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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

For zy<xy<0 to be true either: A:- both x and z are positive and y is negative - in this case then z > x (because zy<xy) B:- both x and z are negative and y is positive - then z < x (because zy<xy).

or in short A:- y < 0 < x < z B:- z < x < 0 < y

Both 1. and 2. tell us that we are looking at situation B. So we need to figure out if |x-z|+|x|=|z| when z < x < 0. in this case |x-z|+|x|=|z|? becomes equivalent to |z|-|x|+|x|=|z| ? the -|x|+|x| cancels so yes it is always true. Try this with some example values to see why it is so.

Both 1. and 2. are SUFFICIENT, so the answer is D.

I used the below method to get the answer D, which is different from bobfirth’s method. Experts’ comments will be highly appreciated.

From the question we already know that zy<xy and both are negative. If y is positive z<x and if y is negative z>x.

Statement 1: z<x, I e. y is positive. So both z and x are negative. Say, x=-2 and z= -3 Plugging in the value |-2-3|+|-3| > |-3| and its true for any negative value of z and x where z<x

Statement 2: y is positive. Did the same and found that the equation can never be equal.

Thank you for your response. I agree answer is - D, but

When you say

"Statement 2: y is positive. Did the same and found that the equation can never be equal"

You are taking the give data zy<xy<0 and the second statment Y> 0 and putting -ve numbers in the equations correct. If thats the case, why do you say that "found that the equation can never be equal"?

if you take the statement zy<xy<0 and y is +ve, that means both x and z are -ve. If you take x = -2 and Z=-3, you end up with 3=3.

@ksrajgmat I was wrong doing the calculation (again). I am just wondering if plugging in the value is the right (quickest) way of doing this sort of problem...... or there is/are any other short cut method...... Dear math gurus: can you please intervene and show us quick method.

The two statements are giving parallel information, hence on its own either statement is sufficient. I recommend using a table and plugging values for the statements.

Instead of plugging in numbers, use the conceptual understanding of mods and inequalities here.

Important points to remember: 1. If ab is negative, either a or b but not both should be negative. 2. If a is positive, |a| = a; if a is negative, |a| = -a 3. If |a| > |b| but a < b, a must be negative.

Given: zy<xy<0 This means zy and xy are both negative. This is possible only if [highlight]either y is negative (and z and x are positive)[/highlight][highlight][/highlight] OR [highlight]both z and x are negative but y is positive[/highlight]. Also, since zy is 'more negative' than xy, the absolute value of z is higher than that of x i.e. [highlight]|z| > |x|[/highlight]

Question: Is |x-z| = |z| - |x|?

Let's look at stmnt 2 first. y > 0 This means that both x and z are negative. We know that if z is negative, |z| = -z. Therefore, the question now is: [highlight]Is |x-z| = x - z[/highlight]? Again, we know that |x-z| is equal to x-z if x-z is positive. So, is x-z positive? We know that x and z are both negative and z is more negative than x (e.g. x = -2 and z = -3). Therefore, x-z will be positive. So the answer to Is |x-z| = x - z? is 'YES' Sufficient.

Stmnt 1: z < x We know from above that |z| > |x|. If still z < x, then z must be negative (which implies that both x and z must be negative). This is exactly same as the situation above in stmnt 2 hence this statement will be sufficient alone.

The answer to this is: you should be able to answer this question?

You don't need a definitive answer, you should just be able to say, yes, with this information you could solve this equation.

and with either A or B, you are able to say, yes you could solve this equation and find the answer. (based on all the clarification posted in the responses) _________________

Thank you Karishma and very interesting way to solve this problem, but little confusing.

Do you see any other simple way to tackle this?

In my opinion, this is the simplest way to tackle it. You can try to plug in numbers if you like though I will not do that because I know it will be very troublesome (and time consuming). I do agree that if you are not very familiar with mods and inequalities, it could be confusing. Questions using both these concepts end up scaring the test takers and are hence considered pretty high level. Then again, if you understand the concepts I discussed under 'Important points to remember', the logic used thereafter will be pretty straight forward. Focus on those points first and understand them well. Then see if the solution is still confusing. _________________

Its really helpful. cleared lot of my confusion. Thanks a lot. The below question is also being discussed under DS of Gmatclub. I am just posting it here so that people like me can connect the confusion and understanding. Appreciate if you discuss a bit using same method:

Thank you Karishma and very interesting way to solve this problem, but little confusing.

Do you see any other simple way to tackle this?

In my opinion, this is the simplest way to tackle it. You can try to plug in numbers if you like though I will not do that because I know it will be very troublesome (and time consuming). I do agree that if you are not very familiar with mods and inequalities, it could be confusing. Questions using both these concepts end up scaring the test takers and are hence considered pretty high level. Then again, if you understand the concepts I discussed under 'Important points to remember', the logic used thereafter will be pretty straight forward. Focus on those points first and understand them well. Then see if the solution is still confusing.

Thank you Karishma.

You are correct. I need to brushup on my basics of mods.

Do you have some websites or PDF documents which you can share to clear the basics of mods and inequalities?

Instead of plugging in numbers, use the conceptual understanding of mods and inequalities here.

Important points to remember: 1. If ab is negative, either a or b but not both should be negative. 2. If a is positive, |a| = a; if a is negative, |a| = -a 3. If |a| > |b| but a < b, a must be negative.

Given: zy<xy<0 This means zy and xy are both negative. This is possible only if [highlight]either y is negative (and z and x are positive)[/highlight][highlight][/highlight] OR [highlight]both z and x are negative but y is positive[/highlight]. Also, since zy is 'more negative' than xy, the absolute value of z is higher than that of x i.e. [highlight]|z| > |x|[/highlight]

Question: Is |x-z| = |z| - |x|?

Let's look at stmnt 2 first. y > 0 This means that both x and z are negative. We know that if z is negative, |z| = -z. Therefore, the question now is: [highlight]Is |x-z| = x - z[/highlight]? Again, we know that |x-z| is equal to x-z if x-z is positive. So, is x-z positive? We know that x and z are both negative and z is more negative than x (e.g. x = -2 and z = -3). Therefore, x-z will be positive. So the answer to Is |x-z| = x - z? is 'YES' Sufficient.

Stmnt 1: z < x We know from above that |z| > |x|. If still z < x, then z must be negative (which implies that both x and z must be negative). This is exactly same as the situation above in stmnt 2 hence this statement will be sufficient alone.

Answer D.

Here is much better analysis of this question by Bunuel: gmatclub. com/forum/absolute-values-101210.html

He points out the peculiarity of this problem and says that "neither of statement is needed to answer the question, stem is enough to do so".

Here is his post:

Bunuel wrote:

This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happen then the answer should be D.

If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)

Look at the inequality \(zy<xy<0\):

We can have two cases:

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\). In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.

And:

B. If \(y>0\) --> when reducing we'll get: \(z<x<0\). In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.

So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that \(z<x\), hence we have case B.

Statement (2) says that \(y>0\), again we have case B.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...