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M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink]
 09 Jun 2012, 08:10
Question Stats:
71% (01:35) correct
28% (01:11) wrong based on 2 sessions
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 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.
To discard other options consider a=−1 and b=2.
Last edited by manulath on 09 Jun 2012, 08:37, edited 1 time in total.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink]
09 Jun 2012, 08:16
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink]
09 Jun 2012, 08:28
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]
09 Jun 2012, 08:29
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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Manager
Joined: 12 May 2012
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GMAT 1: 650 Q51 V25
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink]
09 Jun 2012, 08:33
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]
20 Sep 2012, 11:47
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]
20 Sep 2012, 12:37
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink]
29 Apr 2013, 19:28
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]
29 Apr 2013, 20:58
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 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.
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]
30 Apr 2013, 01:03
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must) [#permalink]
30 Apr 2013, 15:42
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.
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Re: M14, 33 (If ab≠0 and |a|<|b|, which of the following must)
[#permalink]
30 Apr 2013, 15:42
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