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

 It is currently 23 Mar 2019, 22:06 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # 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

Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

meltedcheese wrote:
Would this approach be wrong?

Choose two numbers: a=3, b=2.
a+b>0 and a-b>0 are individually and together fulfilled.

ax+b= 3*x+2=0 => x=-2/3 < 0 => we have shown that both I and II are insufficient.

To get insufficiency you should get both a NO and an YES answers to the question. You have a NO answer, so to get that the statements are insufficient, you should choose numbers which give an YES answer to the question.
_________________
Manager  Joined: 04 Jan 2014
Posts: 78
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

I am getting 2 different equations here.. Please help where I am going wrong..

ax + b = 0

So X=-b/a

1) a+b>0

So, a>-b--- for a>0

1>-b/a
1>x
or -a>-b--- for a<0
1<-x
x<-1

2) a-b>0

a>b---- for a>0
1<b/a
x<-1

-a>b--- for a<0
1<-b/a
1<x
e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2722
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

1
1
Dear sheolokesh

The mistake you made was at the part highlighted in red:

sheolokesh wrote:

1) a+b>0

So, a>-b--- for a>0

1>-b/a
1>x

or -a>-b--- for a<0
1<-x
x<-1

Correct processing of the case a < 0 would be as follow:

From Statement 1,

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

Case: a < 0

Dividing both sides of an inequality with a negative number changes the sign of inequality.

So, dividing both sides of Inequality 1 with a, we get:

1 < $$\frac{-b}{a}$$

Substituting -b/a = x, we get:

1 < x
That is, x > 1

So, from Statement 1, we see that

If a > 0, then x < 1
And, if a < 0, then x > 1

So, we cannot say for sure if x is positive or not.

Similarly, in your analysis of Statement 2, you got confused between the impact of a being positive or negative on the sign of inequality:

sheolokesh wrote:

2) a-b>0

a>b---- for a>0
1<b/a Correct expression: 1 > b/a
x<-1 Correct expression: 1 > -x
Correct expression: -1 < x, that is, x > -1

-a>b--- for a<0 Correct expression: a > b
1<-b/a Correct expression: 1 < b/a (multiplying both sides by negative number changes sign)
1<x Correct expression: 1 < -x
Correct expression: -1 > x, that is, x < -1 (multiplying both sides by negative number changes sign)

It is okay to make mistakes as long as we learn from them. And, the important takeaway from our discussion of this mistake is:

Don't skip steps when multiplying or dividing terms on both sides of an inequality. Because, this step is particularly prone to errors.

Hope our discussion was helpful! Best Regards

Japinder
_________________  Number Properties:- Get 5 free video lessons, 50 practice questions | Algebra:-Get 4 free video lessons, 40 practice questions
Quant Workshop:- Get 100 practice questions | Free Strategy Session:- Key strategy to score 760

Success Stories
Q38 to Q50- Interview call form Wharton | Q35 to Q50 | More Success Stories

Ace GMAT
Articles and Questions to reach Q51 | Question of the week | Tips From V40+ Scorers | V27 to V42:- GMAT 770 | Guide to Get into ISB-MBA

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Manager  B
Joined: 27 Aug 2014
Posts: 73
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

1
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.

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.
_________________
Intern  Joined: 29 Sep 2013
Posts: 6
If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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.

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
Intern  B
Joined: 30 Aug 2016
Posts: 5
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

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 )
Director  S
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 542
Location: India
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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
_________________

GMAT Mentors Manager  S
Joined: 03 Jan 2017
Posts: 145
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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

Intern  B
Joined: 05 Sep 2016
Posts: 13
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

1
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.

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 $$x>1$$.
_________________
Intern  B
Joined: 09 May 2016
Posts: 43
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

1
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.

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.
_________________
BSchool Forum Moderator P
Joined: 05 Jul 2017
Posts: 505
Location: India
GMAT 1: 700 Q49 V36 GPA: 4
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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

_________________
Manager  B
Joined: 27 Aug 2014
Posts: 73
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

Hi Bunuel

Not very sure how you got x>1 region on the numberline.

From stmt 1, I got a-ax>0 or a(1-x)>0 or a(x-1)<0 and on the number line ---(-)----1-----(+)-------- ie x<1.

Pls clarify.

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.

Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

sinhap07 wrote:
Hi Bunuel

Not very sure how you got x>1 region on the numberline.

From stmt 1, I got a-ax>0 or a(1-x)>0 or a(x-1)<0 and on the number line ---(-)----1-----(+)-------- ie x<1.

Pls clarify.

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.

$$a-ax>0$$ --> $$a(1-x)>0$$

So, a and 1-x have the same sign:

Both are positive: $$a>0$$ and $$1-x>0$$, so $$x<1$$

Both are negative: $$a<0$$ and $$1-x<0$$, so $$x>1$$.
_________________
SVP  D
Joined: 26 Mar 2013
Posts: 2099
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

uzzy12 wrote:
If ax + b = 0, is x > 0

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

(1) a + b > 0

Let a = 2 & b = 1.........2+1 > 0...........2x+1=0....x =-1/2 < 0............Answer is NO

Let a = 2 & b =-1.........2 -1 > 0...........2x - 1=0....x =1/2 > 0............Answer is Yes

Insufficient

(2) a - b > 0

Let a = 2 & b = 1.........2 - 1 > 0...........2x+1=0....x =-1/2 < 0............Answer is NO

Let a = 2 & b =-1.........2 + 1 > 0...........2x - 1=0....x =1/2 > 0............Answer is Yes

Insufficient

Combine 1 &2

Use same points above........No clear answer

Insufficient

Manager  B
Joined: 19 Aug 2016
Posts: 84
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

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.

Hi Bunuel

Why are we doing OR in ur statements?

b=-ax. Hence, a-ax>0 then a(1-x)>0. therefore, a>0 or 1-x >0 so 1>x. Thats all..How can we say x<1?
Math Expert V
Joined: 02 Sep 2009
Posts: 53796
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

zanaik89 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.

Hi Bunuel

Why are we doing OR in ur statements?

b=-ax. Hence, a-ax>0 then a(1-x)>0. therefore, a>0 or 1-x >0 so 1>x. Thats all..How can we say x<1?

We have that $$a(1-x)>0$$, so the product of two multiples a and 1 - x is positive. This to be true either both must be positive OR both must be negative.
_________________
Intern  B
Joined: 30 Nov 2017
Posts: 39
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

### Show Tags

1
x = -b/a

x will be greater than 0 only if a and b are of opposite signs.

(1) a + b > 0
Both a and b can be positive, can be negative, or one positive other negative.

(2) a > b
Again, both a and b can be positive, can be negative, or one positive other negative.

Combining,

Adding the two inequalities, we get a > 0

But if a > 0, then b can still be positive or negative. For example, (a,b) can be (2,-1) or (2,1). So, still not sufficient. Hence, E. Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0   [#permalink] 21 May 2018, 00:59

Go to page   Previous    1   2   3    Next  [ 45 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 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®.  