GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 06 Dec 2019, 13:07 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x + y^2 = (x + y ^2)^2, what is the value of y?

Author Message
TAGS:

### Hide Tags

Intern  Joined: 22 Jul 2010
Posts: 26
If x + y^2 = (x + y ^2)^2, what is the value of y?  [#permalink]

### Show Tags

1
17 00:00

Difficulty:   95% (hard)

Question Stats: 47% (02:32) correct 53% (02:25) wrong based on 309 sessions

### HideShow timer Statistics

If x + y^2 = (x + y ^2)^2, what is the value of y?

(1) x = y ^2
(2) xy ^2 = 0
Math Expert V
Joined: 02 Sep 2009
Posts: 59587

### Show Tags

3
3
mitmat wrote:
this is from the same forum and the link to the question and topic for reference is ds-a-maze-of-algebra-1743.html

the question is:

If x + y^2= ( x + y ^2) ^2, what is the value of y?

( 1) x = y ^2
( 2) xy ^2 = 0

I feel the answer is E and i was unable to get to a conclusion from the thread previously posted. Can some one take a look at this?

If x + y^2= ( x + y ^2) ^2, what is the value of y?

(1) $$x=y^2$$. Substitute $$x$$ in $$x+y^2=(x+y^2)^2$$: $$2y^2=4y^4$$ --> $$y^2(2y^2-1)=0$$ --> $$y=0$$ or $$y=\frac{1}{\sqrt{2}}$$ or $$y=-\frac{1}{\sqrt{2}}$$. Not sufficient.

(2) $$xy^2=0$$ --> if $$x=0$$, then in this case: $$x+y^2=(x+y^2)^2$$ becomes: $$y^2=y^4$$ --> $$y^2(y^2-1)=0$$ --> $$y=0$$ or $$y=1$$ or $$y=-1$$. Not sufficient.

(1)+(2) $$x=y^2$$ and $$xy^2=0$$, substitute $$x$$: $$y^4=0$$ --> $$y=0$$. Sufficient.

_________________
##### General Discussion
Director  Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 574
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

### Show Tags

Bunuel wrote:
mitmat wrote:
this is from the same forum and the link to the question and topic for reference is ds-a-maze-of-algebra-1743.html

the question is:

If x + y^2= ( x + y ^2) ^2, what is the value of y?

( 1) x = y ^2
( 2) xy ^2 = 0

I feel the answer is E and i was unable to get to a conclusion from the thread previously posted. Can some one take a look at this?

If x + y^2= ( x + y ^2) ^2, what is the value of y?

(1) $$x=y^2$$. Substitute $$x$$ in $$x+y^2=(x+y^2)^2$$: $$2y^2=4y^4$$ --> $$y^2(2y^2-1)=0$$ --> $$y=0$$ or $$y=\frac{1}{\sqrt{2}}$$ or $$y=-\frac{1}{\sqrt{2}}$$. Not sufficient.

(2) $$xy^2=0$$ --> if $$x=0$$, then in this case: $$x+y^2=(x+y^2)^2$$ becomes: $$y^2=y^4$$ --> $$y^2(y^2-1)=0$$ --> $$y=0$$ or $$y=1$$ or $$y=-1$$. Not sufficient.

(1)+(2) $$x=y^2$$ and $$xy^2=0$$, substitute $$x$$: $$y^4=0$$ --> $$y=0$$. Sufficient.

Bunuel
For Point (2) above, xy2=0, we examine the case where x=0 to find values of y, what about when y=0, which means x-x^2=0 =>
x(x-1)=0; x=0 or x=1
Why did we not examine this sequence?
Math Expert V
Joined: 02 Sep 2009
Posts: 59587

### Show Tags

mainhoon wrote:
Bunuel wrote:
mitmat wrote:
this is from the same forum and the link to the question and topic for reference is ds-a-maze-of-algebra-1743.html

the question is:

If x + y^2= ( x + y ^2) ^2, what is the value of y?

( 1) x = y ^2
( 2) xy ^2 = 0

I feel the answer is E and i was unable to get to a conclusion from the thread previously posted. Can some one take a look at this?

If x + y^2= ( x + y ^2) ^2, what is the value of y?

(1) $$x=y^2$$. Substitute $$x$$ in $$x+y^2=(x+y^2)^2$$: $$2y^2=4y^4$$ --> $$y^2(2y^2-1)=0$$ --> $$y=0$$ or $$y=\frac{1}{\sqrt{2}}$$ or $$y=-\frac{1}{\sqrt{2}}$$. Not sufficient.

(2) $$xy^2=0$$ --> if $$x=0$$, then in this case: $$x+y^2=(x+y^2)^2$$ becomes: $$y^2=y^4$$ --> $$y^2(y^2-1)=0$$ --> $$y=0$$ or $$y=1$$ or $$y=-1$$. Not sufficient.

(1)+(2) $$x=y^2$$ and $$xy^2=0$$, substitute $$x$$: $$y^4=0$$ --> $$y=0$$. Sufficient.

Bunuel
For Point (2) above, xy2=0, we examine the case where x=0 to find values of y, what about when y=0, which means x-x^2=0 =>
x(x-1)=0; x=0 or x=1
Why did we not examine this sequence?

The question is "what is the value of $$y$$". So, we don't care what $$x$$ is.

Statement (2) is: $$xy^2=0$$. This statement would be sufficient if it could give us the single numerical value of $$y$$.

