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Is |a| > |b|? (1) 1/(a − b) > 1/(b − a) (2) a + b < 0

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Bunuel
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Is |a| > |b|? (1) 1/(a − b) > 1/(b − a) (2) a + b < 0 [#permalink] New post   15 Jan 2021, 02:17
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A
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C
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E

Difficulty:

95% (hard)

Question Stats:

38% (02:25) correct 62% (01:59) wrong based on 109 sessions
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Is |a| > |b|?

(1) 1/(a − b) > 1/(b − a)

(2) a + b < 0
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Official Answer
C
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MaryKaaviya
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Re: Is |a| > |b|? (1) 1/(a − b) > 1/(b − a) (2) a + b < 0 [#permalink] New post   16 Jan 2021, 07:30
Answer is C

1) 1/(a-b)>1/(b-a) ==> 2/(a-b)>0

a-b>0 insuff

2) a+b<0
insuff

Combining ==>
case 1: a,b <0
|a|<|b|
case 2: b<0
|a|<|b|

sufficient (C)
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Re: Is |a| > |b|? (1) 1/(a − b) > 1/(b − a) (2) a + b < 0 [#permalink] New post   19 Jan 2021, 10:20
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Bunuel wrote:
Is |a| > |b|?

(1) 1/(a − b) > 1/(b − a)
\(\frac{1}{a-b}-\frac{1}{b-a}>0\)
\(\frac{1}{a-b}-\frac{1}{-(a-b)}>0\)
\(\frac{1}{a-b}+\frac{1}{a-b}>0\)
\(\frac{2}{a-b}>0\)
So a-b>0
Insufficient

(2) a + b < 0
Insufficient


Combined
a-b>0 and a+b<0
So, (a-b)(a+b)<0.......\(a^2-b^2<0.......a^2<b^2...........|a|<|b|\)
Thus, the answer is NO
Sufficient

C


Given statement 1:
\(\frac{1}{a-b}>\frac{1}{b-a}\)

can't we re-arrange the above equation as
(b-a)>(a-b)
2b>2a
b>a
As b>a, which can be used to answer the problem statement i.e. |a| > |b|
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Is |a| > |b|? (1) 1/(a − b) > 1/(b − a) (2) a + b < 0 [#permalink] New post   19 Jan 2021, 17:00
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Hi santosh93 ,

This is a common mistake. This exact expression is a bit special but in general, when you see variables in inequalities we have to be cautious. Here what you did was "multiply both sides by (a - b) then (b - a)" but we don't know if these two terms are positive or negative. We wouldn't know if the inequality should result in ">" or "<", thus we wouldn't be allowed to "move" these terms without knowing the signs.

However, we do know (a - b) and (b - a) have OPPOSITE signs so no matter what the result is \(b - a < a - b\) for this expression and we get \(b < a\) (this is still insufficient alone).

Hope that helps!
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Re: Is |a| > |b|? (1) 1/(a − b) > 1/(b − a) (2) a + b < 0 [#permalink] New post   19 Jan 2021, 17:09
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Bunuel wrote:
Is |a| > |b|?

(1) 1/(a − b) > 1/(b − a)

(2) a + b < 0


Statement 1:

Normally we are not allowed to move variables around in an inequality using multiplication/division but \(a - b\) and \(b - a\) have opposite signs so no matter what this simplifies to \(b - a < a - b\) and \(b < a\). This is still insufficient as we can have \(-5 < 3\) vs \(-1 < 3\).

Statement 2:

(a, b) = (3, -5) vs (a, b) = (-3, -1). Insufficient.

Combined:

We may write statement 2 as \(a < -b\). Combining \(b < a\) from statement 1 we have \(b < a < -b\) which means b must be negative. Since \(a\) is between both b's, \(a\)'s magnitude must be less than that of b, which we can translate to \(|a| < |b|\). Sufficient.

Ans: C
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Re: Is |a| > |b|? (1) 1/(a b) > 1/(b a) (2) a + b < 0 [#permalink] New post   29 Mar 2024, 01:58
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Re: Is |a| > |b|? (1) 1/(a b) > 1/(b a) (2) a + b < 0 [#permalink]
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