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Is n > 6? [#permalink]
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29 Jul 2012, 00:18
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Is n > 6? (1) \(\sqrt{n} > 2.5\) (2) \(n > \sqrt{37}\) The answer to option 2 can be n>6 or n<6. Please explain why option 2 is sufficient. Thanks.
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Re: Is n > 6? [#permalink]
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29 Jul 2012, 10:38
can Sqrt take a negative form . ?



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Re: Is n > 6? [#permalink]
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29 Jul 2012, 14:06
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riteshv wrote: can Sqrt take a negative form . ? Square roots are always positive. If the problem said \(n^2 > 36\), then n could be > or < 6.



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29 Jul 2012, 15:23
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riteshv wrote: can Sqrt take a negative form . ? When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(\sqrt{25}=5\). Even roots have only nonnegative value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\). Hope it helps.
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Re: Is n > 6? [#permalink]
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11 Apr 2013, 23:34
MissionIIM2014 wrote: Is n>6?
1) square root(n)>2.5 2) n>(square root 37)
===============
Typical DS options applicable:
 answers please. 1.\(\sqrt{n}>2,5\) becomes \(n>(2,5)^2\), 2,5^2 (because 25*25=625) is 6,25 so n>6,25. Sufficient 2.\(n>\sqrt{37}\) because \(\sqrt{36}=6,\sqrt{37}>6\), we combine this equations in one: \(n>\sqrt{37}>6\) so n>6. Sufficient IMO D
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Re: Is n > 6? [#permalink]
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12 Apr 2013, 23:16
Hi, Thanks for the response, but got one question. How can you be sure that (square root 37) is only 6...why did not you consider (6).. when (square root 37) is considered as (6) n>6 may not be possible... so B option is not giving an answer with certainty..what do you say? regards Zarrolou wrote: MissionIIM2014 wrote: Is n>6?
1) square root(n)>2.5 2) n>(square root 37)
===============
Typical DS options applicable:
 answers please. 1.\(\sqrt{n}>2,5\) becomes \(n>(2,5)^2\), 2,5^2 (because 25*25=625) is 6,25 so n>6,25. Sufficient 2.\(n>\sqrt{37}\) because \(\sqrt{36}=6,\sqrt{37}>6\), we combine this equations in one: \(n>\sqrt{37}>6\) so n>6. Sufficient IMO D



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Re: Is n > 6? [#permalink]
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12 Apr 2013, 23:45
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MissionIIM2014 wrote: Hi, Thanks for the response, but got one question. How can you be sure that (square root 37) is only 6...why did not you consider (6).. when (square root 37) is considered as (6) n>6 may not be possible... so B option is not giving an answer with certainty..what do you say?
regards
In the GMAT \(\sqrt{25}=5\) and \(\sqrt{36}=6\), the square root of a number is its positive value.
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Re: Is n > 6? [#permalink]
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08 May 2013, 10:55
Zarrolou wrote: MissionIIM2014 wrote: Hi, Thanks for the response, but got one question. How can you be sure that (square root 37) is only 6...why did not you consider (6).. when (square root 37) is considered as (6) n>6 may not be possible... so B option is not giving an answer with certainty..what do you say?
regards
In the GMAT \(\sqrt{25}=5\) and \(\sqrt{36}=6\), the square root of a number is its positive value. Hi, even though it may seems like basic and stupid, but I just want to ask it for clearing my brain. If in GMAT square root is always positive that why cant I take any given number, for eg in the 2nd statement in above equation and say since \(n>\sqrt{37}\) than \(n^2 >37\) and if I again do a Square root n>6 since I can ignore the tive 6. since even squareroot cannot have a negative value. Can I do that? if no, can someone explain me why, and what stops me to do that? can anyone explain me with illustration or an eg?
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Re: Is n > 6? [#permalink]
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08 May 2013, 12:08
nikhil007 wrote: Zarrolou wrote: MissionIIM2014 wrote: Hi, Thanks for the response, but got one question. How can you be sure that (square root 37) is only 6...why did not you consider (6).. when (square root 37) is considered as (6) n>6 may not be possible... so B option is not giving an answer with certainty..what do you say?
regards
In the GMAT \(\sqrt{25}=5\) and \(\sqrt{36}=6\), the square root of a number is its positive value. Hi, even though it may seems like basic and stupid, but I just want to ask it for clearing my brain. If in GMAT square root is always positive that why cant I take any given number, for eg in the 2nd statement in above equation and say since \(n>\sqrt{37}\) than \(n^2 >37\) and if I again do a Square root n>6 since I can ignore the tive 6. since even squareroot cannot have a negative value. Can I do that? if no, can someone explain me why, and what stops me to do that? can anyone explain me with illustration or an eg? Something like n^2 = 36 is different from n= square root 36. In the first case, n can have both positive and negative value because both value satisfy the equation. In the 2nd case, you are explicitly saying that n is positive square root of 36 which is +6. If you try to take the square root of a negative number than that square root takes you to the concept of imaginary numbers. GMAT is not concerned with it and hence you will never see something like n^2 = 36.



