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19 Aug 2022, 04:30
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List A consists of \(n\) integers. One number is removed from list A, and the remaining numbers comprise list B. Is the average (arithmetic mean) of the numbers in list A equal to the average (arithmetic mean) of the numbers in list B?
(1) The sum of the numbers in list A is an odd number.
(2) Exactly half of the numbers in list A is positive.
(1) The sum of the numbers in list A is an odd number.
(2) Exactly half of the numbers in list A is positive.
19 Aug 2022, 04:30
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Official Solution:
List A consists of \(n\) integers. One number is removed from list A, and the remaining numbers comprise list B. Is the average (arithmetic mean) of the numbers in list A equal to the average (arithmetic mean) of the numbers in list B?
Note that for the mean to remain the same, the number removed from list A must be equal to the mean itself (if the number removed from list A is less than the mean of list A, then the mean of list B would be greater than that of list A, and if the number removed from list A is greater than the mean of list A, then the mean of list B would be smaller than that of list A).
(1) The sum of the numbers in list A is an odd number.
This statement is clearly insufficient. For example, if A = {1, 3, 5} and 3 is removed, we'd have a YES answer, but if 1 is removed, we'd have a NO answer.
However, from this statement, we can deduce that \(Sum_A=n*Mean_A=odd\).
(2) Exactly half of the numbers in list A is positive.
This statement alone is also insufficient.
However, from this statement, we can deduce that \(n\) is even: exactly half of the numbers in list A...
(1)+(2) From (2), we deduced that \(n\) is even. Thus, from (1), it follows that \(Sum_A=even*Mean_A=odd\), which means that \(Mean_A\) is NOT an integer! (If \(Mean_A\) were an integer, then \(Sum_A=even*Mean_A\) would have been even, not odd as given in the first statement). Now, since the mean of list A is NOT an integer, we could not have removed a number from A that was equal to the mean because the list contains only integers: List A consists of \(n\) integers... Therefore, the averages of lists A and B cannot be equal. We have a definite NO answer to the question. Sufficient.
Answer: C
List A consists of \(n\) integers. One number is removed from list A, and the remaining numbers comprise list B. Is the average (arithmetic mean) of the numbers in list A equal to the average (arithmetic mean) of the numbers in list B?
Note that for the mean to remain the same, the number removed from list A must be equal to the mean itself (if the number removed from list A is less than the mean of list A, then the mean of list B would be greater than that of list A, and if the number removed from list A is greater than the mean of list A, then the mean of list B would be smaller than that of list A).
(1) The sum of the numbers in list A is an odd number.
This statement is clearly insufficient. For example, if A = {1, 3, 5} and 3 is removed, we'd have a YES answer, but if 1 is removed, we'd have a NO answer.
However, from this statement, we can deduce that \(Sum_A=n*Mean_A=odd\).
(2) Exactly half of the numbers in list A is positive.
This statement alone is also insufficient.
However, from this statement, we can deduce that \(n\) is even: exactly half of the numbers in list A...
(1)+(2) From (2), we deduced that \(n\) is even. Thus, from (1), it follows that \(Sum_A=even*Mean_A=odd\), which means that \(Mean_A\) is NOT an integer! (If \(Mean_A\) were an integer, then \(Sum_A=even*Mean_A\) would have been even, not odd as given in the first statement). Now, since the mean of list A is NOT an integer, we could not have removed a number from A that was equal to the mean because the list contains only integers: List A consists of \(n\) integers... Therefore, the averages of lists A and B cannot be equal. We have a definite NO answer to the question. Sufficient.
Answer: C
01 Apr 2023, 09:03
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This might be a more back of the envelope approach... but my logic was the below:
Prompt: simplify the prompt to: sum(n) / n = sum(n - x)/(n-1) [x being equal to the number that was removed]
- this can be rearranged to: (n-1) * sum(n) = n * sum(n - x)
1. tells us that sum(n) = odd; this doesn't give us much, but we can plug this in: (n-1) * odd = n * sum(n - x)
- We can't get much further than this since x can be either positive or negative and we don't know the denominator
- insufficient
2. tells us that exactly half the numbers in A are positive; this I guess takes some sleuthing but basically you need to know this tells you there is an even amount of numbers in the set.
- this basically tells us: n = even and n-1 = odd
- plugging that in: odd * sum(n) = even * sum(n - x)
- It might be worth mentioning that this point that all we need to prove is if one side is odd and the other is even then they will never be equal
- you can plug in some numbers at this point but eventually you realize we just have no way of knowing.
- insufficient
1+2. This is an easy combination since we have the same simplified equation:
- odd * odd = even * sum(n-x)
- Again you can resort to number picking to confirm, but odd * odd is always odd, and even * odd/even is always even; meaning we have absolute confidence the above equation is not equal.
- sufficient; C
Prompt: simplify the prompt to: sum(n) / n = sum(n - x)/(n-1) [x being equal to the number that was removed]
- this can be rearranged to: (n-1) * sum(n) = n * sum(n - x)
1. tells us that sum(n) = odd; this doesn't give us much, but we can plug this in: (n-1) * odd = n * sum(n - x)
- We can't get much further than this since x can be either positive or negative and we don't know the denominator
- insufficient
2. tells us that exactly half the numbers in A are positive; this I guess takes some sleuthing but basically you need to know this tells you there is an even amount of numbers in the set.
- this basically tells us: n = even and n-1 = odd
- plugging that in: odd * sum(n) = even * sum(n - x)
- It might be worth mentioning that this point that all we need to prove is if one side is odd and the other is even then they will never be equal
- you can plug in some numbers at this point but eventually you realize we just have no way of knowing.
- insufficient
1+2. This is an easy combination since we have the same simplified equation:
- odd * odd = even * sum(n-x)
- Again you can resort to number picking to confirm, but odd * odd is always odd, and even * odd/even is always even; meaning we have absolute confidence the above equation is not equal.
- sufficient; C
