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# If @(n) is defined as the product of the cube root of n and

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If @(n) is defined as the product of the cube root of n and [#permalink]  28 Feb 2011, 02:12
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Difficulty:

35% (medium)

Question Stats:

75% (02:41) correct 25% (02:19) wrong based on 75 sessions
If @(n) is defined as the product of the cube root of n and the positive square root of n, then for what number n does @(n)=50 percent of n?

A. 16
B. 64
C. 100
D. 144
E. 729
[Reveal] Spoiler: OA

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n integer.JPG [ 13.49 KiB | Viewed 1921 times ]

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Last edited by Bunuel on 10 Feb 2014, 00:50, edited 1 time in total.
Renamed the topic and edited the question.
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Re: n [#permalink]  28 Feb 2011, 02:19
for 16, cube root is 2* cube root of 2 and positive square root is 4, so @n = 2*2^(1/3)*4 = 8 *2(1/3) so greater than 8 or greater than 0.5n

Similarly for 64, it is 4*8 = 32 = 0.5*64, so B is correct.

For E, the cube root is 9 and positive square root is 27, so 27*9 is not equal to 0.5*729, so incorrect
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Re: n [#permalink]  28 Feb 2011, 02:27
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If @(n) is defined as the product of the cube root of n and the positive square root of n, then for what number n does @(n)=50 percent of n?

A. 16
B. 64
C. 100
D. 144
E. 729

Given: @(n)=\sqrt[3]{n}*\sqrt[2]{n}. Question: if @(n)=0.5n then n=?

So we have that \sqrt[3]{n}*\sqrt[2]{n}=\frac{1}{2}*n --> 2*\sqrt[3]{n}*\sqrt[2]{n}=n --> take to the 6th power --> 64*n^2*n^3=n^6 --> n=64.

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Re: n [#permalink]  28 Feb 2011, 06:20
n pow(1/3)* n pow(1/2)=0.5n

n pow (5/6)= 0.5n

n pow(1/6)=2

n= 64
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Re: n [#permalink]  28 Feb 2011, 18:51
Expert's post
GMATD11 wrote:
From B and E whts wrong with E

Cube root is the power of 1/3. Square root is the power of 1/2

n^{\frac{1}{3}}*n^{\frac{1}{2}} = \frac{n}{2}
You need to find n. So bring all n's together on one side of the equation and everything else on the other side.

Adding the exponents, n^{\frac{5}{6}} = \frac{n}{2}
Clubbing n's together, 2 = n^{1-\frac{5}{6}}
n = 2^6 = 64

Hence it cannot be 729. If in the question, rather than half, we had a third, answer would have been 729.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Intern Joined: 14 Apr 2011 Posts: 11 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: n [#permalink] 01 Aug 2011, 08:44 the cube root of what integer power of 2 is closest to 50? 1)16 2) 17 3)18 4 ) 19 5) 20 can u pls help me in this by a quicker solution??????? Director Joined: 01 Feb 2011 Posts: 770 Followers: 14 Kudos [?]: 82 [0], given: 42 Re: n [#permalink] 01 Aug 2011, 10:50 n^(5/6) = (1/2)n => n^6-2^6*n^5 = 0 => n =0 or 64. Answer is B. Intern Joined: 27 Feb 2011 Posts: 48 Followers: 0 Kudos [?]: 2 [0], given: 9 Re: n [#permalink] 01 Aug 2011, 12:43 GMATD11 wrote: From B and E whts wrong with E 50% of 729 is not an integer .. whereas @(729) is an integer Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4772 Location: Pune, India Followers: 1115 Kudos [?]: 5054 [1] , given: 164 Re: n [#permalink] 01 Aug 2011, 20:08 1 This post received KUDOS Expert's post sushantarora wrote: the cube root of what integer power of 2 is closest to 50? 1)16 2) 17 3)18 4 ) 19 5) 20 can u pls help me in this by a quicker solution??????? Look at the powers of 2. 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 Since it is an exponential increase, the result increases much more as you go to higher and higher powers. Which powers of 2 are around 50? 2^5 = 32 2^6 = 64 50 is almost in the middle of the two of them but closer to 64. Also, the result increases more with higher powers so I would expect 50 to be almost 2^(5.6) or a little higher. If you find the cube root of 2^18, you will get (2^18)^(1/3) = 2^6 If you find the cube root of 2^17, you will get (2^17)^(1/3) = 2^(5.667) This is the closest. Answer is 17. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: n [#permalink]  01 Aug 2011, 22:35
hi karishma,

seems like the best and easiest ans .. thank you so much .
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Re: If @(n) is defined as the product of the cube root of n and [#permalink]  19 Jul 2014, 08:36
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Re: If @(n) is defined as the product of the cube root of n and   [#permalink] 19 Jul 2014, 08:36
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