GMAT Club World Cup 2022 (DAY 2): List A consists of n integers. One

#1

(1) The sum of the numbers in list A is an odd number
considering that all numbers are different then
list A has even count of numbers and then B will have an odd count of numbers ( 1,2,3,4) ; ( 1,2,3)
or else if numbers are same then list A has odd count of numbers and then B will have an even count of numbers ( 1,1,3) ( 1,3)
for both the cases the avg wont be the same
sufficient
#2
(2) Exactly half of the numbers in list A is positive

A = ( -1,-1,1,1) avg is zero
B + (-1,1,1) avg is not zero
sufficient
option D

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

Given: List A consists of n integers. One number is removed from list A, and the remaining numbers comprise list B.

Asked: Is the average (arithmetic mean) of the numbers in list A equal to the average (arithmetic mean) of the numbers in list B?

Let list A = {a_1, a_2, .... , a_n} and arithmetic mean be M
a_1 + a_2 + ... + a_n = Mn

Let the removed number be a_n

List B = {a_1, a_2, .... , a_(n-1)}
Is a_1 + a_2 + ... + a_(n-1) = a_1 + a_2 + ... + a_(n-1)+ a_n - a_n = Mn - a_n = M(n-1)
Is a_n = M?

(1) The sum of the numbers in list A is an odd number
a_1 + .... + a_n = Mn = odd
M is odd and n is odd.
But nothing is mentioned about removed number a_n or whether it is equal to arithmetic mean M.
NOT SUFFICIENT

(2) Exactly half of the numbers in list A is positive
n = even; Half of the number in list A are positive and half are non-positive.
But nothing is mentioned about removed number a_n or whether it is equal to arithmetic mean M.
NOT SUFFICIENT

(1) + (2)
(1) The sum of the numbers in list A is an odd number
a_1 + .... + a_n = Mn = odd
M is odd and n is odd.
(2) Exactly half of the numbers in list A is positive
n = even; Half of the number in list A are positive and half are non-positive.
But nothing is mentioned about removed number a_n or whether it is equal to arithmetic mean M.
NOT SUFFICIENT

IMO E

Statement 1
Let's say the list is 1, 2, 3, 7, 10 then the mean is 23/5 = 4.6. If I remove 10 from it the mean does not stay the same
Let's say the list is 1, 2, 3, 4, 5 then the mean is 15/5 = 5. If I remove 3 from it the mean stays the same

Statement 2
Let's say the list is 1, 2, 3, -1, -2, -3 then the mean is 0/6 = 0. If I remove any number from it the mean does not stay the same
Let's say the list is 1, 2, 3, 0, -2, -4 then the mean is 0/6 = 0. If I remove 0 from it the mean stays the same

Even with both the statement together, we have the Sum as odd and list A has even numbers
Then we know the mean is Odd/Even which would be a fraction. However, if I remove a number that could be odd or even or 0 from the list I am dividing it by an Odd number and the mean may or may not be the same.

IMHO Option E

Ans:D
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

Statement I :

Case 1

Assume that list A consists of {9,9,9}

Mean (A) = 9

Assume that list B consists of {9,9}

Mean (A) = 9

Both the mean are same - The answer to the question is Yes

Case 2

Assume that list A consists of {1,9,9}

Mean (A) = \frac{19}{3}

Assume that list B consists of {9,9}

Mean (A) = 9

Both the mean are same - The answer to the question is No

So we can eliminate A

Statement II :

We know exactly half of the numbers are positive, therefore the other half is either negative or zero.

If we remove one number from this set (to form Set B), the mean will no longer remain balanced at this position.

Hence knowing this we can conclude that the mean of set A will not be equal to the mean of set B

IMO B

Statement 1 -
Sum of numbers is an odd number.

Case 1 - When all numbers are same
List A can be {1,1,1} and accordingly mean = 1
List B then will become {1,1} and accordingly mean = 1

Case 2 -
When numbers are different
List A can be {1,2,3,4,5} ; mean = 3
List B can become {1,2,3,4} ; mean =2.5

Mean is same in one case and different in another case, hence not sufficient.

Statement 2 -
Half of the terms are positive.
This removes the possibility of the fact that all the numbers are the same. Hence, sufficient.

Choice B is the correct answer.

let's say list A=a1,a2, a3, a4, ...an / ai are integers and i is an indice from 1 to n.

B is the same list of A except for one element so B=a1, A2, A3, ...an-1

AVG(A) =(a1+a2+...an)/n
AVG(B) =(a1+a2+...an-1)/n-1

In order for the two averages to be equal a1+a2+...an)/n = (a1+a2+...an-1)/n-1
=>an=(a1+a2+...an-1)/n-1 (*), this means that the removed element should represent the average of the remaining elements,

Back to statements,

(1) The sum of the numbers in list A is an odd number
We have two cases: the removed number (an) is odd or even
If it is odd, then (a1+a2+...an-1) is even (since a1+a2+...an-1+an is even)
in order that (*) to be correct, n-1 should be even then n should be odd

If it is even, then (a1+a2+...an-1) is odd (since a1+a2+...an-1+an is even)
in order that (*) to be correct, n-1 should be odd then n should be even
nothing to conclude,
(2) Exactly half of the numbers in list A is positive
This simply means that set A contains an even number of elements,
if all the other half are equal to zero,
Then for set B if we remove a Zero, the denominator ( in calculating the AVG) won't change but the numerator will, then AVG(B)> AVG(A)
If we remove another element other than 0, (exemple A(0, 0, 4, 4) B(0, 0, 4), AVG(A)=2 and AVG(B)=1,3)
AVG(A)>AVG(B)
This statement is not sufficiant,

Answer is E

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.

