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If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]
 12 Sep 2011, 17:42
Question Stats:
25% (02:27) correct
75% (01:50) wrong based on 0 sessions
If a does not = 0, is 1/a > a/(b^4 + 3) (1) a^2 = b^2 (2) a^2 = b^4 This comes from the MGMAT advanced quant guide #4-9. Would like to see some tips on how to set this up.
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Re: DS question from MGMAT advanced quant [#permalink]
12 Sep 2011, 23:45
1/a > a/(b^4 + 3) ??? B) a^2 = b^4 thus eqn becomes 1/a > a/(a^2 + 3). if a>0, 3 >0 true. if a<0, 1/c < c/(c^2 + 3) where c=-a and c>0. 3 <0 never true. Insufficient A) a^2 = b^2 if a=1, 1>1/4 true if a=-1, -1>-1/4 false. Insufficient A&B) b^2 = b^4 -> b^2 = 1 or b^2 = 0. thus a=0,+1,-1. a is not equal to 0. if a=1, true, if a=-1,, false IMO
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 00:55
rjdunn03 wrote: If a does not = 0, is 1/a > a/(b^4 + 3) (1) a^2 = b^2 (2) a^2 = b^4 This comes from the MGMAT advanced quant guide #4-9. Would like to see some tips on how to set this up. rjdunn03: List of forums below. Please subscribe to each of these forums separately.
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 01:18
rjdunn03 wrote: If a does not = 0, is 1/a > a/(b^4 + 3) (1) a^2 = b^2 (2) a^2 = b^4 This comes from the MGMAT advanced quant guide #4-9. Would like to see some tips on how to set this up. Just check with a=1, b=1 AND a=-1, b=1 Ans: "E"
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 03:48
rjdunn03 wrote: If a does not = 0, is 1/a > a/(b^4 + 3) (1) a^2 = b^2 (2) a^2 = b^4 This comes from the MGMAT advanced quant guide #4-9. Would like to see some tips on how to set this up. rjdunn03, what is OA ?
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 06:21
Sorry about posting in wrong forum Fluke.
The OA is A. I got this one wrong as well.
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 18:30
Can any experts explain why A is the correct answer here?
Thank you
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 18:34
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rjdunn03 wrote: Can any experts explain why A is the correct answer here?
Thank you OA should be E, not A.
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 19:27
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i got E too.
I think there is a typo in the book. For statement 1, the OE in the book states "b^2 must be positive, so is a". Then they divide by a on both sides. Reading the OE, it seem that the makers of the question meant the first statement in the question to be "a = b^2" instead of "a^2 = b^2".
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 19:38
dreambeliever wrote: i got E too.
I think there is a typo in the book. For statement 1, the OE in the book states "b^2 must be positive, so is a". Then they divide by a on both sides. Reading the OE, it seem that the makers of the question meant the first statement in the question to be "a = b^2" instead of "a^2 = b^2". Yes, I think you are right now that I am looking at the explanation closer. I will ask on the MGMAT forums, but I think you guys are right. This has to be incorrect.
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 19:38
dreambeliever wrote: i got E too.
I think there is a typo in the book. For statement 1, the OE in the book states "b^2 must be positive, so is a". Then they divide by a on both sides. Reading the OE, it seem that the makers of the question meant the first statement in the question to be "a = b^2" instead of "a^2 = b^2". Yes, "a=b^2" will be sufficient. thanks
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 19:44
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Nevermind, I found the error listed on the MGMAT blog. Pg. 106 In "Try-It #4-9," statement 1 is incorrectly written. It should say that "a" is equal to "b^2." (1) a = b^2 http://www.manhattangmat.com/errata-adv-quant-4th.cfmThanks fluke and dreambeliever for your help.
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Re: DS question from MGMAT advanced quant [#permalink]
13 Sep 2011, 22:12
If a does not = 0, is 1/a > a/(b^4 + 3) (1) a^2 = b^2 (2) a^2 = b^4 this might help : 1/a > a/(b^4 + 3) ==> (b^4+3) > a^2 from (1) a = b^2 ==> b= Mod(a) from here, you can say that (b^4+3) > a^2. cz b^2 = a or, b^4 = a^2 (squaring both side) ==> (b^4+X) > a^2 (where x is +ve no) hence, Statement 1 Sufficient. (you can try for any possible values of a & b )
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Re: DS question from MGMAT advanced quant [#permalink]
11 Mar 2012, 06:03
From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...
Then why cant we say that option B alone is also sufficient and hence answer should be D..
To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?
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Re: DS question from MGMAT advanced quant [#permalink]
11 Mar 2012, 11:16
pavanpuneet wrote: From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...
Then why cant we say that option B alone is also sufficient and hence answer should be D..
To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong? THE QUESTION SHOULD READ: If a does not equal to zero, is 1/a > a/(b^4 +3)? Notice that we cannot cross-multiply \frac{1}{a} > \frac{a}{b^4 +3}, since we don't know whether a>0 or a<0, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with b^4 +3, since this expression is always positive: b^4 +3=nonnegative+positive=positive. So, 321kumarsushant's first step is not correct. (1) a=b^2 --> is \frac{1}{b^2} > \frac{b^2}{b^4 +3}? Now, since b^2>0 then here we can safely cross-multiply and the question becomes: is b^4+3>b^4? --> is 3>0? Answer ot this question is YES.Sufficient. (2) a^2=b^4 --> is \frac{1}{a} > \frac{a}{a^2 +3}? We have the same problem: we cannot cross-multiply since we don't know the sign of a. If a=1 then the answer to the question would be YES but if a=-1 then the answer to the question would be NO. Not sufficient. Answer: A.
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]
12 Mar 2012, 02:42
Its clear now! Thanks again for the prompt response Bunuel
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]
22 May 2012, 19:14
Bunuel never ceases to amaze me. His reasoning is so sharp and his methods so efficient, I bet he conquers every problem in under 30 seconds. I have been following his posts extensively hoping to learn to think like he does. Lifelong fan right here.
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Re: If a does not = 0, is 1/a > a/(b^4 + 3)
[#permalink]
22 May 2012, 19:14
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