12 Days of Christmas GMAT Competition - Day 7: Set A consists of disti

Question

12 Days of Christmas GMAT Competition with Lots of Fun

Set A consists of distinct integers. One number is removed from set A, and the remaining numbers comprise set B. Is the average (arithmetic mean) of the numbers in set A equal to the average (arithmetic mean) of the numbers in set B?

(1) The mean of the numbers in A is -2.5.

(2) The number of terms in set B is odd.

https://gmatclub.com/blog/wp-content/uploads/2020/12/12-days-of-christmas.jpg

M36-48

Bunuel · Accepted Answer

Official Solution:

Set A consists of distinct integers. One number is removed from set A, and the remaining numbers comprise set B. Is the average (arithmetic mean) of the numbers in set A equal to the average (arithmetic mean) of the numbers in set B? 
 
Notice that for the mean to remain the same, the number removed from set A must be equal to the mean itself (if the number removed from set A is less than the mean of set A, then the mean of set B would be greater than that of set A and if the number removed from set A is greater than the mean of set A, then the mean of set B would be smaller than that of set A).
 
(1) The average (arithmetic mean) of the numbers in A is -2.5.
 
Since set A consists of distinct  integers , then none of the numbers in the set can be equal to the mean, which is NOT an integer, thus we cannot remove a number from set A such that it equals to the mean, therefore, as discussed above, the mean of set A dose not equal to the mean of set B. Sufficient.
 
(2) The number of terms in set B is odd.
 
This one is clearly insufficient. For example, if A = {1, 2, 3, 6} and 3 is removed we'd have an YES answer but if 6 is removed we'd have a NO answer.

Answer: A

bidskamikaze · Answer

Set A consists of distinct integers. One number is removed from set A, and the remaining numbers comprise set B. Is the average (arithmetic mean) of the numbers in set A equal to the average (arithmetic mean) of the numbers in set B?

(1) The mean of the numbers in A is -2.5.

(2) The number of terms in set B is odd.

(AM = Arithmatic Mean)
 If the numbers in Set A are all integers AND are different, then only way the AM of Set B equals the AM of Set A is that there is a number in the Set A EQUAL to the AM in the Set A and that number itself is removed!

(1) The mean of the numbers in A is -2.5.
This tells us that set A has a non-integer AM. So there is no way AM of Set B = AM of Set A; Hence both Sets will have different AM's.
Sufficient.

(2) The number of terms in set B is odd.
In this case, it is possible that the set contains a number equal to the AM and that number is removed. In this case, AM of Set A = AM of Set B.
OR, in all other cases in which some other number other than AM is removed, Set A and B will have different AM's.
Insufficient.

Answer A.

Gmatfox · Answer

Set A consists of distinct integers. One number is removed from set A, and the remaining numbers comprise set B. Is the average (arithmetic mean) of the numbers in set A equal to the average (arithmetic mean) of the numbers in set B?

(1) The mean of the numbers in A is -2.5.

(2) The number of terms in set B is odd.

A (distinct integers) --> Remove 1 number = B
 Mean A = Mean B Y/N?
 
 1)  Mean A = -2.5 but sum of A is integer --> Number of terms in set A is even
Remove 1 number from A --> Number of terms in set B is odd
--> Sum of B = -2.5 * odd = not integer (Nonsense because B still consists of distinct integers or sum of B must be an integer)
--> Mean A is always different from Mean B --> No
--> Sufficient.
 2)  Number of terms in set B is odd --> Number of terms in set A is even.
Ex: A {1,2,3,6} --> Mean A = 3
 B {1,2,6} --> Mean B = 3
--> Yes 
Ex: A {1,2,3,6} --> Mean A = 3
 B {1,2,3} --> Mean B = 2 
--> No
--> Insufficient
 --> Correct answer: A

dushyanta · Answer

IMO (A)

Set A consists of distinct integers. One number is removed from set A, and the remaining numbers comprise set B. Is the average (arithmetic mean) of the numbers in set A equal to the average (arithmetic mean) of the numbers in set B?

(1) The mean of the numbers in A is -2.5.

(2) The number of terms in set B is odd.

