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# What is the decreasing order from left to right of the following numbe

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Joined: 01 Oct 2017
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WE: Supply Chain Management (Energy and Utilities)
What is the decreasing order from left to right of the following numbe  [#permalink]

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14 Jul 2018, 06:36
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Difficulty:

95% (hard)

Question Stats:

34% (02:49) correct 66% (02:42) wrong based on 31 sessions

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What is the decreasing order from left to right of the following numbers?

$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$

(A) $$\sqrt[4]{3}, \sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}$$
(B) $$\sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}, \sqrt[4]{3}$$
(C) $$\sqrt[5]{6}, \sqrt[15]{45}, \sqrt[4]{3}, \sqrt[10]{12}$$
(D) $$\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}$$
(E) $$\sqrt[4]{3}, \sqrt[15]{45}, \sqrt[5]{6} , \sqrt[10]{12}$$
Senior Manager
Joined: 09 Jun 2014
Posts: 338
Location: India
Concentration: General Management, Operations
Re: What is the decreasing order from left to right of the following numbe  [#permalink]

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14 Jul 2018, 09:50
Good question. No idea how to approach.Any help would be great !!

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Math Expert
Joined: 02 Aug 2009
Posts: 7764
What is the decreasing order from left to right of the following numbe  [#permalink]

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14 Jul 2018, 09:59
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PKN wrote:
What is the decreasing order from left to right of the following numbers?

$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$

(A) $$\sqrt[4]{3}, \sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}$$
(B) $$\sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}, \sqrt[4]{3}$$
(C) $$\sqrt[5]{6}, \sqrt[15]{45}, \sqrt[4]{3}, \sqrt[10]{12}$$
(D) $$\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}$$
(E) $$\sqrt[4]{3}, \sqrt[15]{45}, \sqrt[5]{6} , \sqrt[10]{12}$$

since the roots are 5, 10 and 15 we can easily compare these

1) let's compare $$\sqrt[5]{6}-and-\sqrt[15]{45}$$

$$\sqrt[5]{6}=\sqrt[15]{6^3}=\sqrt[15]{216}......................\sqrt[15]{45}$$............$$216>45$$ so $$\sqrt[5]{6}>\sqrt[15]{45}$$..

2) lets compare $$\sqrt[10]{12}-and-\sqrt[15]{45}$$

$$\sqrt[10]{12}=\sqrt[30]{12^3}=\sqrt[30]{3^3*4^3}................\sqrt[15]{45}=\sqrt[30]{45^2}=\sqrt[30]{3^4*5^2}$$ $$3^45^2>3^34^3$$ hence
$$\sqrt[10]{12}<\sqrt[15]{45}$$

3) finally lets compare $$\sqrt[4]{3}-and-\sqrt[15]{45}$$

$$\sqrt[4]{3}=\sqrt[60]{3^{15}}.................. \sqrt[15]{45}=\sqrt[60]{45^4}=\sqrt[60]{3^8*5^4},$$...
$$3^{15}=3^83^7,3^85^4$$.....3^7>5^4
hence $$\sqrt[4]{3}> \sqrt[15]{45},$$

$$\sqrt[5]{6}> \sqrt[4]{3}> \sqrt[15]{45}> \sqrt[10]{12}$$

D
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Re: What is the decreasing order from left to right of the following numbe  [#permalink]

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14 Jul 2018, 10:17
chetan2u wrote:
PKN wrote:
What is the decreasing order from left to right of the following numbers?

$$\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}$$

(A) $$\sqrt[4]{3}, \sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}$$
(B) $$\sqrt[5]{6}, \sqrt[10]{12}, \sqrt[15]{45}, \sqrt[4]{3}$$
(C) $$\sqrt[5]{6}, \sqrt[15]{45}, \sqrt[4]{3}, \sqrt[10]{12}$$
(D) $$\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}$$
(E) $$\sqrt[4]{3}, \sqrt[15]{45}, \sqrt[5]{6} , \sqrt[10]{12}$$

since the roots are 5, 10 and 15 we can easily compare these

1) let's compare $$\sqrt[5]{6}-and-\sqrt[15]{45}$$

$$\sqrt[5]{6}=\sqrt[15]{6^3}=\sqrt[15]{216}......................\sqrt[15]{45}$$............$$216>45$$ so $$\sqrt[5]{6}>\sqrt[15]{45}$$..

2) lets compare $$\sqrt[10]{12}-and-\sqrt[15]{45}$$

$$\sqrt[10]{12}=\sqrt[30]{12^3}=\sqrt[30]{3^3*4^3}................\sqrt[15]{45}=\sqrt[30]{45^2}=\sqrt[30]{3^4*5^2}$$ $$3^45^2>3^34^3$$ hence
$$\sqrt[10]{12}<\sqrt[15]{45}$$

3) finally lets compare $$\sqrt[4]{3}-and-\sqrt[15]{45}$$

$$\sqrt[4]{3}=\sqrt[60]{3^{15}}.................. \sqrt[15]{45}=\sqrt[60]{45^4}=\sqrt[60]{3^8*5^4},$$...
$$3^{15}=3^83^7,3^85^4$$.....3^7>5^4
hence $$\sqrt[4]{3}> \sqrt[15]{45},$$

$$\sqrt[5]{6}> \sqrt[4]{3}> \sqrt[15]{45}> \sqrt[10]{12}$$

D

Thank you so much Chetan sir!!This is great ..

Posted from my mobile device
Re: What is the decreasing order from left to right of the following numbe   [#permalink] 14 Jul 2018, 10:17
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