Bunuel wrote:

Is \(r > s\) ?

(1) r represents the probability that it will rain on Tuesday.

(2) \(r > 2s\)

OFFICIAL SOLUTION

Is \(r > s\) ? Without looking at the additional information, it's impossible to say, and there's no work that can be done upfront.

On to statement (1)! Rain on Tuesday? Huh? Well if that isn't just the dumbest, randomest, most utterly useless piece of non-information ever...

Statement (1) alone is insufficient.

Statement (2) says that r>2s and at first blush it may seem obvious that if r is more than twice s, then surely r is greater than s itself.

But at this point we should begin to feel like giant suckers. If everything in both statements is that obvious, then what's the point of this problem? What could we be missing? How about negatives. If s<0, then 2s is actually less than s itself, and knowing that r>2s would not establish whether r>s. For instance, if s=−3, then 2s=−6, and if r=−4, then r would be greater than 2s but would not be greater than s.

So actually, it's not obvious that Statement (2) is sufficient; in fact, it's not even true! Statement (2) alone is insufficient.

At this point it might be tempting to go straight to E without even bothering to put the statements together, since statement (1) seemed so incredibly pointless. But that first statement does tell us one critical thing: r cannot be negative; the probability of any event is always a number between 0 and 1.

Now, if s is positive, then 2s>s, and r>2s therefore implies that r>s. If s is 0, then 2s=s, and again r>2s implies that r>s. Finally, if s is negative, then r>s simply based on the fact that r is nonnegative, and any nonnegative number is greater than every negative number. Thus, the two statements together do answer the question by establishing that r>s in every case. C is correct.

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