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Intern  Joined: 17 Dec 2012
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If x and y are integers, what is the value of y ?  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 53% (02:10) correct 47% (02:15) wrong based on 273 sessions

### HideShow timer Statistics If x and y are integers, what is the value of y ?

(1) 2x = y, and x^3 = y^2

(2) y - x = 4, and x^2= 2y

Originally posted by irda on 11 Jun 2014, 03:34.
Last edited by Bunuel on 11 Jun 2014, 03:43, edited 1 time in total.
Edited the OA.
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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If x and y are integers, what is the value of y ?

(1) 2x = y, and x^3 = y^2 --> $$x^3 = (2x)^2$$ --> $$x^2(x-4)=0$$ --> $$x=0$$ and $$y=0$$ OR $$x=4$$ and $$y=8$$. Not sufficient.

(2) y - x = 4, and x^2= 2y --> $$x^2=2(x+4)$$ --> $$x=-2$$ and $$y=2$$ OR $$x=4$$ and $$y=8$$. Not sufficient.

(1)+(2) Common solutions are $$x=4$$ and $$y=8$$. Sufficient.

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Re: If x and y are integers, what is the value of y ?  [#permalink]

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Bunuel wrote:
If x and y are integers, what is the value of y ?

(1) 2x = y, and x^3 = y^2 --> $$x^3 = (2x)^2$$ --> $$x^2(x-4)=0$$ --> $$x=0$$ and $$y=0$$ OR $$x=4$$ and $$y=8$$. Not sufficient.

(2) y - x = 4, and x^2= 2y --> $$x^2=2(x+4)$$ --> $$x=-2$$ and $$y=2$$ OR $$x=4$$ and $$y=8$$. Not sufficient.

(1)+(2) Common solutions are $$x=4$$ and $$y=8$$. Sufficient.

Why statement 1 is not enough ?

as per statement 1, 2x=y i.e. x=y/2 ---1

statement 1 also says, x^3 = y^2, so if we substitute 1 here, we get (y/2)^3 = y^2, solving this we get y = 8.
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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griffins wrote:
Bunuel wrote:
If x and y are integers, what is the value of y ?

(1) 2x = y, and x^3 = y^2 --> $$x^3 = (2x)^2$$ --> $$x^2(x-4)=0$$ --> $$x=0$$ and $$y=0$$ OR $$x=4$$ and $$y=8$$. Not sufficient.

(2) y - x = 4, and x^2= 2y --> $$x^2=2(x+4)$$ --> $$x=-2$$ and $$y=2$$ OR $$x=4$$ and $$y=8$$. Not sufficient.

(1)+(2) Common solutions are $$x=4$$ and $$y=8$$. Sufficient.

Why statement 1 is not enough ?

as per statement 1, 2x=y i.e. x=y/2 ---1

statement 1 also says, x^3 = y^2, so if we substitute 1 here, we get (y/2)^3 = y^2, solving this we get y = 8.

If you read the solution you quote carefully you'll see that there exist two solutions for y, 0 and 8.

I think you reduced (y/2)^3 = y^2 by y^2, which is wrong.

If you divide (reduce) (y/2)^3 = y^2 by y^2 you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND y = 8 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
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If x and y are integers, what is the value of y ?
(1) 2x = y, and x^3 = y^2.
(2) y - x = 4, and x^2 = 2y.
Intern  Joined: 07 Feb 2015
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If x and y are integers, what is the value of y  [#permalink]

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IMO B

Explanation :
Statement 1 : y=2x and x^3=y^2
x^3=4x^2
x^2(x-4)=0
x=0,4
y=0,8
(x,y)=(0,0) or(4,8)
we are getting two values of y so INSUFFICIENT (A cant be the answer )

Statement 2 : y-x=4 and x^2=2y
x^2-2x-8=0
(x-4)(x+2)=0
x=4,-2
y=8,-4
(x,y)=(4,8)or(-2,-4)
again two values of y so INSUFFICIENT
if we combine both statements we get (x,y)=(4,8)
y=8
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Originally posted by masoomdon on 09 Mar 2015, 00:38.
Last edited by masoomdon on 09 Mar 2015, 02:14, edited 1 time in total.
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Re: If x and y are integers, what is the value of y  [#permalink]

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masoomdon wrote:
IMO B

Explanation :
Statement 1 : y=2x and x^3=y^2
x^3=4x^2
x^2(x-4)=0
x=0,4
y=0,8
(x,y)=(0,0) or(4,8)
we are getting two values of y so INSUFFICIENT (A cant be the answer )

Statement 2 : y-x=4 and x^2=2y
x^2-2x-8=0
(x-4)(x+2)=0
x=4,-2
y=8,-4
(x,y)=(4,8)or(-2,-4)
again two values of y so INSUFFICIENT
if we combine both statements we get (x,y)=(4,8)
y=8

Well I guess you want to say C and not B....because your solution (combining both statements) points towards C
Even I got C hence I have posted this question here as the answer given is A which even I feel is wrong....they have cancelled out x in option A rather than subtracting and by cancelling out x we get only one value for x which is 4....

