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If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to

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If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink] New post 10 Dec 2012, 08:01
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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}

m19 q30
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Dec 2012, 08:12, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink] New post 10 Dec 2012, 08:15
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sje12 wrote:
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}

m19 q30


\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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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink] New post 10 Dec 2012, 08:19
Thanks! Very quick and simple way to do that
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink] New post 10 Dec 2012, 08:21
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink] New post 10 Dec 2012, 08:23
I'm not sure which book it is from, however I obtained it from a course instructor in Switzerland (Absolute GMAT, gmat-kurse.ch)
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink] New post 17 Jun 2013, 02:18
Thanks Bunuel
I started with finding the roots & trying to simplify, but ended up nowhere.
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to   [#permalink] 17 Jun 2013, 02:18
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