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Re M1230
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15 Sep 2014, 23:47
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: M1230
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24 Jun 2015, 20:17
Bunuel wrote: 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 \(\{3, 1, 2\}\). III is not necessarily true. The set can contain only zeros.
Answer: A In the explanation for II, it needs to have 4 numbers. But doesnt affect final answer.



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Re: M1230
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25 Jun 2015, 01:42
82vkgmat wrote: Bunuel wrote: 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 \(\{3, 1, 2\}\). III is not necessarily true. The set can contain only zeros.
Answer: A In the explanation for II, it needs to have 4 numbers. But doesnt affect final answer. Edited the typo. Thank you.
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Re M1230
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22 Dec 2015, 02:26
I think this is a poorquality question and the explanation isn't clear enough, please elaborate.



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22 Dec 2015, 02:29



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28 Jun 2016, 15:05
For the same reason we are considering sentence I true, shouldn`t sentence II be true as well? 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? Tks



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30 Jun 2016, 07:21



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07 Sep 2016, 07:58
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: M1230
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14 Dec 2016, 06:25
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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M1230
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Updated on: 05 Dec 2017, 08:57
Bunuel wrote: 82vkgmat wrote: Bunuel wrote: 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 \(\{3, 1, 2\}\). III is not necessarily true. The set can contain only zeros.
Answer: A In the explanation for II, it needs to have 4 numbers. But doesnt affect final answer. Edited the typo. Thank you. @brunell you need to change the official answer in the gmat club test aswell Forget my comment...
Originally posted by cdl1985 on 05 Dec 2017, 08:52.
Last edited by cdl1985 on 05 Dec 2017, 08:57, edited 1 time in total.



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Re: M1230
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24 Mar 2018, 05:01
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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24 Mar 2018, 08:06



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Re M1230
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08 Sep 2018, 00:58
I think this is a poorquality 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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08 Sep 2018, 01:41
Arrgghh Ok got it now. Thanks Bunuel!



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Re M1230
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15 Oct 2018, 06:39
I think this is a poorquality question and I don't agree with the explanation. The question says that all elements of the set are multiplied by ANY constant. This means the constant can be 1 as well. And even in such case the mean will not change.
However, in this case Old mean = New mean and all I, II, III are incorrect.
In addition this is an always true question and not a could be true question. While Statement I is true when Zero is multiplied but not when 1 is multiplied as a constant.



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15 Oct 2018, 06:43



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Re M1230
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17 Nov 2018, 07:05
I think this is a highquality question and I agree with explanation.










