What is the average (arithmetic mean) of x, y and z?

Arithmetic Mean = (x+y+z)/3
From 1:
3x-2y+7z=23; Insufficient
From 2:
4x-3y+5z=5 & -x+6y-2z=58
Adding the two equations,
3x+3y+3z=63
x+y+z=21 --> (x+y+z)/3=7; Sufficient
Answer: B

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What is the average (arithmetic mean) of x, y and z?

(1) 3x – 2y + 7z = 23

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


Transforming the original condition and the question, (x+y+z)/3=? and we just need to know x+y+z. Looking at 2) we have 4x – 3y + 5z = 5 and –x + 6y – 2z = 58, so adding them gives us 3x+3y+3z=63. The condition is sufficient, therefore B is the answer.

KAPLAN OFFICIAL 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).

Nice one.
Here is my solution to this one =>
Mean = Sum/#

Hence to get the mean we need the sum x+y+z

Statement 1=>
clearly not sufficient

Statement 2=>
Adding the two equations we get => 3x+3y+3z=63
Hence x+y+z = 21
Hence Sufficient

Hence B

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