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Is 1/x^5 > y/(y^6+1)?

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Is 1/x^5 > y/(y^6+1)? [#permalink] New post 10 Feb 2011, 13:25
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57% (02:10) correct 43% (01:22) wrong based on 54 sessions
Is 1/x^5 > y/(y^6+1)?

(1) x = y
(2) y > 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 12 Jul 2013, 01:52, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Inequalities [#permalink] New post 10 Feb 2011, 13:52
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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] New post 10 Feb 2011, 13:58
Thank you, Bunuel!!!

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Re: Inequalities [#permalink] New post 12 Feb 2011, 08:40
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] New post 12 Feb 2011, 08:47
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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] New post 12 Feb 2011, 14:06
Bunnel, thank you!!! you had made the inequalities real simple.
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Re: Inequalities [#permalink] New post 14 Feb 2011, 01:18
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] New post 14 Feb 2011, 01:39
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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.

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Re: Inequalities [#permalink] New post 14 Feb 2011, 04:21
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] New post 15 Feb 2011, 07:42
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] New post 15 Feb 2011, 07:56
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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] New post 15 Feb 2011, 08:02
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] New post 01 Mar 2011, 00:24
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] New post 01 Mar 2011, 01:41
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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: inequalities-109031.html#p871109

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.

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

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: Inequalities [#permalink] New post 01 Mar 2011, 02:27
Thanks Bunuel for the brilliant explanation !
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Re: Is 1/x^5 > y/(y^6+1)? [#permalink] New post 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
Re: Is 1/x^5 > y/(y^6+1)?   [#permalink] 07 Feb 2014, 18:55
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