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

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GMAT 1: 660 Q46 V35
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Is |a-b| < |a| +|b| ?  [#permalink]

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New post 03 Aug 2017, 08:35
4
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (01:44) correct 40% (02:02) wrong based on 158 sessions

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

Statement 1: \(\frac{a}{b}\)<0
Statement 2: \(a^2b\)<0
Magoosh GMAT Instructor
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Re: Is |a-b| < |a| +|b| ?  [#permalink]

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New post 03 Aug 2017, 09:52
1
DH99 wrote:
Is |a-b| < |a| +|b| ?

Statement 1: \(\frac{a}{b}\)<0
Statement 2: \(a^2b\)<0

Dear DH99,

I'm happy to respond. :-)

This is a very clever question. Let's think about the prompt.

If a & b have the same sign, then subtracting produces a difference of a smaller absolute value.
\(5 - 3 = 2
(-5) - (-3) = -2\)
These are examples of choices that would produce "yes" answers for the prompt question.

If a & b have opposite signs, the subtracting produces a difference that has the same absolute value as the sum of the absolute values of a & b separately.
\(5 - (-3) = 8\) and \(|8| = 8 = |5| + |-3|\)
\((-5) - 3 = -8\) and \(|-8| = 8 = |-5| + |3|\)
These are examples of choices that would produce "no" answers for the prompt question.

So really, the prompt question is: do a and b have the same sign?

Statement #1: \(\frac{a}{b}\)<0
When is a fraction negative? It's negative when the numerator and denominator have opposite signs. Thus, a & b have opposite signs. We can give a definitive "no" to the prompt question. Because we can give a definitive answer, we know that Statement #1, alone and by itself, must be sufficient.

Statement #2: \(a^2b\)<0
I assume that this was entered correctly, and that DH99 meant \(a^2b\) and not \(a^{2b}\).

Taking this statement at face value, we know \(a^2\) is always positive, regardless of the sign of a, so we know that b just be negative. Thus, we know the sign of b, but the sign of a could be anything. We are not able to give a definitive answer to the prompt. Thus, we know that Statement #2, alone and by itself, must be insufficient.

OA = (A)

Does all this make sense?
Mike :-)
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Re: Is |a-b| < |a| +|b| ?  [#permalink]

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New post 03 Aug 2017, 09:59
1
mikemcgarry wrote:
DH99 wrote:
Is |a-b| < |a| +|b| ?

Statement 1: \(\frac{a}{b}\)<0
Statement 2: \(a^2b\)<0

Dear DH99,

I'm happy to respond. :-)

This is a very clever question. Let's think about the prompt.

If a & b have the same sign, then subtracting produces a difference of a smaller absolute value.
\(5 - 3 = 2
(-5) - (-3) = -2\)
These are examples of choices that would produce "yes" answers for the prompt question.



If a & b have opposite signs, the subtracting produces a difference that has the same absolute value as the sum of the absolute values of a & b separately.
\(5 - (-3) = 8\) and \(|8| = 8 = |5| + |-3|\)
\((-5) - 3 = -8\) and \(|-8| = 8 = |-5| + |3|\)
These are examples of choices that would produce "no" answers for the prompt question.

So really, the prompt question is: do a and b have the same sign?

Statement #1: \(\frac{a}{b}\)<0
When is a fraction negative? It's negative when the numerator and denominator have opposite signs. Thus, a & b have opposite signs. We can give a definitive "no" to the prompt question. Because we can give a definitive answer, we know that Statement #1, alone and by itself, must be sufficient.

Statement #2: \(a^2b\)<0
I assume that this was entered correctly, and that DH99 meant \(a^2b\) and not \(a^{2b}\).

Taking this statement at face value, we know \(a^2\) is always positive, regardless of the sign of a, so we know that b just be negative. Thus, we know the sign of b, but the sign of a could be anything. We are not able to give a definitive answer to the prompt. Thus, we know that Statement #2, alone and by itself, must be insufficient.

OA = (A)

Does all this make sense?
Mike :-)


Yes,thanks mike
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Re: Is |a-b| < |a| +|b| ?  [#permalink]

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New post 24 Aug 2017, 10:03
1
Clearly from A we are getting definite NO.
See the attached pic.
Attachments

IMG_2982.JPG
IMG_2982.JPG [ 905.22 KiB | Viewed 1087 times ]


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Is |a-b| < |a| +|b| ?  [#permalink]

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New post 24 Aug 2017, 11:06
1
DH99 wrote:
Is |a-b| < |a| +|b| ?

Statement 1: \(\frac{a}{b}\)<0
Statement 2: \(a^2b\)<0


Statement 1: implies that either \(a<0\) or \(b<0\)
Putting the values of \(a\) & \(b\) in the inequality as per the scenario
if \(a<0\), then \(|a-b| = |-a-b| = |a+b| = |a| +|b|\). so we get a \(NO\) for the question stem
if \(b<0\), then \(|a-b| = |a-(-b)| = |a+b| = |a| + |b|\). so we get a \(NO\) for the question stem
Hence \(Sufficient\)

[b]Statement 2:/b] implies that \(b<0\) but \(a<0\) or \(a>0\)
Putting the values of \(a\) & \(b\) in the inequality as per the scenario
if both \(a<0\) and \(b<0\), then \(|a-b| = |-a-(-b)| = |-a+b|<|a| + |b|\). so we get a \(YES\) for the question stem
but if \(a>0\) and \(b<0\), then \(|a-b| = |a-(-b)| = |a+b| = |a| + |b|\). so we get a \(NO\) for the question stem
Hence \(Insufficient\)

Option \(A\)
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Is |a-b| < |a| +|b| ?  [#permalink]

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New post 12 Dec 2017, 12:49
\(|a-b| < |a| + |b|\)?

squaring on both sides, (as LHS > 0, RHS > 0, it is safe to square)
=> \(a^2 + b^2 - 2ab < a^2 + b^2 + 2|a||b|\) ?
=> \(-2ab < 2|a||b|\) ?
=> \(-ab < |ab|\) ?
=> question is reduced to \(ab > 0\) ?
=> \(a\) and \(b\) are of same sign?

Let us attack the statements

Statement 1: \(a/b < 0\) => which means \(a\) and \(b\) are of opposite sign, sufficient to answer the question as "NO"
Statement 2: \(a^2 * b < 0\) => \(b\) is negative, but we don't about the sign of a => InSufficient

Answer (A)
GMAT Club Bot
Is |a-b| < |a| +|b| ? &nbs [#permalink] 12 Dec 2017, 12:49
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