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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]

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28 Feb 2011, 03:12
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 10 Feb 2014, 01:50, edited 1 time in total.
Renamed the topic and edited the question.

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28 Feb 2011, 03:19
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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28 Feb 2011, 03:27
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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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28 Feb 2011, 07:20
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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28 Feb 2011, 19:51
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17359 [1], given: 232 Intern Joined: 14 Apr 2011 Posts: 11 Kudos [?]: 1 [0], given: 0 Re: n [#permalink] ### Show Tags 01 Aug 2011, 09:44 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??????? Kudos [?]: 1 [0], given: 0 Director Joined: 01 Feb 2011 Posts: 726 Kudos [?]: 143 [0], given: 42 Re: n [#permalink] ### Show Tags 01 Aug 2011, 11:50 n^(5/6) = (1/2)n => n^6-2^6*n^5 = 0 => n =0 or 64. Answer is B. Kudos [?]: 143 [0], given: 42 Intern Joined: 27 Feb 2011 Posts: 45 Kudos [?]: 4 [0], given: 9 Re: n [#permalink] ### Show Tags 01 Aug 2011, 13:43 GMATD11 wrote: From B and E whts wrong with E 50% of 729 is not an integer .. whereas @(729) is an integer Kudos [?]: 4 [0], given: 9 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17359 [1], given: 232 Location: Pune, India Re: n [#permalink] ### Show Tags 01 Aug 2011, 21:08 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 Get started with Veritas Prep GMAT On Demand for$199

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01 Aug 2011, 23:35
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]

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19 Jul 2014, 09:36
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Re: If (n) is defined as the product of the cube root of n and [#permalink]

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10 Mar 2015, 03:31
VeritasPrepKarishma wrote:
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.

I really need to study these formulas, where do I suggest I go?

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Re: If (n) is defined as the product of the cube root of n and [#permalink]

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10 Mar 2015, 06:05
erikvm wrote:
VeritasPrepKarishma wrote:
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.

I really need to study these formulas, where do I suggest I go?

Here are the basics of exponents and roots:

http://www.veritasprep.com/blog/2011/07 ... eparation/
http://www.veritasprep.com/blog/2011/07 ... ration-ii/
http://www.veritasprep.com/blog/2011/07 ... s-applied/
http://www.veritasprep.com/blog/2011/08 ... -the-gmat/
http://www.veritasprep.com/blog/2011/08 ... exponents/
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Re: If @(n) is defined as the product of the cube root of n and [#permalink]

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10 May 2016, 01:01
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Re: If @(n) is defined as the product of the cube root of n and [#permalink]

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26 Jun 2017, 07:01
Hello from the GMAT Club BumpBot!

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Re: If (n) is defined as the product of the cube root of n and [#permalink]

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28 Jun 2017, 17:01
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GMATD11 wrote:
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

We are given that @(n) is defined as the product of the cube root of n and the positive square root of n.

We need to determine for what number n does @(n) = 50 percent of n.

We see that @n = (∛n)(√n) = (n^(⅓))(n^(½)) = n^(⅓ + ½) = n^(⅚)

We need to determine a value for n when:

n^(⅚) = 0.5n

2n^(⅚) = n

2 = n/n^(⅚)

2 = n^(⅙)

2^6 = n

n = 64

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Re: If (n) is defined as the product of the cube root of n and   [#permalink] 28 Jun 2017, 17:01
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