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

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Intern
Joined: 29 Nov 2012
Posts: 4
If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink]

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Updated on: 10 Dec 2012, 09:12
1
11
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Difficulty:

45% (medium)

Question Stats:

71% (01:14) correct 29% (02:07) wrong based on 265 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. $$\frac{2}{3}$$

m19 q30

Originally posted by sje12 on 10 Dec 2012, 09:01.
Last edited by Bunuel on 10 Dec 2012, 09:12, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 46280
Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink]

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10 Dec 2012, 09:15
3
4
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, 09:23
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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24 Feb 2015, 23:12
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

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

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22 Jan 2018, 22:54
I solved it incorrect with timer but without i did it as below:

- We know that $$10^{10}$$ is a large number so, ignored 7 and 200 from numerator and denominator.
- Now the expression becomes $$\frac{X(X+2)}{X(3X-10)}$$. Cancel out X, left with $$\frac{X+2}{3X-10}$$.
- Apply same logic $$10^{10}$$ is huge, ignored 2 and -10 so eventually the expression becomes $$\frac{1}{3}$$ after cancelling X again. So, Option (B).
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If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to [#permalink]

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23 Jan 2018, 01:04
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

This problem can best be solved by taking $$x^2$$ as common in numerator and $$3x^2$$ in denominator. As $$x$$ is a large number.. anything divided by $$x$$ or $$x^2$$ will be almost negligible provided numerator is also smaller.
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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 Jan 2018, 10:35
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

Plugging 10^10 for x we have:

10^20 + 2(10^10) + 7 for the numerator

3(10^20) - 10(10^10) + 200 for the denominator

Since 10^20 and 3(10^20) are such large values compared to the other terms, we see that the approximate value of the expression is:

10^20/3(10^20) = 1/3

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Re: If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to   [#permalink] 24 Jan 2018, 10:35
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# If x = 10^10, (x^2 + 2x + 7)(3x^2 - 10x + 2) is closest to

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