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

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

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24 Aug 2010, 17:41
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If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0
[Reveal] Spoiler: OA

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24 Aug 2010, 18:00
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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.

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24 Aug 2010, 20:57
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As usual Bunuel's approach is elegant.

Here is how I did it:

x=-b/a; x>0 means a and b have to be different sign. I am usually not good at picking numbers, but in this case if you see a+b>0 and a-b>0 it is clear that 5+2>0 5-2>0 and 5+(-2)>0 5-(-2)>0 hence we cannot say that a and b will have the same sign. Hence E.
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15 Mar 2011, 23:59
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.

you've made things easier bunuel. +1
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17 Mar 2011, 00:09
I've also tried this using numbers:

From (1) we can see that:

let a = 4 b = 6

So a * - 3/2 + b = 0 and x > 0

But if a = -4 and b = 6

then a * 3/2 + 6 = 0 and x < 0

So (1) is not sufficient

From (2) :

If a = 2, b = -2

then a *1 + b = 0 and x > 0

if a = 4 and b = 2 then

a* -1/2 + 2 = 0 whereby x < 0

Combining (1) and (2):

a > 0 , but b may or may not be > 0, so the answer is E
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18 Mar 2011, 14:03
awesome explanations both bunuel and mainhoon --these two approaches would probably serve two different types of students really well. nice work!

there's also a potential blend of the two-- mainhoon's rephrase (are the signs of a and b different) and the stacking/adding of inequalities when evaluating choice c for that same rephrased question (if your head gets turned around when too many numbers are involved)
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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30 Jun 2013, 07:48
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given $$ax=- b$$ then x = $$\frac{-b}{a}$$ question becomes is $$\frac{-b}{a}>0$$

From statement 1

$$a+b > 0$$ so the values for $$(a,b)>> (+,+) , (-,+), (+,-)$$

$$a-b>0$$ so the values for $$(a,b) >> (+,+), (+,-)$$

Combination still 2 cases remain. Hence E

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

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27 Jul 2013, 11:17
fozzzy wrote:
given $$ax=- b$$ then x = $$\frac{-b}{a}$$ question becomes is $$\frac{-b}{a}>0$$

From statement 1

$$a+b > 0$$ so the values for $$(a,b)>> (+,+) , (-,+), (+,-)$$

$$a-b>0$$ so the values for (a,b) >> (+,+), (+,-)

Combination still 2 cases remain. Hence E

Is this method safe?

Actually, all the possible cases are : (+,+) = 7,5 ; (+,-) = 7,-5 ; (-,-) = -5,-7.

Otherwise, IMO it is all good.
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25 Oct 2013, 07:42
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.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

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25 Oct 2013, 08:26
Priya15081 wrote:
Bunuel wrote:
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.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

Why does $$a+b>0$$ imply that either a or b is negative? Why cannot both be positive?
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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12 May 2014, 09:22
I don't understand fossey or bunuel's solution.
Bunuel's soln- I'm kinda confused on how you are marking as both either > and > or < and < .

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

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13 May 2014, 01:16
nivi123 wrote:
I don't understand fossey or bunuel's solution.
Bunuel's soln- I'm kinda confused on how you are marking as both either > and > or < and < .

Hi,

What is the product when 2 positive nos a,b are multiplied. The product is >0
When a positive number and a negative number is multipled the product is <0
When 2 negative nos are multiplied then the product is >0

Now in the given inequality arrived by simplifying the given expression we get that a*(1-x) >0 now this is possible only if either a>0 and (1-x)>0 or a <0 and (1-x) <0

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

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29 Nov 2014, 16:42
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uzzy12 wrote:
If ax + b = 0, is x > 0

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

Guessing numbers helped me.

x = -b/a
Is -b/a > 0?

(1) a > -b
a = 5, b =3 => N
a = 5, b=-3 => Y
NS
(2) a > b
Same numbers as in (1)
NS

(1) + (2)
a > |b|
Same numbers used in (1) and in (2)
NS

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

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30 Nov 2014, 21:24
we have a+b>0
a-b>0
if we add both the equations. a>0
if we subtract both the equations, b>0
then we know both have same signs....
hence can determine, sign of x=-b/a is >0 or not
What is wrong in my explanation?
Bunuel

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

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01 Dec 2014, 03:54
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sudd1 wrote:
we have a+b>0
a-b>0
if we add both the equations. a>0
if we subtract both the equations, b>0
then we know both have same signs....
hence can determine, sign of x=-b/a is >0 or not
What is wrong in my explanation?
Bunuel

For two inequalities, you can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

So, we cannot subtract a+b>0 from a-b>0.

For more check: inequalities-tips-and-hints-175001.html
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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09 Apr 2015, 20:38
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.

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

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10 Apr 2015, 04:19
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.
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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12 Apr 2015, 06:15

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

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

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27 Apr 2015, 21:42
this question is hard.

we should consider/study gmatprep questions carefully because those questions prepresent what gmat test and think.
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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27 Apr 2015, 23:38
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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.

Best Regards

Japinder
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0   [#permalink] 27 Apr 2015, 23:38

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