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:

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

18 Aug 2008, 14:10

zoinnk wrote:

brokerbevo wrote:

Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) \(x^4 \ge 1\) B) \(x^3 \le 27\) C) \(x^2 \ge 16\) D) \(2 \le |x| \le 5\) E) \(2 \le 3x + 4 \le 6\)

E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.

So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further. _________________

Factorials were someone's attempt to make math look exciting!!!

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

18 Aug 2008, 14:34

brokerbevo wrote:

zoinnk wrote:

brokerbevo wrote:

Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) \(x^4 \ge 1\) B) \(x^3 \le 27\) C) \(x^2 \ge 16\) D) \(2 \le |x| \le 5\) E) \(2 \le 3x + 4 \le 6\)

E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.

So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.

D is not the solution because,

\(2 \le |x| \le 5\) Split the inequality, You have two cases, since we are dealing with a mod function, 2<= -x which means, x<= -2 -----(1) 2<=x which means, x>=2 --------(2)

If you analyze this in a number line, you do not get a finite line. You can do the same with the other half of the inequality (with 5) and get a similar result.

The way, I solved E is, 2<= 3x+4 <= 6 -2 <= 3x <=2 -(2/3) <= x <= (2/3) Which gives a finite boundary (line) in a number line and hence E. You can adopt a similar methodology for A, B and C to prove it incorrect. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

18 Aug 2008, 14:55

brokerbevo wrote:

zoinnk wrote:

brokerbevo wrote:

Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) \(x^4 \ge 1\) B) \(x^3 \le 27\) C) \(x^2 \ge 16\) D) \(2 \le |x| \le 5\) E) \(2 \le 3x + 4 \le 6\)

E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.

So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.

The graph for D is disjointed due to the absolute values.

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

18 Aug 2008, 14:57

leonidas wrote:

brokerbevo wrote:

zoinnk wrote:

E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.

So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.

D is not the solution because,

\(2 \le |x| \le 5\) Split the inequality, You have two cases, since we are dealing with a mod function, 2<= -x which means, x<= -2 -----(1) 2<=x which means, x>=2 --------(2)

If you analyze this in a number line, you do not get a finite line. You can do the same with the other half of the inequality (with 5) and get a similar result.

But in D, x IS finite because the boundary of x is as follows: -5 < x < 5 --> x can have no value outside this expression. For example, x can not equal -6, -97, 22, or 9 ---> x is finite in D ??? _________________

Factorials were someone's attempt to make math look exciting!!!

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

18 Aug 2008, 15:14

leonidas wrote:

brokerbevo wrote:

zoinnk wrote:

So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.

D is not the solution because,

\(2 \le |x| \le 5\) Split the inequality, You have two cases, since we are dealing with a mod function, 2<= -x which means, x<= -2 -----(1) 2<=x which means, x>=2 --------(2)

If you analyze this in a number line, you do not get a finite line. You can do the same with the other half of the inequality (with 5) and get a similar result.

But in D, x IS finite because the boundary of x is as follows: -5 < x < 5 --> x can have no value outside this expression. For example, x can not equal -6, -97, 22, or 9 ---> x is finite in D ???

The question stem says that- "single line segment of finite length" Solving D, -5 <= x <= 5 and 2 <= x <= -2 If you plot these on the number line, you will get two line segments satisfying the conditions: -5 <= x <= -2 and 2 <= x<= 5 Hence, not D. Sorry, I take back what I said earlier about not making a finite line segment for D. It actually gives two finite line segments. Hope this helps. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

18 Aug 2008, 15:35

leonidas wrote:

leonidas wrote:

zoinnk wrote:

The question stem says that- "single line segment of finite length" Solving D, -5 <= x <= 5 and 2 <= x <= -2 If you plot these on the number line, you will get two line segments satisfying the conditions: -5 <= x <= -2 and 2 <= x<= 5 Hence, not D. Sorry, I take back what I said earlier about not making a finite line segment for D. It actually gives two finite line segments. Hope this helps.

Ahh yes. That makes sense now. I wasn't taking into account the fact that -5 < x < 5 includes 1 and -1 (for example) which does not satisfy \(2 \le |x| \le 5\). Its definitely 2 different line segments. _________________

Factorials were someone's attempt to make math look exciting!!!

Re: Which of the following inequalities has a solution set that, [#permalink]

Show Tags

20 Sep 2014, 15:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Which of the following inequalities has a solution set that, [#permalink]

Show Tags

20 Sep 2014, 16:55

Expert's post

1

This post was BOOKMARKED

brokerbevo wrote:

Which of the following inequalities has a solution set that when graphed on the number line, is a single line segment of finite length?

A. x^4 ≥ 1 B. x^3 ≤ 27 C. x^2 ≥ 16 D. 2≤ |x| ≤ 5 E. 2 ≤ 3x+4 ≤ 6

The key words in the stem are: "a single line segment of finite length"

Now, answer choices A, B, and C can not be correct answers as solutions sets for these exponential functions are not limited at all (>= for even powers and <= for odd power) and thus can not be finite (x can go to + or -infinity for A and C and x can got to -infinity for B). As for D: we have that absolute value of x is between two positive values, thus the solution set for x (because of absolute value) will be two line segments which will be mirror images of each other.

Answer: E.

Just to demonstrate:

A. x^4 >= 1 --> \(x\leq{-1}\) or \(x\geq{1}\): two infinite ranges;

B. x^3 <= 27 --> \(x\leq{3}\): one infinite range;

C. x^2 >= 16 --> \(x\leq{-4}\) or \(x\geq{4}\): two infinite ranges;

D. 2 <= |x| <= 5 --> \(-5\leq{x}\leq{-2}\) or \(2\leq{x}\leq{5}\): two finite ranges;

E. 2 <= 3x+4 <= 6 --> \(-2\leq{3x}\leq{2}\) --> \(-\frac{2}{3}\leq{x}\leq{\frac{2}{3}}\): one finite range.

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...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...