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# Is root (x-3)^2 = 3-x ? (1) x ≠ 3 (2) x | x | > 0

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Is root (x-3)^2 = 3-x ? (1) x ≠ 3 (2) x | x | > 0 [#permalink]

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02 Aug 2007, 09:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0
_________________

cool

VP
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02 Aug 2007, 16:36
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

B.
Reduced question: Is |x-3| = 3-x ?
The statement can only be true when x<3.

(1) x is not equal to 3, but it can be less than 3 or more than 3. INSUFFICIENT.

(2) -x|x|>0, can only be true when x<0; thus, less than 3.
SUFFICIENT.
Director
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02 Aug 2007, 22:29
bkk145 wrote:
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

B.
Reduced question: Is |x-3| = 3-x ?
The statement can only be true when x<3>0, can only be true when x<0; thus, less than 3.
SUFFICIENT.

HI bkk145,
I am considering it this way..

Is (x-3) = (3-x) or, ----------1
-(x-3) = (3-x)
i.e (3-x) = (3-x) ----------2

So can't we say that St1 is suff. given that we have an equation 2 on LHS that is equal to RHS. i.e eq'n 2 is always true...
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03 Aug 2007, 07:51
Amit05 wrote:
bkk145 wrote:
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

B.
Reduced question: Is |x-3| = 3-x ?
The statement can only be true when x<3>0, can only be true when x<0; thus, less than 3.
SUFFICIENT.

HI bkk145,
I am considering it this way..

Is (x-3) = (3-x) or, ----------1
-(x-3) = (3-x)
i.e (3-x) = (3-x) ----------2

So can't we say that St1 is suff. given that we have an equation 2 on LHS that is equal to RHS. i.e eq'n 2 is always true...

Amit, what's LHS and RHS? confused...
Director
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03 Aug 2007, 08:02
bkk145 wrote:
Amit05 wrote:
bkk145 wrote:
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

B.
Reduced question: Is |x-3| = 3-x ?
The statement can only be true when x<3>0, can only be true when x<0; thus, less than 3.
SUFFICIENT.

HI bkk145,
I am considering it this way..

Is (x-3) = (3-x) or, ----------1
-(x-3) = (3-x)
i.e (3-x) = (3-x) ----------2

So can't we say that St1 is suff. given that we have an equation 2 on LHS that is equal to RHS. i.e eq'n 2 is always true...

Amit, what's LHS and RHS? confused...

LHS = Left hand side of the equation
RHS = Right hand side of the equation.

The equation here is 2.
VP
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03 Aug 2007, 08:50
Amit05 wrote:
bkk145 wrote:
Amit05 wrote:
bkk145 wrote:
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

B.
Reduced question: Is |x-3| = 3-x ?
The statement can only be true when x<3>0, can only be true when x<0; thus, less than 3.
SUFFICIENT.

HI bkk145,
I am considering it this way..

Is (x-3) = (3-x) or, ----------1
-(x-3) = (3-x)
i.e (3-x) = (3-x) ----------2

So can't we say that St1 is suff. given that we have an equation 2 on LHS that is equal to RHS. i.e eq'n 2 is always true...

Amit, what's LHS and RHS? confused...

LHS = Left hand side of the equation
RHS = Right hand side of the equation.

The equation here is 2.

If I understand you correctly, you can't consider the equations separately. Both equations must be true and they are only true for certain values of x, not all.

For example, for equation 2, you can pick any value of x and it will work. Great!, However, this is not the case for equation 1. I think of it like when you have two ranges of x values. Say, first range is 1<x<10, second range is 2<x<5. If you want to find valid range of x that is true for both, you must pick smaller one.

Hope this helps.
Director
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04 Aug 2007, 06:48
Ok .. Got it .. here I am with my explanation:

As bkk145 mentioned, The question asks whether |x-3| = 3 - x or not ?

Now, |x-3| = (x -3 ) i.e When (x-3) is +ve i.e (x-3) > 0 or
|x-3| = -(x-3) i.e When (x-3) is -ve i.e (x-3) < 0

So as asked in the Q stem, |x-3| = 3 - x only when |x-3| = - (x-3) or in other words when x-3 is -ve which is possible only if x < 3.

As per the St1, x ≠ 3 which implies x can be <3> 3. Hence St-1 is in-suff.

St2 : – x | x | > 0
=> This statement is only true |x| = -x since -x * -x = x ^2 > 0. If |x| = x then -x * x < 0 which negates st2.

Therefore x < 0. Hence X < 3. Hence we can conclude that (x - 3) <0> |x-3| = 3 - x. hence St2 is suff to answer this question.

The another method to solve this question is by using the distance logic.

Since we are asked whether |x-3| = 3 - x which in other words means whether distance between 3 and x = -ve of distance between x and 3. And the same logic can be applied here that distance only be a negative number x < 3.

