Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Attend a Veritas Prep GMAT Class for Free. With free trial classes you can work with a 99th percentile expert free of charge. Learn valuable strategies and find your new favorite instructor; click for a list of upcoming dates and teachers.
Join us for a debate among the MBA experts from Stratus Admissions on how these elite MBA programs are different from each other, what are their strengths & weaknesses, and how to decide which one is better for you.
Are you struggling to achieve your target GMAT score? Most students struggle to cross GMAT 700 because they lack a strategic plan of action. This Free Strategy Webinar, which will empower you to create a well-defined study plan to score 760+ on GMAT.
To score Q50 on GMAT Quant, you would need a strategic plan of action which takes your strengths and weakness into consideration. Attend this free workshop to attempt a supervised quant quiz and gain insights that can help you save 35+ hours.
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
27 Apr 2019, 08:10
Top Contributor
Expert Reply
2
Bookmarks
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
RELATED VIDEO FROM MY COURSE
_________________
If you enjoy my solutions, I think you'll like my GMAT prep course.
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
27 Apr 2019, 08:11
1
Bookmarks
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).
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
27 Apr 2019, 09:38
1
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
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
07 May 2019, 17:36
Expert Reply
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.
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
16 May 2019, 13:21
Expert Reply
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
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
06 Nov 2019, 20:52
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.
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
20 Dec 2019, 14:33
1
Expert Reply
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.
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
20 Dec 2019, 21:57
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.
Re: Is the average (arithmetic mean) of the numbers x, y, and z greater th
[#permalink]
29 Jan 2020, 19:11
1
Expert Reply
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