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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, 10:24
Awesomely explained by @frozzy
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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?
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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] 27 Apr 2015, 21:42

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