Thanks bkk145. It was a simple concept but I getting lost in my confusion. I went though the doc again and spent some time. Modulus and inequality problems are the hot favs of GMAT test takers.
Senior Manager
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07 Aug 2007, 12:41
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

Just one simple concept will give you the right answer quickly:

The function sqrt(x) is positive; that is, sqrt(x)>0 for any x >=0. With this in mind: sqrt((x-3)^2)=3-x if and only if 3>x. So, what the question is actually asking is, "is 3>x?" Just by simple inspection you'll notice that only (2) has enough data to solve the problem, therefore B.
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07 Aug 2007, 13:21
Manager
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07 Aug 2007, 14:08
Amit05 wrote:
Ok .. Got it .. here I am with my explanation:

As bkk145 mentioned, The question asks whether |x-3| = 3 - x or not ?

Now, |x-3| = (x -3 ) i.e When (x-3) is +ve i.e (x-3) > 0 or
|x-3| = -(x-3) i.e When (x-3) is -ve i.e (x-3) < 0

So as asked in the Q stem, |x-3| = 3 - x only when |x-3| = - (x-3) or in other words when x-3 is -ve which is possible only if x < 3.

As per the St1, x ≠ 3 which implies x can be <3> 3. Hence St-1 is in-suff.

St2 : – x | x | > 0
=> This statement is only true |x| = -x since -x * -x = x ^2 > 0. If |x| = x then -x * x < 0 which negates st2.

Therefore x < 0. Hence X < 3. Hence we can conclude that (x - 3) <0> |x-3| = 3 - x. hence St2 is suff to answer this question.

The another method to solve this question is by using the distance logic.

Since we are asked whether |x-3| = 3 - x which in other words means whether distance between 3 and x = -ve of distance between x and 3. And the same logic can be applied here that distance only be a negative number x < 3.

Thanks bkk145. It was a simple concept but I getting lost in my confusion. I went though the doc again and spent some time. Modulus and inequality problems are the hot favs of GMAT test takers.

Not that I wanna be rude, but is this Jimmy Kimmel with a turban on the picture??
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07 Aug 2007, 14:46
I haven't tried anything with Statement 2 yet, but I believe you have enough data from statement 1 to get the answer.

Quote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3

(x-3)^2 = 3-x

x^2 - 6x + 9 = 3-x
x^2 - 5x + 6 = 0
(x-3)(x-2) = 0

x is not = to 3 so x must = 2.

then you can just plug 2 back into the original equation.

what does this root mean though? is asking if the square root of the left side = right side? if so, won't finding 2 still be sufficient?
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07 Aug 2007, 14:46
I got down to BCE but don't get statement #2
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08 Aug 2007, 03:25
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

Stat 1: Insuff
If x=2 then LHS=RHS and answer is yes
If x=5 then LHS>RHS and answer is no

Stat 2: Suff
-x|x|>0

The only way this expression can be greater than 0 is if |x|=-x because -ve*-ve will give us a =ve

For |x| to be equal to -x, x has to be less than 0. Therefore, stat. 2 tells us is that x<0

LHS=RHS for all values x<0.
I tried x=-1,-2,-5
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08 Aug 2007, 03:30
Andr359 wrote:
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

Just one simple concept will give you the right answer quickly:

The function sqrt(x) is positive; that is, sqrt(x)>0 for any x >=0. With this in mind: sqrt((x-3)^2)=3-x if and only if 3>x. So, what the question is actually asking is, "is 3>x?" Just by simple inspection you'll notice that only (2) has enough data to solve the problem, therefore B.

Andr359, can you pls. explain how we know that 3-x is greater than 0? If 3-x is less than 0 then the concept above will not be helpful right?
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08 Aug 2007, 04:21
GK_Gmat wrote:
Andr359 wrote:
jet1445 wrote:
11. Is root (x-3)^2 = 3-x ?
(1) x ≠ 3
(2) – x | x | > 0

Just one simple concept will give you the right answer quickly:

The function sqrt(x) is positive; that is, sqrt(x)>0 for any x >=0. With this in mind: sqrt((x-3)^2)=3-x if and only if 3>x. So, what the question is actually asking is, "is 3>x?" Just by simple inspection you'll notice that only (2) has enough data to solve the problem, therefore B.

Andr359, can you pls. explain how we know that 3-x is greater than 0? If 3-x is less than 0 then the concept above will not be helpful right?

It will be helpful because u switch to imaginay number after that... which is out of the GMAT ... So, yes, it cannot be that Sqrt( A ) < 0 so we conclude :

Sqrt((x-3)^2) >= 0
<=> 3-x >= 0
<=> x =< 3 : the question is boiled down to this

BTW, excellent explanation from Andr359
Re: Equations   [#permalink] 08 Aug 2007, 04:21
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