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

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There is a set of beads, each of which is painted either red [#permalink]

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New post 30 Aug 2013, 02:06
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Question Stats:

47% (02:14) correct 53% (01:44) wrong based on 216 sessions

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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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Joined: 31 Jan 2013
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Schools: ISB '15
WE: Consulting (Energy and Utilities)
Re: probability question from Princeton review [#permalink]

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New post 30 Aug 2013, 03:12
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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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Re: There is a set of beads, each of which is painted either red [#permalink]

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New post 02 Jun 2014, 08:53
I don't get it. Bunuel could you solve alternatively?

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

Thanks!
Cheers
J :)

Kudos [?]: 704 [0], given: 355

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Joined: 06 Sep 2013
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Kudos [?]: 704 [0], given: 355

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Re: There is a set of beads, each of which is painted either red [#permalink]

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New post 02 Jun 2014, 08:53
I don't get it. Bunuel could you solve alternatively?

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

Thanks!
Cheers
J :)

Kudos [?]: 704 [0], given: 355

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Re: There is a set of beads, each of which is painted either red [#permalink]

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New post 02 Mar 2016, 05:17
Hello from the GMAT Club BumpBot!

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Re: There is a set of beads, each of which is painted either red [#permalink]

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New post 15 Jun 2017, 03:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: There is a set of beads, each of which is painted either red [#permalink]

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New post 16 Jun 2017, 03:07
I got B right away as the only way the prob of only Blue beads > split Red ones is if there were more blue than red beads.
But can someone help do it algebraically as I'd like to see it efficiently laid out using algebra. Thank you.

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Re: There is a set of beads, each of which is painted either red   [#permalink] 16 Jun 2017, 03:07
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