From (2):
Either $$y=0$$ OR $$x=0$$. $$y=0$$ is obvious solution and we should concentrate on another option, which is $$x=0$$, to see whether this case can give the same or different solutions of $$y$$ and thus to determine whether this statement is sufficient.

$$x=0$$ gives 3 values of $$y$$: 0, 1, and -1, so we can say that this statement is not sufficient.

Hope it's clear.
_________________
Director  Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 574
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

### Show Tags

So for (2) itself, y=0; x=0 or 1 is a valid solution. With x=0, we end up with y=0,1,-1 and thus if we went for (1)+(2), we can see that x=0,y=0 is the only solution that is common across all cases. In this case it so happens that the obvious solution (0,0) is what we end up with, however, in the general case (say for another problem), would you still not check the y=0 solutions for x, because the questions asks for the value of not x? I wanted to know the thought process. Thanks

Another question - when quote an answer in the forum, I see the window flicker and move up and down randomly (I have to try hard to keep the window and the space I am typing in in view.. Anything I am doing wrong here? Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 59587

### Show Tags

mainhoon wrote:
So for (2) itself, y=0; x=0 or 1 is a valid solution. With x=0, we end up with y=0,1,-1 and thus if we went for (1)+(2), we can see that x=0,y=0 is the only solution that is common across all cases. In this case it so happens that the obvious solution (0,0) is what we end up with, however, in the general case (say for another problem), would you still not check the y=0 solutions for x, because the questions asks for the value of not x? I wanted to know the thought process. Thanks

Another question - when quote an answer in the forum, I see the window flicker and move up and down randomly (I have to try hard to keep the window and the space I am typing in in view.. Anything I am doing wrong here? Thanks

If I'm asked to find y and say I get that y=some value, then why should I find the value of another unknown?

_________________
Senior Manager  Joined: 21 Oct 2013
Posts: 408
If x + y^2 = (x + y^2)^2, what is the value of y?  [#permalink]

### Show Tags

If x + y^2 = (x + y^2)^2, what is the value of y?

(1) x = y^2
(2) xy^2 = 0

OE
(1): Substitute y^2 for x into x + y^2 = (x + y^2)^2 → y^2 + y^2 = (y^2 + y^2)^2 → 2y^2 = 4y^4
As y may be equal to 0, cannot divide by y^2
2y^2 = 4y^4 → 4y^4 – 2y^2 = 0
Factor out 2y2 from left side of equation to give 2y^2(2y^2 – 1) = 0
→ either 2y^2 = 0 or 2y^2 – 1 = 0.
When product of a group of terms is 0, at least one of the terms must be 0
If 2y^2 = 0 → y = 0
If (2y^2 – 1) = 0 → 2y^2 = 1 → y = ±√(1/2)
→ y = 0 or ±√(1/2)
Insufficient

(2): Either x = 0 or y = 0
If x = 0, substitute 0 for x into x + y^2 = (x + y^2)^2
→ 0 + y^2 = (0 + y^2)^2 → y^2 = (y^2)^2 → y^2 = y^4 → y^4 – y^2 = 0 → y^2(y^2 – 1) = 0
If y^2 = 0 → y = 0
If y^2 – 1 = 0 → y^2 = 1 & y = ±1
If x = 0, → y = –1, 0, or 1
Insufficient

Combined: From (1) y = 0 or ±√(1/2). From (2) y = –1, 0, or 1 → y = 0
Sufficient

Hi, I want to know if we have more simple way to solve this question, please.
Math Expert V
Joined: 02 Sep 2009
Posts: 59587
Re: If x + y^2 = (x + y^2)^2, what is the value of y?  [#permalink]

### Show Tags

1
goodyear2013 wrote:
If x + y^2 = (x + y^2)^2, what is the value of y?

(1) x = y^2
(2) xy^2 = 0

OE
(1): Substitute y^2 for x into x + y^2 = (x + y^2)^2 → y^2 + y^2 = (y^2 + y^2)^2 → 2y^2 = 4y^4
As y may be equal to 0, cannot divide by y^2
2y^2 = 4y^4 → 4y^4 – 2y^2 = 0
Factor out 2y2 from left side of equation to give 2y^2(2y^2 – 1) = 0
→ either 2y^2 = 0 or 2y^2 – 1 = 0.
When product of a group of terms is 0, at least one of the terms must be 0
If 2y^2 = 0 → y = 0
If (2y^2 – 1) = 0 → 2y^2 = 1 → y = ±√(1/2)
→ y = 0 or ±√(1/2)
Insufficient

(2): Either x = 0 or y = 0
If x = 0, substitute 0 for x into x + y^2 = (x + y^2)^2
→ 0 + y^2 = (0 + y^2)^2 → y^2 = (y^2)^2 → y^2 = y^4 → y^4 – y^2 = 0 → y^2(y^2 – 1) = 0
If y^2 = 0 → y = 0
If y^2 – 1 = 0 → y^2 = 1 & y = ±1
If x = 0, → y = –1, 0, or 1
Insufficient

Combined: From (1) y = 0 or ±√(1/2). From (2) y = –1, 0, or 1 → y = 0
Sufficient

Hi, I want to know if we have more simple way to solve this question, please.

Merging similar topics. Please refer to the solution above.
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 13722
Re: If x + y^2 = (x + y ^2)^2, what is the value of y?  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: If x + y^2 = (x + y ^2)^2, what is the value of y?   [#permalink] 27 Nov 2019, 13:59
Display posts from previous: Sort by

# If x + y^2 = (x + y ^2)^2, what is the value of y?  