Is |xy|>x^2y^2

This topic has expert replies
Source: — Data Sufficiency |

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

absolute value

by GMATGuruNY » Sun Oct 09, 2016 3:48 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Needgmat wrote:Is |xy|>x^2y^2 ?

1) 0 < x^2 < 1/4

2) 0 < y^2 < 1/9
Statement 1:
If x=1/10 and y=1, then |xy| > x²y².
If x=1/10 and y=0, then |xy| = x²y².
INSUFFICIENT.

Statement 2:
If y=1/10 and x=1, then |xy| > x²y².
If y=1/10 and x=0, then |xy| = x²y².
INSUFFICIENT.

Statements combined:
If x=1/10 and y=1/10, then |xy| > x²y².
If x=1/10 and y=-1/10, then |xy| > x²y².
If x=-1/10 and y=-1/10, then |xy| > x²y².
The cases above illustrate the following:
Since x and y must be positive or negative fractions between -1 and 1, |xy| > x²y².
SUFFICIENT.

The correct answer is C.
Last edited by GMATGuruNY on Sun Sep 02, 2018 2:58 am, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

absolute value

by GMATGuruNY » Sun Oct 09, 2016 4:17 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Needgmat wrote:Is |xy|>x^2y^2 ?

1) 0 < x^2 < 1/4

2) 0 < y^2 < 1/9

Alternate approach:

Since both sides of the question stem are NONNEGATIVE, we can square the inequality:
|xy|² > (x²y²)²
x²y² > x�y�.

Since the resulting inequality is valid only if xy is NONZERO, x²y² > 0.
Thus, we can safely divide both sides by x²y²:
(x²y²)/(x²y²) > (x�y�)/(x²y²)
1 > x²y².

Question stem, rephrased:
Is x²y² < 1?

Statement 1:
No information about y².
INSUFFICIENT.

Statement 2:
No information about x².
INSUFFICIENT.

Statements combined:
Since 0<x²<1/4 and 0<y²<1/9, x²y² < 1.
SUFFICIENT.

The correct answer is C.
Last edited by GMATGuruNY on Sun Sep 02, 2018 2:57 am, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Legendary Member
Posts: 712
Joined: Fri Sep 25, 2015 4:39 am
Thanked: 14 times
Followed by:5 members

by Mo2men » Wed Aug 29, 2018 5:58 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

GMATGuruNY wrote:
Needgmat wrote:Is |xy|>x^2y^2 ?

1) 0<x^2<1/4

2) 0<y^2<1/9

Alternate approach:

Since both sides of the question stem are NONNEGATIVE, we can square the inequality:
|xy|² > (x²y²)²
x²y² > x�y�.

Since the resulting inequality is valid only if xy is NONZERO, x²y² > 0.
Thus, we can safely divide both sides by x²y²:
(x²y²)/(x²y²) > (x�y�)/(x²y²)
1 > x²y².

Question stem, rephrased:
Is x²y² < 1?

Statement 1:
No information about y².
INSUFFICIENT.

Statement 2:
No information about x².
INSUFFICIENT.

Statements combined:
Since 0<x²<1/4 and 0<y²<1/9, x²y² < 1.
SUFFICIENT.

The correct answer is C.
Dear GMATGuru

After combining the two statements, can I rephrase the question as follow:

Since both sides of the question are positives numbers and neither x nor y can take zero value, So we can safely divide both sides by |xy|

The question becomes: Is |xy| < 1 ?

Is the above valid ?

Thanks

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

absolute value

by GMATGuruNY » Sun Sep 02, 2018 2:56 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Mo2Men wrote:
After combining the two statements, can I rephrase the question as follow:

Since both sides of the question are positives numbers and neither x nor y can take zero value, So we can safely divide both sides by |xy|

The question becomes: Is |xy| < 1 ?

Is the above valid ?

Thanks
Looks good!
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pm
Thanked: 59 times
Followed by:33 members

by fskilnik@GMATH » Mon Sep 03, 2018 7:57 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Needgmat wrote:Is |xy|>x^2y^2 ?

1) 0 < x^2 < 1/4

2) 0 < y^2 < 1/9
\[\,{\left( {xy} \right)^2}\,\,\mathop < \limits^? \,\,\,\left| {xy} \right|\]
\[\left( 1 \right){\kern 1pt} \,\,\,0 < {x^2} < \frac{1}{4}\,\,\,\,\left\{ \begin{gathered}
\,\,Take\,\,\left( {x\,;\,y} \right) = \left( {\frac{1}{3}\,\,;\,\,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\,\,Take\,\,\left( {x\,;\,y} \right) = \left( {\frac{1}{3}\,\,;\,\,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right){\kern 1pt} \,\,\,0 < {y^2} < \frac{1}{9}\,\,\,\,\left\{ \begin{gathered}
\,\,Take\,\,\left( {x\,;\,y} \right) = \left( {0\,\,;\,\,\frac{1}{4}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\,\,Take\,\,\left( {x\,;\,y} \right) = \left( {1\,\,;\,\,\frac{1}{4}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\end{gathered} \right.\,\,\]
\[\left( {1 + 2} \right){\kern 1pt} \,\,\,\]
\[\left. \begin{gathered}
0 < {x^2} < \frac{1}{4}\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,\mathop \Rightarrow \limits^{\sqrt {} } \,\,\,\,0 < \left| x \right| < \frac{1}{2} \hfill \\
\mathop \Rightarrow \limits^{0\, < \,\,\left| x \right|\, < \,\,1} \,\,0 < {\left| x \right|^{\,2}} < \left| x \right|\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\mathop \Rightarrow \limits^{{{\left| x \right|}^{\,2}} = \,\,{x^2}} \,\,\,\,\,0 < {x^2} < \left| x \right|\,\,\,\,\,\,\,\,\,\, \hfill \\
0 < {y^2} < \frac{1}{9}\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,\mathop \Rightarrow \limits^{\sqrt {} } \,\,\,\,0 < \left| y \right| < \frac{1}{3} \hfill \\
\mathop \Rightarrow \limits^{0\, < \,\,\left| y \right|\, < \,\,1} \,\,0 < {\left| y \right|^{\,2}} < \left| y \right|\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\mathop \Rightarrow \limits^{{{\left| y \right|}^{\,2}} = \,\,{y^2}} \,\,\,\,\,0 < {y^2} < \left| y \right| \hfill \\
\end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,0 < {\left( {xy} \right)^2} < \left| {xy} \right|\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

The above follows the notations and rationale taught in the GMATH method.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br