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If the average (arithmetic mean) of set A is 10,000 and the average (a

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If the average (arithmetic mean) of set A is 10,000 and the average (arithmetic mean) of set B is 10,000, what is the range of set A and set B combined?

1) The range of set A is 6,000
2) The range of set B is 3,000
[Reveal] Spoiler: OA

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If the average (arithmetic mean) of set A is 10,000 and the average (a [#permalink]

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New post 08 Mar 2017, 04:13
MathRevolution wrote:
If the average (arithmetic mean) of set A is 10,000 and the average (arithmetic mean) of set B is 10,000, what is the range of set A and set B combined?

1) The range of set A is 6,000
2) The range of set B is 3,000



Range is defined as highest no. of set - Lowest no of set

(1) No info of total terms in set A...not suff
(2)No info of total terms in set B...not suff

Ans E
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If the average (arithmetic mean) of set A is 10,000 and the average (a [#permalink]

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New post 08 Mar 2017, 07:21
vivek2k70 wrote:
please explain more on this.


Since both individual options are insuff
combined we can have:-

Let set A={ 7000,10000,10000,10000,13000}
set B={ 8500,10000,10000,10000,11500}
Range=6000
or

set A ={6850,10100,10100,10100,12850}
set B= {6800,10100,10100,10100,12900}

Range = 3000

or u can take any such sets....

Thus E
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Re: If the average (arithmetic mean) of set A is 10,000 and the average (a [#permalink]

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New post 10 Mar 2017, 02:11
==> If you modify the original condition and the question, from range= Max-min, you get set A: Ra =Ma-ma and set B: Rb=Mb-mb. Then, there are 6 variables and 2 equations, and in order to match the number of variables to the number of equations, there must be 4 more equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), the answer becomes E as well. Also, there is no relationship between the average and the range, and thus the answer is definitely E.

Answer: E
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Re: If the average (arithmetic mean) of set A is 10,000 and the average (a   [#permalink] 10 Mar 2017, 02:11
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