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M19-27

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M19-27 [#permalink]

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New post 16 Sep 2014, 01:06
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\(\frac{1}{\sqrt{51}}\) is between:

A. 0 and \(\frac{1}{9}\)
B. \(\frac{1}{9}\) and \(\frac{1}{7}\)
C. \(\frac{1}{7}\) and \(\frac{1}{6}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{5}\) and \(\frac{1}{2}\)
[Reveal] Spoiler: OA

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Re M19-27 [#permalink]

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New post 16 Sep 2014, 01:06
Official Solution:

\(\frac{1}{\sqrt{51}}\) is between:

A. 0 and \(\frac{1}{9}\)
B. \(\frac{1}{9}\) and \(\frac{1}{7}\)
C. \(\frac{1}{7}\) and \(\frac{1}{6}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{5}\) and \(\frac{1}{2}\)

Because \(\sqrt{49} = 7 \lt \sqrt{51} \lt \sqrt{81} = 9\), \(\frac{1}{\sqrt{51}}\) is between \(\frac{1}{9}\) and \(\frac{1}{7}\).

Answer: B
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M19-27 [#permalink]

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New post 16 Jan 2017, 14:38
I might be being stupid here but is it not also between 0 and 1/9?

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Re: M19-27 [#permalink]

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New post 16 Jan 2017, 15:32
sefwow wrote:
I might be being stupid here but is it not also between 0 and 1/9?


No, Because \(\frac{1}{\sqrt{51}}\) is bigger than \(\frac{1}{9}\).

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Re: M19-27 [#permalink]

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New post 29 Aug 2017, 16:04
The key to this question is to not be too worried about knowing the precise square root or decimal of a number, but to be able to approximate and detect what is unnecessary in the problem, which in this case the problem tries to throw off the test taker by adding in reciprocals.

The question simply asks to determine what two numbers is 1/\(\sqrt{51}\) between. To make the problem simpler, dumb down the question and focus only on \(\sqrt{51}\) and not 1/\(\sqrt{51}\). We know that \(\sqrt{51}\) is between \(\sqrt{49}\) and \(\sqrt{64}\); therefore,
7 < \(\sqrt{51}\) < 8.

Since 1/\(\sqrt{51}\) will yield a smaller fraction due to a larger number in the denominator, 1/\(\sqrt{51}\) is smaller than 1/7 and larger than 1/8.

Looking at the answer choice, the only number smaller than 1/8 is 0 and 1/9. This leaves us with answer choices A and B.

Because we deduced that 1/\(\sqrt{51}\) is smaller than 1/7 and larger than 1/8, we know that 1/\(\sqrt{51}\) is larger than 1/9 but smaller than 1/7. The answer is B.

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Re: M19-27   [#permalink] 29 Aug 2017, 16:04
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