BTG set 550-700

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BTG set 550-700

by Night reader » Thu Jan 06, 2011 4:40 pm
What is the value of x?

(1) Sqrt (x^4) = 9

(2) Sqrt (x^2) = -x

[spoiler]I inferred everything possible here and decided with E, but OA suggests C; yet explanation did not seem reasonable to me, as 0 is one possible value for st.(2) and we may not deduce that combined st.(1&2) is |x|=3 => x=-3. I think this problem has ambiguity.
[/spoiler]

(2) Sqrt (x^2) = -x ............ :( is this possible with -ve, I guess only 0 here would be right fit...
Source: — Data Sufficiency |

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by anshumishra » Thu Jan 06, 2011 5:21 pm
Night reader wrote:What is the value of x?

(1) Sqrt (x^4) = 9

(2) Sqrt (x^2) = -x

[spoiler]I inferred everything possible here and decided with E, but OA suggests C; yet explanation did not seem reasonable to me, as 0 is one possible value for st.(2) and we may not deduce that combined st.(1&2) is |x|=3 => x=-3. I think this problem has ambiguity.
[/spoiler]

(2) Sqrt (x^2) = -x ............ :( is this possible with -ve, I guess only 0 here would be right fit...
x = ?

Statement 1:
sqrt(x^4) = 9
=> |x^2| = 9
=> x = +3 or -3 - Insufficient

Statement 2:
sqrt(x^2) = -x
=> |x| = -x
=> x = -ve - Insufficient

Combining 1 and 2 :
x = -3

Hence , C
Thanks
Anshu

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by Night reader » Thu Jan 06, 2011 5:25 pm
anshumishra wrote:
Night reader wrote:What is the value of x?

(1) Sqrt (x^4) = 9

(2) Sqrt (x^2) = -x

[spoiler]I inferred everything possible here and decided with E, but OA suggests C; yet explanation did not seem reasonable to me, as 0 is one possible value for st.(2) and we may not deduce that combined st.(1&2) is |x|=3 => x=-3. I think this problem has ambiguity.
[/spoiler]

(2) Sqrt (x^2) = -x ............ :( is this possible with -ve, I guess only 0 here would be right fit...
x = ?

Statement 1:
sqrt(x^4) = 9
=> |x^2| = 9
=> x = +3 or -3 - Insufficient

Statement 2:
sqrt(x^2) = -x
=> |x| = -x
=> x = -ve - Insufficient

Combining 1 and 2 :
x = -3

Hence , C
Anshu, root-3rd power (x) OR x^1/3 could be -ve not x^1/2 ok? :( and x^2 can't be equal to -ve; |x| can be x OR -x. So it's not clear yet

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by Night reader » Thu Jan 06, 2011 5:29 pm
let me correct myself a bit: x^1/2 could be negative but Not in GMAT.

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by anshumishra » Thu Jan 06, 2011 5:31 pm
Night reader wrote: Anshu, root-3rd power (x) OR x^1/3 could be -ve not x^1/2 ok? :( and x^2 can't be equal to -ve; |x| can be x OR -x. So it's not clear yet
It can be :
sqrt(-3*-3) = sqrt(9) = 3 = -(-3)
=> sqrt(x^2) = -x
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by anshumishra » Thu Jan 06, 2011 5:34 pm
Night reader wrote:let me correct myself a bit: x^1/2 could be negative but Not in GMAT.
Actually it is not -ve. When we append a -ve sign before x and since x is negative, the sqrt becomes positive.
I always use :

sqrt(x^2) = |x| , as it resolves lots of confusion.
Thanks
Anshu

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by Night reader » Thu Jan 06, 2011 5:39 pm
anshumishra wrote:
Night reader wrote:let me correct myself a bit: x^1/2 could be negative but Not in GMAT.
Actually it is not -ve. When we append a -ve sign before x and since x is negative, the sqrt becomes positive.
I always use :

sqrt(x^2) = |x| , as it resolves lots of confusion.
nice, the same here for sqrt(a^2)=|a|, but had to scrub the memory rust from my -ve-ve old literacy. Thanks

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by Anurag@Gurome » Thu Jan 06, 2011 6:35 pm
I just want to add only one small correction in Anshu's solution.
anshumishra wrote:Statement 2:
sqrt(x^2) = -x
=> |x| = -x
=> x = -ve - Insufficient
Actually |x| = -x implies x ≤ 0, i.e. x is non-positive.
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by Night reader » Thu Jan 06, 2011 8:11 pm
Anurag@Gurome wrote:I just want to add only one small correction in Anshu's solution.
anshumishra wrote:Statement 2:
sqrt(x^2) = -x
=> |x| = -x
=> x = -ve - Insufficient
Actually |x| = -x implies x ≤ 0, i.e. x is non-positive.
correct, we should imply x is non-positive, not -ve; hence x can be 0 or -ve
thanks Anurag

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by anshumishra » Thu Jan 06, 2011 8:23 pm
Night reader wrote:
Anurag@Gurome wrote:I just want to add only one small correction in Anshu's solution.
anshumishra wrote:Statement 2:
sqrt(x^2) = -x
=> |x| = -x
=> x = -ve - Insufficient
Actually |x| = -x implies x ≤ 0, i.e. x is non-positive.
correct, we should imply x is non-positive, not -ve; hence x can be 0 or -ve
thanks Anurag
That's right. Hopefully wouldn't miss it the next time.
Thanks
Anshu

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