Set A - distinct integers

One number removed to make it Set B

For us to make mean of Set A = mean of set B,  the number removed should equal to mean of A/ B

1) The mean of numbers in A is -2.5

Set of Integers with mean =-2.5 
In this case we would have to remove -2.5 from the Set A to keep the mean consistent in Set B
This cannot be the case as the Set A has all integers, we cannot remove a fraction from it.

Sufficient

(2) The number of terms in set B is odd.
No mention of anything else. B could have anything or any number of digits with mean anywhere

Insufficient

Nishant1795 · Answer

The Mean of Set A and B will be equal if the extra integer in A is equal to its mean. Let's evaluate the 2 statements:

1. Sufficient. 
The mean of A = -2.5. Since both sets include distinct  integers  and the mean is not an integer, hence statement 1 is sufficient.

2. Insufficient.
This statement can be true if Set A consists of (1,2,3,6) with mean = 3 and Set B consists of (1,2,6) with mean = 3. 
This statement can be false if Set A consists of (1,2,3,4) with mean = 2.5 and Set B consists of (1,2,3) with mean = 2.

Hence the answer is Option A.

plaverbach · Answer

Set A consists of distinct integers. One number is removed from set A, and the remaining numbers comprise set B. Is the average (arithmetic mean) of the numbers in set A equal to the average (arithmetic mean) of the numbers in set B?

(1) The mean of the numbers in A is -2.5.

(2) The number of terms in set B is odd.

\frac{A-x}{n-i} = \frac{A}{n} 
xn=A
x= \frac{A}{n}

1. Since x is an integer, x cannot be -2.5, sufficient!
2. From this we know that n is even and A is even, but that is not sufficient.

Ans: A

CluelessLearner · Answer

(1) The mean of the numbers in A is -2.5.
Mean isn't an integer so it's definitely not one of the numbers in the set. Removing any one will change the average, so  Sufficient

(2) The number of terms in set B is odd.
Even if it's odd, removing a term from any side will change the average.  Not Sufficient

A.

ashu2526 · Answer

So option A.
We can only determine after knowing both the sentence 
There be in short there will be no numbers of such set

Explanation:
Mean is -2.5 (2.5 means no must have been a remainder of 2 when divided by the total integers in the set)
Questions already says one digit has been moved from A to B
Now if A is even - 1 = odd
odd/odd will never get 2.5 because number will be of form X0 and no odd*5 = any perfect no

So A is even 
if A is even then possible number of integers in A 2,4,6,8,10 because each has 0 in last when multiplied by 5
Now lets see which is possible

ATQ after removing 1 digit from A, B will have same average as A had
N/1 = not possible
N/3 = X.333333
N/5 = not possible
N/7 = to possible
........

Answer A, alone enough to tell no such set will occur

zhanbo · Answer

I chose (A). This is a rather tricky question!

We need to bear in mind that Set A (and therefore set B) consists of  distinct integers . Failing to appreciate its significance will lead us to choose other options.

In order to answer the question whether the average of A is equal to the average of B, we just need to know whether the removed number is equal to the average of A. 
> If the removed number happens to be the average of A, Set A and Set B will have the same average. 
> Otherwise, Set A and Set B will have different average.

(1) The average of the numbers in A is -2.5.
Because the average is not an integer, the removed number cannot happen to be the average of A. 
Set A and Set B have different average. 
SUFFICIENT.

(2) The number of terms in set B is odd.
So the number of terms in set A is even. 
2.1 It should be easy to find instances in which Set A and Set B have different average.
For example, Set A = {1, 2, 3, 4}. 
2.2 But it is also possible to find some instances in which Set A and Set B have the same average.
For example, Set A = {2, 4, 6, 12}. After we remove 6, Set B = {2, 4, 12}
Set A and B have the same average. 
INSUFFICIENT

So, we have to choose (A). In actual test, I think I will get it wrong.

Poojita · Answer

(AM = Arithmatic Mean)
 If the numbers in Set A are all integers AND are different, then only way the AM of Set B equals the AM of Set A is that there is a number in the Set A EQUAL to the AM in the Set A and that number itself is removed!

Could you help me understand how you inferred this part?

Poojita · Answer

For us to make mean of Set A = mean of set B,  the number removed should equal to mean of A/ B  
I did'nt quite understand this part
Please help

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

12 Days of Christmas GMAT Competition - Day 7: Set A consists of disti

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

12 Days of Christmas GMAT Competition - Day 7: Set A consists of disti

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards