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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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2 00:00

Difficulty:   55% (hard)

Question Stats: 62% (02:28) correct 38% (03:09) wrong based on 21 sessions

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If $$\sqrt{x} = 12$$, $$\sqrt{y}$$ = 8, $$\sqrt{z} = 6$$ 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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Abhi077 wrote:
If $$\sqrt{x} = 2$$, $$\sqrt{y}$$ = 8, $$\sqrt{z} = 6$$ 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, $$\sqrt{y}$$ = 8
i.e.$$y^{1/8} = 8$$

i.e. $$y = 8^8$$

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

Also, $$\sqrt{z} = z^{1/6} = 6$$

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 = $$2^{96}$$

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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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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
_________________ 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   