I am waiting for some expert to reply and confirm...thanks.
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Re: If x and y are integers, what is the value of y  [#permalink]

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fireinbelly wrote:
masoomdon wrote:
IMO B

Explanation :
Statement 1 : y=2x and x^3=y^2
x^3=4x^2
x^2(x-4)=0
x=0,4
y=0,8
(x,y)=(0,0) or(4,8)
we are getting two values of y so INSUFFICIENT (A cant be the answer )

Statement 2 : y-x=4 and x^2=2y
x^2-2x-8=0
(x-4)(x+2)=0
x=4,-2
y=8,-4
(x,y)=(4,8)or(-2,-4)
again two values of y so INSUFFICIENT
if we combine both statements we get (x,y)=(4,8)
y=8

Well I guess you want to say C and not B....because your solution (combining both statements) points towards C
Even I got C hence I have posted this question here as the answer given is A which even I feel is wrong....they have cancelled out x in option A rather than subtracting and by cancelling out x we get only one value for x which is 4....

I am waiting for some expert to reply and confirm...thanks.

yeah that was a mistake i meant (C)
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If x and y are integers, what is the value of y ?  [#permalink]

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If x and y are integers, what is the value of y ?
(1)$$2x = y$$, and $$x^3 = y^2$$
(2) $$y - x = 4$$, and $$x^2 = 2y$$
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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C.

Statement 1: Solving statement 1, we get x = 0 or x = 4. INSUFFICIENT.

Statement 2: Solving statement 2, we get x = 4 or x = 0. INSUFFICIENT.

1+2 : x =4. SUFFICIENT.
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St1:

y = 2x
y^2 = 4x^2

Plugging this in the other equation:

x^3 = y^2
x^3 = 4x^2
x^3 - 4x^2 = 0
x^2(x - 4) = 0
x can be 0 or 4 -> y can be 0 or 8.
Insufficient.

St2:

y = x + 4

Plugging this in the other equation:

x^2 = 2y
x^2 = 2x + 8
x^2 - 2x - 8 = 0
(x + 2)(x - 4) = 0
x can be -2 or 4 -> y can be 8 or 2.
Insufficient.

St1 + St2:

x = 4.
y = 8.
Sufficient.

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Re: If x and y are integers, what is the value of y ?  [#permalink]

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BrainLab are you sure OA is A.

I also getting C.
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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abhimahna wrote:
BrainLab are you sure OA is A.

I also getting C.

Hi,

yes, the official answer is A, but I also don't agree with it. You can not cancel out a variable if you are not 100% sure that this variable is not equal to zero.
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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BrainLab wrote:
abhimahna wrote:
BrainLab are you sure OA is A.

I also getting C.

Hi,

yes, the official answer is A, but I also don't agree with it. You can not cancel out a variable if you are not 100% sure that this variable is not equal to zero.

Yes, I never cancelled any variable. This is the Basic rule of DS which I am aware of. But my concern is how can the answer be A.

May be we need to wait for some experts to help !!
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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BrainLab wrote:
If x and y are integers, what is the value of y ?
(1)$$2x = y$$, and $$x^3 = y^2$$
(2) $$y - x = 4$$, and $$x^2 = 2y$$

Merging topics. Please refer to the discussion above.

P.S. The correct answer is C, not A.
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If x and y are integers, what is the value of y?  [#permalink]

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If x and y are integers, what is the value of y?

1.) $$2x = y$$, and $$x^3 = y^2$$

2.) $$y-x = 4$$, and $$x^2 = 2y$$

Source: Jeff Sackman Algebra Challenge
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Re: If x and y are integers, what is the value of y?  [#permalink]

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I keep getting C. OA is A though. Plz correct me where I’m going wrong on this one.

Here is how I approach this:

From Condition 1 I get:

x^3 = (2x)^2
x^3 = 4x^2
x^3 - 4x^2 = 0

(x^2)(x-4) = 0

x could be either 0 or 4.

Plug into 2x = y, y could be either 0 or 8. Hence Insuff.

From Condition 2:

x^2 = 2(4+x)
x^2 = 8 + 2x
x^2 - 2x - 8 = 0
(x-4)(x+2) = 0

Hence x could be either -2 or 4.

Plug into y - x = 4, y could be either 2 or 8. Hence Insuff.

Taking both together y can be 8 for both conditions, so y will be 8. So C.
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Re: If x and y are integers, what is the value of y ?  [#permalink]

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Scyzo wrote:
If x and y are integers, what is the value of y?

1.) $$2x = y$$, and $$x^3 = y^2$$

2.) $$y-x = 4$$, and $$x^2 = 2y$$

Source: Jeff Sackman Algebra Challenge

Merging topics. Please refer to the discussion above.

P.S. The correct answer is C, not A.
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