 + 12/12 + 16/12)(C+D)
Sum = (37/12)(C+D)

The MEAN of those 6 terms = (37/12)(C+D)/6 = (37/72)(C+D)
The MEDIAN of those 6 terms = (C+D)/2 = (1/2)(C+D) = (36/72)(C+D)

Comparing the median and the mean now, we can see that the median will ALWAYS be less than the mean. Thus, the answer to the question is ALWAYS NO.
Combined, SUFFICIENT.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

rsarashi wrote:Given the ascending set of positive integers { a, b, c, d, e, f}, is the median greater than the mean?

(1) a + e = (3/4)( c + d)

(2) b + f = (4/3)( c + d)

OAC

We see that each statement by itself is not sufficient. While statement 1 does not tell anything about b and f, statement 2 does not tell anything about a and e.

Let's combine both the statements.

We know that

Mean = Sum of {a, b, c, d, e, f} divided by 6 = (a+b+c+d+e+f)/6 and
Median = Mean of the middle-most values = (c+d)/2

Adding both the statements, we get

a+f+b+e = (3/4 + 4/3) (c+d)

Adding (c+d) both the side, we get

a+b+c+d+e+f = (3/4 + 4/3 + 1) (c+d)

Sum of all the six numbers = 37/12 (c+d)

Diving both the sides by 6, we get

Sum of all the six numbers/6 = [37/12]*[(c+d)/6]

Mean = [37/36]*[(c+d)/2]

Mean = 37/36*Median

=> Mean > Median. Sufficient.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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