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# If a and b are non-zero numbers, is a > b ?

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Director
Joined: 02 Oct 2017
Posts: 728
If a and b are non-zero numbers, is a > b ?  [#permalink]

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15 Nov 2018, 04:54
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Difficulty:

85% (hard)

Question Stats:

37% (01:29) correct 63% (01:22) wrong based on 92 sessions

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If a and b are non-zero numbers, is a > b ?

(1) |a| = b

(2) |b|*a < |a*b|

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Manager
Joined: 02 Aug 2015
Posts: 117
Re: If a and b are non-zero numbers, is a > b ?  [#permalink]

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15 Nov 2018, 09:54
push12345 wrote:
If a and b are non-zero numbers, is a > b ?

(1) |a| = b

(2) |b|*a < |a*b|

Statement 1 - There are two cases possible: a is negative or a is positive. Either way A is not greater than B. Sufficient.

Statement 2 - A is negative, else the given inequality will fail. B can either be positive or negative, so statement not sufficient to answer.

Cheers!
Director
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
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Location: India
If a and b are non-zero numbers, is a > b ?  [#permalink]

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16 Nov 2018, 01:59
Hi,

Given,

“a” and “b” are non-zero numbers.

Question:

Is a > b?

Definition of modulus:

|x| = x, if x ≥ 0

|x| = -x, if x < 0

Statement I is sufficient:

|a| = b

a = b, if a > 0

or

a = -b, if a < 0

In either case, “a” is less than or equal to “b”.

So, the answer to the question is always NO.

So sufficient.

Statement II is insufficient:

|b|*a < |a*b|

Modulus is a non-negative function.

So, right hand side(|a*b|) is always is positive.

So, the statement hold true is, only if “a” is negative.

But we don’t know anything about “b”, it could be either positive or negative.

If a = -2 and b = -3, then “a” is greater than “b”, so answer to the question is YES.

If a = -2 and b = 3, then “a” is less than “b”, so answer to the question is NO.

So insufficient.

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Re: If a and b are non-zero numbers, is a > b ?  [#permalink]

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16 Nov 2018, 05:06
push12345 wrote:
If a and b are non-zero numbers, is a > b ?

(1) |a| = b

(2) |b|*a < |a*b|

$$a,b\,\, \ne 0\,\,\,\left( * \right)$$

$$a\,\,\mathop > \limits^? \,\,b$$

$$\left( 1 \right)\,\,\,\left| a \right|\,\,\, = b\,\,\,\, \Rightarrow \,\,\,\,\,\,a\,\,\mathop > \limits^? \,\,\left| a \right|\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\left[ {\,x \le \left| x \right|\,\,{\rm{for}}\,\,{\rm{all}}\,\,{\rm{values}}\,\,{\rm{of}}\,\,x\,} \right]$$

$$\left( 2 \right)\,\,a\left| b \right|\,\, < \,\,\left| {ab} \right| = \left| a \right| \cdot \left| b \right|\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left| b \right|\,\, > \,\,0\,\,\left( * \right)} \,\,\,a < \left| a \right|\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a < 0\,\,\,$$

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

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

Regards,
Fabio.
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Intern
Joined: 22 Aug 2018
Posts: 10
Re: If a and b are non-zero numbers, is a > b ?  [#permalink]

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24 Nov 2018, 01:45
push12345 wrote:
If a and b are non-zero numbers, is a > b ?

(1) |a| = b

(2) |b|*a < |a*b|

Hi guys,

You all point out that the problem with statement 2 is that we can't say if b is positive or not.

In my opinion this is not the only problem, because if for example a is -2 (we know for sure that a is negative) and |b| is 1 we could say that a is always < b

So the problem with statement 2 is not only related to sign of b

Statement 1 is clearly sufficient.

Ans: A
Re: If a and b are non-zero numbers, is a > b ? &nbs [#permalink] 24 Nov 2018, 01:45
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