M12-30

Question

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. 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

Bunuel · Answer

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. 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. 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

Aswath Kumar · Answer

If the constant multiplied is equal to 1, then option 1 also fails.

I am not sure if I am missing something. A constant can be any real integer right?

Bunuel · Answer

The mean of four integers will not change if all the integers are multiplied by ANY constant. Of course the mean won't change if you multiply all the integers by 1 and in this case the mean could be any number but if we want the mean not to change when we multiply by ANY constant than the mean must be 0. Check for more here: https://gmatclub.com/forum/the-mean-of- ... 70142.html

Akanksha007 · Answer

I think this is a  high-quality  question and I agree with explanation.

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation. A simple but a very good concept question

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

gokusan · Answer

Hi Bunuel, in the example you have given above, I interpreted it as the largest members being 1 and 2 and the smallest members being -3 and 0. So the sum still comes to be zero which is also what I did during practice. Would it make more sense to say sum of the largest and smallest member is zero, instead of saying members?

Bunuel · Answer

When referring to the "sum of the largest and smallest members," it typically means considering only one largest number and one smallest number from the list, not multiple. I've never seen any example of the opposite. So, I think the wording is fine as it is.

7seven7 · Answer

I think this is a  high-quality  question and the explanation isn't clear enough, please elaborate.

As the question states "the average of a list of four integers remains unchanged when all the integers are multiplied by any constant", then doesn't it mean that it should be true for all constants or for any one constant ??

As, I assumed that it should be true for all constants and so, the only set is {0,0,0,0}

If we assume a set like in solution i.e. {-3,0,1,2} multiplying it buy anyhing other than -1/0/1 would cause the avg to change and so, I assumed that this set is not applicable as per question stem.

Kindly help in explaining why "multiplied by any constant" is taken to be true if multiplying with any 1 is enough (as I assumed it as must be true).

Bunuel · Answer

If you multiply each term of the set {-3, 0, 1, 2} by ANY constant, the mean will remain the same. For example:

• If you multiply each term by -2, the new set becomes {6, 0, -2, -4}, and the mean remains 0.
• If you multiply each term by π, the new set becomes {-3π, 0, π, 2π}, and the mean is still 0.

IIMA-ABHISHEK · Answer

"bunnel" 
For the second statement, the sum of largest and smallest members should be zero but as per your example it would be -3+2= (-1) and not zero. Will you please explain.

Bunuel · Answer

You are missing a point. The second statement is "The sum of the largest and smallest members of the list is 0."

In the example {-3, 0, 1, 2}, the largest number is 2 and the smallest is -3. So the sum of the largest and smallest is 2 + (-3) = -1, which is not 0.

This shows that the second statement is  not necessary  true. A list with a mean of 0 can still have the largest and smallest numbers not adding up to 0.

luisdicampo · Answer

Deconstructing the Question  
The average of four integers remains unchanged when all numbers are multiplied by any constant.
We are asked which statements must be true.

Step-by-step  
Let the original average be  m .

After multiplying all numbers by a constant  k , the new average becomes:
 k·m

Since the average remains unchanged:
 k·m = m

Rewriting:
 (k - 1)m = 0

This must hold for any  k , so the only possibility is:
 m = 0

Evaluate the statements:
I. The average is 0 → must be true.
II. Largest + smallest = 0 → not required.
III. Must contain both positive and negative integers → not required (e.g., all zeros).

Answer: I only

totamofficiis · Answer

I like the solution - it’s helpful.

M12-30

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