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Rank those three in order from smallest to biggest.

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Rank those three in order from smallest to biggest.  [#permalink]

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New post 26 Oct 2016, 21:53
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Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I

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Rank those three in order from smallest to biggest.  [#permalink]

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New post Updated on: 28 Oct 2016, 20:06
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Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Raise all the options to the power of 4.

I. 2^4* 5^2 = 400

II. 3^4*2^2 = 324

III. 401

II < I < III

C

Edited calculation error


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Originally posted by acegmat123 on 27 Oct 2016, 00:32.
Last edited by acegmat123 on 28 Oct 2016, 20:06, edited 2 times in total.
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Re: Rank those three in order from smallest to biggest.  [#permalink]

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New post 26 Oct 2016, 22:21
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


IMO C
2 sqrt5= 2*2.236= 4.472
3 sqrt2= 3* 1.414= 4.242
(401)^(1/4) = 4.475 (1, 16, 81, 256, 625) === (401)^(1/4) will fall between 256 and 625
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Rank those three in order from smallest to biggest.  [#permalink]

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New post 27 Oct 2016, 02:10
acegmat123 wrote:
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Raise all the options to the power of 4.

I. 2* 5^2 = 50

II. 3*2^2 = 12

III. 401

II < I < III

C

1 is 16*25= 400
2 is 81*4=324
3 is 401

Liked your style of solution though :) i went the conventional way. Took me a bit more time than this solution :o :-D :-D :-D
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Re: Rank those three in order from smallest to biggest.  [#permalink]

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New post 27 Oct 2016, 07:27
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Put all numbers into their most basic square root forms.

I. \(2\sqrt{5}\) = \(\sqrt{4*5} = \sqrt{20}\)

II. \(3\sqrt{2}\)
= \(\sqrt{9*2}\) = \(\sqrt{18}\)

III. \(\sqrt[4]{401}\). This one is a bit more difficult. To take the 4th root is to take the square root twice. The square root of 401 is slightly more than 20, so the 4th root of 401 will be slightly more than \(\sqrt{20}\).

Thus, our ordering from smallest to largest is II,I,III.

Answer: C
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Re: Rank those three in order from smallest to biggest.  [#permalink]

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New post 12 Dec 2016, 09:04
speedilly wrote:
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Put all numbers into their most basic square root forms.

I. \(2\sqrt{5}\) = \(\sqrt{4*5} = \sqrt{20}\)

II. \(3\sqrt{2}\)
= \(\sqrt{9*2}\) = \(\sqrt{18}\)

III. \(\sqrt[4]{401}\). This one is a bit more difficult. To take the 4th root is to take the square root twice. The square root of 401 is slightly more than 20, so the 4th root of 401 will be slightly more than \(\sqrt{20}\).

Thus, our ordering from smallest to largest is II,I,III.

Answer: C


For the last step I would suggest to bring \(\sqrt{20}\) to \(2\sqrt{5}\)[/b], which is a little bit smaller than I
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Re: Rank those three in order from smallest to biggest.  [#permalink]

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New post 22 Jan 2018, 07:17
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


I. \((2\sqrt{5})^2\) = 20 & \(20^2\) = \(400\)
II.\((3\sqrt{2})2\) = 18 & \(18^2\) = \(324\)
III.\((\sqrt[4]{401})^4\) = \(401\)

So, Ascending order of arrangement will be II, I, III , answer will be (C)
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Re: Rank those three in order from smallest to biggest.  [#permalink]

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New post 22 Jan 2018, 07:30
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I



Best way to answer such questions is APPROXIMATION

I. \(2\sqrt{5}\) = 2*2.25 = 4.5


II. \(3\sqrt{2}\) = 3*1.41 = 4.2


III. \(\sqrt[4]{401}\) = \(\sqrt{20}\) = 4.49


Answer: Option C
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Rank those three in order from smallest to biggest.  [#permalink]

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New post 22 Jan 2018, 08:27
Bunuel wrote:
Rank those three in order from smallest to biggest.

I. \(2\sqrt{5}\)

II. \(3\sqrt{2}\)

III. \(\sqrt[4]{401}\)


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I


Best way to answer this is \(a^4 = 400, b^4 = 162, c^4 = 401\).. Now if we have doubt in this try\(a^4 - b^4 = (a-b)(a+b)(a^2+b^2)\).. Similarly,\(c^4 - a^4, c^4 - b^4\)
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Rank those three in order from smallest to biggest. &nbs [#permalink] 22 Jan 2018, 08:27
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