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another one

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Source: — Data Sufficiency |
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Re: another one

by logitech » Sun Nov 30, 2008 10:10 pm
muzali wrote:Would appreciate a detailed explanation.
Muzali,

You can rephrase this question as:

Is X<3

St 1)

X is not 3. Insuf

2) -x > 0 since absolute is +

x<0

well if x is less than 0, it is less than 3 so SUF

I would go with B.
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Re: another one

by bacali » Sun Nov 30, 2008 10:27 pm
logitech wrote:
muzali wrote:Would appreciate a detailed explanation.
Muzali,

You can rephrase this question as:

Is X<3

St 1)

X is not 3. Insuf

2) -x > 0 since absolute is +

x<0

well if x is less than 0, it is less than 3 so SUF

I would go with B.
wait doesn't root(x-3)^2 imply |x-3| rather than just x-3???
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by bacali » Sun Nov 30, 2008 10:30 pm
I'll go with D.
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by austin » Sun Nov 30, 2008 10:37 pm
Is x-3 = 3-x????
or is x = 3???

1. x!= 3 Sufficient

2. -x|x| > 0
|x| is always positive. If -x|x| > 0 then x has to be negative.
eg. x = -3; -(-3)|-3| >0.. so x cannot be 3..

D
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by logitech » Sun Nov 30, 2008 10:53 pm
Yes

root(x-3)^2 imply |x-3|

X>3 it is : X-3

X<3 it is : 3-X

so The question asks whether x is less than 3. And it is B!



:lol:
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Re: another one

by sudhir3127 » Sun Nov 30, 2008 10:55 pm
muzali wrote:Would appreciate a detailed explanation.

NOTE : We should know that a square root cannot be negative in GMAT . Our aim is to prove that LHS = RHS . we have to solve the equation to get a value

Statement 1

Nothing is told whether X is postive/negative .thus insufficent .


Statment 2.

-Xmod(x) >0

For the above to be true X has to be negative .

thus when X is negative

sqrt [(x-3)^2] = 3-x

put X = -4

sqrt( 49) = 3-(-4)
7 =7

this is possible only when X<0 and B tells us that,

Hope its clear. do let me know if u have any doubts..
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by logitech » Sun Nov 30, 2008 11:00 pm
Guys do not leave this question BEFORE you understand why B is the correct answer. Know your absolute values!
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by austin » Sun Nov 30, 2008 11:00 pm
Agree.. my mistake...
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by muzali » Mon Dec 01, 2008 9:50 am
Thank you all. Want to know a good resource to refer to regarding absolute values. Could you suggest me some? Thanks.
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by logitech » Mon Dec 01, 2008 10:09 am
https://www.purplemath.com/modules/absolute.htm
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by dzinyo00 » Mon Dec 01, 2008 12:35 pm
Not so sure of ans. B

Given sqrt((x-3)^2) you know that (x-3)^2 is positive so you can go ahead and simplify and then equate to the other side.

i.e (x-3)^(2*1/2) = 3-x --> rephrases to be is x = 3?

Ans option A, in my opinion should be correct.
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by pbanavara » Mon Dec 01, 2008 7:01 pm
dzinyo00 wrote:Not so sure of ans. B

Given sqrt((x-3)^2) you know that (x-3)^2 is positive so you can go ahead and simplify and then equate to the other side.

i.e (x-3)^(2*1/2) = 3-x --> rephrases to be is x = 3?

Ans option A, in my opinion should be correct.
I guess solving the equation gives only one of the possible values, doesn't necessarily mean that's the only value. x=3 solves the equation if you substitute the value back but so does x=-4. X being negative ensures that the equation is always true for negative values.

Hope this helps.

- pradeep
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