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

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

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

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

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"

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?

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.

Given ax + b = 0, hence x = -b/a, & is x > 0?

Statement 1:

a + b > 0

a = 1, b = 0, x = 0
a = -1, b = 2, x = 2 > 0

Statement 1 alone is Not Sufficient.

Statement 2:

a - b > 0

a = 1, b = 0, x = 0
a =1, b = -1, x = 1 > 0

Statement 2 alone is Not Sufficient.

Combining both statements,
a = 1, b = 0, x = 0
a = 2, b = -1, x = 1/2

Combing both statements is Not Sufficient.

Answer E.


Thanks,
GyM

I love this question. Probably, because one can solve it under 1 minute if you are clear on the rules of positive & negative (in inequality).

The question is saying the following: ax + b =0 => b=-ax or => -b/a = x

So the rephrased question is not to check x > 0 but is -b/a > 0. Now this has two implications as follows:-

1. Case I: If a > 0, b < 0 [Note: ',' indicates and]
2. Case II: If a <0, b > 0

So another deep down will give us the following phrasing:- Is Case I or Case II true (If either one of them is true, then you can see we will get our answer to the question)

Now, Statement A: a + b > 0 i.e. we don't the 'sign' of either a or b: both could be positive; one could be large positive, one could be small negative. So clearly, Not Sufficient

Moving on to statement B: a - b > 0 i.e. same as the above statement only with a minor tweak, hence not sufficient

Combining both the statement: Now inequalities can be added in the same direction; therefore, we get the following:-
2a > 0 => a > 0
Do you think this information is enough to answer the question?. The answer is a resounding 'No'. Why is that?
Well, the statement only gives us the sign of a; however, we don't know whether b is either > 0 or less < 0 or even equal to 0.

Kudos, if you liked the solution

Regards,
Abheek

Target question: Is x > 0

Given: ax + b = 0

Statement 1: a + b > 0
At this point, we have 1 equation, 1 inequality, and THREE variables.
Even if we had 2 EQUATIONS and 3 variables, we probably wouldn't be able to make any conclusions about whether x is positive or negative.
Given this, let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: a - b > 0
This is the same scenario as statement 1, so let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

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 \(x>1\).

A video explanation can be found here:
https://www.youtube.com/watch?v=cYPgNTx-XK8&t=5s

For DS questions, always ask "What would I need to know in order to answer this question?" In this case, isolate x:

ax + b = 0

ax = -b

x = -b/a

Whether x is positive depends upon the signs of a and b -- The real question is, "What are the signs of a and b?"

(1) either both numbers are positive, or one is positive and one is negative. Insifficient.

(2) same problem. Insufficient.

Together, if you add the two inequalities you'll have 2a > 0, which means a > 0, but that's only half of what we need to know...

Answer E.

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

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.

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

Solution:

The easiest method to solve a DS question depends on the question. For certain questions (especially for statements which are not sufficient to provide an answer), the question can be easily solved by plugging in numbers. Other questions require you to use algebraic manipulations to arrive at the correct answer. In either case, you can use a combination of both methods; i.e. use algebraic manipulations to get some information about the statements and plug in numbers only after that so that you don’t waste time trying unnecessary cases.

Let me show what I mean on one of the examples you provided.

In the first example, assuming that a is non-zero, we can write x = -b/a. This expression tells us that x is -1 times the ratio of b to a; i.e. if b and a have the same sign then x is negative and if b and a have opposite signs, then x is positive. When plugging in numbers for a and b, this information tells me to only look for examples where a and b have the same and opposite signs, so that if I tried a = 1 and b = -1, I won’t waste time trying a = -1 and b = 1 since I know both will produce a positive x.

Statement One Alone:

For an example where a and b have the same sign, we can simply take a = b = 1. In this case, x + 1 = 0 and hence, x = -1 i.e. x > 0 is not true.

For an example where a and b have opposite signs, we can take a = 2 and b = -1. In this case, 2x - 1 = 0 and hence, x = 1/2 i.e. x > 0 is true.

Statement one alone is not sufficient.

Statement Two Alone:

For an example where a and b have the same sign, we can take a = 2 and b = 1. In this case, 2x + 1 = 0 and hence, x = -1/2 i.e. x > 0 is not true.

For an example where a and b have opposite signs, we can take a = 1 and b = -1. In this case, x - 1 = 0 and hence, x = 1 i.e. x > 0 is true.

Statement two alone is not sufficient.

Statements One and Two Together:

Notice that if a + b > 0 and a - b > 0 are true at the same time, then adding the inequalities side by side, we determine that 2a > 0; i.e. a > 0.

If a = 2 and b = 1, then 2x + 1 = 0 and hence, x = -1/2. Notice that a = 2 and b = 1 satisfy both a + b > 0 and a - b > 0. In this case, x < 0.

If a = 2 and b = -1, then 2x - 1 = 0 and hence, x = 1/2. Notice that a = 2 and b = - 1 satisfy both a + b > 0 and a - b > 0. In this case, x > 0.

Statements one and two together is not sufficient.

Answer: E

One glance at the statements is enough to tell that this question is more about a and b and less about x. The equation given as data just furthers the point.

ax + b = 0. Since the question stem asks if x>0, we keep x on one side of the equation.
x = - \(\frac{b }{ a}\).

Therefore, the question stem can now be rephrased as “Is –\(\frac{b }{ a}\) > 0?”.

-(\(\frac{b}{a}\)) will be positive if (\(\frac{b}{a}\)) is negative; If (\(\frac{b}{a}\)) has to be negative, b and a should be of opposite signs.

The question can be rephrased to “Are a and b of opposite signs?”. You will see that the statements now make a lot more sense because the question is also about the signs of a and b.

From statement I alone, a+b > 0.

Note that a sum or two variables is never enough to tell you a lot about the signs, compared to a product or a division of terms.
At the most, a+b>0 can tell you that both numbers cannot be ZERO and both cannot be negative at the same time. That’s about it!

Statement I alone is insufficient to say if a and b, are of opposite signs. Answer option A and D can be eliminated. Possible answer options are B, C and E.

From statement II alone, a-b>0.

A similar thing can be said about a difference of terms.
At the most, a-b>0 tells us that both numbers are not ZERO. Nothing more, nothing less!

Statement II alone is insufficient to answer the question. Answer option B can be eliminated. Possible answer options are C and E.

Combining statements I and II, we have the following:

From statement I, a + b > 0; from statement II, a – b > 0.

Since both inequalities have the same sign, they can be added. In fact, GMAT expects you to know this property of inequalities and use it to combine statements.

Adding the two inequalities, we have
2a > 0 or a > 0.

The combination of statements tells us that a is positive but provides us with no information about b.
The combination of statements is insufficient. Answer option C can be eliminated.

The correct answer option is E.

Hope that helps!
Aravind B T

Dear sheolokesh

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



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:



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

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

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 \(x>1\).

you've made things easier bunuel. +1