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Is x^2 - y^2 > 0?

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Is x^2 - y^2 > 0?  [#permalink]

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New post 21 Mar 2017, 18:52
1
4
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

37% (02:28) correct 63% (02:25) wrong based on 101 sessions

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Is \(x^2-y^2>0\) ?

(1) \(x^2-y^2>(x - y)\)
(2) \(x^2-y^2<(x + y)\)

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Re: Is x^2 - y^2 > 0?  [#permalink]

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New post 09 Jan 2019, 11:30
3
Here's how I solved this - please critique!

Is \(x^{2}\)-\(y^{2}\)>0?

Rewriting the stem: Is (x-y)(x+y)>0?

Statement 1) \(x^{2}\)-\(y^{2}\)>(x-y)

Rewriting: (x-y)(x+y)>(x-y)

If (x-y)>0
Dividing both sides by (x-y) we are left with (x+y)>1.

If (x-y)<0
Dividing both sides by (x-y) we are left with (x+y)<1.

Because we don't know if (x-y) is positive, negative or zero, Statement 1 is insufficient.


Statement 2) \(x^{2}\)-\(y^{2}\)<(x+y)

Rewriting: (x-y)(x+y)<(x+y)

If (x+y)>0
Dividing both sides by (x+y) we are left with (x-y)<1.

If (x+y)<0
Dividing both sides by (x+y) we are left with (x-y)>1.

Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient.

Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient.

E is the answer.
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Is x^2 - y^2 > 0?  [#permalink]

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New post 10 Jan 2019, 00:53
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ocelot22 wrote:
hazelnut wrote:
Is \(x^2-y^2>0\) ?

(1) \(x^2-y^2>(x - y)\)
(2) \(x^2-y^2<(x + y)\)


Notice that x^2-y^2 factors to (x+y)(x-y) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(x-y)

then the question becomes ab>0 Yes/NO?

(1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=-1/2 a=1/2, then ab<0 NO.

Since we get a Yes and a No. (1) is NS

(2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=-1 b=2, then ab<0 No Since we get a Yes and a No NS

(1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=-1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E.


Hi ocelot22

Nice solution. However, I would take your concept in simplicity by the following:

ab>b
ab<a
-----------Subtract the inequalities as different signs

ab -ab > b -a

0 > b -a

a > b....Form here it is easy to judge quickly

If a = 10 & b= 4...Then answer is Yes

If a = -10 & b = 4....Then answer No

Answer: E

hazelnut
Can you please post Veritas solution
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Re: Is x^2 - y^2 > 0?  [#permalink]

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New post 09 Jan 2019, 13:05
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hazelnut wrote:
Is \(x^2-y^2>0\) ?

(1) \(x^2-y^2>(x - y)\)
(2) \(x^2-y^2<(x + y)\)


Notice that x^2-y^2 factors to (x+y)(x-y) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(x-y)

then the question becomes ab>0 Yes/NO?

(1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=-1/2 a=1/2, then ab<0 NO.

Since we get a Yes and a No. (1) is NS

(2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=-1 b=2, then ab<0 No Since we get a Yes and a No NS

(1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=-1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E.
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Re: Is x^2 - y^2 > 0?  [#permalink]

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New post 22 Mar 2017, 02:01
ziyuen wrote:
Is \(x^2-y^2>0\) ?

(1) \(x^2-y^2>(x - y)\)
(2) \(x^2-y^2<(x + y)\)


Hi

In other words "Is \(x^2 > y^2\) ?

(1) \(x^2-y^2>(x - y)\)

\(x(x - 1) > y(y - 1)\)

We can have \(x>y>0\) -----> \(3>2>0\)

\(3*2>2*1\) --------> \(6^2 > 2^2\) Yes.

Or \(0<x<y\) ---------> \(0<\frac{1}{3}<\frac{1}{2}\)

\(\frac{1}{3}*(-\frac{2}{3}) "?" \frac{1}{2} * (-\frac{1}{2})\) -------> \(-\frac{2}{9} > -\frac{1}{4}\) ----------> \((\frac{1}{3})^2 < (\frac{1}{2})^2\) No. Insufficient.

