Given that $$p=996$$, which of the following best approximates the value of $$\frac{p}{3p - 1}$$?

Incorrect.
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__Plug In__ $$p = 1{,}000$$ so that $$\frac{p}{(3p - 1)}$$ becomes
>$$\displaystyle \frac{1{,}000}{(3{,}000 - 1)}$$
>$$\displaystyle = \frac{1{,}000 }{ 2{,}999}$$,
which is not equal to $$\frac{1}{9}$$.
Hence, eliminate this answer choice.

Incorrect.
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__Plug In__ $$p = 1{,}000$$ so that $$\frac{p}{3p - 1}$$ becomes
>$$\displaystyle \frac{1{,}000}{3{,}000 - 1} = \frac{1{,}000 }{ 2{,}999}$$,
which is not equal to $$\frac{1}{8}$$. Hence, eliminate this answer choice.

Incorrect.
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__Plug In__ $$p = 1{,}000$$ so that $$\frac{p}{3p - 1}$$ becomes
>$$\displaystyle \frac{1{,}000}{3{,}000 - 1} = \frac{1{,}000 }{ 2{,}999}$$,
which is not equal to $$\frac{1}{4}$$. Hence, eliminate this answer choice.

Incorrect.
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__Plug In__ $$p = 1{,}000$$ so that $$\frac{p}{3p - 1}$$ becomes
>$$\displaystyle \frac{1{,}000}{3{,}000 - 1} = \frac{1{,}000 }{ 2{,}999}$$,
which is not equal to $$\frac{1}{6}$$. Hence, eliminate this answer choice.

Correct.
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Incorrect.
[[Snippet]]
__Plug In__ $$p = 1{,}000$$ so that $$\frac{p}{3p - 1}$$ becomes
>$$\displaystyle \frac{1{,}000}{3{,}000 - 1} = \frac{1{,}000 }{ 2{,}999}$$,
which is not equal to $$\frac{1}{2}$$. Hence, eliminate this answer choice.

Did you use 1,000 instead of 996?

And why wouldn't you?

Oh, but there's a bunch of other stuff you can do if you are a masochistic person that seems much more enjoyable than trying to calculate $$\frac{996}{3(996)-1}$$.
You are not a human calculator, and the GMAT does not test your ability to become one. Use approximations to cut through the hurdle of a long calculation, then aggressively __POE__ answers that are out of the __Ballpark__.

The number 1,000 is very close to 996 and gives you a number that's a pleasure to work with.
>$$\displaystyle \frac{1{,}000}{3(1{,}000)-1}$$
Of course, for the sake of __Ballparking__ we can ignore the "1" in the denominator and get
>$$\displaystyle \frac{1{,}000}{3{,}000} = \frac{1}{3}$$.
By __Ballparking__, you get to a quick and painless estimation of the
correct answer. In our case, all other answers are out of the __Ballpark__.

$$\frac{1}{8}$$

$$\frac{1}{6}$$

$$\frac{1}{4}$$

$$\frac{1}{3}$$

$$\frac{1}{2}$$

Yes, I did.

No, I did not.

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