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

Question

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

chetan2u · Accepted Answer

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

BrentGMATPrepNow · Answer

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 RELATED VIDEO https://www.youtube.com/watch?v=MyG1K3ee69w

DavidTutorexamPAL · Answer

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).

Posted from my mobile device

ZoltanBP · Answer

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

ScottTargetTestPrep · Answer

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

EMPOWERgmatRichC · Answer

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

Final Answer:

Show Spoiler

A

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

uchihaitachi · Answer

The question asks us is 
 \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.

Answer - A

kri93 · Answer

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

EMPOWERgmatRichC · Answer

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

Rocknrolla21 · Answer

Hello, experts!

Why hasn't anyone considered negative integers as an option for Statement 2?

I considered x=-6, z=-4, and y=-2. In this case, 2z = -8 is not greater than -4.

X+Y+Z/3 = (-6)+(-2)+(-4)/3

-12/3 = -4

Thanks!

EMPOWERgmatRichC · Answer

Hi Rocknrolla21,

TESTing VALUES is a great way to approach this question - and you can absolutely use negative integers when working through this prompt. In certain DS prompts, that type of 'thoroughness' in your thinking (meaning that you're considering more than just positive integers) is essential to proving what the correct answer is. With this prompt, it's not necessary though - and some minor tweaks to the positive numbers that you might TEST is all that's needed to prove that Fact 2 is Insufficient.

With a bit of work, we can rewrite the given question as:

Is (X+Y) > 2Z?

(2) X < Z < Y

With this inequality, we can TEST VALUES and track the results. The first example is arguably one of the easiest that we could use:
IF.... 
X=1, Z=2, Y=3, then (1+3) is NOT greater than (2)(2), so the answer to the question is NO.

Since we now have a "NO" answer, if there is a way to get a "YES" answer, then we'd be done working here. Simply changing the value of Y is all that's needed:
IF....
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

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

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=L1IvXvM6rDY

GMATinsight · Answer

Answer: Option A

Video solution by  GMATinsight

https://www.youtube.com/watch?v=el7wi9Jii3Y

Suruchim12 · Answer

Hi Experts,  EMPOWERgmatRichC

Can someone help me understand, if I were to take the logical approach and not simplify the question to x+y >2z

Then for St. (2) : if we are told that y is the largest no. and x is the smallest, invariably putting z as the middle doesn't z then become the mean for a 3 number sequence. And therefore, the mean is not greater than z and sufficient to answer the question.

I understand that's not OA but if someone can explain the flaw in the above deduction pls! TIA

chetan2u · Answer

z in middle means ‘z’ is the median, but you cannot say anything about mean. 
1) 1<2<3….Here z is equal to the mean, 2.
2) 1<3<4….Here, mean is 8/3 or 2.67, and z>2.67
3) 1<2<6… Here, mean is 9/3 or 3, and z<3

Suruchim12 · Answer

I get where I went wrong. The mean and median is equal ONLY in  evenly spaced sets  and since we do not have that info we can't assume it and therefore the mean can be different.

EMPOWERgmatRichC · Answer

Hi Suruchim12, Since chetan2u has already answered your question (with some examples), I won't rehash any of that work here. In the broader sense though, you have to be careful about the 'assumptions' that you put on to any DS question that you work through. In most of these prompts, there will be an 'obvious' answer - and you stated one in your explanation (re: Z is in the 'middle' of X and Y). However, that is NOT what X < Z < Y actually means (this means that the value of Z falls somewhere between and X and Y; maybe it's the exact 'middle', but it might not be). DS questions are designed to test a variety of different skills (far more than just 'math' skills), including the thoroughness of your thinking. Going forward, if you find yourself thinking that the given information in Fact 1 or Fact 2 leads to just one solution, take a moment to try anything else that you can think of that 'fits' what you're told. It won't take very long to do that work and those few extra seconds might help you to avoid missing out on some easy points. GMAT assassins aren't born, they're made, Rich Contact Rich at: Rich.C@empowergmat.com

Deepam10 · Answer

On St2, can we do as below without testing values?

Given x<z<y
Subtract Z from all
x-z<z-z<y-z or 
x<0<y
and here we do not have anything for z hence not suff.  Bunuel

Bunuel · Answer

No. When subtracting z you get:

x - z < z - z < y - z
x - z < 0 < y - z

Which is not the same as x < 0 < y.

I'd say the most straightforward method to see that the second statement is insufficient is to recognize that the question essentially asks if the average of three numbers (x, y, and z) is more than the middle number (z), which obviously can't be determined.

Hope it helps.

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

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