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Is x > 6 ?

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akxshay
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Is x > 6 ? [#permalink] New post  13 Mar 2016, 15:28
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Is x > 6 ?

1) \(\sqrt{x} > 2.5\)

2) \(x > \sqrt{37}\)
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Is x > 6 ? [#permalink] New post  13 Mar 2016, 15:58
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akxshay wrote:
I understand that 1) is SUFFICIENT because x > 6.25 hence x > 6, but not sure why 2) is SUFFICIENT ?

37^1/2 can have two values -6.08 and +6.08. If I consider +6.08 then it is sufficient, but what about the negative value ?

If I take the negative value then x > -6.08 which means x is not > 6. Hence the answer should be A), but the OA is D)

I am sure I am making some mistake, kindly help me to understand if my interpretation of taking a -ve is wrong and why ?


The mistake you are making is that \(x^2 = 36\) has 2 roots \(\pm 6\) (as it is a second degree equation in x) but \(x=\sqrt{36}\) only has 1 solution = +6 as it is a linear equation in x

Thus, from statement 2, \(x > \sqrt{37}\) means that \(x > 6.aaa\) and NOT \(x > - 6.aaa\)

Remember this rule for GMAT.

Thus both statements are sufficient.

D is the correct answer.

Hope this helps.
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akxshay
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Re: Is x > 6 ? [#permalink] New post  13 Mar 2016, 15:33
I understand that 1) is SUFFICIENT because x > 6.25 hence x > 6, but not sure why 2) is SUFFICIENT ?

37^1/2 can have two values -6.08 and +6.08. If I consider +6.08 then it is sufficient, but what about the negative value ?

If I take the negative value then x > -6.08 which means x is not > 6. Hence the answer should be A), but the OA is D)

I am sure I am making some mistake, kindly help me to understand if my interpretation of taking a -ve is wrong and why ?
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akxshay
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Re: Is x > 6 ? [#permalink] New post  13 Mar 2016, 16:14
Engr2012 wrote:
The mistake you are making is that \(x^2 = 36\) has 2 roots \(\pm 6\) (as it is a second degree equation in x) but \(x=\sqrt{36}\) only has 1 solution = +6 as it is a linear equation in x

Thus, from statement 2, \(x > \sqrt{37}\) means that \(x > 6.aaa\) and NOT \(x > - 6.aaa\)

Remember this rule for GMAT.

Thus both statements are sufficient.

D is the correct answer.

Hope this helps.



Got it, Thanks a lot ! It does make sense.

Cheers ! :)
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Re: Is x > 6 ? [#permalink] New post  13 Mar 2016, 22:01
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akxshay wrote:
I understand that 1) is SUFFICIENT because x > 6.25 hence x > 6, but not sure why 2) is SUFFICIENT ?

37^1/2 can have two values -6.08 and +6.08. If I consider +6.08 then it is sufficient, but what about the negative value ?

If I take the negative value then x > -6.08 which means x is not > 6. Hence the answer should be A), but the OA is D)

I am sure I am making some mistake, kindly help me to understand if my interpretation of taking a -ve is wrong and why ?


The square root function cannot give negative result: \(\sqrt{nonnegative \ value}\geq{0}\). For example, \(\sqrt{4}=2\) (not +2 and -2). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.
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Re: Is x > 6 ? [#permalink] New post  21 Sep 2019, 09:45
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akxshay wrote:
Is x > 6 ?

1) \(\sqrt{x} > 2.5\)

2) \(x > \sqrt{37}\)


Is x > 6 ?

1) \(\sqrt{x} > 2.5\)
x>6.25>6
x>6
SUFFICIENT

2) \(x > \sqrt{37}\)
\(x>\sqrt{37}>\sqrt{36}=6\)
x>6
SUFFICIENT

IMO D
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Re: Is x > 6 ? [#permalink] New post  08 Oct 2019, 10:30
akxshay wrote:
Is x > 6 ?

1) \(\sqrt{x} > 2.5\)

2) \(x > \sqrt{37}\)


Asked: Is x > 6 ?

1) \(\sqrt{x} > 2.5\)
x > 6.25 > 6
SUFFICIENT

2) \(x > \sqrt{37}\)
x > \sqrt{37} > \sqrt{36} = 6
x> 6
SUFFICIENT

IMO D
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sampad
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Re: Is x > 6 ? [#permalink] New post  31 Dec 2019, 22:20
i am really speechless after seeing this solution. sqrt of 37 is only 6.abc not -6.abc. But for x^2=25 x will have 2 values +5 and -5. So can anyone please explain how we are getting the values of x. is it not the sqrt of 25?
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Basim2016
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Re: Is x > 6 ? [#permalink] New post  09 Jan 2020, 03:11
Dear Sampad,

X^2 has 2 solutions, becuase it has the power of 2. Kindly read earlier /above posts.
Surely, you will be clear after reading above posts

Regards,
Basim
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