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).
_________________

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
_________________

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.

Hi Kzivrev

choice D also have I before II

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

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 \(\sqrt{5/9}\)

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
_________________

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________