S1: "The sum of the numbers in list A is an odd number"

Consider example
A={2,3} => Mean = 2.5
B=(2} => Mean =2.
So means are NOT EQUAL in this case

again consider another example
A={2,3,4} => Mean = 3
B={2,4} => Mean =3
So means are EQUAL in this case.

So statement is INSUFFICIENT


S2: "Exactly half of the numbers in list A is positive "

Here number of elements in set A must be even.

Now, consider example,
A={0,0,4,12} => Mean=4
B={0,0,12} => Mean=4
So means are EQUAL in this case.

Consider example,
A={-1,5} => Mean = 2
B={5} => Mean 5
So means are NOT EQUAL in this case.


S1 + S2 :
"(1) The sum of the numbers in list A is an odd number
(2) Exactly half of the numbers in list A is positive"

From 2, number of elements in A is even, and sum of the elements is odd. Algebraically,
Let sum of set A elements SA and sum of B elements SB, and number of elements of set A = n+1 and that of set B=n
For Mean to be EQUAL,
SA/(n+1)=SB/n [n<>0]
OR n*SA=(n+1)*SB
Now n is ODD, SA is ODD. Thus n*SA=ODD
and n+1 is EVEN. Thus (n+1)*SB=EVEN
As EVEN is NOT EQUAL TO ODD, mean of these 2 cannot be EQUAL

So together SUFFICIENT

ANS C

1) 1 is insufficient because - A and D out
if we consider list A as [4,3,0,0]
and list B as [4,3,0]
Averages are not equal
Now for list A=[2,3,4]
and list B =2,4
Averages are equal
2) 2 is insufficient because - B is out
As per the statement Number of terms in the list A is even
if we consider list A as [4,3,0,0]
and list B as [4,3,0]
Averages are not equal
Now for list A =[16,4,0,-4]
list B = [16,-4,0]
Averages are equal
Combining 1 and 2
Sum(A) - odd
Number of terms in list A - even -n
Number of terms in list B - odd - n-1
Sum(A)/n=Sum(B)/(n-1)
Sum(A)*(n-1)=Sum(B)*n
Num as n is even thus the RHS of the equation is even while the LHS is always odd as both the terms are odd
Thus this is not possible.The average cannot be same.
C is the correct choice

E
1. a=15/3 =5 B remove 5 -> 10/2 =5 or remove 3 -> 12/2=6 NS
2. Half of the numbers are positive but we don't know which numbers are removed NS
1 and 2 together are NS

Answer: B

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
Let A = (1,1,1,1,1)
=> B = (1,1,1,1)
AM of A = 5/5 = 1
AM of B = 4/4 =1
AM of A = AM of B.

Consider another set A = (1,2,3,4,5) and B = (1,2,3,4)
AM of A = (1+2+3+4+5)/5 = 3
AM of B = ((1+2+3+4)/4 = 2.5
AM of A not equal to AM of B

Insufficient.

(2) Exactly half of the numbers in list A is positive
Let A = (-1,1) and B = (1)
AM of A and B will be different.
As per the statement, list A should consist of even numbers as exactly half of them are negative.
=> 0 will not be part of the list.
If we remove one of the number to get list B then it will consist of odd numbers.
In this case, AM of A cannot be equal to AM of B.
Sufficient.

Statement 2 alone is sufficient.

Imo C

Statement 1: Sum of the numbers in list A is odd
Let's take a set with minimum numbers
A = {0,1,2} and B = {0,2} then the average is 1 for both set
A = {0,1,2} and B = {0,1} then the average is not the same.
Insufficient

Statement 2: Exactly half of the numbers in list A is positive
Take set
A= {-1,0,1} and B = {-1,0} then the average is not same.
A= {-1,0,1} and B = {-1,1} then the average is same.
Insufficient

Combining both statements
Take set A- {-1,0,1,3} Average is 3/4 = 3/4
Take B -{-1,1,3} Average is 3/3 =1
or Take B -{-1,0,3} Average is 2/3
The average for any such sets will be different.

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

Let the numbers be 5, 5, 7, 8, and 10. Their sum is an odd number 35 and average is 7. If I take out 7 from the set and calculate the average, it’s still 7. Yes answer. However, if I take any other number, the average will change. So, No answer.
Insufficient.



(2) Exactly half of the numbers in list A is positive

This statement informs that the number of integers set A has is EVEN. So, let the numbers be -3, 0, 1, and 2 whose average is 0. If I exclude 0 from the set, the mean is still 0. We have Yes answer. If I take out any other number, the answer is No.
Insufficient.



1+2) So we know that the average for set A looks as ODD/EVEN which ends anyway with decimal …,5 (for example; 2,5 or 3,5 or 100,5 that ends anyway with 5)

If we exclude any number from set A, we will have either ODD/ODD or EVEN/ODD whose decimal representation never ends with 5. So, averages are not equal.


So C

Statement A

We don't get any information about B

Statement B

We don't get any information about B

Answer is E

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

In this case the means of set A and set B will never be equal. Let's take an example:

Set A {0,0,0,1} Mean = 1/4

Set B {0,0,1} Mean = 1/3

Sufficient

(2) Exactly half of the numbers in list A is positive

In this case the means of set A and set B will never be equal. Let's take an example:

Set A {0,-1} Mean = -1/2

Set B {-1} Mean = -1

Suficient

IMO Option D