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If ab ≠ 0, is ab > a/b ?

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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 05 Jan 2016, 21:45
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If ab ≠ 0, is ab > a/b ?

(1) |b| > 1

(2) ab + a/b > 0

Modify the original condition and the question and multiply b^2 on the both equations, which becomes ab^3>ab? --> ab^3-ab>0? -->ab(b^2-1)>0?. There are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) &2), |b|>1 -> b^2>1, ab and a/b get to have the same sign in 2). That is, ab>0 -> a/b>0(this is divided by b^2 and as b^2 is a positive number, direction of the sign doesn’t change.) So, 2) becomes ab>0, which is yes and sufficient. Therefore, the answer is C.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 09 Jan 2016, 09:01
Bunuel wrote:
If ab ≠ 0, is ab > a/b ?

Is \(ab > \frac{a}{b}\)? --> is \(ab - \frac{a}{b}>0\)? --> \(\frac{a(b^2-1)}{b}>0\)? --> \(\frac{a}{b}*(b^2-1)>0\)?

(1) |b| > 1 --> both sides are non-negative, thus we can square: \(b^2>1\) --> \(b^2-1>0\). We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) \(b^2-1>0\) and \(\frac{a}{b} > 0\), thus their product \(\frac{a}{b}*(b^2-1)>0\). Sufficient.

Answer: C.

Hope it's clear.


\(ab > \frac{a}{b}\)

Could someone explain if instead of moving the a/b to the left side, we multiple by b on both sides?

\(ab > \frac{a}{b}\) Multiple by b on both sides
\(ab^2 > a\) Then divide by a
\(b^2 > 1\)

I assume this operation is not allowed, by can someone explain to me why? Could it be because we do not know the sign of b?
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 09 Jan 2016, 09:13
ZaydenBond wrote:


Could someone explain if instead of moving the a/b to the left side, we multiple by b on both sides?

\(ab > \frac{a}{b}\) Multiple by b on both sides
\(ab^2 > a\) Then divide by a
\(b^2 > 1\)

I assume this operation is not allowed, by can someone explain to me why? Could it be because we do not know the sign of b?


Its because of the most important rule in inequalities. You need to reverse the sign of the inequality when you multiple by a negative quantity.

In the question at hand, you do not know whether b>0. If it is, then yes you can multiply the inequality by 'b' but if b<0 then you must reverse the sign of inequalilty. You can also remember this rule by the following example.

You know 2>1 but what about -2 and -1? -2<-1 or in the other words, when you are given 2>1 and you want to multiple throughout by -1, you get -2 < -1 (with reversed sign).

This is the reason why it is much less time consuming to bring a/b to the other side than to multiply throughout by 'b' as you do not know the sign of 'b'

Hope this helps.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 02 Feb 2017, 16:49
gmatquant25 wrote:
If \(ab ≠ 0\), is \(ab > \frac{a}{b}\)?

(1) \(|b| > 1\)

(2) \(ab + \frac{a}{b} > 0\)


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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Re: If ab ≠ 0, is ab > a/b ? &nbs [#permalink] 24 Sep 2018, 06:54

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