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

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.

Similar questions to practice:
which-of-the-following-is-closest-to-10180-1030-a-110224.html
the-value-of-10-8-10-2-10-7-10-3-is-closest-to-which-of-95082.html
tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029229

Hope it helps.
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I'm not sure which book it is from, however I obtained it from a course instructor in Switzerland (Absolute GMAT, gmat-kurse.ch)

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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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).

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.

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

Answer: B
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