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If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12

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If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 26 Dec 2016, 10:59
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If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post Updated on: 27 Dec 2016, 03:51
Bunuel wrote:
If |x – y| = 10, what is the value of x?

(1) x > 0
(2) y = 12


Question: If absolute value of difference of x and y is 10, then find "x"?

Statement -1 is insufficient as it does not gives us the value of "y".

Statement - 2 is insufficient as x can be 2 or 22.

Even combining both we don't get a single value for "x"
So IMO (E)
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Originally posted by QZ on 26 Dec 2016, 11:16.
Last edited by QZ on 27 Dec 2016, 03:51, edited 2 times in total.
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Re: If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 27 Dec 2016, 03:02
Distance between x and y on the number line is 10, what is x

1 is obviously insuff

From 2

X could be 22 or 2

Both

Still insuff

E


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Re: If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 07 Jan 2017, 02:03
Hi,
I don´t understand why statement 2 is also insuficient.

Statement 2: y=12

x-12=10
x=10+12
x= 22

Suff.
Thanks!!
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If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post Updated on: 09 Jan 2017, 02:51
lauramo wrote:
Hi,
I don´t understand why statement 2 is also insuficient.

Statement 2: y=12

x-12=10
x=10+12
x= 22

Suff.
Thanks!!


The concept is an "absolute value" is the distance between two points on a number line and therefore it is always greater than or equal to zero. It can never be negative as distance can never be negative.

For this question, LHS is an absolute value so it can't be negative and hence x can be 2 & 22 both.

|2-12| = |-10| = 10

|22-12| = |10| = 10

So there are two values of x satisfying this equation, and hence the answer is E.

For Y/N questions we need one and only one value to make the statement sufficient.

Hope this helps.
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Originally posted by QZ on 07 Jan 2017, 02:29.
Last edited by QZ on 09 Jan 2017, 02:51, edited 1 time in total.
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Re: If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 07 Jan 2017, 04:21
Thank you very much for the response.

I see why x can have two values: 2 or 10.
However what I still don´t understand is why we cannot resolve this inequality as a regular equation to obtain only one value.

Thanks.
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If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 07 Jan 2017, 12:06
lauramo wrote:
Thank you very much for the response.

I see why x can have two values: 2 or 10.
However what I still don´t understand is why we cannot resolve this inequality as a regular equation to obtain only one value.

Thanks.



It can be solved as a regular inequality if there is no modulus sign.

If \(x - y = 10\) then as per statement 2 x can only have a single value as 22.

But for\(|x – y| = 10\), we need to take care of absolute value on both sides of "0"on number line.

\(|x – y| = (x - y) = 10\) , putting y = 12 we get x - 12 = 10 and \(x = 22\)

\(|x – y| = -(x - y) = 10\), putting y = 12 we get,

\(-x + 12 = 10\)

\(-x =10 - 12\)

\(-x = -2\)

\(x = 2\)

If I say absolute value of x is 5 then it means distance of x from 0 on the number line is 5 units and these 5 units will be on both sides of "0".
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Re: If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 09 Jan 2017, 02:15
Of course!!
Thank you so much for your patience!
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Re: If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12 [#permalink]

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New post 01 Sep 2017, 15:20
Bunuel wrote:
If |x – y| = 10, what is the value of x?

(1) x > 0
(2) y = 12


Let's break this down

Stmnt 1

No info about y and x and y could take on infinite values

Stmnt 2

two ways to break down abs. value

Pos
x-12 =10
x =22

Neg
-lx-12l =10
-x +12 =10
-x= -2
x =2

insuff

Stmnt 1 and 2

Even if we know that x is greater than 0 there are two values that fit that criteria insuff

E
Re: If |x – y| = 10, what is the value of x? (1) x > 0 (2) y = 12   [#permalink] 01 Sep 2017, 15:20
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