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# M28-03

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Math Expert
Joined: 02 Sep 2009
Posts: 53020
M28-03  [#permalink]

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16 Sep 2014, 00:28
00:00

Difficulty:

35% (medium)

Question Stats:

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

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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) $$x(y-3)=0$$

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Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re M28-03  [#permalink]

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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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Concentration: Entrepreneurship, Strategy
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Re: M28-03  [#permalink]

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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
Intern
Joined: 07 Feb 2017
Posts: 6
M28-03  [#permalink]

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7334
Re: M28-03  [#permalink]

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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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

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Re: M28-03  [#permalink]

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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.
Manager
Joined: 02 Jan 2016
Posts: 123
Re: M28-03  [#permalink]

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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 ?
Math Expert
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Posts: 53020
M28-03  [#permalink]

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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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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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Joined: 09 Mar 2018
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M28-03  [#permalink]

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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

NEVER assume things.
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: M28-03  [#permalink]

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

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