It is currently 20 Oct 2017, 15:24

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0

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

Hide Tags

Manager
Manager
avatar
B
Joined: 27 Aug 2014
Posts: 78

Kudos [?]: 2 [0], given: 3

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 23 Jul 2015, 00:19
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel

If I find from the question stem itself that x is negative i.e. x=-b/a, in this case, can we say we dont require any additional statements to prove it and hence, E?

Kudos [?]: 2 [0], given: 3

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129030 [0], given: 12187

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 23 Jul 2015, 02:34
sinhap07 wrote:
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel

If I find from the question stem itself that x is negative i.e. x=-b/a, in this case, can we say we dont require any additional statements to prove it and hence, E?


No.

1. You cannot write x = -b/a from ax + b = 0, because a can be 0 and division by 0 is not allowed.
2. Even if we had x = -b/a, minus sign does not indicate that x is negative. If b/a is negative itself then x = -b/a = -(negative) = positive.

Hope it's clear.
_________________

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

Kudos [?]: 129030 [0], given: 12187

Intern
Intern
avatar
Joined: 29 Sep 2013
Posts: 9

Kudos [?]: 5 [0], given: 11

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 14 Aug 2015, 15:20
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Thanks for your answer Bunuel, great as always. My approach was:

ax + b = 0 implies that either:
ax = negative and b is positive {a = positive, x = negative or x = positive, a = negative}
or
ax = positive and b is negative {a = negative, x = negative or x = positive, a = positive}
or
ax = 0 and b = 0

1) a + b > 0
Insufficient because no conclusion can be reached on sign of a or b

2) a - b > 0
which means that a > b
if a > b, then: b must be negative or zero. this means that we cannot reach conclusion on x.
if b is negative, then ax = positive. This means that a = negative, x = negative or a = positive, x = positive
if b is zero, then ax = 0, which means that x = 0 as a > b and cannot be 0.
x still has wide range of possibilities.

1 + 2, a does not help in answering whether b must be negative or zero or tell us what sign a should be. Thus, question raised in 2 is still not solved. Therefore E

Kudos [?]: 5 [0], given: 11

Manager
Manager
avatar
Joined: 31 Jul 2014
Posts: 147

Kudos [?]: 54 [0], given: 373

GMAT 1: 630 Q48 V29
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 02 Sep 2015, 06:59
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

Kudos [?]: 54 [0], given: 373

Manager
Manager
avatar
Joined: 31 Jul 2014
Posts: 147

Kudos [?]: 54 [0], given: 373

GMAT 1: 630 Q48 V29
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 02 Sep 2015, 07:00
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

Kudos [?]: 54 [0], given: 373

Manager
Manager
avatar
Joined: 31 Jul 2014
Posts: 147

Kudos [?]: 54 [0], given: 373

GMAT 1: 630 Q48 V29
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 02 Sep 2015, 07:01
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

Kudos [?]: 54 [0], given: 373

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129030 [0], given: 12187

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 02 Sep 2015, 07:13
anupamadw wrote:
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking


Please read the whole thread: if-ax-b-0-is-x-0-1-a-b-0-2-a-b-99749.html#p1449499
_________________

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

Kudos [?]: 129030 [0], given: 12187

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 16636

Kudos [?]: 273 [0], given: 0

Premium Member
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 23 Sep 2016, 10:27
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 273 [0], given: 0

Intern
Intern
avatar
B
Joined: 30 Aug 2016
Posts: 5

Kudos [?]: [0], given: 92

Location: Italy
Schools: HEC Dec"18 (I)
GMAT 1: 700 Q47 V41
GPA: 3.83
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 23 Dec 2016, 03:11
Another way:

we know that x = - b/a

Is x>0?

for x to be positive we must know signs of b and a:
if b/a > 0 (i.e. they have the same sign) then x<0
if b/a < 0 (i.e. one pos one neg) then x>0
In both cases we'd be able to answer.

1. a + b > 0. Clearly NS because they can be both positive, both negative or pos/neg
2. a -b > 0. Clearly NS because they can be both positive, both negative or pos/neg

1.+2. We only know that their sum is positive and one is bigger than the other: they can be both positive, both negative or pos/neg

E

(this was my first post, hello GMATClub :) )

Kudos [?]: [0], given: 92

Senior Manager
Senior Manager
avatar
B
Joined: 05 Jan 2017
Posts: 435

Kudos [?]: 59 [0], given: 15

Location: India
Premium Member
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 23 Feb 2017, 05:59
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Prompt analysis
ax+b=0 or x =-b/a

Super set
The answer will be either yes or no.

