"The only way this absolute value difference would be less than 1/2 is if it is zero"
AustinKL wrote:
Assume that all positive integers are colored either red or blue. If a set S consisting entirely of positive integers is chosen, where R is the number of red integers chosen and B is the number of blue integers chosen, the discrepancy of set S is \(|R-B|\). If set Q is such a set, is its discrepancy greater than \(\frac{1}{2}\) ?
(1) \(\frac{(R)}{(R+B)}\) < \(\frac{(B)}{(R+B)}\)
(2) \(R<B\)
Dear
AustinKL,
I'm happy to respond.
Like many
Veritas questions, this is a great question. It has the feel of a GMAT Quant question: it looks as if it would require extensive complicated calculations, but when we unpack what it is really asking, it's very simple.
Let's think about this. R & B are integers, because they are counts. The difference (R - B) could be any integer, positive or negative or zero. The absolutely value |R - B| could be any positive integer or zero. The only way this absolute value difference would be less than 1/2 is if it is zero. If it is zero, then it must be true that R = B, the number of reds equals the number of blues. That's what the prompt question is really asking. If we know that R = B, then we can give a definitive "yes" answer. If we know for a fact that R and B are not equal, then we can give a definitive no answer.
Statement #1: \(\frac{R}{(R + B)}\) < \(\frac{B}{(R + B)}\)
Multiply both side by (R + B), and we get R < B. They are not equal, so we can give a definitive "no" to the prompt question. Because we can give a definitive answer, this statement is
sufficient.
Forget statement #1 now.
Statement #2: \(R < B\)
This is a direct statement that R and B are not equal. Again, we can give a definitive "no" to the prompt question. Again, because we can give a definitive answer, this statement is
sufficient.
Each statement, on its own, is sufficient. OA =
(D)Does all this make sense?
Mike
close
64% (01:50) correct 
