GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Aug 2018, 03:55

Close

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
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.

Close

Request Expert Reply

Confirm Cancel

Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
User avatar
S
Joined: 24 Oct 2016
Posts: 72
GMAT 1: 670 Q46 V36
CAT Tests
Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1  [#permalink]

Show Tags

New post 02 Aug 2018, 10:47
2
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

53% (01:37) correct 47% (01:49) wrong based on 30 sessions

HideShow timer Statistics

Is x > 0?


(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

_________________

Everything You Need To Know About Verb Tenses

Manager
Manager
User avatar
S
Joined: 24 Oct 2016
Posts: 72
GMAT 1: 670 Q46 V36
CAT Tests
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1  [#permalink]

Show Tags

New post 02 Aug 2018, 10:50
1
dabaobao wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1



(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. 
We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are: 

x + 3 = +(4x – 3)       and      x + 3 = –(4x – 3) 
x + 3 = 4x – 3                       x + 3 = –4x + 3 
6 = 3x                                  5x = 0 
2 = x                                    x = 0 

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them. 

|(2) + 3| = 4(2) – 3    and      |(0) + 3| = 4(0) – 3 
|5| = 8 – 3                            |3| = –3 
5 = 5                                    3 = –3 

Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive. 

(2) SUFFICIENT: Here, again, we must consider the two possible solutions for the absolute value expression. They are: 

x + 1 = +(2x – 1)       and      x + 1 = –(2x – 1) 
x + 1 = 2x – 1                       x + 1 = –2x + 1 
2 = x                                    3x = 0 
                                            x = 0 

Once again, we need to verify that both solutions are valid. We need to plug x = 2 and x = 0 into the original equation: 

|(2) + 1| = 2(2) – 1    and      |(0) + 1| = 2(0) – 1 
|3| = 4 – 1                            |1| = –1 
3 = 3                                    1 = –1 

One again, 2 is the only valid solution and we can determine that x is positive. 

The correct answer is D. (OE: Manhattan Prep)
_________________

Everything You Need To Know About Verb Tenses

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47981
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1  [#permalink]

Show Tags

New post 03 Aug 2018, 00:05
2
dabaobao wrote:
dabaobao wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1



(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. 
We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are: 

x + 3 = +(4x – 3)       and      x + 3 = –(4x – 3) 
x + 3 = 4x – 3                       x + 3 = –4x + 3 
6 = 3x                                  5x = 0 
2 = x                                    x = 0 

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them. 

|(2) + 3| = 4(2) – 3    and      |(0) + 3| = 4(0) – 3 
|5| = 8 – 3                            |3| = –3 
5 = 5                                    3 = –3 

Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive. 

(2) SUFFICIENT: Here, again, we must consider the two possible solutions for the absolute value expression. They are: 

x + 1 = +(2x – 1)       and      x + 1 = –(2x – 1) 
x + 1 = 2x – 1                       x + 1 = –2x + 1 
2 = x                                    3x = 0 
                                            x = 0 

Once again, we need to verify that both solutions are valid. We need to plug x = 2 and x = 0 into the original equation: 

|(2) + 1| = 2(2) – 1    and      |(0) + 1| = 2(0) – 1 
|3| = 4 – 1                            |1| = –1 
3 = 3                                    1 = –1 

One again, 2 is the only valid solution and we can determine that x is positive. 

The correct answer is D. (OE: Manhattan Prep)


This question has MUCH easier solution.

Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) \(|x + 1| = 2x – 1\). The same here. LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

Answer: D.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1 &nbs [#permalink] 03 Aug 2018, 00:05
Display posts from previous: Sort by

Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.