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Math Expert V
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Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 77% (01:40) correct 23% (02:05) wrong based on 516 sessions

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Is the average (arithmetic mean) of the numbers x, y, and z greater than z ?

(1) z − x < y − z
(2) x < z < y

DS69502.01
OG2020 NEW QUESTION

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Is the average (arithmetic mean) of the numbers x, y, and z greater than z ?

We are looking for whether $$\frac{x+y+z}{3}>z......x+y+z>3z.......x+y>2z$$

(1) z − x < y − z
Suff

(2) x < z < y
We do not know the values..
Insuff

A
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Posts: 4219
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Bunuel wrote:
Is the average (arithmetic mean) of the numbers x, y, and z greater than z?

(1) z − x < y − z
(2) x < z < y

Target question: Is the average (arithmetic mean) of the numbers x, y, and z greater than z?
This is a good candidate for rephrasing the target question.
Rewrite the question as "Is (x + y + z)/3 > z?"
Multiply both sides by 3 to get: "Is x + y + z > 3z?"
Subtract z from both sides to get: "Is x + y > 2z?"

REPHRASED target question: Is 2z less than x + y?

Statement 1: z − x < y − z
Add z to both sides to get: 2z − x < y
Add x to both sides to get: 2z < x + y
PERFECT!
The answer to the REPHRASED target question is YES, 2z IS less than x+y
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: x < z < y
There are several values of x, y and z that satisfy statement 2. Here are two:
Case a: x = 0, y = 3 and z = 1. In this case, 2z = 2(1) = 2 and x + y = 0 + 3 = 3. So, the answer to the REPHRASED target question is YES, 2z IS less than x+y
Case b: x = 0, y = 3 and z = 2. In this case, 2z = 2(2) = 4 and x + y = 0 + 3 = 3. So, the answer to the REPHRASED target question is NO, 2z is NOT less than x+y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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1
While our research found that in most DS questions the fastest way to solve the question is the Logical approach, in questions like this one, in which both the given information and the statements can be further simplified, the Precise approach might be faster.
The questions, translated into an algebraic expression, is whether (x+y+z)/3 > z, or, is x +y+z>3z, which can be simplified to: x+y>2z?
By 'moving' the negative variables in statement (1) to the other side of the inequality we get 2z < x+y, which is exactly what we were looking for.
Statement (2) on the other hand tells is that one variable is larger than z while the other is smaller, but we don't know how large are the differences (if y was far greater than z, their sum could have been greater than 2z. Also, if both x and y were greater than z, then their sum would have definitely been greater than 2z).
So statement (2) is not sufficient, while statement (1) is.

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Bunuel wrote:
Is the average (arithmetic mean) of the numbers x, y, and z greater than z ?

(1) z − x < y − z
(2) x < z < y

DS69502.01
OG2020 NEW QUESTION

The original question: Is $$\frac{x+y+z}{3}>z$$ ?
The rephrased question: Is $$x+y+z>3z$$ ? $$\implies$$ Is $$x+y>2z$$ ? $$\implies$$ Is $$z<\frac{x+y}{2}$$ ?

In words: Is $$z$$ less than the average of the numbers $$x$$ and $$y$$ ?

1) We know that $$z-x<y-z$$, which we can rearrange.

$$2z<x+y$$

$$z<\frac{x+y}{2}$$

Thus, the answer to the rephrased question is a definite Yes. $$\implies$$ Sufficient

2) We know that $$x<z<y$$ and can test possible cases. If $$x=1$$, $$z=1.5$$, and $$y=2$$, then the answer to the rephrased question is No. However, if $$x=1$$, $$z=1.4$$, and $$y=2$$, then the answer to the rephrased question is Yes. Thus, we can't get a definite answer to the rephrased question. $$\implies$$ Insufficient

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Bunuel wrote:
Is the average (arithmetic mean) of the numbers x, y, and z greater than z ?

(1) z − x < y − z
(2) x < z < y

DS69502.01
OG2020 NEW QUESTION

We need to determine:

Is (x + y + z)/3 > z ?

Is x + y + z > 3z ?

Is x + y > 2z ?

Statement One Alone:

z − x < y − z

Manipulating the inequality, we have:

2z < y + x

We see that the question has been answered. Statement one alone is sufficient to answer the question.

Statement Two Alone:

x < z < y

Statement two is not sufficient. If x = 1, z = 2, y = 3, then the average of x, y, and z is not greater than z. However if x = 1, z = 2, and y = 100, then the average is x, y, and z is greater than z.

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Hi All,

We're asked if the average (arithmetic mean) of the numbers X, Y, and Z is GREATER than Z. This is a YES/NO question and can be approached with a mix of Arithmetic and TESTing VALUES. To start, we can 'rewrite' the question a bit:

Is (X+Y+Z)/3 > Z?
Is (X+Y+Z) > 3Z?
Is (X+Y) > 2Z?

By comparison, this is an easier question to answer than what we were initially given.

(1) Z - X < Y - Z

With Fact 1, we can rewrite the inequality as:
2Z < X + Y
This Fact tells us that (X+Y) IS greater than 2Z, so the answer to the question is clearly YES.
Fact 1 is SUFFICIENT

(2) X < Z < Y

With this inequality, we can TEST VALUES and track the results.
IF....
X=1, Z=2, Y=3, then (1+3) is NOT greater than (2)(2), so the answer to the question is NO.
X=1, Z=2, Y=4, then (1+4) IS greater than (2)(2), so the answer to the question is YES.
Fact 2 is INSUFFICIENT

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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$$\frac{x+y+z}{3}$$ > z
Rephrasing it, we get is x+y > 2z

Statement 1 -
we get 2z < x + y
Sufficient

Statement 2 -
The average lies between the min and max number.
So, we know that avg lies between x and y.
Now, it can be either greater than, less than or equal to z.
Insufficient.

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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Is (x+y+z)/3 > z
Is x+y+z > 3Z
Is x+y > 2Z

1) Y+X > 2Z: sufficient
2) Insuff.   Take the example:
x=1 z = 1.2 y = 1.3.
X+Y is not greater than 2Z

Likewise you can prove X+Y > 2Z

Therefore statement 2 is insufficient.
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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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How can we multiply x+y+z/3>z......x+y+z>3z without knowing the signs?
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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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1
kri93 wrote:
How can we multiply x+y+z/3>z......x+y+z>3z without knowing the signs?

Hi kri93,

The issue that you're referring to is only relevant when you're multiplying or dividing an inequality by a NEGATIVE number. Here, we're multiplying both sides by POSITIVE 3, so we don't have worry about the 'direction' of the inequality.

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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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EMPOWERgmatRichC wrote:
kri93 wrote:
How can we multiply x+y+z/3>z......x+y+z>3z without knowing the signs?

Hi kri93,

The issue that you're referring to is only relevant when you're multiplying or dividing an inequality by a NEGATIVE number. Here, we're multiplying both sides by POSITIVE 3, so we don't have worry about the 'direction' of the inequality.

GMAT assassins aren't born, they're made,
Rich

Thank you for the clarification:) Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th   [#permalink] 20 Dec 2019, 22:57
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