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# If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2)

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Intern
Status: GMAT on july 13.
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If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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Updated on: 24 May 2018, 21:34
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95% (hard)

Question Stats:

38% (01:29) correct 62% (01:31) wrong based on 112 sessions

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If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

Originally posted by Viserion99 on 22 May 2018, 22:28.
Last edited by Bunuel on 24 May 2018, 21:34, edited 2 times in total.
Edited the OA.
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Joined: 28 Nov 2017
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Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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23 May 2018, 00:27
Viserion99 wrote:
If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

Even though official answer is $$E$$, I will go with $$C$$.
Because, $$(1)$$ and $$(2)$$ taken together, results in $$n=25$$. This is sufficient to answer the question.
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Tulkin.

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Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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23 May 2018, 02:29
Tulkin987 wrote:
Viserion99 wrote:
If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

Even though official answer is $$E$$, I will go with $$C$$.
Because, $$(1)$$ and $$(2)$$ taken together, results in $$n=25$$. This is sufficient to answer the question.
Yes.
Thanks for the reassurance.
Even, I went with C.

Sent from my Moto G (5S) Plus using GMAT Club Forum mobile app
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Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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24 May 2018, 05:59
2
Viserion99 wrote:
If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

The data is sufficient if we can determine whether the equation holds good with the information in the statements.

n is a positive integer.
Properties of perfect squares. The difference between squares of consecutive positive integers will increase by 2.
For instance, 2^2 - 1^2 = 3; 3^2 - 2^2 = 5; 4^2 - 3^2 = 7; 5^2 - 4^2 = 9; 6^2 - 5^2 = 11 and so on.

Statement 1: n is a perfect square.
If n = 4; $$\sqrt{4+11} \neq \sqrt{4} + 1$$
If n = 25; $$\sqrt{25+11} = \sqrt{25} + 1$$

Cannot determine whether the equality will hold good. Statement 1 alone is not sufficient.

Statement 2: n + 11 is a perfect square
If n = 14; n + 11 = 25, which is a perfect square. $$\sqrt{14+11} \neq \sqrt{14} + 1$$
If n = 25; n + 11 = 36, which is a perfect square. $$\sqrt{25+11} = \sqrt{25} + 1$$

Combining the two statements: n is a perfect square and (n + 11) is also a perfect square.
If the equation were to hold good, the two numbers n and (n + 11) will have to be squares of two consecutive numbers.
From the properties of difference between square of positive integers, there can exist only one set of values where the difference between squares of consecutive numbers is 11.
The difference between 6^2 and 5^2 is 11 and we cannot find any other set of two consecutive integers that will satisfy this condition. i.e., the only value that satisfies is when n = 25 and (n + 11) = 36.

Manager
Joined: 26 Dec 2017
Posts: 96
Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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24 May 2018, 06:26
$$\sqrt{n+11} = \sqrt{n} + 1$$?
Squaring on both ides and solve for n then we will get n=25.So we need to get a unique value 25 from A or b or both
(1) n is a perfect square -25,200,4,9,...
(2) n + 11 is a perfect square.-25,14,5,....

n=25 is only perfect square with n+11 a perfect square so unique and C is the answer.
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Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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24 May 2018, 16:06
Viserion99 wrote:
If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

Dear Bunuel,

I have got same answer C as the above. Can you change the OA?
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Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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10 Jun 2018, 07:57
Viserion99 wrote:
If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

The data is sufficient if we can determine whether the equation holds good with the information in the statements.

n is a positive integer.
Properties of perfect squares. The difference between squares of consecutive positive integers will increase by 2.
For instance, 2^2 - 1^2 = 3; 3^2 - 2^2 = 5; 4^2 - 3^2 = 7; 5^2 - 4^2 = 9; 6^2 - 5^2 = 11 and so on.

Statement 1: n is a perfect square.
If n = 4; $$\sqrt{4+11} \neq \sqrt{4} + 1$$
If n = 25; $$\sqrt{25+11} = \sqrt{25} + 1$$

Cannot determine whether the equality will hold good. Statement 1 alone is not sufficient.

