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# M19-30

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Math Expert
Joined: 02 Sep 2009
Posts: 42247

Kudos [?]: 132677 [0], given: 12331

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16 Sep 2014, 01:07
Expert's post
7
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Difficulty:

35% (medium)

Question Stats:

71% (01:10) correct 29% (01:56) wrong based on 111 sessions

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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

_________________

Kudos [?]: 132677 [0], given: 12331

Math Expert
Joined: 02 Sep 2009
Posts: 42247

Kudos [?]: 132677 [0], given: 12331

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16 Sep 2014, 01:07
Expert's post
3
This post was
BOOKMARKED
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, $$\frac{10^{20} + 2*10^{10} + 7}{3*10^{20} - 10*10^{10} + 200} \approx \frac{10^{20}}{3*10^{20}}=\frac{1}{3}$$.

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Intern
Joined: 17 Jul 2015
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Schools: Booth '18 (II)

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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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Intern
Joined: 22 Nov 2014
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22 Dec 2016, 02:32
can we also take x=10..i am getting same answer

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Math Expert
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Posts: 42247

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22 Dec 2016, 02:38
yaash wrote:
can we also take x=10..i am getting same answer

In this case yes because the options are widespread enough.
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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 $$\frac{1}{3}$$.

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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# M19-30

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