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M02 #19

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M02 #19 [#permalink] New post 26 Dec 2008, 17:28
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Is x \gt y^2 ?

1. x \gt y + 5
2. x^2 - y^2 = 0

[Reveal] Spoiler: OA
C

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Re: DS- m02 [#permalink] New post 27 Dec 2008, 11:22
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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] New post 27 Dec 2008, 14:13
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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] New post 27 Dec 2008, 20:09
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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] New post 28 Dec 2008, 18:13
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] New post 28 Dec 2008, 20:23
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] New post 28 Dec 2008, 21:42
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] New post 10 Jan 2009, 21:49
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x-ALI-x 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] New post 20 Jan 2009, 01:57
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x-ALI-x 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] New post 25 Jan 2009, 01:18
x-ALI-x wrote:
botirvoy wrote:
Stmt2: /x/=/y/
Again, x can or cannot be +ve.


how did you come up with this?

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.
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Re: M02 - 19: Do we need to verify this extra bit or not? [#permalink] New post 27 Mar 2009, 05:28
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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] New post 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.

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

Thanks!

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Re: M02 - 19: Do we need to verify this extra bit or not? [#permalink] New post 10 Apr 2009, 22:56
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] New post 11 Apr 2009, 13:04
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] New post 23 Sep 2010, 07:05
my opinion:

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
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Re: M02 #19 [#permalink] New post 23 Sep 2010, 07:09
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cnrnld wrote:
my opinion:

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.

Answer: C.

Hope it's clear.
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Re: M02 #19 [#permalink] New post 24 Sep 2010, 03:32
It means both Pico. you should always look at that as (X+Y)(X-Y)=0. So X=+-Y or X = /Y/
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Re: M02 #19 [#permalink] New post 25 Sep 2010, 04:00
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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] New post 08 Nov 2010, 21:54
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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Re: M02 #19   [#permalink] 08 Nov 2010, 21:54
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