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

Question

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

(1) |a| = b

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

Diwakar003 · Answer

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.

So answer is A.

Cheers!

KrishnakumarKA1 · Answer

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.

So the answer is A.

fskilnik · Answer

a,b\,\, e 0\,\,\,\left( * ight) a\,\,\mathop > \limits^? \,\,b \left( 1 ight)\,\,\,\left| a ight|\,\,\, = b\,\,\,\, \Rightarrow \,\,\,\,\,\,a\,\,\mathop > \limits^? \,\,\left| a ight|\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{ m{NO}}} ight angle \,\,\,\,\,\,\,\,\,\left \left( 2 ight)\,\,a\left| b ight|\,\, < \,\,\left| {ab} ight| = \left| a ight| \cdot \left| b ight|\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left| b ight|\,\, > \,\,0\,\,\left( * ight)} \,\,\,a < \left| a ight|\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a < 0\,\,\, \left\{ \matrix{ \,{ m{Take}}\,\,\left( {a,b} ight) = \left( { - 1,1} ight)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{ m{NO}}} ight angle \,\, \hfill \cr \,{ m{Take}}\,\,\left( {a,b} ight) = \left( { - 1, - 2} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{ m{YES}}} ight angle \,\, \hfill \cr} ight.\,\,\,\, This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.

Manfo · Answer

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

bumpbot · Answer

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

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

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