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:

(1) x < x^2 so x can be less than -1 and greater than 1 ! insufficient! (2) x < x^3 so x is greater than -1 so it can be negative or positive insufficent!

(1) and (2) only said than x is greater than -1 so x can be positive or negative.... insufficient!

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Question: is \(x>0\)?

When considering statements together we can see that YES answer is easy to get: \(x=2>0\), so answer YES --> \((x=2)<(x^2=4)<(x^3=8)\);

Now, for NO answer we should pick negative \(x\) --> ANY negative \(x\) would be less than \(x^2\) as \(x^2\) would be positive, so satisfying (1) is easy. Next, in order negative \(x\) to be less than \(x^3\), it shoul be a fraction in the range (-1,0) --> \(x=-\frac{1}{2}<0\), so answer NO --> \((x=-\frac{1}{2})<(x^3=-\frac{1}{8})<(x^2=\frac{1}{4})\).

I'm assuming you are trying to divide both sides by x. The problem with an inequality is that you don't know if x is positive or negative. If it is negative, you need to change the sign to greater than in this case.

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

Can I deduce as following : 1) \(x < x^2\) holds true for x > 1 and x < 0 Thus, not sufficient

2) \(x < x^3\) holds true for x > 0 and 0 < x < 1 Not sufficient

No unique value is determined when combining 1 & 2. Thus, E.

Please correct me if I am incorrect in the above logic.

You are correct.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.
_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

So you can not divide both parts of inequality \(x<x^2\) by \(x\) as you don't know the sign of this unknown: if \(x>0\) you should write \(1<x\) BUT if \(x<0\) you should write \(1>x\).

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

Can I deduce as following : 1) \(x < x^2\) holds true for x > 1 and x < 0 Thus, not sufficient

2) \(x < x^3\) holds true for x > 0 and 0 < x < 1 Not sufficient

No unique value is determined when combining 1 & 2. Thus, E.

Please correct me if I am incorrect in the above logic.

You are correct.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.

No your ranges for (2) are not correct, it should be: \(-1<x<0\) or \(x>1\). Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution:

Is \(x>0\)?

(1) \(x<x^2\) --> \(x*(x-1)>0\) --> \(x<0\) or \(x>1\) --> ---------0----1----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\) --> ----(-1)----(0)----(1)----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(1)+(2) Intersection of the ranges from (1) and (2) is: ----(-1)----(0)----(1)---- Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.

No your ranges for (2) are not correct, it should be: \(-1<x<0\) or \(x>1\). Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution:

Is \(x>0\)?

(1) \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\) --> ---------0----1----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\) --> ----(-1)----(0)----(1)----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(1)+(2) Intersection of the ranges from (1) and (2) is: ----(-1)----(0)----(1)---- Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

Answer: E.

Hope it helps.[/quote]

Hi Bunuel,

Could you please explain why you're considering only the negative factors while considering \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\).. From \(x*(1-x)>0\) can't we consider both x and (x-1) to be positive also ?

Similarly for \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\), I seem to be missing your logic.

Could you please explain why you're considering only the negative factors while considering \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\).. From \(x*(1-x)>0\) can't we consider both x and (x-1) to be positive also ?

Similarly for \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\), I seem to be missing your logic.

Thanks

Actually, both cases were considered.

\(x<x^2\) --> \(x*(x-1)>0\) x>0 and x-1>0 (x>1) --> x>1; x<0 and x-1<0 (x<1) --> x<0.

Thus, \(x<x^2\) holds true for \(x<0\) and \(x>1\).

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

So you can not divide both parts of inequality \(x<x^2\) by \(x\) as you don't know the sign of this unknown: if \(x>0\) you should write \(1<x\) BUT if \(x<0\) you should write \(1>x\).

Hope it helps.

Hi Bunnuel

With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52 TO make things easier i ll post the question

If X and Y are positive , Is X<10<Y? 1. X<Y and XY = 100 2. X^2 <100< Y^2

In the explanation part 2nd stem is directly reduced to X<10<Y

Isnt your explanation contradictory to above expalantion of OG Pls help in getting my concept right.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

So you can not divide both parts of inequality \(x<x^2\) by \(x\) as you don't know the sign of this unknown: if \(x>0\) you should write \(1<x\) BUT if \(x<0\) you should write \(1>x\).

Hope it helps.

Hi Bunnuel

With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52 TO make things easier i ll post the question

If X and Y are positive , Is X<10<Y? 1. X<Y and XY = 100 2. X^2 <100< Y^2

In the explanation part 2nd stem is directly reduced to X<10<Y

Isnt your explanation contradictory to above expalantion of OG Pls help in getting my concept right.

First of all, they are not reducing, they are taking square root. Also, notice that the stem says that x and y are positive numbers.

But is it ok to take square root,considering signs are unknown in above scenario.

If we were not told that x and y are positive numbers, then from x^2<100<y^2, we would have |x|<10<|y| (since \(\sqrt{x^2}=|x|\)).

Generally:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...