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M21-02

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What is the median of set \(S =\{a - b, b - a, a + b\}\) ?


(1) The mean of set \(S\) is equal to \(a + b\).

(2) The range of set \(S\) is equal to \(2b\).
[Reveal] Spoiler: OA

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(1) The mean of set \(S\) is equal to \(a + b\). Given that \(mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b\), which leads to \(a+b=0\). Now, if \(a+b=0\), then \(a-b\) and \(b-a\) are either both zeros (if \(a=b=0\)) or have different signs (if \(a \ne b\)). In any case the median of \(S\) is \(a+b=0\). Sufficient.

(2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.


Answer: A
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Re: M21-02 [#permalink]

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New post 04 Oct 2014, 06:26
In option 2, The range (largest - Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, b-a, a-b ?

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cjhande wrote:
In option 2, The range (largest - Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, b-a, a-b ?


It's explained in the solution above:

If a = b = 0, then S = {0, 0, 0} --> the range = 0 - 0 = 0 = 2b.
If a = 0 and b = 1, then S = {-1, 1, 1} --> the range = 1 - (-1) = 2 = 2b.
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New post 25 Jul 2016, 05:32
a+b - (a-b) = 2b, which is true for a=b=0 as well. Hence that makes b-a as the median. From first option I got a+b as the median. What am I missing here?

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New post 05 Sep 2016, 09:56
Option (1) states mean is a+b, which can be inferred from the question stem.

How does this lead a+b to 0? Could someone please clarify?

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New post 07 Sep 2016, 01:41
Bunuel wrote:
manhasnoname wrote:
Option (1) states mean is a+b, which can be inferred from the question stem.

How does this lead a+b to 0? Could someone please clarify?


By simplifying this: \(mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b\)



doesnt simlifying leads to:

a+B/3=a+b ????????????

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Manmeet0 wrote:
Bunuel wrote:
manhasnoname wrote:
Option (1) states mean is a+b, which can be inferred from the question stem.

How does this lead a+b to 0? Could someone please clarify?


By simplifying this: \(mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b\)



doesnt simlifying leads to:

a+B/3=a+b ????????????


Yes, but we can simplify further:

(a + b)/3 = a + b;

a + b = 3(a + b);

2(a + b) = 0;

a + b = 0.
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New post 31 Dec 2016, 13:49
I worked the second statement as following:

Range= Highest Number - Lowest Number
R=2b
In order for R to be 2b, the only possible combination is (b+a)-(a-b).
Thus, (a-b)≤(b-a)<(b+a) or (a-b)<(b-a)≤(b+a)
So b-a is the median.

Where am I wrong?
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arven wrote:
I worked the second statement as following:

Range= Highest Number - Lowest Number
R=2b
In order for R to be 2b, the only possible combination is (b+a)-(a-b).
Thus, (a-b)≤(b-a)<(b+a) or (a-b)<(b-a)≤(b+a)
So b-a is the median.

Where am I wrong?


(2) The range of set \(S\) is equal to \(2b\). If \(a=b=0\), then \(median=0\) but if \(a=0\) and \(b=1\), then \(median=1\). Not sufficient.
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M21-02 [#permalink]

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New post 11 Feb 2017, 12:15
Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.

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New post 11 Feb 2017, 12:22
brooklyndude wrote:
Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.


A set, by definition, is a collection of elements without any order. (While, a sequence, by definition, is an ordered list of terms.)
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Re: M21-02 [#permalink]

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New post 17 Aug 2017, 08:37
Did not understand the inference of 1st option.

How the median of the set is a+b ??

looking at the set median is b-a .

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New post 19 Aug 2017, 05:24
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.

as explained even if a=b=0 or a=1, b=0, mean is b-a.

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New post 19 Aug 2017, 05:37
nitin083 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.

as explained even if a=b=0 or a=1, b=0, mean is b-a.


The point is that in data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.
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Re: M21-02   [#permalink] 19 Aug 2017, 05:37
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