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Re: Is n > 6? [#permalink]
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08 May 2013, 14:19
[/quote] Bluelagoon wrote: you are explicitly saying that n is positive square root of 36 which is +6. If you try to take the square root of a negative number than that square root takes you to the concept of imaginary numbers. GMAT is not concerned with it and hence you will never see something like n^2 = 36. Yes I am explicitly saying that because for \(\sqrt{36}\) as per Gmat 6 shouldn't be an option other wise if your are given statement like x= \(\sqrt{36}\) you will have to consider 2 roots +6, whereas we say that even square root will only have positive number as a answer in Gmat, coz 6 is imaginary
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Re: Is n > 6? [#permalink]
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08 May 2013, 19:56
nikhil007 wrote: Bluelagoon wrote: you are explicitly saying that n is positive square root of 36 which is +6. If you try to take the square root of a negative number than that square root takes you to the concept of imaginary numbers. GMAT is not concerned with it and hence you will never see something like n^2 = 36. Yes I am explicitly saying that because for \(\sqrt{36}\) as per Gmat 6 shouldn't be an option other wise if your are given statement like x= \(\sqrt{36}\) you will have to consider 2 roots +6, whereas we say that even square root will only have positive number as a answer in Gmat, coz 6 is imaginary  6 is NOT imaginary, SQUARE ROOT (or any even root) of a negative integer is imaginary (imaginary numbers are out of scope for GMAT). Guys, could you please point me to the instructions where we are asked to consider only positive roots for numbers?



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Re: Is n > 6? [#permalink]
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09 May 2013, 02:47
anilisanil wrote: nikhil007 wrote: Bluelagoon wrote: you are explicitly saying that n is positive square root of 36 which is +6. If you try to take the square root of a negative number than that square root takes you to the concept of imaginary numbers. GMAT is not concerned with it and hence you will never see something like n^2 = 36. Yes I am explicitly saying that because for \(\sqrt{36}\) as per Gmat 6 shouldn't be an option other wise if your are given statement like x= \(\sqrt{36}\) you will have to consider 2 roots +6, whereas we say that even square root will only have positive number as a answer in Gmat, coz 6 is imaginary  6 is NOT imaginary, SQUARE ROOT (or any even root) of a negative integer is imaginary (imaginary numbers are out of scope for GMAT). Guys, could you please point me to the instructions where we are asked to consider only positive roots for numbers? Any nonnegative real number has a unique nonnegative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root. When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(\sqrt{25}=5\). Even roots have only nonnegative value on the GMAT.
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Re: Is n > 6? [#permalink]
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05 Dec 2014, 17:19
Bunuel wrote: Is n > 6?
(1) \(\sqrt{n} > 25\) > \(n>25^2\). Sufficient.
(2) \(n > \sqrt{37}\). Since \(\sqrt{37}> \sqrt{36}\), then \(n > \sqrt{37}>\sqrt{36}\) > \(n>6\). Sufficient.
Answer: D. Hi Bunuel, My question showed stated one as "2.5" not "25'. I didn't what the (2.5) ^2 was so I did 2^2 = 4 and (1/2)^2 = 1/4, therefore, less than 5. How are we supposed to square something like 2.5?



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Is n > 6? (Please help me understand why the answer is that.) [#permalink]
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29 Jul 2015, 09:58
Is n > 6? (1) \(\sqrt{n} > 2.5\) (2) \(n> \sqrt{37}\) Could someone please explain why the answer is D?
My thought is this: In statement (2), \(\sqrt{36}\) can be either 6 or 6. Because 6 is possible, statement (2) shouldn't be sufficient to answer the question.



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