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If 12th root of x is equal to 12, 8th root of y equals to 8

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If 12th root of x is equal to 12, 8th root of y equals to 8  [#permalink]

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New post Updated on: 13 Apr 2019, 00:55
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If \(\sqrt[12]{x} = 12\), \(\sqrt[8]{y}\) = 8, \(\sqrt[6]{z} = 6\) then which of the following is the median of\(\sqrt{x^{12}}\) , \(\sqrt{y^8}\) and\(\sqrt{z^6}\)

A) \(8^{98}\)

B)\(2^{96}\)

C)\(6^{18}\)

D)\(8^{10}\)

E)\(4^{12}\)

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Originally posted by Abhi077 on 13 Apr 2019, 00:01.
Last edited by Abhi077 on 13 Apr 2019, 00:55, edited 2 times in total.
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If 12th root of x is equal to 12, 8th root of y equals to 8  [#permalink]

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New post Updated on: 13 Apr 2019, 04:44
Abhi077 wrote:
If \(\sqrt[12]{x} = 2\), \(\sqrt[8]{y}\) = 8, \(\sqrt[6]{z} = 6\) then which of the following is the median of\(\sqrt{x^{12}}\) , \(\sqrt{y^8}\) and\(\sqrt{z^6}\)

A) \(8^{98}\)

B)\(2^{96}\)

C)\(6^{18}\)

D)\(8^{10}\)

E)\(4^{12}\)


\(x^{1/12} = 12\)
i.e.\(x = 12^{12}\)

i.e. \(\sqrt{x^{12}}\) \(= x^6 = 12^{72}\)


Now, \(\sqrt[8]{y}\) = 8
i.e.\(y^{1/8} = 8\)

i.e. \(y = 8^8\)

i.e. \(\sqrt{y^8} = y^4 = 8^{32} = 2^{96}\)

Also, \(\sqrt[6]{z} = z^{1/6} = 6\)

i.e. \(z = 6^6\)

i.e. \(\sqrt{z^6} = z^3 = 6^{18}\)


\(\sqrt{z^6} = 6^{18}\) is the smallest number

and \(12^{72} = 2^{144}*3^{72}\) is even bigger than \(2^{96}\)

So median = Middle term in increasing order of these numbers = \(2^{96}\)



Answer: Option B
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Originally posted by GMATinsight on 13 Apr 2019, 00:51.
Last edited by GMATinsight on 13 Apr 2019, 04:44, edited 1 time in total.
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Re: If 12th root of x is equal to 12, 8th root of y equals to 8  [#permalink]

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New post 13 Apr 2019, 00:58
GMATinsight i'm sorry there was an error in typing. it's 12th root of x = 12 and not 2. I have corrected it. answer is B
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Re: If 12th root of x is equal to 12, 8th root of y equals to 8   [#permalink] 13 Apr 2019, 00:58

If 12th root of x is equal to 12, 8th root of y equals to 8

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