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Assume that all positive integers are colored either red or blue. If a

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Assume that all positive integers are colored either red or blue. If a  [#permalink]

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New post 22 Feb 2017, 22:58
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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) \(R<B\)

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Assume that all positive integers are colored either red or blue. If a  [#permalink]

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New post 23 Feb 2017, 12:15
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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 absolute 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 :-)
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Re: Assume that all positive integers are colored either red or blue. If a  [#permalink]

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New post 31 Jul 2017, 10:30
Can you please explain this

"The only way this absolute value difference would be less than 1/2 is if it is zero"


mikemcgarry wrote:
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 :-)
Magoosh GMAT Instructor
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Joined: 28 Dec 2011
Posts: 4485
Re: Assume that all positive integers are colored either red or blue. If a  [#permalink]

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New post 31 Jul 2017, 11:25
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gps5441 wrote:
Can you please explain this

"The only way this absolute value difference would be less than 1/2 is if it is zero"

Dear gps5441,

I'm happy to respond. :-)

I'm going to ask you to think about the context of that statement:
R & B are integers, because they are counts. The difference (R - B) could be any integer, positive or negative or zero. The absolute 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.

The final statement is a consequence of all the previous statements. Which statements in this argument do you not understand?

Mike :-)
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Mike McGarry
Magoosh Test Prep


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Re: Assume that all positive integers are colored either red or blue. If a   [#permalink] 31 Jul 2017, 11:25
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