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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
1: if y = +ve, yes. If y is -ve but <-2 (approx.), then not.
2: x^2 - y^2 = 0 x^2 = y^2
suff. no matter x is +ve or -ve, y^2 always>x. So x > y^2 is not true.
Hence B.
Sorry it was overlooked and, in fact, is more difficult than it initially looked.. It should be C not B cuz if x and y are fraction, x may or may not be > y^2.
Using st 2 only: suppose if x = -0.5 and y = 0.5, x < y^2. if x = 0.5 and y = 0.5, x > y^2.
From 1 and 2: y must be < -2.50 & x must be > 2.50 accordingly. In any case, x < y^2. Therefore, it is C. _________________
thanks gmattiger for the reply. can u pls explain how did u come up with the 2.5 value. "y must be < -2.50 & x must be > 2.50 accordingly. "
Also, can we solve this w/o plugging in numbers.?
From 1 and 2, i could figure out that x is positive and y is negative.
Thanks for ur help.
thats the point here.
since x^2 = y^2 and x > y+5, y < -2.5 and x > 2.5 because if y = lets say -2, x > 3. In that case x^2 cannot be equal to y^2. x^2 would be equal to y^2 only when lxl = lyl. if y is not <-2.5, then it violates the given information that lxl = lyl or x^2 = y^2.
Stmt1: x>y+5 x can take both +ve and -ve: try x=7, y=1; and x=-1, y=-7 Insuff.
Stmt2: /x/=/y/ Again, x can or cannot be +ve.
1&2 together: Let's see what kind of numbers x and y can take. a) x=7, y=-7 b) x=-7, y=7 c) x=-7, y=-7 Only (a) satisfies both Stmt 1and 2. The answer to the question is "No".
x^2-y^2=0 => x^2=y^2; now you take square root of both sides. But because x and y are variables and we dont know there signs, we need to express them in absolute forms. Use numbers to check it.
Re: M02 - 19: Do we need to verify this extra bit or not? [#permalink]
27 Mar 2009, 05:28
1
This post received KUDOS
First, (see the red part) if we use both statements, it's clear that Y is negative and X is positive but their absolute values are the same (e.g. \(X=3\) and \(Y=-3\)). It's also clear that \(X \not\gt Y^2\). Second, (see the green part) this is what the OE tells us. If we use both statements, we see from S1 that \(X > Y\) and from S2 -> \(|X|=|Y|\). This is only true if we use BOTH statements together. Using both statements we see that we can answer the question.
Does it answer your question?
asthanap wrote:
I checked the solution to this question. It is say as X is +ve and Y is -ve therefore x>y2 is false. I think this conlusion is valid only for y <-2.5 or X >2.5.
As the absolute value of Y is > 2.5 hence X can never be greater than Y^2
Can someone verify my understanding? I raised this question to know whether this extra consideration is necessary.
Re: M02 - 19: Do we need to verify this extra bit or not? [#permalink]
27 Mar 2009, 06:02
Will upload smaller images. Actually, I wanted to verify whether it make sense to do extra work that I mentioned in my analysis. This took some extra time, and I was not sure whether it was wroth spending time on this extra verification. It seems not required. Thanks for your help.
It means +5 is (y +5) is not adding any value. This can be any positive number.
dzyubam wrote:
Can you please post smaller images? It gets too big in my screen.
Re: M02 - 19: Do we need to verify this extra bit or not? [#permalink]
27 Mar 2009, 07:01
Any positive number as long as it's X is 5 points greater. We're using S1 together with S2 in order to prove that Y is negative. If we didn't use S1 and S2 together then Y could be positive.
If you're not sure about the answer than it's a good to idea to do extra verification. It depends on the situation though.
asthanap wrote:
Will upload smaller images. Actually, I wanted to verify whether it make sense to do extra work that I mentioned in my analysis. This took some extra time, and I was not sure whether it was wroth spending time on this extra verification. It seems not required. Thanks for your help.
It means +5 is (y +5) is not adding any value. This can be any positive number.
dzyubam wrote:
Can you please post smaller images? It gets too big in my screen.
statement 2: x^2-y^2=0 we get 1) x=y, or x=-y; 2) x=y=0
then x>y^2 is false, statement 2 alone is sufficient.
choice B
Is \(x>y^2\)?
(1) \(x>y+5\) --> \(x-y>5\). Clearly insufficient, for example: if \(x=1\) and \(y=-10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient.
(2) \(x^2-y^2=0\) --> \((x-y)(x+y)=0\) --> so either \(x-y=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient.
(1)+(2) As from (1) \(x-y>5\neq{0}\), then from (2) must be true that \(x+y=0\) --> so \(x=-y\) --> substitute \(x\) in (1) --> \(-y-y>5\) --> \(y<-\frac{5}{2}<0\), as \(x=-y\), then \(x>\frac{5}{2}>0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x>y^2\)" is NO. Sufficient.
To elaborate more as \(x=-y>0\), the only chance for \(x>y^2\) to hold true (or which is the same for \(x>x^2\) to hold true) would be if \(x\) is fraction (\(0<x<1\)). For example if \(x=\frac{1}{2}\) and \(y=-\frac{1}{2}\) then \(x=\frac{1}{2}>y^2=\frac{1}{4}\). But the fact that \(x>\frac{5}{2}>0\) rules out this option.
From #1: x > y + 5 => x > y^2 for y = 1 but x maybe < y^2 for y=100 Hence #1 by itself is insufficient.
From #2: x^2 - y^2 = 0 => x^2 = y^2 => |x| = |y|
so x> y^2 for all 0<y <1 but x =y^2 for y=1 and x<y^2 for -1 < (y=x)< 0
and hence #2 is also inconclusive and hence insufficient by itself.
Both together: x > y + 5 & |x| = |y|
This is only possible if y is negative and x is positive such that |x| = |y| furthermore since x > y + 5 it can be written as |y| > y + 5 => |y| - y > 5 => |y| > 5/2 => |y| > 2.5 => |y| < y^2 => x < y^2
Hence both together are sufficient to conclude. Hence C.