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If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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26 Dec 2012, 04:31
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If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\) (1) k > 3n (2) n + k > 3n
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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26 Dec 2012, 04:36




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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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27 Jun 2014, 01:36
Bunuel wrote: If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?
Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" > is \(k>3n\)?
(1) k > 3n. Sufficient.
(2) n + k > 3n > \(k>2n\). Not sufficient.
Answer: A. Hi Bunuel, I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive. What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 > \(\sqrt{x}\) = 3 or 3 Same thought or n! Please help me to clarify, thank you so much!
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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27 Jun 2014, 01:53
LucyDang wrote: Bunuel wrote: If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?
Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" > is \(k>3n\)?
(1) k > 3n. Sufficient.
(2) n + k > 3n > \(k>2n\). Not sufficient.
Answer: A. Hi Bunuel, I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive. What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 > \(\sqrt{x}\) = 3 or 3 Same thought or n! Please help me to clarify, thank you so much! 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, \(\sqrt{9}=3\), NOT +3 or 3. In contrast, the equation \(x^2=9\) has TWO solutions, +3 and 3. 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: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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27 Jun 2014, 04:06
Bunuel wrote: LucyDang wrote: Bunuel wrote: If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?
Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" > is \(k>3n\)?
(1) k > 3n. Sufficient.
(2) n + k > 3n > \(k>2n\). Not sufficient.
Answer: A. Hi Bunuel, I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive. What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 > \(\sqrt{x}\) = 3 or 3 Same thought or n! Please help me to clarify, thank you so much! 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, \(\sqrt{9}=3\), NOT +3 or 3. In contrast, the equation \(x^2=9\) has TWO solutions, +3 and 3. 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. I got it, thank you!!
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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19 Aug 2014, 06:01
Bunuel wrote: If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?
Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" > is \(k>3n\)?
(1) k > 3n. Sufficient.
(2) n + k > 3n > \(k>2n\). Not sufficient.
Answer: A. How do you get to the conclusion that \(\sqrt{n+k}>2\sqrt{n}\)? Even if k > 3n it can be either n+k > 4n or n+k < 4n, since we don't know more about n. Thanks .



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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19 Aug 2014, 07:06
tobiasfr wrote: Bunuel wrote: If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?
Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" > is \(k>3n\)?
(1) k > 3n. Sufficient.
(2) n + k > 3n > \(k>2n\). Not sufficient.
Answer: A. How do you get to the conclusion that \(\sqrt{n+k}>2\sqrt{n}\)? Even if k > 3n it can be either n+k > 4n or n+k < 4n, since we don't know more about n.Thanks . Not sure how you get the above... Anyway, the question asks whether \(\sqrt{n+k}>2\sqrt{n}\)? After algebraic manipulations shown in my solution the question becomes: is \(k>3n\)? The first statement answers this question, which makes it sufficient.
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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09 Dec 2016, 13:39
Hi,
If it were not given that n and k are positive integers, then taking the underroot of this expression (Underoort n+k > 2 underroot n) would have resulted in absolute value form as l n + k l > 4 l n l. Am I right?
Just wanted to make sure that I get the concept right.
Thanks for your help.
Regards, H



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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12 Mar 2017, 05:54
HarveyKlaus wrote: Hi,
If it were not given that n and k are positive integers, then taking the underroot of this expression (Underoort n+k > 2 underroot n) would have resulted in absolute value form as l n + k l > 4 l n l. Am I right?
Just wanted to make sure that I get the concept right.
Thanks for your help.
Regards, H Nope. It still would have given the positive value, however, we won't be sure of the sign of the n and k if the statement "they are +ve integers" isn't mentioned. \sqrt{10036} \sqrt{64} = 8



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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23 Mar 2017, 11:54
Careless mistake! Can someone share a link on how to reduce careless mistake please? Frustrating to miss this kind of easy questions
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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23 Mar 2017, 22:44



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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23 Mar 2017, 22:57
Thanks a lot Bunnuel. Very useful
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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19 Jul 2017, 18:58
Hi Bunuel,
I couldn't understand the 2nd part.
The objective is the check if k>3n (after simplification). From option 2 we find K>2n , so it's also sufficient to answer that K is not greater than 3n.
Please explain.



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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08 Aug 2017, 22:38
Bunuel wrote: IWilWin wrote: Hi Bunuel,
I couldn't understand the 2nd part.
The objective is the check if k>3n (after simplification). From option 2 we find K>2n , so it's also sufficient to answer that K is not greater than 3n.
Please explain. Say a question asks: is x > 3? We got that x > 2. Is this sufficient to answer the question? No, because if x= 2.5, then the answer is NO but if x = 4, then the answer is YES. Perfect Explanation, This is what I was looking for.



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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11 Oct 2017, 10:46
Bunuel wrote: IWilWin wrote: Hi Bunuel,
I couldn't understand the 2nd part.
The objective is the check if k>3n (after simplification). From option 2 we find K>2n , so it's also sufficient to answer that K is not greater than 3n.
Please explain. Say a question asks: is x > 3? We got that x > 2. Is this sufficient to answer the question? No, because if x= 2.5, then the answer is NO but if x = 4, then the answer is YES. Bunuel : I always have a confusion in questions such as these... We know that on simplifying the question stem we get k>3n The second premise is reduced to k>2n Now my understanding is that the second premise is sufficient for us to say that k is not greater than 3n, hence answer choice D. When it itself deduces to k>2n isn't it enough for us to answer the original question stem whether k>3n? In this case No k is not greater than 3n



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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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11 Oct 2017, 20:03
avaneeshvyas wrote: Bunuel wrote: IWilWin wrote: Hi Bunuel,
I couldn't understand the 2nd part.
The objective is the check if k>3n (after simplification). From option 2 we find K>2n , so it's also sufficient to answer that K is not greater than 3n.
Please explain. Say a question asks: is x > 3? We got that x > 2. Is this sufficient to answer the question? No, because if x= 2.5, then the answer is NO but if x = 4, then the answer is YES. Bunuel : I always have a confusion in questions such as these... We know that on simplifying the question stem we get k>3n The second premise is reduced to k>2n Now my understanding is that the second premise is sufficient for us to say that k is not greater than 3n, hence answer choice D. When it itself deduces to k>2n isn't it enough for us to answer the original question stem whether k>3n? In this case No k is not greater than 3nNo. Say a question asks is x > 3? (1) x > 2. Is this sufficient to answer the question whether x is greater than 3? No. If x = 100, then we'd have an YES answer to the question but if x = 2.5 (so if x is any number from 2 to 3), then we'd have a NO answer to the question. In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".
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Re: If n and k are positive integers, is (n + k)^(1/2) > 2n^(1/2)?
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25 Nov 2017, 23:17
We know that on simplifying the question stem we get k>3n The second premise is reduced to k>2n Now my understanding is that the second premise is sufficient for us to say that k is not greater than 3n, hence answer choice D. When it itself deduces to k>2n isn't it enough for us to answer the original question stem whether k>3n? In this case No k is not greater than 3n[/quote]
No. Say a question asks is x > 3?
(1) x > 2. Is this sufficient to answer the question whether x is greater than 3? No. If x = 100, then we'd have an YES answer to the question but if x = 2.5 (so if x is any number from 2 to 3), then we'd have a NO answer to the question.
In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".[/quote]
This explanation..... this is going to save lots of lives on the GMAT. Thank you Bunuel!



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