Translation
In order to find the exact answer, we need:
1# exact value of a and b.
2# any equation to find the a and b
3# any equation to find the exact or the range for -b/a

Statement analysis
St 1: a +b >0 or -b/a<1 or x <1, therefore x can be x can be greater than 0 or less than 0. INSUFFICIENT
St 2: a -b >0 or b/a<1 or -x<1 or x>-1.therefore x can be x can be greater than 0 or less than 0. INSUFFICIENT

St 1 & st2: -1 < x < 1. therefore x can be x can be greater than 0 or less than 0. INSUFFICIENT
Option E

Kudos [?]: 59 [0], given: 15

Manager
Manager
avatar
S
Joined: 03 Jan 2017
Posts: 197

Kudos [?]: 9 [0], given: 4

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 28 Mar 2017, 10:07
from the beginning we can make up an equation to x=-b/a
1)a+b>0 or a-ax>0. a(1-x)>0 not enough data
2) a-b>0 or a-ax>0. a(1+x)>0 not enough data
1+2) let's sum the equation, because they have the same signs. a>0, from this point we can try the other numbers:
1-x>0, x<1
1+x>0, x>-1
this gives us no answer if x is positive or negative

Answer is E

Kudos [?]: 9 [0], given: 4

Intern
Intern
avatar
B
Joined: 05 Sep 2016
Posts: 17

Kudos [?]: [0], given: 10

GMAT ToolKit User
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 01 Apr 2017, 09:50
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) -->either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel I have a doubt in equality:
a>0 and 1−x>0, so x<1 OR a<0 and 1−x<0, so x>1. not able to get how did a>0 turned to a<0 and x<1 to 1-x<0

Kudos [?]: [0], given: 10

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129030 [1], given: 12187

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 01 Apr 2017, 10:05
1
This post received
KUDOS
Expert's post
jamescath wrote:
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) -->either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel I have a doubt in equality:
a>0 and 1−x>0, so x<1 OR a<0 and 1−x<0, so x>1. not able to get how did a>0 turned to a<0 and x<1 to 1-x<0


We have \(a(1-x)>0\). For a product of two numbers to be positive they must have the same sign. Two scenarios:

1. Both are positive:
\(a>0\) and \(1-x>0\).
\(1-x>0\) is the same as \(x<1\).

2. Both are negative:
\(a<0\) and \(1-x<0\).
\(1-x<0\) is the same as \(x>1\).
_________________

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

Kudos [?]: 129030 [1], given: 12187

Intern
Intern
avatar
B
Joined: 09 May 2016
Posts: 47

Kudos [?]: 10 [0], given: 3

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 17 Jun 2017, 03:03
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.



Bunuel Please help me in understanding this by (1)+(2) we came up to \(a>0\), now from question we have \(-b=ax\) which means product of ax is negative and by that logic would not x be <0 as we have got \(a>0\).... Please help me in understanding flaw in my logic

Kudos [?]: 10 [0], given: 3

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129030 [0], given: 12187

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 17 Jun 2017, 04:04
KARISHMA315 wrote:
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.



Bunuel Please help me in understanding this by (1)+(2) we came up to \(a>0\), now from question we have \(-b=ax\) which means product of ax is negative and by that logic would not x be <0 as we have got \(a>0\).... Please help me in understanding flaw in my logic


The highlighted part is not correct.

From \(b=-ax\) knowing that \(a>0\), does not necessarily mean that x is negative. Consider x = -1/2 and x = 1/2. Notice that we know nothing about b, we don't know whether it's positive or negative.
_________________

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

Kudos [?]: 129030 [0], given: 12187

Manager
Manager
avatar
S
Joined: 05 Jul 2017
Posts: 153

Kudos [?]: 30 [0], given: 162

Location: India
Concentration: Entrepreneurship, Technology
GMAT 1: 700 Q49 V36
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

New post 31 Aug 2017, 23:34
Hey Bunuel

I just wanted to check if my approach is correct. Can you let me know? Below is how I solved it

Given:-

Statement 1:-
a+b>0
a-ax>0
a(1-x)>0
i. Either a>0 , then x can be >0 or <0
ii. OR x<1 , x can be <0 or >0
iii. OR a>0 and x<1, we still can't say anything about about x

Insufficient


Statement 2:-
a-b>0
a+ax>0
a(1+x)>0
i. Either a>0 , then x can be >0 or <0
ii. OR x>-1 , x can be <0 or >0
iii. OR a>0 and x>-1, we still can't say anything about about x

If I combine both statements, I can't conclude anything about x

Answer E

Kudos [?]: 30 [0], given: 162

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0   [#permalink] 31 Aug 2017, 23:34

Go to page   Previous    1   2   [ 36 posts ] 

Display posts from previous: Sort by

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0

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


GMAT Club MBA Forum Home| About| Terms and Conditions| 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®.