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M28-03

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M28-03  [#permalink]

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New post 16 Sep 2014, 00:28
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

64% (00:46) correct 36% (00:56) wrong based on 263 sessions

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Re M28-03  [#permalink]

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New post 16 Sep 2014, 00:28
1
Official Solution:


Given \(y = \frac{x^2 - y^2}{x-y}\). Simplify: \(y = \frac {(x - y)(x+y)}{x-y}\). Reduce by x-y: \(y=x+y\). So, we have that \(x=0\).

(1) \(x+y-3=0\). Since \(x=0\), then \(y=3\). Sufficient.

(2) \(x(y-3)=0\). Since \(x=0\), then \(y\) can take ANY value. Not sufficient.


Answer: A
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Re: M28-03  [#permalink]

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New post 14 Jun 2017, 06:49
If x≠yx≠y and y=x2−y2x−yy=x2−y2x−y, then what is the value of yy?
(1) x+y−3=0x+y−3=0
(2) x(y−3)=0

I am not so sure for statement 2.
If x(y−3)=0
They x=0 or y=3
If x=0, they y=0. As per question If x≠y, so y cannot be 0.
Not sure whats wrong with my logic.
Any expert can explain.
ThankS
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New post 19 Jul 2017, 08:01
Given y=x2−y2x−yy=x2−y2x−y. Simplify: y=(x−y)(x+y)x−yy=(x−y)(x+y)x−y. Reduce by x-y: y=x+yy=x+y. So, we have that x=0x=0.

(1) x+y−3=0x+y−3=0. Since x=0x=0, then y=3y=3. Sufficient.

(2) x(y−3)=0x(y−3)=0. Since x=0x=0, then y can take ANY value. Not sufficient.

regarding point two .
i did not understand why Y can take any value ? .
(y-3) has to be equal to zero ?
thanks in advance
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Re: M28-03  [#permalink]

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New post 19 Jul 2017, 08:14
1
ashimd wrote:
Given y=x2−y2x−yy=x2−y2x−y. Simplify: y=(x−y)(x+y)x−yy=(x−y)(x+y)x−y. Reduce by x-y: y=x+yy=x+y. So, we have that x=0x=0.

(1) x+y−3=0x+y−3=0. Since x=0x=0, then y=3y=3. Sufficient.

(2) x(y−3)=0x(y−3)=0. Since x=0x=0, then y can take ANY value. Not sufficient.

regarding point two .
i did not understand why Y can take any value ? .
(y-3) has to be equal to zero ?
thanks in advance



Hi statement 2 says x(y-3)=0..
If x=0, y-3 can be anything including 0 because 0*(y-3) will always be 0 irrespective of value of y-3..
But if x is Not EQUAL to 0, y-3 has to be 0...
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Re: M28-03  [#permalink]

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New post 11 Oct 2018, 06:53
This is a great question!

I chose D, because my approach for the 2) was

=X(Y-3)=0
= Xy-3x = 0
= xy = 3x
y=3

But I found why I was wrong.
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Re: M28-03  [#permalink]

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New post 15 Nov 2018, 18:49
I did exactly the same way... hanamana

=X(Y-3)=0
= Xy-3x = 0
= xy = 3x
y=3


Is it because we know from stem that X = 0 and 0/0 is undefined ?

But its not explicitly given in the question stem ?

can you me your opinion on this ? and Bunuel chetan2u Sirs ?
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New post 15 Nov 2018, 22:19
hero_with_1000_faces wrote:
I did exactly the same way... hanamana

=X(Y-3)=0
= Xy-3x = 0
= xy = 3x
y=3


Is it because we know from stem that X = 0 and 0/0 is undefined ?

But its not explicitly given in the question stem ?

can you me your opinion on this ? and Bunuel chetan2u Sirs ?


From the stem (after we simplified given equation \(y = \frac{x^2 - y^2}{x-y}\)) we got that x = 0. Since we cannot divide by 0, then you cannot divide xy = 3x by x to get y = 3. Basically for (2) we have that 0(y - 3) = 0. Now, 0(y - 3) is 0 regardless of the value of y.
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Re M28-03  [#permalink]

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New post 26 Jan 2019, 20:22
I think this is a poor-quality question and I don't agree with the explanation. For statement 2, if you know one of the values in an equation, you substitute and find the other value. (Basic of GMAT math). this explanation seems to violate that
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M28-03  [#permalink]

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New post 26 Jan 2019, 22:44
Bunuel wrote:
If \(x\neq{y}\) and \(y = \frac{x^2 - y^2}{x-y}\), then what is the value of \(y\)?


(1) \(x+y-3=0\)

(2) \(x(y-3)=0\)


y = x-y * x+y / x-y

y = x + y

from this we can say x =0 but y can be any value, integer ,fraction or a number

(1) \(x+y-3=0\)
From here we get a definite value as y = 3
Sufficient

(2) \(x(y-3)=0\)[/quote]
Now, i didn't realize this initially

0 * (y-3) = 0

y can be any value > 3

3.1, 6 , 9, these values will satisfy the statement

But will give multiple values for the question what is the value of y

A

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Re: M28-03  [#permalink]

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New post 27 Jan 2019, 01:50
itsrahulmohan wrote:
I think this is a poor-quality question and I don't agree with the explanation. For statement 2, if you know one of the values in an equation, you substitute and find the other value. (Basic of GMAT math). this explanation seems to violate that


Have you read this?
(2) \(x(y-3)=0\). Since \(x=0\), then \(y\) can take ANY value. Not sufficient.

We want of find the value of y. Can you?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M28-03   [#permalink] 27 Jan 2019, 01:50
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