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

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Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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Updated on: 15 Oct 2018, 07:21
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Question Stats:

49% (01:44) correct 51% (02:10) wrong based on 116 sessions

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

(1) ab< 0

(2) a^b < 0

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Originally posted by push12345 on 15 Oct 2018, 07:14.
Last edited by Bunuel on 15 Oct 2018, 07:21, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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15 Oct 2018, 08:40
(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.
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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15 Oct 2018, 10:29
push12345 wrote:
Is |a - b| < |a| + |b| ?

(1) ab< 0

(2) a^b < 0

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

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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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15 Oct 2018, 11:52
push12345 wrote:
Is |a - b| < |a| + |b| ?

(1) ab< 0

(2) a^b < 0

$$\left| {a - b} \right|\,\,\mathop < \limits^? \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,ab\,\,\mathop > \limits^? \,\,\,0$$

(*) This equivalence will be PROVED at the end of this post. Ignore this proof if you don´t like math!

$$\left( 1 \right)\,\,ab < 0\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

$$\left( 2 \right)\,\,\,{a^b} < 0\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {a,b} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,\,{{\left( { - 1} \right)}^1} = - 1\,\,\,} \right] \hfill \cr \,{\rm{Take}}\,\,\left( {a,b} \right) = \left( { - 1, - 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,\,{{\left( { - 1} \right)}^{ - 1}} = {1 \over {{{\left( { - 1} \right)}^1}}} = - 1\,\,\,} \right]\,\, \hfill \cr} \right.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

POST-MORTEM:

$$\left( * \right)\,\,\,\left\{ \matrix{ \,\left( i \right)\,\,\,\,\left| {a - b} \right|\,\, < \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\,\, \Rightarrow \,\,\,\,ab > 0 \hfill \cr \,\left( {ii} \right)\,\,\,\,ab > 0\,\,\,\, \Rightarrow \,\,\,\,\left| {a - b} \right|\,\, < \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left| {a - b} \right|\,\, \ge \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\, \Rightarrow \,\,\,\,\,ab \le 0 \hfill \cr} \right.\,$$

$$\left( i \right)\,\,\,\,\left| {a - b} \right|\,\, < \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{squaring}}} \,\,\,\,{\left( {a - b} \right)^2} < \,\,\,{a^2} + 2\left| {ab} \right| + {b^2}\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\, - ab < \left| {ab} \right|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,ab > 0$$

$$\left( {ii} \right)\,\,\,\,\left| {a - b} \right|\,\, \geqslant \,\,\,\left| a \right| + \left| b \right|\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{squaring}}} \,\,\,\,{\left( {a - b} \right)^2} \geqslant \,\,\,{a^2} + 2\left| {ab} \right| + {b^2}\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\, - ab \geqslant \left| {ab} \right|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,ab \leqslant 0\,\,\,$$
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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15 Oct 2018, 22:43
GMATinsight wrote:
push12345 wrote:
Is |a - b| < |a| + |b| ?

(1) ab< 0

(2) a^b < 0

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

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?
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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16 Oct 2018, 04:43
1
MrCleantek wrote:
GMATinsight wrote:
push12345 wrote:
Is |a - b| < |a| + |b| ?

(1) ab< 0

(2) a^b < 0

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

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?

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
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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17 Oct 2018, 23:08
GMATinsight wrote:

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

Agreed! I will now remember to not open modulus by default. Thank you..
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0  [#permalink]

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17 Oct 2018, 23:41
2
2
push12345 wrote:
Is |a - b| < |a| + |b| ?

(1) ab< 0

(2) a^b < 0

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
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Re: Is |a - b| < |a| + |b| ? (1) ab< 0 (2) a^b < 0   [#permalink] 17 Oct 2018, 23:41
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