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# Is x > 6 ?

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Intern
Joined: 29 Mar 2015
Posts: 12
Location: India
GPA: 3.23
WE: Information Technology (Computer Software)
Is x > 6 ?  [#permalink]

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13 Mar 2016, 15:28
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Difficulty:

25% (medium)

Question Stats:

70% (00:55) correct 30% (01:02) wrong based on 241 sessions

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Is x > 6 ?

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

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

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Thanks,

Akki

Intern
Joined: 29 Mar 2015
Posts: 12
Location: India
GPA: 3.23
WE: Information Technology (Computer Software)
Re: Is x > 6 ?  [#permalink]

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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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Akki

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Is x > 6 ?  [#permalink]

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13 Mar 2016, 15:58
3
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.

Hope this helps.
Intern
Joined: 29 Mar 2015
Posts: 12
Location: India
GPA: 3.23
WE: Information Technology (Computer Software)
Re: Is x > 6 ?  [#permalink]

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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.

Hope this helps.

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

Cheers !
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Akki

Math Expert
Joined: 02 Sep 2009
Posts: 50042
Re: Is x > 6 ?  [#permalink]

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13 Mar 2016, 22:01
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]

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21 Nov 2017, 08:27
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is x > 6 ? &nbs [#permalink] 21 Nov 2017, 08:27
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