uggh a tough one

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uggh a tough one

by sana.noor » Thu Aug 29, 2013 8:47 pm
The table below represents three sets of numbers with their respective medians, means and standard deviations. The third set, Set [A+B], denotes the set that is formed by combining Set A and Set B.

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If X - Y > 0 and L - M = 0, then which of the following must be true?

I. Z > N
II. R > M
III. Q > R

I only
II only
III only
I and II only
none

OA is None
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by [email protected] » Thu Aug 29, 2013 11:51 pm
Hi sana.noor,

What is the source of this question? It's WAY too involved for a statistics question and doesn't follow the typical GMAT patterns for Roman Numeral questions.

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by sana.noor » Fri Aug 30, 2013 12:05 am
manhattan word problem question bank...:( i am quite upset since i have solved manhattan question bank. i thought i am really doing well in quant but i guess i am wrong..:(
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by rishianand7 » Fri Aug 30, 2013 12:35 am
Im not sure if my solution takes into account all the possible cases, but given the time constraint on the GMAT, this is what I would have done about this problem:

Let Set A be : 1,2,3,4,5
Let Set B be : 6,7,8,9,10

Then A+B will be : 1,2,3,4,5,6,7,8,9,10

Note : Since the mean and median are equal for both sets according to the question, I have taken 2 evenly spaced sets to solve the question.

Now for the Sets above:

Set A: X=3, Y=3, Z=1

Set B: L=8, M=8, N=1

Set C: Q=5.5, R=5.5, S=1

From the values above, its clear that the answer is "None"

PS : Manhattan questions are really hard, but are good for practice.

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by vipulgoyal » Fri Aug 30, 2013 12:40 am
I found the detailed solution in my old notes, hope it"ll help



If X - Y > 0, then X > Y and the median of A is greater than the mean of set A. If L - M
= 0, then L = M and the median of set B is equal to the mean of set B.
I. NOT NECESSARILY: According to the table, Z > N means that the standard deviation
of set A is greater than that of set B. Standard deviation is a measure of how close the
terms of a given set are to the mean of the set. If a set has a high standard deviation, its
terms are relatively far from the mean. If a set has a low standard deviation, its terms
are relatively close to the mean.
Recall that a median separates the set into two as far as the number of terms. There is
an equal number of terms both above and below the median. If the median of a set is
greater than the mean, however, the terms below the median must collectively be
farther from the median than the terms above the median. For example, in the set {1,
89, 90}, the median is 89 and the mean is 60. The median is much greater than the
mean because 1 is much farther from 89 than 90 is.
Knowing that the median of set A is greater than the mean of set A just tells us that the
terms below set A's median are further from the median than the terms above set A's
median. This does not necessarily imply that the terms, overall, are further away from
the mean than in set B, where the terms below the median are the same distance from
the median as the terms above it. In fact, a set in which the mean and median are
equal can have a very high standard deviation if the terms are both far below the mean
and far above it.
II. NOT NECESSARILY: According to the table, R > M implies that the mean of set [A +
B] is greater than the mean of set B. This is not necessarily true. When two sets are
combined to form a composite set, the mean of the composite set must either
be between the means of the individual sets or be equal to the mean of both
of the individual sets. To prove this, consider the simple example of one member
sets: A = [3], B = [5], A + B = [3, 5]. In this case the mean of A + B is greater than the
mean of A and less than the mean of B. We could easily have reversed this result by
reversing the members of sets A and B.
III. NOT NECESSARILY: According to the table, Q > R implies that the median of the set
[A + B] is greater than the mean of set [A + B]. We can extend the rule given in
statement II to medians as well: when two sets are combined to form a composite
set, the median of the composite set must either be between the medians of
the individual sets or be equal to the median of one or both of the individual
sets. While the median of set A is greater than the mean of set A and the median of set
B is equal to the mean of set B, set [A + B] might have a median that is greater or less
than the mean of set [A + B]. See the two tables for illustration:
Set Median Mean Result
A 1, 3, 4 3 2.67 Median > Mean
B 4, 5, 6 5 5 Median = Mean
A + B 1, 3, 4, 4, 5, 6 4 3.83 Median > Mean
Set Median Mean Result
A 1, 3, 3, 4 3 2.75 Median > Mean
B 10, 11, 12 11 11 Median = Mean
A + B 1, 3, 3, 4, 10, 11, 12 4 6.29 Median < Mean
Therefore none of the statements are necessarily true and the correct answer is E.