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M14-33

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M14-33  [#permalink]

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New post 16 Sep 2014, 00:54
1
4
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

62% (02:00) correct 38% (01:44) wrong based on 58 sessions

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Math Expert
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Re M14-33  [#permalink]

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New post 16 Sep 2014, 00:54
2
Official Solution:

If \(ab \ne 0\) and \(|a| \lt |b|\), which of the following must be negative?

A. \(\frac{a}{b} - \frac{b}{a}\)
B. \(\frac{a - b}{a + b}\)
C. \(a^b - b^a\)
D. \(a \frac{b}{a - b}\)
E. \(\frac{b - a}{b}\)


\(|a| \lt |b|\) means that \(a^2 \lt b^2\), which can be written as \(a^2 - b^2 \lt 0\) or as \((a - b)(a + b) \lt 0\). So, \(a - b\) and \(a + b\) have the opposite signs, which means that \(\frac{a - b}{a + b}\) will always be negative.

To discard other options consider \(a=-1\) and \(b=2\).


Answer: B
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Re: M14-33  [#permalink]

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New post 10 Nov 2014, 04:58
1
Bunnel,

option e: (b-a)/b = 1 - (a/b)
Case i: if either a or b is negative , a/b is positive , so the 1 - (a/b) is positive
case ii: if both are of the same sign, since |a|<|b|, 0< a/b < 1 => 1 - (a/b) is always positive
Hence option e.

What am I missing? :shock:

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Re: M14-33  [#permalink]

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New post 10 Nov 2014, 05:42
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Re: M14-33  [#permalink]

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New post 10 Nov 2014, 05:49
2
:wall

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Re: M14-33  [#permalink]

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New post 10 Nov 2015, 04:15
|a|<|b| is possible in below scenarios; (I don't like using smart numbers, but in that case I could not avoid using them)

a=2 b=3
a=-2 b=-3
a=2 b=-3
a=-2 b=3

If you try each of them individually for each choice, you will get negative result for the equation in choice B.
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Re M14-33  [#permalink]

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New post 20 Aug 2016, 04:53
I think this is a high-quality question and I agree with explanation.
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Re M14-33  [#permalink]

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New post 21 Feb 2019, 08:12
I think this is a high-quality question and I agree with explanation.
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Re M14-33  [#permalink]

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New post 21 Aug 2019, 13:33
I think this is a high-quality question and I agree with explanation.
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Re: M14-33  [#permalink]

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New post 10 Sep 2019, 23:52
I think this is a high-quality question and I agree with explanation.
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Re: M14-33   [#permalink] 10 Sep 2019, 23:52
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