Is 1/x^5 > y/(y^6+1)? (1) x = y (2) y > 0 : Data Sufficiency (DS)

gmatjon · 2011-02-10T13:25:29-08:00

Is \frac{1}{x^5} > \frac{y}{(y^6+1)} ? (1) x = y (2) y > 0 Untitled.jpg

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12 Feb 2011, 09:40

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Bunuel

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 $x=y=0$?

Bunuel

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12 Feb 2011, 09:47

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amod243

Bunuel

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 cannot be zero.

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12 Feb 2011, 15:06

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Bunnel, thank you!!! you had made the inequalities real simple.

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14 Feb 2011, 02:18

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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

Bunuel

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14 Feb 2011, 02:39

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bigtooth81

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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14 Feb 2011, 05:21

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Bunuel

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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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15 Feb 2011, 08:42

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Bunuel

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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15 Feb 2011, 08:56

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mariyea

Bunuel

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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15 Feb 2011, 09:02

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Bunuel

mariyea

Bunuel

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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01 Mar 2011, 01:24

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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

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.

Bunuel

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01 Mar 2011, 02:41

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gmat1220

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

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: https://gmatclub.com/forum/inequalities- ... ml#p871109

You cannot 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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01 Mar 2011, 03:27

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Thanks Bunuel for the brilliant explanation !

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07 Feb 2014, 19:55

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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

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08 Jul 2019, 07:05

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Bunuel

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.

Bunuel Could you please tell me how you cross multiplied the equations? Y^5 (LHS) as well as y (RHS) are negative here. Also, could you please share some resource to learn cross multiplication in inequalities. Thanks!

Bunuel

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08 Jul 2019, 22:58

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Bunuel

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.

Bunuel Could you please tell me how you cross multiplied the equations? Y^5 (LHS) as well as y (RHS) are negative here. Also, could you please share some resource to learn cross multiplication in inequalities. Thanks!

It could also be possible that a and c are both negative and c is more negative than a. I jus do not get how to approach such questions. Is there a frame of mind one should be in? Bunuel

Does this question have a verified source?[/quote]

Sorry, not sure what you are trying to say there but below are materials that could help solving inequalities questions:

9. Inequalities

Theory
Inequalities Made Easy!
Inequalities: Tips and hints
Arithmetic with Inequalities
Wavy Line Method Application - Complex Algebraic Inequalities
Solving Quadratic Inequalities: Graphic Approach
Inequalities - Quadratic Inequalities
Graphic approach to problems with inequalities
Inequalities trick
INEQUATIONS(Inequalities) Part 1
INEQUATIONS(Inequalities) Part 2

Questions
Inequality and absolute value questions from my collection
DS Questions
PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.

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09 Jul 2019, 10:03

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gmatjon

Is $\frac{1}{x^5} > \frac{y}{(y^6+1)}$?

(1) x = y

(2) y > 0

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Untitled.jpg

#1
would be valid for $\frac{1}{x^5} > \frac{y}{(y^6+1)}$ when +ve and not sufficient when -ve
#2
y>0 ; x not given
insufficient
from 1 &2
x& y are +ve so $\frac{1}{x^5} > \frac{y}{(y^6+1)}$ is sufficeint
IMO C

Is 1/x^5 > y/(y^6+1)? (1) x = y (2) y > 0

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