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18 Mar 2022, 06:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
37% (02:16) correct
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wrong
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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.
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
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) None
Are You Up For the Challenge: 700 Level Questions
1.PNG [ 1.69 KiB | Viewed 5320 times ]
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
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) None
Are You Up For the Challenge: 700 Level Questions
Attachment:
1.PNG [ 1.69 KiB | Viewed 5320 times ]
21 Mar 2022, 18:30
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Bunuel
Breaking Down the Info:
I. This means the Std of set A is greater than that of set B, which can never be concluded since these sets are independent.
II. If Y < M, then R < M. If Y > M, then R > M. Again, the two original sets are independent, so we can't conclude anything about Y vs M, and R vs M.
III. We have X > Y and L = M. The only conclusion for the new median (Q) is that it must be between X and L, inclusive. We can also say that for R, it must be between Y and M, conclusive. Thus there is no clear relation between Q and R.
Answer: E
22 May 2026, 11:28
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The key idea here is:
The three sets are independent.
Knowing something about:
* mean vs median in Set A
* mean vs median in Set B
does NOT force a relationship between:
* standard deviations
* combined median
* combined mean
Question:
Set A:
* median = X
* mean = Y
* std dev = Z
Set B:
* median = L
* mean = M
* std dev = N
Combined set:
* median = Q
* mean = R
Given:
X − Y > 0 → X > Y
L − M = 0 → L = M
Need statements that MUST be true.
---
First understand what:
X > Y
means.
Median > Mean.
That usually means left-skewed distribution.
But it tells NOTHING definite about standard deviation.
Similarly:
L = M
means symmetric/balanced in some way.
Again nothing definite about spread.
So statement I is suspicious.
---
I. Z > N
“Std dev of A greater than std dev of B”
No connection.
You can easily make:
* Set A tightly packed
* Set B wildly spread
while still keeping:
* median > mean in A
* median = mean in B
So NOT must.
Eliminate I.
---
II. R > M
R is combined mean.
M is mean of Set B.
Combined mean must lie between the two means.
But we do NOT know whether Y > M or Y < M.
Since R is weighted average of Y and M:
R=\frac{n_AY+n_BM}{n_A+n_B}
R could be:
* greater than M
* equal to M
* less than M
depending on Y.
So II not must.
---
III. Q > R
Q = combined median
R = combined mean
Again no guarantee.
Even if one set has median > mean, the other may offset it.
Combined set could have:
* Q > R
* Q = R
* Q < R
Nothing fixed.
So III also not must.
Hence:
None must be true.
Answer: (E) None
Extra's
Because mean–median difference and standard deviation measure completely different things.
* Mean vs median → tells you about skew/asymmetry
* Standard deviation → tells you about spread
A distribution can be:
* highly skewed but tightly packed
OR
* barely skewed but extremely spread out
Example 1: small SD, noticeable mean-median gap
Numbers:
1, 5, 5, 5, 5
Median = 5
Mean = 21/5 = 4.2
Gap = 0.8
But spread is not huge.
---
Example 2: enormous SD, zero mean-median gap
Numbers:
−1000, 0, 0, 0, 1000
Median = 0
Mean = 0
Gap = 0
But SD is massive.
So mean-median gap does NOT determine SD.
Even among sets where median > mean:
Set A:
1,5,5,5,5
Set B:
−1000,100,100,100,100
For Set B:
Median = 100
Mean = −600/5 = −120
Huge median-mean gap and huge SD.
But you can also make:
4,5,5,5,5
Median = 5
Mean = 4.8
Still median > mean, but tiny SD.
So:
“median > mean”
only tells direction of skew,
not how spread out numbers are.
The three sets are independent.
Knowing something about:
* mean vs median in Set A
* mean vs median in Set B
does NOT force a relationship between:
* standard deviations
* combined median
* combined mean
Question:
Set A:
* median = X
* mean = Y
* std dev = Z
Set B:
* median = L
* mean = M
* std dev = N
Combined set:
* median = Q
* mean = R
Given:
X − Y > 0 → X > Y
L − M = 0 → L = M
Need statements that MUST be true.
---
First understand what:
X > Y
means.
Median > Mean.
That usually means left-skewed distribution.
But it tells NOTHING definite about standard deviation.
Similarly:
L = M
means symmetric/balanced in some way.
Again nothing definite about spread.
So statement I is suspicious.
---
I. Z > N
“Std dev of A greater than std dev of B”
No connection.
You can easily make:
* Set A tightly packed
* Set B wildly spread
while still keeping:
* median > mean in A
* median = mean in B
So NOT must.
Eliminate I.
---
II. R > M
R is combined mean.
M is mean of Set B.
Combined mean must lie between the two means.
But we do NOT know whether Y > M or Y < M.
Since R is weighted average of Y and M:
R=\frac{n_AY+n_BM}{n_A+n_B}
R could be:
* greater than M
* equal to M
* less than M
depending on Y.
So II not must.
---
III. Q > R
Q = combined median
R = combined mean
Again no guarantee.
Even if one set has median > mean, the other may offset it.
Combined set could have:
* Q > R
* Q = R
* Q < R
Nothing fixed.
So III also not must.
Hence:
None must be true.
Answer: (E) None
Extra's
Because mean–median difference and standard deviation measure completely different things.
* Mean vs median → tells you about skew/asymmetry
* Standard deviation → tells you about spread
A distribution can be:
* highly skewed but tightly packed
OR
* barely skewed but extremely spread out
Example 1: small SD, noticeable mean-median gap
Numbers:
1, 5, 5, 5, 5
Median = 5
Mean = 21/5 = 4.2
Gap = 0.8
But spread is not huge.
---
Example 2: enormous SD, zero mean-median gap
Numbers:
−1000, 0, 0, 0, 1000
Median = 0
Mean = 0
Gap = 0
But SD is massive.
So mean-median gap does NOT determine SD.
Even among sets where median > mean:
Set A:
1,5,5,5,5
Set B:
−1000,100,100,100,100
For Set B:
Median = 100
Mean = −600/5 = −120
Huge median-mean gap and huge SD.
But you can also make:
4,5,5,5,5
Median = 5
Mean = 4.8
Still median > mean, but tiny SD.
So:
“median > mean”
only tells direction of skew,
not how spread out numbers are.
