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Is 1/x^5 > y/(y^6+1)? [#permalink]
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10 Feb 2011, 14:25
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Is 1/x^5 > y/(y^6+1)? (1) x = y (2) y > 0
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Last edited by Bunuel on 12 Jul 2013, 02:52, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Inequalities [#permalink]
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10 Feb 2011, 14:52
Is 1/x^5 > y/(y^6+1)? (1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases: A. \(y<0\) > cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO. B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES. (2) y>0. Clearly insufficient as no info about \(x\). (1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient. Answer: C.
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Re: Inequalities [#permalink]
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10 Feb 2011, 14:58
Thank you, Bunuel!!!
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Re: Inequalities [#permalink]
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12 Feb 2011, 09:40
Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\) > cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C. HI Bunuel.. in S1: \(x=y\) , you considered 2 cases , \(y<0\) and \(y>0\) but why you did not consider \(x=y=0\)?
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Re: Inequalities [#permalink]
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12 Feb 2011, 09:47
amod243 wrote: Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\) > cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C. HI Bunuel.. in S1: \(x=y\) , you considered 2 cases , \(y<0\) and \(y>0\) but why you did not consider \(x=y=0\)? x is in denominator it can not be zero.
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Re: Inequalities [#permalink]
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12 Feb 2011, 15:06
Bunnel, thank you!!! you had made the inequalities real simple.



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Re: Inequalities [#permalink]
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14 Feb 2011, 02:18
Hi guys, Can we do it this way 1/x^5>y/(y^6 + 1) => X^5<(Y^6+1)/Y S1. Y^5 < Y^5 + 1/Y > Insufficient S2 Y>0 > Insufficient S1 + S2 > 1/Y >0 > Y^5 < Y^5 + 1/Y > C is the answer
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Re: Inequalities [#permalink]
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14 Feb 2011, 02:39



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Re: Inequalities [#permalink]
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14 Feb 2011, 05:21
Bunuel wrote: bigtooth81 wrote: Hi guys,
Can we do it this way
1/x^5>y/(y^6 + 1) => X^5<(Y^6+1)/Y
S1. Y^5 < Y^5 + 1/Y > Insufficient S2 Y>0 > Insufficient
S1 + S2 > 1/Y >0 > Y^5 < Y^5 + 1/Y > C is the answer You can not cross multiply as you don't know the signs of x^5 and y (y^6+1 is positive). For example if x=1 and y=1 then: 1/x^5=1>1/2=y/(y^6+1) but x^5=1>2=(y^6+1)/y (> not < as you wrote). Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.Got you. Thanks bro
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Re: Inequalities [#permalink]
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15 Feb 2011, 08:42
Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\)> cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C. In A, are you asking us to consider y being negative or did you derive it. Consideration would make sense to me, but I'm just trying to learn for the future. Sorry to be asking all this simple questions, but I just need to know. Thank you Bunuel! Mari
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Re: Inequalities [#permalink]
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15 Feb 2011, 08:56
mariyea wrote: Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\)> cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C. In A, are you asking us to consider y being negative or did you derive it. Consideration would make sense to me, but I'm just trying to learn for the future. Sorry to be asking all this simple questions, but I just need to know. Thank you Bunuel! Mari We should determine whether \(\frac{1}{y^5}>\frac{y}{y^6+1}\) is true. You can do this algebraically or with number plugging (for example test y=1 to get YES and then y=1 to get NO). As for algebraic approach: we should somehow simplify \(\frac{1}{y^5}>\frac{y}{y^6+1}\) (as it looks kind of ugly) to get the answer. For this we consider two cases: \(y<0\) and \(y>0\) (y=0 is not possible as y is in denominator and we know that division by zero is undefined). As we proceed, we get NO answer for an assumption \(y<0\) and we get YES answer for an assumption \(y>0\). Hope it's clear.
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Re: Inequalities [#permalink]
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15 Feb 2011, 09:02
Bunuel wrote: mariyea wrote: Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\)> cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C. In A, are you asking us to consider y being negative or did you derive it. Consideration would make sense to me, but I'm just trying to learn for the future. Sorry to be asking all this simple questions, but I just need to know. Thank you Bunuel! Mari We should determine whether \(\frac{1}{y^5}>\frac{y}{y^6+1}\) is true. You can do this algebraically or with number plugging (for example test y=1 to get YES and then y=1 to get NO). As for algebraic approach: we should somehow simplify \(\frac{1}{y^5}>\frac{y}{y^6+1}\) (as it looks kind of ugly) to get the answer. For this we consider two cases: \(y<0\) and \(y>0\) (y=0 is not possible as y is in denominator and we know that division by zero is undefined). As we proceed, we get NO answer for an assumption \(y<0\) and we get YES answer for an assumption \(y>0\). Hope it's clear. Yes! It is very clear now, thanks! I was just wondering whether you considered y ve... now I understand. Thank you so much for your patience!
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Re: Inequalities [#permalink]
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01 Mar 2011, 01:24
Hi Bunuel If it would have been Is 1/x^5 > 1/(y^6+1) ? Then can I rephrase it as Is x^5 < y^6+1 ? irrespective of the signs. Am I correct? Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\) > cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C.



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Re: Inequalities [#permalink]
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01 Mar 2011, 02:41
gmat1220 wrote: Hi Bunuel If it would have been Is 1/x^5 > 1/(y^6+1) ? Then can I rephrase it as Is x^5 < y^6+1 ? irrespective of the signs. Am I correct? Bunuel wrote: Is 1/x^5 > y/(y^6+1)?
(1) x=y > is \(\frac{1}{y^5}>\frac{y}{y^6+1}\)? Two cases:
A. \(y<0\) > cross multiply and as for negative \(y\): \(y^5<0\) and \(y^6+1>0\) flip the sign (because of negative \(y^5\)). The question becomes: is \(y^6+1<y^6\)? > is \(1<0\)? In this case the answer would be NO.
B. \(y>0\) > cross multiply and as for positive \(y\) both \(y^5\) and \(y^6+1\) are positive remain the sign. The question becomes: is \(y^6+1>y^6\)? > is \(1>0\)? In this case the answer would be YES.
(2) y>0. Clearly insufficient as no info about \(x\).
(1)+(2) As from (2) \(y>0\) then we have case B and the answer is YES. Sufficient.
Answer: C. Check this: inequalities109031.html#p871109You can not cross multiply as you don't know the sign of x^5. Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.
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Re: Inequalities [#permalink]
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01 Mar 2011, 03:27
Thanks Bunuel for the brilliant explanation !



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Re: Is 1/x^5 > y/(y^6+1)? [#permalink]
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Re: Is 1/x^5 > y/(y^6+1)? [#permalink]
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07 Feb 2014, 19:55
Is 1/x^5 > y/(y^6+1)? or is x^5 < (y^6+1)/y? (taking reciprocals changes the sign of inequality)
(1) x = y > is x^ 5 < (x^6 + 1)/x or is x^5 < x^5 + 1/x? put x = 1, yes. Put x = 1, No. (2) y > 0 > no info on x. NOT sufficient
Combining, x = y > 0. For +ve values, x^5 always less than x^5 +1/x. example: integer values. Put x = 2; Yes. noninteger values: put 1/2. 1/x on the RHS will make it RHS a bigger number.
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Re: Is 1/x^5 > y/(y^6+1)?
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