If |x−y|=y, what is the value of x?

Question

(1) xy>0
(2) y=6

chetan2u · Accepted Answer

Hi,

lets see what info can we get from Q 
|x−y|=y
 inference:  
1)y is positive
2) x= 0 or x=2y

lets see the statement 
 (1) xy>0 
this tells us that xis NOT equal to 0..
we already know both x and y are positive
so x=2y..
we do not know the NUMERIC value of y..
Insuff

(2) y=6 
this tells us either x=2y=12 or x=0..
two values
Insuff

Combined
X is not 0, so x=12
Suff

C

Mo2men · Answer

|x−y| means the LHS is Positive.....Thus, y is positive

1) xy>0

Let x =2, y = 1

Let x = 4, y = 2

Insufficient

2) y=6

|x−6|=6

x = 12 or x =-6

Insufficient

Combine 1 & 2

y is positive and according to Statement 1, x must be positive

We get one solution x =12

Answer: C

GMATYoda · Answer

Using the two-case rule for absolute values, you can restate the information in the stimulus as:

Then performing the algebra in each case, the first tells you that x = 2y, and the second tells you that x = 0. 
Those are then the two possibilities for x: either x = 0, or it equals 2y.

The correct answer is C.

fskilnik · Answer

VERY beautiful problem, Bunuel. (Kudos!)

? = x

\left| {x - y} ight| = y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,y \geqslant 0\,\,\,\,\,\,\,\,{	ext{AND}}\,\,\,\,\,\,\,\left\{ \begin{gathered}
 \,\,x = 0\,\,\,,\,\,\,y \geqslant 0\,\,\,{	ext{free}} \hfill \
 \,\,{	ext{OR}}\,\,\,\, \hfill \
 0 
e x\,\,,\,\,\,{	ext{0}}\,\,\mathop {	ext{ < }}\limits^{\left( * ight)} \,\,{	ext{dist}}\left( {x,y} ight) = {	ext{dist}}\left( {y,0} ight)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,0 < y < x = 2y \hfill \ 
\end{gathered} ight.

\left( * ight)\,\,\,0 
e x = y\,\,\,\,\, \Rightarrow \,\,\,\,\,0 = \left| {x - y} ight| = y = x\,\,\,\,{	ext{impossible}}\,\,\,

\left( 1 ight)\,\,\,xy > 0\,\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {2,1} ight)\,\,\,\, \Rightarrow \,\,\,{	ext{?}}\,\,{	ext{ = }}\,\,{	ext{2}}\,\, \hfill \
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {4,2} ight)\,\,\,\, \Rightarrow \,\,\,{	ext{?}}\,\,{	ext{ = }}\,\,{	ext{4}}\,\, \hfill \ 
\end{gathered} ight.

\left( 2 ight)\,\,\,y = 6\,\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {0,6} ight)\,\,\,\, \Rightarrow \,\,\,{	ext{?}}\,\,{	ext{ = }}\,\,{	ext{0}}\,\, \hfill \
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {12,6} ight)\,\,\,\, \Rightarrow \,\,\,{	ext{?}}\,\,{	ext{ = }}\,\,{	ext{12}}\,\, \hfill \ 
\end{gathered} ight.

\left( {1 + 2} ight)\,\,\,\,\,\left\{ \begin{gathered}
 \,x 
e 0 \hfill \
 \,y = 6 \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = x = 2y = 12\,\,\,\,\, \Rightarrow \,\,\,\,\,{	ext{SUFF}}.

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

sahilvermani · Answer

(1) xy > 0
So, both x and y are positive or both x and y are negative.

But from the question statement (|x−y|=y) we know that y is positive.
So, basically both x and y are positive.
But from this, we cannot find the value of x.
Insufficient.

(2)

|x-6| = 6

x > 6
x - 6 = 6
x = 12

or

x < 6
-x + 6 = 6
x = 0

So, two solutions (x=12 or x=0) for x. Hence, insufficient.

Combining, since xy >0, we know that x cannot be 0. So, the only value possible is x=12.

Sufficient.
C.

natheodore · Answer

If |x−y|=y, what is the value of x?

(1) xy>0
(2) y=6

For (1)

both x and y has to be positive or negative. However, y cannot be negative here, else it will not be equal to LHS. Insufficient

For (2)

|x-6|=6

x=0 or x=12. Insufficient

Using (1) and (2),

x= 12. Answer is C

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If |x−y|=y, what is the value of x?

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