It is currently 29 Oct 2020, 17:42

### GMAT Club Daily Prep

#### 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

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

# |3 + 3x| < –2x

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 13718
GRE 1: Q160 V160
Followers: 305

Kudos [?]: 3552 [1] , given: 12677

|3 + 3x| < –2x [#permalink]  08 Aug 2018, 16:31
1
KUDOS
Expert's post
00:00

Question Stats:

64% (01:07) correct 35% (01:20) wrong based on 92 sessions
$$|3 + 3x| < –2x$$

 Quantity A Quantity B $$|x|$$ $$4$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E
[Reveal] Spoiler: OA

_________________

- Do you want all the GRE Prep Club resources at a glance (sort of ) ?? Look at GRE Prep Club Official LinkTree page
- Do you want all the GRE Prep Club FREE resources in one spot ?? Look at Free GRE Materials - Where to get it!! (2020)
- Do you want all the FREE GRE tests on earth ?? Look at FREE GRE Practice Tests [Collection] - New Edition (2021)
- Do you want to know the GRE score in the Top 25 BS in 7 continents ?? Look at here The GRE average score at Top 25 Business Schools 2020 Ed.
- Need Practice 20 Free GRE Quant Tests available for free with 20 Kudos
- GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Intern
Joined: 10 Aug 2018
Posts: 33
Followers: 0

Kudos [?]: 24 [2] , given: 0

Re: |3 + 3x| < –2x [#permalink]  11 Aug 2018, 12:29
2
KUDOS
We solve the inequality:

|3+3x| < –2x

We have two possible options:

3+3x < -2x

and

3+3x > 2x

In both cases, we move x terms to the left side and constant terms over to the right side:

5x < -3

and

x > -3

So x is constrained between -5/3 and -3. In both cases, |x| < 4. So the answer is B.
Intern
Joined: 02 Oct 2019
Posts: 9
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: |3 + 3x| < –2x [#permalink]  13 Nov 2019, 18:18
Hi,

I'm having trouble with this one. I selected D as the answer. My reasoning:

When we solve for an absolute value of a variable, we are accounting for both scenarios of the bars reversing x and keeping x as it is. So, when we're given any inequality or equation with absolute value and variables, we have to solve for both scenarios. That makes perfect sense.

But it also follows from this reasoning that both scenarios can't be true--it's only one or the other. For this problem in particular, it could be that x > -3 and not less than -3/5 or it could be that x is less than -3/5 but not greater than -3.

Say that it were only true that x > -3. X could well be 4 or more, making D the correct answer.

Why is this incorrect? Could it be that this is a faulty question by Manhattan? Since this is highly unlikely I'm guessing I'm just missing a detail, so I would love it if someone could point it out for me!

Thank you.
Intern
Joined: 03 Nov 2019
Posts: 11
Followers: 0

Kudos [?]: 12 [2] , given: 3

Re: |3 + 3x| < –2x [#permalink]  14 Nov 2019, 07:52
2
KUDOS
So, you have good reasoning up to the point where you solve for both scenarios:
X < -3/5
AND
X > -3
Now here's where you are getting it a bit twisted: BOTH CONDITIONS MUST BE MET, NOT ONLY ONE.
So, values of X go from -3 to -3/5 (not including). Or in mathematical terms: X = (-3,-3/5)

I hope this helps!

P.S: I have no idea why I can't quote the comment of einalemjs...
Attachments

Possible values of X.jpg [ 9.87 KiB | Viewed 958 times ]

Intern
Joined: 03 Nov 2019
Posts: 11
Followers: 0

Kudos [?]: 12 [0], given: 3

Re: |3 + 3x| < –2x [#permalink]  14 Nov 2019, 07:53
[

Intern
Joined: 02 Oct 2019
Posts: 9
Followers: 0

Kudos [?]: 3 [2] , given: 4

Re: |3 + 3x| < –2x [#permalink]  14 Nov 2019, 09:20
2
KUDOS
Why is it that both scenarios must be true? I thought when you take an absolute value, you are either reversing the negative or leaving it as it was... but it can't be both negative and positive at once. Meaning, while it's useful to know both possible scenarios, it's not actually possible that they both occur simultaneously, since it can only be one OR the other. So what is the reasoning behind that?

It's not like finding the roots of a quadratic or accounting for positive/negative roots, where both are correct. It's a way of accounting for uncertainty in absolute value, but in this case it is either one case or the other, because the starting value could not have been both cases at once.

Given all of this, answer choice D makes the most sense, since it's possible that the |x| is less than, equal to, or greater than 4.

I really don't see how this is faulty reasoning. I hope someone can explain!

EDiT: I think I understand now, from doing more questions. It seems there's nothing wrong with my reasoning--the variables can't be both cases at once. But we're not precisely talking about the sign of the variable, but rather the output of the absolute value sign. Further, since we're dealing with inequalities, the values can satisfy both conditions. In the context of absolute value, since the resulting quantity will always be positive for the same inputs, the variable must satisfy the inequalities of both positive and negative values.

Thanks! Resolved.
Posted from my mobile device
Re: |3 + 3x| < –2x   [#permalink] 14 Nov 2019, 09:20
Display posts from previous: Sort by