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If x=y^2

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If x=y^2  [#permalink]

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New post 02 Mar 2018, 18:04
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A
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C
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E

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Question Stats:

71% (00:57) correct 29% (00:59) wrong based on 17 sessions

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

(1) x = 4

(2) x + y = 2

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If x=y^2  [#permalink]

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New post 02 Mar 2018, 21:31
pushpitkc niks18 Hatakekakashi
amanvermagmat Bunuel chetan2u

Any another approach than number picking here?

I tried below, but could not get to OE:

St 1. From x = 4, we get y = +/- 2
Question asks: y-x or y - \(y^2\) or y * (y-1) = ?
since we have two values of y from st 1 , this st will not lead
to UNIQUE ans and hence is insuff.

St 2: x+y =2
\(y^2\) + y = 2
y * (y+1) =2
How to proceed from here?
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Re: If x=y^2  [#permalink]

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New post 02 Mar 2018, 22:19
y^2 + y = 2
y^2 + y - 2 = 0
Solve this quadratic equation and again you will get two values of y as:

y=1 and y=-2

Combine the two statements and you have your answer

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If x=y^2  [#permalink]

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New post 02 Mar 2018, 22:29
Thanks amanvermagmat

Is my analysis of St 1 correct or you have better approach?
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If x=y^2  [#permalink]

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New post 02 Mar 2018, 22:38
Hey adkikani

I think another way of going about this(at least with regards to statement 2 should be)

St2 : This can be rewritten as y = 2 - x
Therefore, the value y - x will take is 2 - x - x = 2 - 2x.
Since we don't know the value of x, we cannot come to a conclusive answer.

But then on combining the information from both the statements,
we have a unique solution for y - x = 2 - 2(4) = -6

That's why the answer should be Option C. Hope this helps!
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Re: If x=y^2  [#permalink]

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New post 03 Mar 2018, 02:06
adkikani wrote:
If x = \(y^2\), what is the value of y – x?

(1) x = 4

(2) x + y = 2


This question is discussed here: if-x-y-2-what-is-the-value-of-y-x-1-x-4-2-x-y-206393.html Hope it helps.

Please search before posting and name topics properly. The name of this topic should have been: "If x = y^2, what is the value of y – x? (1) x = 4 (2) x + y = 2" NOT "If x=y^2". Such short names mess up similar topics search block. Thank you.

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Re: If x=y^2 &nbs [#permalink] 03 Mar 2018, 02:06
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