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705-805 Level|   Roots|      
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Bunuel
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Rank the following quantities in order, from smallest to biggest. 13 Mar 2015, 07:44
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44% (01:34) correct 56% (01:26) wrong based on 685 sessions

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Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) III, I, II
(E) III, II, I
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A
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Bunuel
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Rank the following quantities in order, from smallest to biggest. 15 Mar 2015, 22:08
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Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

First of all, clearly \(\sqrt{2/3}=\sqrt{\frac{4}{9}}<\sqrt{\frac{5}{9}}\)

So, II is bigger than I. Now, what about III? When we take higher order roots, the values move closer to one. If the number starts larger than one, then higher and higher roots make it smaller, closer to one. If the number starts between 0 and 1, then higher and higher roots make it larger, closer to one. Therefore, III is larger than II. From smallest to biggest, I, II, III.

Answer = (A).
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Rank the following quantities in order, from smallest to biggest. 13 Mar 2015, 09:10
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Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

in positive numbers , we should remember that square root of a fraction is always greater than the fraction...that is \(\sqrt{x/y}\)> \(x/y\), where x<y.... and square or higher power of a fraction will keep reducing the value with higher powers...
\(2/3\)=\(\sqrt{4/9}\)<\(\sqrt{5/9}\)<5\(\sqrt{4/9}\)
so ans A..l,ll,lll
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Rank the following quantities in order, from smallest to biggest. 09 Apr 2015, 18:07
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well to figure out the first two expresion it is easy to determine that II > I, so since it is asked to make smalest to bigest, we can conclude that I needs to be before II, looking in the answer choices only A has I before II so safely we can pick A as final answer. Im sure if the answer choices had different setup the outcome would be much harder to determine if we are not sure how to evaulate III.
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Rank the following quantities in order, from smallest to biggest. 09 Apr 2015, 22:33
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Hi Kzivrev

choice D also have I before II
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Rank the following quantities in order, from smallest to biggest. 10 Apr 2015, 16:14
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HI Naina1, you are totlay correct, I didnt see that one , I guess was just lucky to get the correct answer, I was doing it fast, and I do understant the concept behind the root fraction. I hope to be lucky on the D-day :)
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Rank the following quantities in order, from smallest to biggest. 11 Dec 2017, 20:39
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Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

Hi Bunuel

Can you please help me to clarify the doubt.

From Stmnt 1 & 2 when we compare the numerator of fraction then \(\sqrt{5}\) > 2. However, \(\sqrt{5/9}\) can be written as \(\sqrt{5}\)/3 or \(\sqrt{5}\) / -3 when base is +ve \(\sqrt{5}/\) 3 > 2/3 but when base is -ve \(\sqrt{5}/\) -3 < 2/3.

Please advise how do we decide on the base of \(\sqrt{5/9}\)
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Rank the following quantities in order, from smallest to biggest. 11 Dec 2017, 20:49
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rahul16singh28
Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

Hi Bunuel

Can you please help me to clarify the doubt.

From Stmnt 1 & 2 when we compare the numerator of fraction then \(\sqrt{5}\) > 2. However, \(\sqrt{5/9}\)can be written as \([fraction]5[/fraction]\)/ 3 or \(\sqrt{5}\) / -3 when base is +ve \(\sqrt{5}/\) 3 > 2/3 but when base is -ve \(\sqrt{5}/\) -3 < 2/3.

Please advise how do we decide on the base of \sqrt{5/9}

Hi..
Square root is always positive..
So √9= 3 only..
But square is where you look at both + & -
X^2=9.. x=3 or -3
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Rank the following quantities in order, from smallest to biggest. 22 Apr 2018, 00:56
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Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

2/3 = \(\sqrt{\frac{4}{9}}\) <\(\sqrt{\frac{5}{9}}\)
So, I < II

Now since 0<5/9<1.. So powers will reduce its value and roots will increase its value.
Square root will be lees than cube root will be less than 4th root will be less than 5th root.. and so on.. Th value will keep on increasing towards 1.

Hence, \(\sqrt{\frac{5}{9}}\) < \(\sqrt[5]{\frac{5}{9}}\)

So, II < III

Hence, I < II < III.
Answer A
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Rank the following quantities in order, from smallest to biggest. 28 Jul 2023, 10:31
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Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9\)

III. \(\sqrt[5]{\frac{5}{9\)

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

Kudos for a correct solution.


We have: 2/3, (5/9)^0.5 and (5/9)^0.2

Since 5/9 < 1: Higher the power, lower will be the value. Thus, (5/9)^0.5 < (5/9)^0.2

Comparing 2/3 and (5/9)^0.5: Squaring both, we get 4/9 and 5/9. Since 4/9 < 5/9, we have: 2/3 < (5/9)^0.5

Thus: 2/3 < (5/9)^0.5 < (5/9)^0.2, i.e. I < II < III

Answer A
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Rank the following quantities in order, from smallest to biggest. 30 May 2025, 02:11
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didn't get the explanation provided please if you could elaborate how III is bigger?
Bunuel
Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

First of all, clearly \(\sqrt{2/3}=\sqrt{\frac{4}{9}}<\sqrt{\frac{5}{9}}\)

So, II is bigger than I. Now, what about III? When we take higher order roots, the values move closer to one. If the number starts larger than one, then higher and higher roots make it smaller, closer to one. If the number starts between 0 and 1, then higher and higher roots make it larger, closer to one. Therefore, III is larger than II. From smallest to biggest, I, II, III.

Answer = (A).
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Bunuel
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Rank the following quantities in order, from smallest to biggest. 30 May 2025, 10:04
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shubhim20
didn't get the explanation provided please if you could elaborate how III is bigger?
Bunuel
Bunuel
Rank the following quantities in order, from smallest to biggest.

I. 2/3

II. \(\sqrt{\frac{5}{9}}\)

III. \(\sqrt[5]{\frac{5}{9}}\)

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

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

First of all, clearly \(\sqrt{2/3}=\sqrt{\frac{4}{9}}<\sqrt{\frac{5}{9}}\)

So, II is bigger than I. Now, what about III? When we take higher order roots, the values move closer to one. If the number starts larger than one, then higher and higher roots make it smaller, closer to one. If the number starts between 0 and 1, then higher and higher roots make it larger, closer to one. Therefore, III is larger than II. From smallest to biggest, I, II, III.

Answer = (A).

Here’s the key idea: when you take a root of a number between 0 and 1, the result is larger than the original number. And the higher the root, the closer it gets to 1.

For example, \(\sqrt{\frac{1}{2}} < \sqrt[3]{\frac{1}{2}} < \sqrt[4]{\frac{1}{2}} < \sqrt[5]{\frac{1}{2}}...\)

So, \(\sqrt{\frac{5}{9}}\), which is about 0.75, is less than \(\sqrt[5]{\frac{5}{9}}\), which is about 0.9.
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