|
|
GMAT 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.
20 May 2012, 19:02
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (02:12) correct
44%
(02:12)
wrong
based on 2695
sessions
History
Date
Time
Result
Not Attempted Yet
If x^2 = y^2, is true that x > 0 ?
(1) x = 2y+1
(2) y <= -1
(1) x = 2y+1
(2) y <= -1
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
21 May 2012, 00:38
Kudos
Bookmarks
If x^2 = y^2, is true that x>0?
\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).
(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.
(2) y<= -1. Clearly insufficient.
(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.
Answer: C.
Hope it's clear.
\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).
(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.
(2) y<= -1. Clearly insufficient.
(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.
Answer: C.
Hope it's clear.
18 Jun 2012, 17:16
Kudos
Bookmarks
The way I usually approach these problems is with plugging in examples, like -1, 0, and 1 to see when the equations hold true.
1)
x = 2y +1
Here, we actually don't need to do anything. Obviously we can't know if x>0 if we don't know the value of y.
2) y<=-1
Alone, this obviously tells us nothing about x! Insufficient. Answer is either C or E.
1+2)
Let's plug a few values of y that are <= -1 into the equation from situation #1 to see how they affect x:
x=2y+1
x=2(-1)+1 = -1
x=2(-2)+1 = -3
x=2(-3)+1 = -5
Obviously, this will continue as a series.
Therefore, we clearly know that x will never be >0 and therefore, C is the answer.
1)
x = 2y +1
Here, we actually don't need to do anything. Obviously we can't know if x>0 if we don't know the value of y.
2) y<=-1
Alone, this obviously tells us nothing about x! Insufficient. Answer is either C or E.
1+2)
Let's plug a few values of y that are <= -1 into the equation from situation #1 to see how they affect x:
x=2y+1
x=2(-1)+1 = -1
x=2(-2)+1 = -3
x=2(-3)+1 = -5
Obviously, this will continue as a series.
Therefore, we clearly know that x will never be >0 and therefore, C is the answer.
