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M14, 33 (If ab≠0 and |a|<|b|, which of the following must)

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M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 09 Jun 2012, 07:10
This post is edited to rectify the incorrect choice.

If ab≠0 and |a|<|b|, which of the following must be negative?

A. (a/b)− (b/a)
B. (a−b)/(a+b)
C. a^b−b^a
D. a(b/(a−b))
E. (b−a)/b

[Reveal] Spoiler:
OA: B


OE:
[Reveal] Spoiler:
Solution
|a|<|b| means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.

To discard other options consider a=−1 and b=2.

Last edited by manulath on 09 Jun 2012, 07:37, edited 1 time in total.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 09 Jun 2012, 07:16
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If ab\neq{0} and |a|<|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. \frac{ab}{a-b}

E. \frac{b-a}{b}

|a|<|b| means that a^2<b^2 --> a^2-b^2<0 --> (a-b)(a+b)<0, so a-b and a+b have the opposite signs, which means that \frac{a-b}{a+b} will always be negative.

Answer: B.

To discard other options consider a=-1 and b=2.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 09 Jun 2012, 07:28
manulath wrote:
If ab≠0 and |a|<|b|, which of the following must be negative?

A. (a/b)− (b/b)
B. (a−b)/(a+b)
C. a^b−b^a
D. a(b/(a−b))
E. (b−a)/b


With respect to the Choice A : (a/b)− (b/b)

as ab≠0, what ever the value of b,
b/b is always 1
choice A reduces to a/b-1

as |a|<|b| ......... a/b will always be less than 1

scenario 1: a>0, b>0
clearly a/b <1

scenario 2: a<0, b<0
in a/b, the negative signs will cancel each other
Hence once again a/b<1

scenario 3 and 4: a<0, b>0 or a>0,b<0
in a/b, one value is +ve and other is -ve
Hence a/b will be -ve
that is to say a/b<0
or to say a/b<1

As we see that in all the above cases a/b < 1
in choice A: (a/b)-(b/b) will always be -ve

Where did I went wrong?

Even if I put the values give in OE

manulath wrote:
OE: Solution
|a|<|b| means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.

[highlight]To discard other options consider a=−1 and b=2[/highlight]


A. (a/b)− (b/b) => (-1/2) - (2/2) => -1/2 - 1 = -3/2 a negative value
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 09 Jun 2012, 07:29
Expert's post
manulath wrote:
manulath wrote:
If ab≠0 and |a|<|b|, which of the following must be negative?

A. (a/b)− (b/b)
B. (a−b)/(a+b)
C. a^b−b^a
D. a(b/(a−b))
E. (b−a)/b


With respect to the Choice A : (a/b)− (b/b)

as ab≠0, what ever the value of b,
b/b is always 1
choice A reduces to a/b-1

as |a|<|b| ......... a/b will always be less than 1

scenario 1: a>0, b>0
clearly a/b <1

scenario 2: a<0, b<0
in a/b, the negative signs will cancel each other
Hence once again a/b<1

scenario 3 and 4: a<0, b>0 or a>0,b<0
in a/b, one value is +ve and other is -ve
Hence a/b will be -ve
that is to say a/b<0
or to say a/b<1

As we see that in all the above cases a/b < 1
in choice A: (a/b)-(b/b) will always be -ve

Where did I went wrong?

Even if I put the values give in OE

manulath wrote:
OE: Solution
|a|<|b| means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.

[highlight]To discard other options consider a=−1 and b=2[/highlight]


A. (a/b)− (b/b) => (-1/2) - (2/2) => -1/2 - 1 = -3/2 a negative value


There is a typo in option A. It should read a/b-b/a.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 09 Jun 2012, 07:33
Bunuel wrote:
A. \frac{a}{b} - \frac{b}{a}




Okay it seems that question I got was incorrect.
I have send the screenshot to you as pm.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 20 Sep 2012, 10:47
Hi Bunuel,

Could you please explain as to how did you know that a= -1 and b = 2 would make all the other answers invalid?was it just hit and trial or was there any specific reason for choosing these two values?

Thanks.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 20 Sep 2012, 11:37
Expert's post
negative wrote:
Hi Bunuel,

Could you please explain as to how did you know that a= -1 and b = 2 would make all the other answers invalid?was it just hit and trial or was there any specific reason for choosing these two values?

Thanks.


With a certain algebraic manipulations with the expression in the stem we can get that B is the correct answer. So, the example which discard other options are just to illustrate that they are wrong, and yes you can find the proper numbers for that by trial and error.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 29 Apr 2013, 18:28
I thought |x| = \sqrt{x^2}? How do you arrive at |a| = a^2?
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 29 Apr 2013, 19:58
manulath wrote:
This post is edited to rectify the incorrect choice.

If ab≠0 and |a|<|b|, which of the following must be negative?

A. (a/b)− (b/a)
B. (a−b)/(a+b)
C. a^b−b^a
D. a(b/(a−b))
E. (b−a)/b

[Reveal] Spoiler:
OA: B


OE:
[Reveal] Spoiler:
Solution
|a|<|b| means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.

To discard other options consider a=−1 and b=2.

I've also done the way Bunuel explained ........................... as
Given .......... |a|<|b| i.e., a^2<b^2
Therefore, a^2-b^2<0 & therefore, (a-b)(a+b)<0 & ultimately divide both sides with (a+b)
Hence, , I got ... \frac{a-b}{a+b} < 0
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 30 Apr 2013, 00:03
Expert's post
youngkacha wrote:
I thought |x| = \sqrt{x^2}? How do you arrive at |a| = a^2?


a^2<b^2 is obtained by squaring |a|<|b| (we can safely square |a|<|b| since both sides of the inequality are non-negative).

Hope it's clear.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink] New post 30 Apr 2013, 14:42
Bunuel wrote:
youngkacha wrote:
I thought |x| = \sqrt{x^2}? How do you arrive at |a| = a^2?


a^2<b^2 is obtained by squaring |a|<|b| (we can safely square |a|<|b| since both sides of the inequality are non-negative).

Hope it's clear.


Oh okay, I see now. This was a tough one for me.

Thank you.
Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must)   [#permalink] 30 Apr 2013, 14:42
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