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11 Sep 2008, 22:10
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A
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Difficulty:
75%
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Question Stats:
49% (01:41) correct
51%
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wrong
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If the average (arithmetic mean) of a list of four integers remains unchanged when all the integers are multiplied by any constant, which of the following statements must be true?
I. I. The average (arithmetic mean) of the list is 0.
II. The sum of the largest and smallest members of the list is 0.
III. The list 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
I. I. The average (arithmetic mean) of the list is 0.
II. The sum of the largest and smallest members of the list is 0.
III. The list 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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29 Apr 2023, 14:48
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Official Solution:
If the average (arithmetic mean) of a list of four integers remains unchanged when all the integers are multiplied by any constant, which of the following statements must be true?
I. I. The average (arithmetic mean) of the list is 0.
II. The sum of the largest and smallest members of the list is 0.
III. The list 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
Since the mean remains unchanged when all the integers of the list are multiplied by any constant, it must also remain unchanged when all the integers are multiplied by 0. If we multiply all integers by 0, we get a list containing only 0's: {0, 0, 0, 0}, which has a mean of 0. Therefore, the mean of the original list must also be 0.
I. I. The average (arithmetic mean) of the list is 0. As discussed earlier, this statement must be true.
II. The sum of the largest and smallest members of the list is 0. This statement is not necessarily true. We can have a list with a mean of 0, where the sum of the largest and smallest members is not 0. For example, consider the list {-3, 0, 1, 2}.
III. The list contains both positive and negative integers. This statement is not necessarily true, as the list can consist of only zeros.
Answer: A
If the average (arithmetic mean) of a list of four integers remains unchanged when all the integers are multiplied by any constant, which of the following statements must be true?
I. I. The average (arithmetic mean) of the list is 0.
II. The sum of the largest and smallest members of the list is 0.
III. The list 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
Since the mean remains unchanged when all the integers of the list are multiplied by any constant, it must also remain unchanged when all the integers are multiplied by 0. If we multiply all integers by 0, we get a list containing only 0's: {0, 0, 0, 0}, which has a mean of 0. Therefore, the mean of the original list must also be 0.
I. I. The average (arithmetic mean) of the list is 0. As discussed earlier, this statement must be true.
II. The sum of the largest and smallest members of the list is 0. This statement is not necessarily true. We can have a list with a mean of 0, where the sum of the largest and smallest members is not 0. For example, consider the list {-3, 0, 1, 2}.
III. The list contains both positive and negative integers. This statement is not necessarily true, as the list can consist of only zeros.
Answer: A
General Discussion
Archit143
Joined: 21 Sep 2012
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Schools: Cox '16 (S) AGSM '16 Tulane '16 (S) McCombs '16 (S) Neeley '16 (S) Jones '16 (S) ISB '15 (D) MIT-LGO '16 (M)
GMAT 1: 410 Q35 V11
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Schools: Cox '16 (S) AGSM '16 Tulane '16 (S) McCombs '16 (S) Neeley '16 (S) Jones '16 (S) ISB '15 (D) MIT-LGO '16 (M)
GMAT 3: 630 Q45 V31
Posts: 720
03 Feb 2013, 15:27
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I have a doubt in this question
If we consider a set {1,2,3,4}, when multiplied with a constant we take that constant as 1 and multiply resulting set will contain the same numbers as the original. so how can the answer be I only.....
If we consider a set {1,2,3,4}, when multiplied with a constant we take that constant as 1 and multiply resulting set will contain the same numbers as the original. so how can the answer be I only.....
