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Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink ]
20 Dec 2009, 06:50
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Is

\sqrt{(x-3)^2}=3-x ?

(1)

x\neq{3} (2) -x|x| > 3

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Ans: E. First statement is not sufficient because if x=4, it may or may not be true. LHS = +1 or -1 while RHS = -1. Second statement leads us to -3^1/2 < x < 0. For all the values between 0 and -3^1/2, which leads to |LHS| = RHS.

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jan4dday wrote:

Is \sqrt{(x-3)^2}=3-x ? (1) x not = 3 (2) -x|x| > 3 will give OA as soon as 1st few replies come in

Remember:

\sqrt{x^2}=|x| .

\sqrt{(x-3)^2}=|x-3| . So the question becomes is

|x-3|=3-x .

When

x>3 , then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When

x\leq{3} , then

LHS=|x-3|=-x+3=3-x=RHS , hence in this case equation holds true.

Basically question asks is

x\leq{3} ?

(1) x is not equal to 3. Clearly insufficient.

(2)

-x|x| > 3 , basically this inequality implies that

x<0 , hence

x<3 . Sufficient.

Answer: B.

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Re: How to solve this Question...I have no clue... [#permalink ]
27 Feb 2010, 20:47

is \sqrt{(x-3)^2} = 3-x? square root of a number is positive. if x<=3 .. 3-x is always positive if x>3 ... 3-x is negative st 1) x!=3 . not sufficient as we dont know if x>3 or x<3 st 2) -x |x| > 0 |x| is always greater than 0, so x has to be negative for the expression to be > 0 if x is negative, then \sqrt{(x-3)^2} = 3-x B

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Re: How to solve this Question...I have no clue... [#permalink ]
27 Feb 2010, 20:58

Thank you very much.

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Re: How to solve this Question...I have no clue... [#permalink ]
17 Mar 2011, 02:58

Just a slight correction I think... St2 is saying x<-sqrt of 3 which is still a sufficient condition.

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I got x<-1.732 from statement 2

This satisfies the x<3 requisite

Hence B

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Re: How to solve this Question...I have no clue... [#permalink ]
17 Mar 2011, 22:07

sqrt((x-3)^2) = |x-3| = x-3 if x > 3 or 3-x if x < 3

(1) is not enough

(2) -x|x| is positive then

So -x is also +ve and hence x is -ve

so x < 3, hence answer is B

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Re: How to solve this Question...I have no clue... [#permalink ]
18 Mar 2011, 19:49

I'm a bit confused by this problem Why can I not square both sides of the statement to get rid of the radical? I will then have (x-3)^2=(3-x)^2 Statement 1 and 2 both will not apply in this case and the answer is then E. However if I choose Bunuels method of doing it, I still dont get how B is the answer. Statement 2 basically says that x=-2,-3....and so on and so forth will satisfy the statement. So basically all negative numbers equal to or less than 2 will satisfy the condition...

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Is square_root of (x-3)^2 = 3 - x ? [#permalink ]
27 Jun 2012, 11:08

Is

\sqrt{(x-3)^2} = 3 - x ?

(1) x is not equal to 3

(2)

-x |x| > 0 According to my reasoning:

\sqrt{(x-3)^2} = |x-3| So, the question is:

Is

|x-3| = 3 -x ?

In other words:

Is

x - 3 < 0 ?

Is

x < 3 ?

So, let's see the clues:

(1) x is not equal to 3

INSUFFICIENT.

(2)

-x |x| > 0 Based on this info we know that x is not zero.

So,

-x > 0 x < 0 SUFFICIENT.

IMO, the OA is :

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Re: Is square_root of (x-3)^2 = 3 - x ? [#permalink ]
27 Jun 2012, 11:15

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Bunuel wrote:

jan4dday wrote:

Is \sqrt{(x-3)^2}=3-x ? (1) x not = 3 (2) -x|x| > 3 will give OA as soon as 1st few replies come in

Remember:

\sqrt{x^2}=|x| .

\sqrt{(x-3)^2}=|x-3| . So the question becomes is

|x-3|=3-x .

When

x>3 , then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When

x\leq{3} , then

LHS=|x-3|=-x+3=3-x=RHS , hence in this case equation holds true.

Basically question asks is

x\leq{3} ?

(1) x is not equal to 3. Clearly insufficient.

(2)

-x|x| > 3 , basically this inequality implies that

x<0 , hence

x<3 . Sufficient.

Answer: B.

Dear Bunuel,

is it possible in this question to solve by taking conditions x>0 and x<0 for |x-3| = 3-x instead of x>3 ?

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Re: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink ]
01 Jul 2012, 12:07

B it is!!Good explanation though!

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DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink ]
21 Nov 2012, 19:11

I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help

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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink ]
21 Nov 2012, 21:02

arvindbhat1887 wrote:

I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help

The question is

Is

\sqrt{(x-3)^2}= 3-x ?

(1) x<>3

(2) -x|x|>0

\sqrt{y} can only be positive or 0.

So,The question is basically asking whether 3-x >= 0

1) Insufficient. If x=1, true, If x=10, false.

2) From this we get that x is negative. So -x is postivie. So, 3 + (-x) will always be a positive number. Sufficient.

Answer is hence B.

Kudos Please... If my post helped.

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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink ]
22 Nov 2012, 04:48

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Bunuel wrote:

jan4dday wrote:

Is \sqrt{(x-3)^2}=3-x ? (1) x not = 3 (2) -x|x| > 3 will give OA as soon as 1st few replies come in

Remember:

\sqrt{x^2}=|x| .

\sqrt{(x-3)^2}=|x-3| . So the question becomes is

|x-3|=3-x .

When

x>3 , then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When

x\leq{3} , then

LHS=|x-3|=-x+3=3-x=RHS , hence in this case equation holds true.

Basically question asks is

x\leq{3} ?

(1) x is not equal to 3. Clearly insufficient.

(2)

-x|x| > 3 , basically this inequality implies that

x<0 , hence

x<3 . Sufficient.

Answer: B.

(2) Actually

-x|x| > 3 , implies that

x<-sqrt(3) , but still, as you mentioned, consequently implies that

x<3 . Sufficient.

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Hi Bunuel, In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}? Please help me understand this concept. Regards, Ravi

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email2vm wrote:

Hi Bunuel, In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}? Please help me understand this concept. Regards, Ravi

I got it now!

|x-3| = (3-x) ===> |3-x| = (3-x) ===>which is only possible when (3-x) >=0 (i.e +ve)

(like |x| =x only when x>=0)

so we have 3-x >=0 ====> x<=3

Also for option B

-x|x| > 3

==> x|x| < 3

so |x| is always positive but so make it less than 3 , x must be negative. i.e. x<0