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Does (x + a)^2 = y^2?

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Does (x + a)^2 = y^2? [#permalink]

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New post 15 Jan 2015, 06:02
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Question Stats:

77% (00:41) correct 23% (01:08) wrong based on 101 sessions

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Re: Does (x + a)^2 = y^2? [#permalink]

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New post 17 Jan 2015, 01:54
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(1) We can subtract a from both sides of the equation and square it and we get to the equation in the question stem. So statement 1 is sufficient for any values x, y and a.

(2) This statement seems to contradict the equation in the question stem, so we are inclined to conclude that the statement is sufficient to answer the question. However, because the only equation we get is \((y+2a)^2=y^2\). However, there might be values for y and a that make the equation true. If a=0, then \((y+2a)^2=y^2\). This statement is therefore not sufficient.

Answer A.
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Re: Does (x + a)^2 = y^2? [#permalink]

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New post 17 Jan 2015, 02:05
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Does \((x + a)^2 = y^2\)?

(1) x = y - a
=> x=y-a
=> x-a=y
==> squaring both sides => \((x+a)^2 = y^2\) sufficient

(2) x = y + a

=> x-a=y
=> squaring both sides => \((x-a)^2 = y^2\) insufficient

Hence ans is A!

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Re: Does (x + a)^2 = y^2? [#permalink]

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New post 17 Jan 2015, 06:18
Here we go -


St1: x = y - a

Q : (x + a)^2 = y^2
Substitute the value given in st1

( y - a + a)^2 = y ^ 2

Clearly statement 1 is sufficient

St2: x = y + a

Substitute
(y + 2a) ^ 2
No information is given about a...
so it may or may not be equal to y^2

Clearly option A is right

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Re: Does (x + a)^2 = y^2? [#permalink]

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New post 21 Nov 2017, 10:01
Bunuel wrote:
Does (x + a)^2 = y^2?

(1) x = y - a

(2) x = y + a

Kudos for a correct solution.



Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.
We can modify the original condition and question as follows.

\((x + a)^2 = y^2\)
\(⇔ (x + a)^2 - y^2 = 0\)
\(⇔ (x + a + y )(x+a-y) = 0\)
\(⇔ x = -y - a\) or \(x = y - a\)

Therefore, the condition 1) only is sufficient.
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Re: Does (x + a)^2 = y^2? [#permalink]

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New post 21 Nov 2017, 12:02
from Statement 1
x= y-a
so put the value in main equation
(x+a)^2
(Y-a+a)^2
Y^2
same as RHS
sufficient
from statement 2 :
x = y + a
put in the equation (x+a)^2=y^2
we will get (y+2a)^2
This can be equal to y^2 when a= 0
and for some values it will not be equal to Y^2
hence we get more than one possibility.
therefore insufficient.

answer : A

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Re: Does (x + a)^2 = y^2?   [#permalink] 21 Nov 2017, 12:02
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