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

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What is the average (arithmetic mean) of x, y and z?  [#permalink]

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07 Sep 2015, 04:28
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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

Kudos for a correct solution.

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Re: What is the average (arithmetic mean) of x, y and z?  [#permalink]

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07 Sep 2015, 06:19
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Bunuel wrote:
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

Kudos for a correct solution.

Solution: We need to find mean i.e $$\frac{x+y+z}{3}$$.
In order to find this we need either values of x,y and z or value of x+y+z

Statement1 : We have 3 variables and 1 equation. So, we cannot find x,y and z. Finding x+y+z is also impossible here.
So, insufficient.

Statement2 : There is a small trap here. We have 3 variables and 2 equations,we cant find x,y and z individually. But adding those two equations we get the value of x+y+z. This is enough to find the mean.
So, sufficient.

Option B
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Re: What is the average (arithmetic mean) of x, y and z?  [#permalink]

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07 Sep 2015, 06:32
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Bunuel wrote:
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

Kudos for a correct solution.

Basically, we need to find x + y + z

Statement 1) Insufficient. We cannot find the sum of x + y + z through any mathematical operation.

Statement 2 ) Add both the equations and you get x + y + z = 21 . Sufficient
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What is the average (arithmetic mean) of x, y and z?  [#permalink]

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07 Sep 2015, 21:50
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Arithmetic Mean = (x+y+z)/3
From 1:
3x-2y+7z=23; Insufficient
From 2:
4x-3y+5z=5 & -x+6y-2z=58
3x+3y+3z=63
x+y+z=21 --> (x+y+z)/3=7; Sufficient
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What is the average (arithmetic mean) of x, y and z?  [#permalink]

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08 Sep 2015, 07:48
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

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.
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Re: What is the average (arithmetic mean) of x, y and z?  [#permalink]

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14 Sep 2015, 07:40
Bunuel wrote:
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

Kudos for a correct solution.

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).

Attachment:

Screen-shot-2010-11-09-at-12.28.24-PM.png [ 9.96 KiB | Viewed 15375 times ]

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Re: What is the average (arithmetic mean) of x, y and z?  [#permalink]

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21 Dec 2016, 17:20
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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Re: What is the average (arithmetic mean) of x, y and z?  [#permalink]

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14 Apr 2019, 08:00
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Re: What is the average (arithmetic mean) of x, y and z?   [#permalink] 14 Apr 2019, 08:00
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# What is the average (arithmetic mean) of x, y and z?

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