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12 Jul 2022, 08:00
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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
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
M38-07
(1) The sum of the numbers in list A is an odd number
(2) Exactly half of the numbers in list A is positive
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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M38-07
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19 Aug 2022, 04:31
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Bunuel
GMAT CLUB Official Explanation:
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?
Notice 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 one is clearly insufficient. For example, if A = {1, 3, 5} and 3 is removed we'd have an YES answer but if 1 is removed we'd have a NO answer.
Notice though, from this statement it follows that \(Sum_A=n*Mean_A=odd\).
(2) Exactly half of the numbers in list A is positive.
So what? Not sufficient.
Notice though, from this statement it follows that \(n\) is even: exactly half of the numbers in list A...
(1)+(2) From (2) we got 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, then we could not have removed a number from A, which was equal to the mean because the list has only integers in it: 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
General Discussion
12 Jul 2022, 08:58
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Bunuel
List A : (n) integers; List B: (n-1) integers {1 integer is removed from List A and that makes the contents of List B}
Is Avg[A] = Avg
[b]If you examine closely, average of A and B will only be the same if the number removed from A (to make the content of List B) is the actual mean of List A
For example: If List A = [1,2,3,4,5], Avg=3, and if you remove 3 from this list and List B =[1,2,4,5], its average is also 3. This is the only way the 2 averages remain same if 1 element from A is removed to form B
(1) The sum of the numbers in list A is an odd number
Let us look at some examples
List A = [1,2,3,4,5]: Sum=15, Avg=3
Take 3 out of A => List B = [1,2,4,5]: Sum=12, Avg=3 YES, SAME AVERAGE
But Take 5 out of A => List B = [1,2,3,4]: Sum=10, Avg=2.5 NO, DIFFERENT AVERAGE
NOT SUFFICIENT
(2) Exactly half of the numbers in list A is positive
Number of elements in List A have to be even now since half are positive and half need to be non positive
Again, examples to help process of elimination
List A=[-2,-1,2,9]: Sum=8, Avg=2
If List B=[-2,-1,9]: Sum=6, Avg=2 YES
If List B=[-2,-1,2]: Sum=-1, Avg=-1/3 NO
NOT SUFFICIENT
Statement 1 and Statement 2 combined
From Statement 1, we get that sum of terms has to be odd in List A, and from statement 2, we get that number of terms in List A has to be even and half of them need to be positive
So, if number of terms are even and sum is odd then average will NEVER BE AN INTEGER because ODD/EVEN is never an integer
And mean of both Lists can only be same if the integer taken out from List A is the same as the mean of List A but if the mean of List A is not an integer, then irrespective of what integer you take out from A, its mean will never equal mean of List B
SUFFICIENT
Answer - C
12 Jul 2022, 18:35
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Bunuel
Statement (1) alone:
If A is {0,1,2} and B is {0,1}, the average of A is not equal to the average of B.
If A is {0,1,2} and B is {0,2}, the average of A is equal to the average of B.
Do we have enough information from statement (1) alone to know whether the average of A is equal to the average of B? No. BCE.
Statement (2) alone:
If A is {-3,0,1,2} and B is {0,1,2}, the average of A is not equal to the average of B.
If A is {-3,0,1,2} and B is {-3,1,2}, the average of A is equal to the average of B.
Do we have enough information from statement (2) alone to know whether the average of A is equal to the average of B? No. CE.
Statements (1) and (2) together:
We know from the words "exactly half" that we must have an even number of elements in A. And since the sum of A is odd, we know that we have an odd number of odd elements (which means that we also have an odd number of even elements, which means that we have an even total number of elements). The average of A is therefore an odd divided by an even. The average of B will either be an odd divided by an odd or an even divided by an odd. Either way, it's not going to equal the average of A.
Do statements (1) and (2) together give us enough information to know whether the average of A is equal to the average of B? Yes. C.
Answer choice C.
