## GMAT Sample Problem: Data Sufficiency Averages

While you may often see the topic of averages come up in Problem Solving word problems, you can just as easily see averages within an algebraic Data Sufficiency question, such as the following example.

## Sample Problem:

What is the average (arithmetic mean) of

x, yandz?(1) 3x – 2y + 7z = 23

(2) 4x – 3y + 5z = 5 and –x + 6y – 2z = 58

## Solution:

In this data sufficiency question we are asked to find the average of three unknowns. Remember that when asked to find an average, you need to find the sum of the terms divided by the number of terms. In this case we would need to know the sum of x + y + z and divide it by 3. The key to remember, is that we do not need to know x, y and z individually, only their sum. As long as we can do this, we will be able to find the average.

Statement 1 tells us 3x – 2y + 7z = 23. From this statement we are unable to determine x, y and z individually and we are also unable to find the sum of x, y and z directly. Statement 1 is, therefore, insufficient.

Statement 2 tells us 4x – 3y + 5z = 5 and –x + 6y – 2z = 58. At first this statement looks insufficient, as we have three equations and two variables, meaning that we are unable to solve for x, y and z. However, if we add these two equations together we get:

If we divide 3x + 3y + 3z = 63 by 3, we are left with x + y + z = 21. As we know the sum of x, y and z, statement 2 is sufficient.

Since statement 2 is sufficient and statement 1 is not, we do not need to check if the statements are sufficient together. Our answer must be (B).

## 7 comments

Michael Feres on November 21st, 2010 at 4:39 pm

Bret

How are you. Big fan of the site. I take the gmat in two weeks and have a full two weeks off of work to finish up my study prep. I am feeling more confident day by day. I am almost finished with the Data Sufficiency practice questions in the OG 12th edition. I am having trouble with question 128 and the explanation in the back of the book.

A school administrator will assign each student in a group of n students to one of m classrooms. If 3<M<13<N, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?

(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.

(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.

If you can please provide as much detail as possible in the description of the answer.

Thank you

Bakhtior on November 22nd, 2010 at 12:51 am

M can equal 4,5,6,7,8,9,10,11,12. Lets assume that answer 1 is sufficient, then all 3n number can be divided by these numbers. If n=14, it is correct, n=15, n=16 correct, but if n=17 or 19, in general n=prime numbers it will be incorrect. 13 is prime number.

Both are insufficient.

Maybe I couldn't catch the meaning of problem, simply this is my solution out of consideration.

Aditya on November 22nd, 2010 at 2:55 am

This problem would be from a easy category...I think one would get this kinda problem for a score of around 40 in quant..

Just my two cents...

Michael Feres on November 22nd, 2010 at 6:26 am

B is sufficient

Ash on November 23rd, 2010 at 12:43 am

I guess the answer is B.

What we are trying to do here is to see that m divides n. If this is possible then there will be equal number of students in each class equal to the quotient.

i) if we consider this statement then this means that 3N/m is an integer. Now N can be any big number while m can vary from 4 to 12. Notice that in this range 6, 9 and 12 are included hence depending on N which can be 17 or 21 suppose N may or may not be divisible by n.

Insufficient

ii) here since 13 is not in the range of values of M. Implies that n is divisible by m

Sufficient.

Hence, B

Neil on November 30th, 2010 at 6:00 pm

Sorry still very confusing. You mentioned that "here since 13 is not in the range of values of M. Implies that n is divisible by m

Sufficient."

How could this happen? If 13 is not in the rang of M, the answer should be unclear because we don't have the standard to judge the final answer?

Suppose now we are saying there are only 4 or 5 classrooms now(limited to two), and I want you to judge if we can put 13n students in these classroom and you know N >13. How can u make sure if we can put them in or not?

Thanks.

Ash on November 30th, 2010 at 10:27 pm

@Neil

M cannot is smaller than 13 and can only have integer values from 4-12.

Again 13 is a prime number and for 13N students to fit in M classrooms N must be divisible by M, since M cannot have a value of 13.

So N is some value which is divsible by M because M cannot be 13. N can be any big number say(=50) while 13N = 13*50 = 650 while M=5.

SO the point to note here is that 13N is divisible by M =>

N is divisible by M. since M cannot be equal to 13.