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

(1) a + b > 0
(2) a - b > 0
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Re: GMAT PREP DS  [#permalink]

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New post 24 Aug 2010, 18:00
53
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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.

Answer: E.
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Re: GMAT PREP DS  [#permalink]

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New post 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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Re: GMAT PREP DS  [#permalink]

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New post 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.

Answer: E.

you've made things easier bunuel. +1
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Re: GMAT PREP DS  [#permalink]

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New post 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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Re: GMAT PREP DS  [#permalink]

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New post 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: GMATPREP - Please explain my flaw  [#permalink]

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New post 14 Apr 2011, 10:59
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sonnco wrote:
ax + b = 0, is x > 0?
1. a + b > 0
2. a - b > 0
How is the OA answer right?! Your input is greatly appreciated.


\(ax+b=0\)
\(ax=-b\)
\(x=-\frac{b}{a}\)

\(Is \hspace{3} x > 0\)
\(Is \hspace{3} -\frac{b}{a} > 0\)
\(Is \hspace{3} \frac{b}{a} < 0\)

OR

Does a and b have opposite signs?

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

It doesn't tell us anything about sign.
a=5
b=2
a>-b as 5>-2
Same sign. No.

a=5
b=-2
a>-b as 5>2
Opposite signs. Yes.

Not Sufficient.

2. \(a - b > 0\)
\(a>b\)

It doesn't tell us anything about sign.
a=5
b=2
a>b as 5>2
Same sign. No.

a=5
b=-2
a>b as 5>-2
Opposite signs. Yes.

Combining both;

a > |b|

Using same sample set:
a=5, b=2, No.
a=5, b=-2, Yes.
Not Sufficient.

Ans: "E"
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Re: GMATPREP - Please explain my flaw  [#permalink]

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New post 14 Apr 2011, 11:27
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sonnco wrote:
ax + b = 0, is x > 0?

1. a + b > 0
2. a - b > 0



You can look at this geometrically - because the equation is that of a line, the question is set up so that you can look at it as a coordinate geometry question. The equation y = ax + b is the equation of a line with slope a and y-intercept b. When we plug in y=0, we are finding the x-intercept of the line. So the question is just asking if the x-intercept of the line y=ax + b is positive. In a diagram, that would be true if our line crosses the x-axis to the right of the origin (0,0).

Now, using both statements, if a > -b and a > b, then a > |b|. So a is clearly positive, and our line has a positive slope and is climbing as you move to the right. We can now just draw two different scenarios on the coordinate plane, one where b is positive and one where b is negative. If you choose, say, b = 1 and a = 5, then if you quickly sketch your line you'll see that the x-intercept of y = ax + b is negative, and if b=-1 and a = 5 then if you quickly sketch your line you'll see that the x-intercept of y=ax+b is positive, so not sufficient, and the answer is E.
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Re: GMATPREP - Please explain my flaw  [#permalink]

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New post 14 Apr 2011, 22:08
(1) Let a = 2, b = 6


2 * -3 + 6 = 0, x = -3 < 0

Let a = -2, b = 6

-2 * 3 + 6 = 0, x = 3 > 0


(1) is insufficient

(2)

Let a = 6 b = -3

6 * 1/2 - 3 = 0, x = 1/2 > 0


Let a = -3 b = -4

-3 * -4/3 - 4 = 0, x = -4/3 < 0

(2) is insufficient


(1) and (2)

2a > 0 => a > 0

So b can be -ve or +ve and |a| > |b|

So x can be +ve or -ve so that ax + b = 0

Let a = 6, b = 5 or -5

6 * -5/6 + 5 = 0, x < 0

6 * 5/6 - 5 = 0, x > 0

Answer - E
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Re: GMATPREP - Please explain my flaw  [#permalink]

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New post 14 Apr 2011, 22:46
sonnco wrote:
ax + b = 0, is x > 0?

1. a + b > 0
2. a - b > 0


x = -b/a. The problem boils down to. ---- Do we know sign of a and b??

1) + 2).
a + b > 0
a - b > 0
Adding we get a > 0. Also absolute value of a is greater than absolute value of b. However nothing can be concluded about the sign of b. Hence E.
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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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New post 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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New post 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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Re: GMAT PREP DS  [#permalink]

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New post 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.

Answer: E.


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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Re: GMAT PREP DS  [#permalink]

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New post 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.

Answer: E.


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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New post 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 < .
Can Somebody please explain this
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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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New post 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 < .
Can Somebody please explain this


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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New post 29 Nov 2014, 16:42
1
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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New post 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?
some please help...
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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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New post 01 Dec 2014, 03:54
1
1
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?
some please help...
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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New post 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 &nbs [#permalink] 09 Apr 2015, 20:38

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