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(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.

(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 \(x=y=0\)? _________________

Working without expecting fruit helps in mastering the art of doing fault-free action !

(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. _________________

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. _________________

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 _________________

"Life is like a box of chocolates, you never know what you'r gonna get"

(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 _________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

(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 \(y>0\).

(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! _________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

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.

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.

You 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. _________________

Re: Is 1/x^5 > y/(y^6+1)? [#permalink]
07 Feb 2014, 09:04

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is 1/x^5 > y/(y^6+1)? [#permalink]
07 Feb 2014, 18:55

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. non-integer values: put 1/2. 1/x on the RHS will make it RHS a bigger number.

C

gmatclubot

Re: Is 1/x^5 > y/(y^6+1)?
[#permalink]
07 Feb 2014, 18:55

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