Is |a - b|

Question

Is |a - b| < |a| + |b| ?

(1) ab< 0

(2) a^b < 0

chetan2u · Accepted Answer

Two ways...

1)   Logical inference  
The right side is adding two absolute values so that is the MAX value possible. The Left side has the DISTANCE between the two values. So if they are on  same side of 0 , it will be less and of   on either side of 0  , it will be again MAX possible and thus equal to right hand side.
So   we are looking for the relative signs of a and b, whether they are of Same sign or different  
(1) ab<0
So both a and b are of different signs and thus both sides will be EQUAL
Ans NO
Sufficient
(2)a^b<0
a is surely <0 but B could be anything
Insufficient
A

2)    Algebraic  
Square both sides
 |a+b|^2<(|a|+|b|)^2.......2ab<2|a||b|.......2|a||b|-2ab>0 
When is this possible -  when both a and b are of opposite signs 
Thus   we are looking for the relative signs of a and b, whether they are of Same sign or different  
Rest will be same as above
A

dvishal387 · Answer

(1) ab< 0 => a or b is negative.
If a<0 and b>0 then |-a-b|=a+b=|a+b|
We get a definite answer no to the question so sufficient.

(2) a^b < 0=> a is negative
but we don't know what the sign of b will be so not sufficient
So A is the correct choice.

GMATinsight · Answer

Question  :Is |a - b| < |a| + |b| ?

On the right side of equation we will always see the sum of absolute values of a and b
But on the left hand side the |a - b| to be smaller, it's important to have the same sign because in case of opposite signs of a and b, left side of equation will be same as the right side of the equation

i.e. for the condition of the question to be true it's must for a and b to have same sign therefore,

Question REPHRASED  : Do a and b have the same sign ?

Statement 1 : ab< 0

i.e. a and b have opposite sign only then the product of a and b will be negative hence

SUFFICIENT

Statement 2 : a^b < 0 
 (-1)^1 < 0 
Also,  (-1)^{-1} < 0 
i.e. and b may have the same sign as well as opposite signs hence

NOT SUFFICIENT

Answer: Option A

fskilnik · Answer

\left| {a - b} ight|\,\,\mathop < \limits^? \,\,\,\left| a ight| + \left| b ight|\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * ight)} \,\,\,\,\,ab\,\,\mathop > \limits^? \,\,\,0 (*) This equivalence will be PROVED at the end of this post. Ignore this proof if you don´t like math! \left( 1 ight)\,\,ab < 0\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{ ext{NO}}} ight angle \,\,\,\, \Rightarrow \,\,\,\,{ ext{SUFF}}. \left( 2 ight)\,\,\,{a^b} < 0\,\,\,\left\{ \matrix{ \,{ m{Take}}\,\,\left( {a,b} ight) = \left( { - 1,1} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{ m{NO}}} ight angle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left \hfill \cr \,{ m{Take}}\,\,\left( {a,b} ight) = \left( { - 1, - 1} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{ m{YES}}} ight angle \,\,\,\,\,\,\,\,\,\,\,\,\,\left \,\, \hfill \cr} ight. This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. POST-MORTEM: \left( * ight)\,\,\,\left\{ \matrix{ \,\left( i ight)\,\,\,\,\left| {a - b} ight|\,\, < \,\,\,\left| a ight| + \left| b ight|\,\,\,\,\,\,\, \Rightarrow \,\,\,\,ab > 0 \hfill \cr \,\left( {ii} ight)\,\,\,\,ab > 0\,\,\,\, \Rightarrow \,\,\,\,\left| {a - b} ight|\,\, < \,\,\,\left| a ight| + \left| b ight|\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left| {a - b} ight|\,\, \ge \,\,\,\left| a ight| + \left| b ight|\,\,\,\,\, \Rightarrow \,\,\,\,\,ab \le 0 \hfill \cr} ight.\, \left( i ight)\,\,\,\,\left| {a - b} ight|\,\, < \,\,\,\left| a ight| + \left| b ight|\,\,\,\,\,\mathop \Rightarrow \limits^{{ ext{squaring}}} \,\,\,\,{\left( {a - b} ight)^2} < \,\,\,{a^2} + 2\left| {ab} ight| + {b^2}\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\, - ab < \left| {ab} ight|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,ab > 0 \left( {ii} ight)\,\,\,\,\left| {a - b} ight|\,\, \geqslant \,\,\,\left| a ight| + \left| b ight|\,\,\,\,\,\mathop \Rightarrow \limits^{{ ext{squaring}}} \,\,\,\,{\left( {a - b} ight)^2} \geqslant \,\,\,{a^2} + 2\left| {ab} ight| + {b^2}\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\, - ab \geqslant \left| {ab} ight|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,ab \leqslant 0\,\,\,

MrCleantek · Answer

I got the explanation but I messed up my solution when I removed the modulus with + and - signs for each of them! I mean when we remove modulus then there are 2 possibilities right? |a| => a and -a. Why we did not do this?

GMATinsight · Answer

Hi  MrCleantek

Just a suggestion that I share with many of my students...  Be logical first and use maths second ... So I tend to avoid opening Modulus till it's not needed.

And the need to open modulus will be seen in less than 10% questions of inequality

MrCleantek · Answer

Agreed! I will now remember to not open modulus by default. Thank you..

GMATGuruNY · Answer

Since both sides of the question stem are NONNEGATIVE, we can square the inequality: (|a-b|)^2 < (|a| + |b|)^2 a^2 + b^2 - 2ab < a^2 + b^2 + 2|ab| -ab < |ab| The resulting inequality is true only if ab > 0. Question stem, rephrased: Is ab > 0? Statement 1: The answer to the rephrased question stem is NO. SUFFICIENT. Statement 2: If a=-1 and b=1, the answer to the rephrased question stem is NO. If a=-1 and b=-1, the answer to the rephrased question stem is YES. INSUFFICIENT. A

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Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0

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