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There is a set of beads, each of which is painted either red

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Intern
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Joined: 26 Aug 2013
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There is a set of beads, each of which is painted either red [#permalink] New post 30 Aug 2013, 01:06
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A
B
C
D
E

Difficulty:

  55% (medium)

Question Stats:

45% (02:38) correct 54% (01:27) wrong based on 83 sessions
There is a set of beads, each of which is painted either red or blue. The beads are split, with each bead being cut in half. Then they are merged, where all the halves are randomly reassembled back into the same number of beads that there were to begin with. This results in r red beads, b blue beads, and p beads that are half red and half blue. Before the split and merge, were there more red beads than blue beads?

(1) after the split and merge, the probability of picking a bead that is only red is less than the probability of picking a bead that is at least half blue.
(2) After the split and merge, the probability of picking a bead that is only blue is greater than the probability of picking a bead that is at least half red.

[Reveal] Spoiler:
Answer is B.
Well, I got the answer but it took me so long to solve it.
Can someone help me to solve this kind of questions quicker?
[Reveal] Spoiler: OA
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Intern
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Joined: 31 Jan 2013
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WE: Consulting (Energy and Utilities)
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Re: probability question from Princeton review [#permalink] New post 30 Aug 2013, 02:12
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Consider the Number of Red and Blue beads as R and B respectively and, after splitting, the number of beads as R1,B1 and P(mixed)

We need to prove R>B
From the given information, We could say
R+B = R1+B1+P
Or
R = R1+P/2
B = B1+P/2

1) Given R1<B1+P -
R1<B+P/2 or R1+P/2<B+P
R<B+P - from this equation, there is no way we can interpret about R or B. Hence not Sufficient

2) Given B1>R1+P -
From this we could definitely say B>B1>R1+P
B> (R1+P/2)+P/2
So we can definitely say,
B>R
hence St2 is sufficent to answer the question.

I hope this helps
/SW
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Joined: 06 Sep 2013
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Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
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Re: There is a set of beads, each of which is painted either red [#permalink] New post 02 Jun 2014, 07:53
I don't get it. Bunuel could you solve alternatively?

Question is to express half/half red and blue balls algebraically

Thanks!
Cheers
J :)
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User avatar
Joined: 06 Sep 2013
Posts: 1629
Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
WE: Corporate Finance (Investment Banking)
Followers: 10

Kudos [?]: 127 [0], given: 254

GMAT ToolKit User
Re: There is a set of beads, each of which is painted either red [#permalink] New post 02 Jun 2014, 07:53
I don't get it. Bunuel could you solve alternatively?

Question is to express half/half red and blue balls algebraically

Thanks!
Cheers
J :)
Re: There is a set of beads, each of which is painted either red   [#permalink] 02 Jun 2014, 07:53
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There is a set of beads, each of which is painted either red

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