Statement 2: n + 11 is a perfect square
If n = 14; n + 11 = 25, which is a perfect square. $$\sqrt{14+11} \neq \sqrt{14} + 1$$
If n = 25; n + 11 = 36, which is a perfect square. $$\sqrt{25+11} = \sqrt{25} + 1$$

Combining the two statements: n is a perfect square and (n + 11) is also a perfect square.
If the equation were to hold good, the two numbers n and (n + 11) will have to be squares of two consecutive numbers.
From the properties of difference between square of positive integers, there can exist only one set of values where the difference between squares of consecutive numbers is 11.
The difference between 6^2 and 5^2 is 11 and we cannot find any other set of two consecutive integers that will satisfy this condition. i.e., the only value that satisfies is when n = 25 and (n + 11) = 36.

Dear Bunuel,

Shouldn't the answer be E? I see a lot of my friends have said C, considering that n=25 and n+11 = 36, would solve the problem.

I say that because \sqrt{36} would be +/- 6 and not just =6. Similarly, the \sqrt{25}, will be +/-5 and not just +5.
Given this, I believe the ans to be E. I note that the restriction to be a positive integer is for n, not on \sqrt{n}.

Requesting your expertise! Thanks a lot.
Math Expert
Joined: 02 Sep 2009
Posts: 46207
Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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10 Jun 2018, 10:19
jayantbakshi wrote:
Viserion99 wrote:
If n is a positive integer greater than 1, is $$\sqrt{n+11} = \sqrt{n} + 1$$?

(1) n is a perfect square
(2) n + 11 is a perfect square.

The data is sufficient if we can determine whether the equation holds good with the information in the statements.

n is a positive integer.
Properties of perfect squares. The difference between squares of consecutive positive integers will increase by 2.
For instance, 2^2 - 1^2 = 3; 3^2 - 2^2 = 5; 4^2 - 3^2 = 7; 5^2 - 4^2 = 9; 6^2 - 5^2 = 11 and so on.

Statement 1: n is a perfect square.
If n = 4; $$\sqrt{4+11} \neq \sqrt{4} + 1$$
If n = 25; $$\sqrt{25+11} = \sqrt{25} + 1$$

Cannot determine whether the equality will hold good. Statement 1 alone is not sufficient.

Statement 2: n + 11 is a perfect square
If n = 14; n + 11 = 25, which is a perfect square. $$\sqrt{14+11} \neq \sqrt{14} + 1$$
If n = 25; n + 11 = 36, which is a perfect square. $$\sqrt{25+11} = \sqrt{25} + 1$$

Combining the two statements: n is a perfect square and (n + 11) is also a perfect square.
If the equation were to hold good, the two numbers n and (n + 11) will have to be squares of two consecutive numbers.
From the properties of difference between square of positive integers, there can exist only one set of values where the difference between squares of consecutive numbers is 11.
The difference between 6^2 and 5^2 is 11 and we cannot find any other set of two consecutive integers that will satisfy this condition. i.e., the only value that satisfies is when n = 25 and (n + 11) = 36.

Dear Bunuel,

Shouldn't the answer be E? I see a lot of my friends have said C, considering that n=25 and n+11 = 36, would solve the problem.

I say that because \sqrt{36} would be +/- 6 and not just =6. Similarly, the \sqrt{25}, will be +/-5 and not just +5.
Given this, I believe the ans to be E. I note that the restriction to be a positive integer is for n, not on \sqrt{n}.

Requesting your expertise! Thanks a lot.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
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Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2) [#permalink]

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10 Jun 2018, 21:54
1
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.[/quote]

Thanks so much Bunuel, this was very helpful! Very kind of you.

Much appreciated
Re: If n is a positive integer greater than 1, is (n + 11)^(1/2) = n^(1/2)   [#permalink] 10 Jun 2018, 21:54
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