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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] New post 28 Feb 2011, 02:12
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71% (02:40) correct 29% (02:19) wrong based on 113 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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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] New post 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] New post 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.

Answer: B.
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Re: n [#permalink] New post 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] New post 28 Feb 2011, 18:51
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GMATD11 wrote:
From B and E whts wrong with E


Just follow the rules of exponents. Answer will follow.
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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Re: n [#permalink] New post 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???????
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Re: n [#permalink] New post 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.
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Re: n [#permalink] New post 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
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Re: n [#permalink] New post 01 Aug 2011, 20:08
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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.
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Re: n [#permalink] New post 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] New post 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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