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Is \(x \gt y^2\) ? 1. \(x \gt y + 5\) 2. \(x^2  y^2 = 0\) Source: GMAT Club Tests  hardest GMAT questions



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Re: DS m02 [#permalink]
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27 Dec 2008, 12:22
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jackychamp wrote: Is x > y^2?
1. x>y+5 2. X^2  Y^2 = 0 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.
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Re: DS m02 [#permalink]
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27 Dec 2008, 15:13
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what if x=1/10 and y=1/10
hence x^2 = y^2
and x > y^2.
so, its true



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Re: DS m02 [#permalink]
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27 Dec 2008, 21:09
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I am getting E
plot the following:
x = y^2 (parabola, only exists in I and II quadrents) > the region to the right of the parabola is the required area in the question
y = x  5 (we'll convert it to inequality by shading the area below the line since y < x  2)
y^2 = x^2 gives two lines > y = x and y = x > since it's in equation the solution lies ON the two lines.. i.e. we don't shade any areas
now B clearly lies inside as well as outside the required region > insuff
A is a maybe too
Combining A and B gives a maybe as well > E



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Re: DS m02 [#permalink]
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28 Dec 2008, 19:13
GMAT TIGER wrote: jackychamp wrote: Is x > y^2?
1. x > y+5 2. X^2  Y^2 = 0 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.
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Re: DS m02 [#permalink]
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28 Dec 2008, 21:23
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.



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Re: DS m02 [#permalink]
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28 Dec 2008, 22:42
jackychamp wrote: 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. hth.
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Re: M02Q19 [#permalink]
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10 Jan 2009, 22:49
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xALIx wrote: Is \(x> y^2\) ?
1. \(x > y + 5\) 2. \(x^2  y^2 = 0\)
Please explain. 1: if x is 7, and y = 1, yes. if x = 10 and y = 4, no. not sufff... 2: x = + or  y if x is +ve fraction, yes and vice versa. not suff. 1&2: y < 2.50 and x > 2.50. So in that case: x is not greater than y^2. Hence C.
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Re: M02Q19 [#permalink]
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20 Jan 2009, 02:57
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xALIx wrote: Is \(x> y^2\) ?
1. \(x > y + 5\)
2. \(x^2  y^2 = 0\)
Please explain. Stem: \(x>y^2\)? => x=+ve; y=+ve; ve 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". Hence, C.



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Re: M02Q19 [#permalink]
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25 Jan 2009, 02:18
xALIx wrote: botirvoy wrote: Stmt2: /x/=/y/ Again, x can or cannot be +ve.
how did you come up with this? x^2y^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.



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Re: M02  19: Do we need to verify this extra bit or not? [#permalink]
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27 Mar 2009, 06:28
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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.
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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.
Thanks!



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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.
Thanks!
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Re: M02  19: Do we need to verify this extra bit or not? [#permalink]
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10 Apr 2009, 23:56
Guys, sorry .. i want to clarify this too. Doesn't x^2=y^2 mean that x=y or x=y? Thanks !!



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Re: M02  19: Do we need to verify this extra bit or not? [#permalink]
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11 Apr 2009, 14:04
piccolino wrote: Guys, sorry .. i want to clarify this too. Doesn't x^2=y^2 mean that x=y or x=y? Thanks !! Basically, yes. You have to be careful with 0 in this situation though (0=0 as it's neither negative nor potisive).
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Re: M02 #19 [#permalink]
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23 Sep 2010, 08:05
my opinion:
statement 2: x^2y^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



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Re: M02 #19 [#permalink]
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23 Sep 2010, 08:09
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cnrnld wrote: my opinion:
statement 2: x^2y^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) \(x>y+5\) > \(xy>5\). 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) \(x^2y^2=0\) > \((xy)(x+y)=0\) > so either \(xy=0\) 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) \(xy>5\neq{0}\), then from (2) must be true that \(x+y=0\) > so \(x=y\) > substitute \(x\) in (1) > \(yy>5\) > \(y<\frac{5}{2}<0\), 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. Answer: C. Hope it's clear.
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Re: M02 #19 [#permalink]
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24 Sep 2010, 04:32
It means both Pico. you should always look at that as (X+Y)(XY)=0. So X=+Y or X = /Y/



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Re: M02 #19 [#permalink]
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Is x > y^2 ?
1. x > y + 5 2. x^2  y^2 = 0
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.
HTH.



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Re: M02 #19 [#permalink]
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08 Nov 2010, 22:54
Thanks Bunuel. You are awesome! I suggest that you replace the online OE for this question on 1)+2) with what you posted here.
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