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Why in this case it's wrong to square all and remove the square root?
In this question https://gmatclub.com/forum/2-240047.html I saw that it's feasible... why in this case it isn't?
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MvArrow
Why in this case it's wrong to square all and remove the square root?
In this question https://gmatclub.com/forum/2-240047.html I saw that it's feasible... why in this case it isn't?


Let a= \(\frac{\sqrt{21}}{6}\)

If you want to remove the square root, square both sides

\(a^2 =\frac{21}{6^2}\) = \(\frac{21}{36} = \frac{7*3}{12*3} = \frac{7}{12}\)

Taking square root, we get a = \(\sqrt{\frac{7}{12}}\)(Option D)
Hope it helps!
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MvArrow
Which of the following is equivalent to \(\frac{\sqrt{21}}{6}\)?

A. \(\frac{\sqrt{7}}{2}\)

B. 7/12

C. \(\sqrt{\frac{7}{2}}\)

D. \(\sqrt{\frac{7}{12}}\)

E. \(\frac{7\sqrt{3}}{2}\)

Why in this case it's wrong to square all and remove the square root?
In this question https://gmatclub.com/forum/2-240047.html I saw that it's feasible... why in this case it isn't?

You can do this way but you should square the options too to compare apples with apples. When you square, the question becomes: which of the following is equivalent to \((\frac{\sqrt{21}}{6})^2=\frac{21}{36}=\frac{7}{12}\). Option D squared gives exactly this value: \((\sqrt{\frac{7}{12}})^2=\frac{7}{12}\)

Answer: D.

Hope it helps.
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Bunuel
Which of the following is equivalent to \(\frac{\sqrt{21}}{6}\)?

A. \(\frac{\sqrt{7}}{2}\)

B. 7/12

C. \(\sqrt{\frac{7}{2}}\)

D. \(\sqrt{\frac{7}{12}}\)

E. \(\frac{7\sqrt{3}}{2}\)

\(\frac{\sqrt{21}}{6}\)
\(\frac{\sqrt{7*3}}{\sqrt{3*3*2*2}}\)
\(\frac{\sqrt{7}}{\sqrt{3*2*2}}\)
\(\frac{\sqrt{7}}{\sqrt{12}}\)

Answer D.
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Bunuel
MvArrow
Which of the following is equivalent to \(\frac{\sqrt{21}}{6}\)?

A. \(\frac{\sqrt{7}}{2}\)

B. 7/12

C. \(\sqrt{\frac{7}{2}}\)

D. \(\sqrt{\frac{7}{12}}\)

E. \(\frac{7\sqrt{3}}{2}\)

Why in this case it's wrong to square all and remove the square root?
In this question https://gmatclub.com/forum/2-240047.html I saw that it's feasible... why in this case it isn't?

You can do this way but you should square the options too to compare apples with apples. When you square, the question becomes: which of the following is equivalent to \((\frac{\sqrt{21}}{6})^2=\frac{21}{36}=\frac{7}{12}\). Option D squared gives exactly this value: \((\sqrt{\frac{7}{12}})^2=\frac{7}{12}\)

Answer: D.

Hope it helps.

Great! Thanks! I din't think about it
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Bunuel
Which of the following is equivalent to \(\frac{\sqrt{21}}{6}\)?

A. \(\frac{\sqrt{7}}{2}\)

B. 7/12

C. \(\sqrt{\frac{7}{2}}\)

D. \(\sqrt{\frac{7}{12}}\)

E. \(\frac{7\sqrt{3}}{2}\)


We can also ballpark...no issues with applying square root rules and what we need to consider when we square things.

\(\sqrt{21}\) is somewhere between 4 and 5. Let's call it 4.5. \(\frac{4.5}{6}=0.75\)

Let's check the answer choices.

A. \(\sqrt{7}\) is somewhere between 2 and 3. Either way, it's greater than 2, so this answer choice is greater than 1. Eliminate.

B. Hmmm, it's between 0.5 and 1, so it's close, but we want 0.75 and this is just barely more than 0.5. Eliminate.

C. \(\sqrt{3.5}\) is greater than 1. Eliminate.

D. 7/12 is just a little bigger than 0.5, so this is between 0.5 and 1, and it's close to the middle of that range. Keep it.

E. \(3.5*\sqrt{3}\) is greater than 1. Eliminate.

We only have one that's close enough to be correct.

Answer choice D.


You might think that a solution like this is "not as good" as doing the "real" math. I'll ask you WHY? What about the "real" math makes it better? Which method gets to the correct answer? Both. Which method introduces a lower probability of making a careless mistake with a fraction? I'd argue that ballparking is safer. Do business schools see your scratch paper for the test and analyze whether you did the "real" math, or do they just see a score? Sooooo, what about the "real" math makes it better on the GMAT? No bonus points for doing it that way; a right answer is a right answer! Ballparking is a very useful skill.
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