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

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Senior SC Moderator
Joined: 14 Nov 2016
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Assume that all positive integers are colored either red or blue. If a  [#permalink]

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22 Feb 2017, 22:58
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Difficulty:

55% (hard)

Question Stats:

64% (01:50) correct 36% (01:36) wrong based on 121 sessions

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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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Joined: 28 Dec 2011
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Assume that all positive integers are colored either red or blue. If a  [#permalink]

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23 Feb 2017, 12:15
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3
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
_________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
##### General Discussion
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Re: Assume that all positive integers are colored either red or blue. If a  [#permalink]

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31 Jul 2017, 10:30

"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
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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31 Jul 2017, 11:25
1
gps5441 wrote:

"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
_________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Re: Assume that all positive integers are colored either red or blue. If a   [#permalink] 31 Jul 2017, 11:25
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