If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to : GMAT Problem Solving (PS)
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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]

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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]

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

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Hope it helps.
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink]

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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]

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10 Dec 2012, 08:21
sje12 wrote:
Thanks! Very quick and simple way to do that

May I ask, where did you take this question from?
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink]

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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]

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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]

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24 Feb 2015, 20:59
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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink]

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24 Feb 2015, 22:12
Hi All,

The phrasing in this question ("is closest to") is meant to hint that we can avoid an exact calculation.

From the answer choices, we can see that the denominator is going to be bigger than the numerator, so we have to think about how those two values really relate to one another....

Since X = 10^10, we know that we're dealing with a BIG number

The numerator gives us X^2 and the denominator gives us 3(X^2)

(10^10)^2 = 10^20 which is A LOT BIGGER than 10^10

Here they are, for context:

10^10 = 10,000,000,000
10^20 = 100,000,000,000,000,000,000
3(10^20) = 300,000,000,000,000,000,000

10^20 and 3(10^20) are so much bigger than the other "elements" in the numerator and denominator that those other elements are "negligible" (by comparison) to the overall calculation.

This means that we're basically dealing with (X^2 + a little)/(3X^2 - a little). That fraction is approximately 1/3

[Reveal] Spoiler:
B

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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to   [#permalink] 24 Feb 2015, 22:12
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# If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to

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