(2) \(x^2-y^2<(x + y)\)

\(x(x - 1) < y(y + 1)\)

\(x=2, y=4\) ----> \(2*1 < 4*5\) -----> \(2^2 < 4^2\) No.

\(x=\frac{1}{2}, y=\frac{1}{3}\) ---> \(\frac{1}{2}*(-\frac{1}{2}) "?" \frac{1}{3}*\frac{4}{3}\) -----> \(-\frac{1}{4} < \frac{4}{9}\) -------> \((\frac{1}{2})^2 > (\frac{1}{4})^2\) Yes. Insufficient.

(1)&(2) \(x^2 - y^2 > x - y\) & \(x^2 - y^2 < x + y\)

or \(y^2 - x^2 < y - x\) & \(x^2 - y^2 < x + y\)

Adding we'll get: 0 < 2y ----> y>0. But we already have different answers when y is positive (x can be >0, <0, integer or a fraction). Insufficient.

Answer E.

What's the source of the question?
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Re: Is x^2 - y^2 > 0?  [#permalink]

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New post 09 Jan 2019, 20:13
emockus wrote:
Here's how I solved this - please critique!

Is \(x^{2}\)-\(y^{2}\)>0?

Rewriting the stem: Is (x-y)(x+y)>0?

Statement 1) \(x^{2}\)-\(y^{2}\)>(x-y)

Rewriting: (x-y)(x+y)>(x-y)

If (x-y)>0
Dividing both sides by (x-y) we are left with (x+y)>1.

If (x-y)<0
Dividing both sides by (x-y) we are left with (x+y)<1.

Because we don't know if (x-y) is positive, negative or zero, Statement 1 is insufficient.


Statement 2) \(x^{2}\)-\(y^{2}\)<(x+y)

Rewriting: (x-y)(x+y)<(x+y)

If (x+y)>0
Dividing both sides by (x+y) we are left with (x-y)<1.

If (x+y)<0
Dividing both sides by (x+y) we are left with (x-y)>1.

Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient.

Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient.

E is the answer.


Thanks for the explanation
I dont think we can divide by x-y. since its no where mentioned that x is not equal to y
similarly for the next given statement x can be equal to -y.
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Re: Is x^2 - y^2 > 0?  [#permalink]

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New post 24 May 2019, 05:05
Bunuel, could you provide your insight on this question? I'm stuck on why E is the answer.

Per Statement 1, when x-y >0, x + y >1 (x=3, y=2) and |x|>|y|. When x-y<0, x+y<1 (x=(-2.5), y=3), and |x|<|y|. NS.

Per Statement 2, when x+y>0, x-y<1 (x=3, y=2.5) and |x|>|y|. When x+y<0, x-1>1 (x=(-2), y=(-4), and |x|<|y|. NS.

x=3 and y = 2, works for both statements. I can't find values that work for |x|<|y| fr both statements.
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Re: Is x^2 - y^2 > 0?  [#permalink]

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New post 25 May 2019, 08:52
hazelnut wrote:
Is \(x^2-y^2>0\) ?

(1) \(x^2-y^2>(x - y)\)
(2) \(x^2-y^2<(x + y)\)


Hi, I answered E using the following approach.
(1)
\(x^2-y^2>(x - y)\)
\((x + y)(x - y)>(x - y)\)
\((x + y)>1\)
Insufficient.

(2)
\(x^2-y^2<(x + y)\)
\((x + y)(x - y)<(x + y)\)
\((x - y)<1\)
Insufficient.

Together both statements are still insufficient as the possibility for (x - y) can range from positive fraction < 1 to unlimited negatives.
Let me know if that makes sense.
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Re: Is x^2 - y^2 > 0?   [#permalink] 25 May 2019, 08:52
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