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M19-30

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M19-30 [#permalink]

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New post 16 Sep 2014, 01:07
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If \(x = 10^{10}\), \(\frac{x^2 + 2x + 7}{3x^2 - 10x + 200}\) is closest to:

A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{5}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 01:07
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Official Solution:

If \(x = 10^{10}\), \(\frac{x^2 + 2x + 7}{3x^2 - 10x + 200}\) is closest to:

A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{5}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)


\(\frac{x^2 + 2x + 7}{3x^2 - 10x + 200} = \frac{1 + \frac{2}{x} + \frac{7}{x^2}}{3 - \frac{10}{x} + \frac{200}{x^2}}\). When \(x\) is large, \(\frac{2}{x}\), \(\frac{7}{x^2}\), \(\frac{10}{x}\), and \(\frac{200}{x^2}\) are small and the fraction is very close to \(\frac{1}{3}\).

Alternative Explanation

\(\frac{x^2 + 2x + 7}{3x^2 - 10x + 200}=\frac{10^{20} + 2*10^{10} + 7}{3*10^{20} - 10*10^{10} + 200}\).

Note that we need approximate value of the given expression. Now, since \(10^{20}\) is much larger number than \(2*10^{10} + 7\), then \(2*10^{10} + 7\) is pretty much negligible in this case. Similarly \(3*10^{20}\) is much larger number than \(-10*10^{10} + 200\), so \(-10*10^{10} + 200\) is also negligible in this case.

So, \(\frac{10^{20} + 2*10^{10} + 7}{3*10^{20} - 10*10^{10} + 200} \approx \frac{10^{20}}{3*10^{20}}=\frac{1}{3}\).


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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M19-30 [#permalink]

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New post 11 Sep 2015, 14:52
I think this is a high-quality question and I agree with explanation.
However, the exponents are not clear. I thought they are actually 3 and not 2.

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New post 22 Dec 2016, 02:32
can we also take x=10..i am getting same answer

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New post 22 Dec 2016, 02:38

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M19-30 [#permalink]

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New post 07 Sep 2017, 14:28
Bunuel wrote:
Official Solution:

If \(x = 10^{10}\), \(\frac{x^2 + 2x + 7}{3x^2 - 10x + 200}\) is closest to:

A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{5}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)


\(\frac{x^2 + 2x + 7}{3x^2 - 10x + 200} = \frac{1 + \frac{2}{x} + \frac{7}{x^2}}{3 - \frac{10}{x} + \frac{200}{x^2}}\). When \(x\) is large, \(\frac{2}{x}\), \(\frac{7}{x^2}\), \(\frac{10}{x}\), and \(\frac{200}{x^2}\) are small and the fraction is very close to \(\frac{1}{3}\).



Answer: B



please explain the first part/ method. How did you solve it.

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M19-30   [#permalink] 07 Sep 2017, 14:28
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