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

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

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New post 26 Apr 2019, 02:45
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Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th  [#permalink]

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New post 26 Apr 2019, 05:00
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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
Add z on both sides....2z-x<y.....y+x>2z...
Suff

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

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

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New post 27 Apr 2019, 09:10
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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

Answer: A

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

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New post 27 Apr 2019, 09:11
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.
The correct answer is (A).

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

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New post 27 Apr 2019, 10:38
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

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

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New post 07 May 2019, 18:36
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.

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

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New post 16 May 2019, 14:21
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] 16 May 2019, 14:21
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