GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 24Get an Extra 30% Off Target Test Prep Plans
08:00 PM EDT
-11:59 PM EDT
Grab 30% off any Target Test Prep GRE plan during our Flash Sale and join Brian and many other students who have used Target Test Prep to score high on the GRE. Just enter the coupon code FLASH30 at checkout to save big.
x2 > y2 and x > –|y|
[#permalink]
02 Sep 2018, 07:32
02 Sep 2018, 07:32
3
Expert Reply
5
Bookmarks
00:00
Question Stats:
68% (01:12) correct
31% (00:52) wrong based on 120 sessions
Hide Show timer Statistics
\(x^2 > y^2\) and \(x > -|y|\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
y |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: x2 > y2 and x > –|y|
[#permalink]
03 Sep 2018, 10:26
03 Sep 2018, 10:26
5
1
Bookmarks
sandy wrote:
\(x^2 > y^2\) and \(x > -|y|\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
y |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Given \(x^2 > y^2\) and \(x > -|y|\)
From \(x^2 > y^2\) => \(x^2\) - \(y^2\) > 0 => ( x + y ) ( x - y ) > 0.
Here ( x + y ) ( x - y ) > 0 either both should be +ve or both -ve to be greater than 0.
But from \(x > -|y|\) we get ( x + y ) > 0 ( Note here |y| is converted to y )
Since we came to know that ( x + y ) > 0 then must be ( x - y ) > 0
So ( x - y ) > 0 => x > y.
A.
General Discussion
Re: x2 > y2 and x > –|y|
[#permalink]
03 Sep 2018, 15:40
03 Sep 2018, 15:40
2
if \(x^2>y^2\), then \(|x|>|y|\)
if x>0 y>0, then x>y since \(|x|>|y|\);
if x>0 y<0, then x>y. Because \(x>-|y|\) and \(-|y|=-(-y)=y\);
x<0 y>0 is not possible. Because \(x>-|y|\) and \(-|y|=-y\), so \(x>-y\), which means \(|x|<|y|\), but we already know \(|x|>|y|\), so this is a contradiction;
x<0 y<0 is also not possbile, Because \(x>-|y|\) and \(-|y|=-(-y)=y\) therefore x>y but then \(|x|<|y|\), and we already know \(|x|>|y|\), so this is a contradiction;
if x>0 y>0, then x>y since \(|x|>|y|\);
if x>0 y<0, then x>y. Because \(x>-|y|\) and \(-|y|=-(-y)=y\);
x<0 y>0 is not possible. Because \(x>-|y|\) and \(-|y|=-y\), so \(x>-y\), which means \(|x|<|y|\), but we already know \(|x|>|y|\), so this is a contradiction;
x<0 y<0 is also not possbile, Because \(x>-|y|\) and \(-|y|=-(-y)=y\) therefore x>y but then \(|x|<|y|\), and we already know \(|x|>|y|\), so this is a contradiction;
