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edit: I should add, though, that because this is a DS question, we can stop long before we reach the answer.
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Re: OG C DS 131 Is 5^k less than 1000? a. 5^(k+1) >3000 b. [#permalink]

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23 Nov 2011, 10:55

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Simple Solution: 1) One variable. Inequality. Cannot determine the value of k -> Insufficiant 2) One Vasriable. One euality Equation. I can determine the value of k -> Sufficient

Hence B. I would not calculate anything. Time to solve - < 10 Secs.
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Re: OG C DS 131 Is 5^k less than 1000? a. 5^(k+1) >3000 b. [#permalink]

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23 Nov 2011, 11:12

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@iamgame I agree with your theory for statement 2 but I disagree for statement 1

Just because it is inequality and one variable you can't dismiss it. If after simplification you got the statement 1 as \(5^k > 1200\), it would have been sufficient.

We have to be careful in generalizing that rule. This is especially dangerous for high level inequality problems.

Simple Solution: 1) One variable. Inequality. Cannot determine the value of k -> Insufficiant 2) One Vasriable. One euality Equation. I can determine the value of k -> Sufficient

Hence B. I would not calculate anything. Time to solve - < 10 Secs.

If you're approaching DS questions in that way, you won't get very many of them right, unfortunately. For example, if you saw this question:

Is 5^k < 125?

1) k < 3 2) 5^k = 5*5^(k-1)

then Statement 1 is sufficient, since if k is less than 3, then 5^k is less than 5^3 = 125. Statement 2 gives an equation, but it is not sufficient, since it is always true - it gives you no information at all about the value of k. So in this example, the statement with the inequality *is* sufficient, and the statement with the equation is *not* sufficient. There are countless similar examples that you can find among official questions.
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(1) 5^(k+1) > 3,000 --> \(5^k>600\) --> if \(k=4\) then the answer is YES: since \(600<(5^4=625)<1,000\) but if \(k=10\), for example, then the answer is NO. Not sufficient.

(2) 5^(k-1) = 5^k - 500 --> we can solve for k and get the single numerical value of it, hence this statement is sufficient. Just to illustrate: \(5^k-5^{k-1}=500\) --> factor out \(5^{k-1}\): \(5^{k-1}(5-1)=500\) --> \(5^{k-1}=125\) --> \(k-1=3\) --> \(k=4\). Sufficient.

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