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M12-30

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M12-30  [#permalink]

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New post 16 Sep 2014, 00:47
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A
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  75% (hard)

Question Stats:

47% (01:38) correct 53% (01:35) wrong based on 96 sessions

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The mean of four integers will not change if all the integers are multiplied by any constant. What is always true about this set of numbers?

I. The mean of the set is 0

II. The sum of the largest member and the smallest member of the set is 0

III. The set contains both positive and negative integers


A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

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New post 16 Sep 2014, 00:47
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Official Solution:


The mean of four integers will not change if all the integers are multiplied by any constant. What is always true about this set of numbers?

I. The mean of the set is 0

II. The sum of the largest member and the smallest member of the set is 0

III. The set contains both positive and negative integers


A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III


If the constant is 0, the mean of the new set is 0. Thus, the mean of the original set has to be 0 as well.

II is not necessarily true. Consider \(\{-3, 0, 1, 2\}\).

III is not necessarily true. The set can contain only zeros.


Answer: A
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Re M12-30  [#permalink]

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New post 22 Dec 2015, 03:26
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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Re: M12-30  [#permalink]

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New post 22 Dec 2015, 03:29
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Re: M12-30  [#permalink]

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New post 07 Sep 2016, 08:58
edutorrescasana wrote:
Considering "mean of four integers will not change if all the integers are multiplied by any constant", shouldnt all the numbers be 0 and then the sum of the largest and smallest be 0?


All numbers being zero is a valid set, but not the only valid set. The question asks "What is always true..."

Possible sets of 4 integers whose mean does not change if all a integers are multiple by any constant:
0, 0 ,0 0
-2, -1, -2, 2
-3, 0, 1, 2

The largest and smallest integers in the last example do not equal zero.
There are not positive and negative integers in the first set.
All have a mean of 0.
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Re: M12-30  [#permalink]

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New post 14 Dec 2016, 07:25
It is a very interesting question, because we may consider 1 also as the mean of the set, but it is discarded once you realize that the constant that multiplies the mean can be positive or negative. In this case, 0 remains as the only mean.

Statements II and III were easy to discard thereafter.
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Re: M12-30  [#permalink]

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New post 24 Mar 2018, 06:01
If the constant multiplied is equal to 1, then option 1 also fails.

I am not sure if I am missing something. A constant can be any real integer right?
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New post 24 Mar 2018, 09:06
Aswath Kumar wrote:
If the constant multiplied is equal to 1, then option 1 also fails.

I am not sure if I am missing something. A constant can be any real integer right?


The mean of four integers will not change if all the integers are multiplied by ANY constant.

Of course the mean won't change if you multiply all the integers by 1 and in this case the mean could be any number but if we want the mean not to change when we multiply by ANY constant than the mean must be 0.

Check for more here: https://gmatclub.com/forum/the-mean-of- ... 70142.html
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Re M12-30  [#permalink]

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New post 08 Sep 2018, 01:58
I think this is a poor-quality question and I don't agree with the explanation. What if the set was {1,1,1,1} and the constant was 1. The above assumption of the set and constant will not have a mean of 0. The question doesn't specify whether the integers are "distinct".
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New post 08 Sep 2018, 02:05
sumitkar007 wrote:
I think this is a poor-quality question and I don't agree with the explanation. What if the set was {1,1,1,1} and the constant was 1. The above assumption of the set and constant will not have a mean of 0. The question doesn't specify whether the integers are "distinct".


Please re-read the question and solution more carefully:

The mean of four integers will not change if all the integers are multiplied by ANY constant. What is always true about this set of numbers?
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Re: M12-30  [#permalink]

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New post 08 Sep 2018, 02:41
Arrgghh Ok got it now. Thanks Bunuel!
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Re M12-30  [#permalink]

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New post 17 Nov 2018, 08:05
I think this is a high-quality question and I agree with explanation.
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Re M12-30  [#permalink]

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New post 27 Mar 2019, 07:50
I don't agree with the explanation. what if the constant is 1?
The mean of the set will still be the same.
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New post 27 Mar 2019, 07:52
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Re: M12-30   [#permalink] 27 Mar 2019, 07:52
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