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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.

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".

Re: M02 - 19: Do we need to verify this extra bit or not? [#permalink]

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27 Mar 2009, 06:28

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.

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.

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.

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]

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27 Mar 2009, 07:02

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]

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27 Mar 2009, 08:01

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.