GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 12:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 26 Aug 2013
Posts: 11
There is a set of beads, each of which is painted either red  [#permalink]

### Show Tags

30 Aug 2013, 02:06
2
11
00:00

Difficulty:

85% (hard)

Question Stats:

53% (02:23) correct 47% (02:26) wrong based on 259 sessions

### HideShow timer Statistics

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.

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?
Intern
Joined: 31 Jan 2013
Posts: 17
Schools: ISB '15
WE: Consulting (Energy and Utilities)
Re: probability question from Princeton review  [#permalink]

### Show Tags

30 Aug 2013, 03:12
1
1
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
Intern
Joined: 07 Dec 2016
Posts: 36
Re: There is a set of beads, each of which is painted either red  [#permalink]

### Show Tags

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.
Senior Manager
Joined: 02 Apr 2014
Posts: 468
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
There is a set of beads, each of which is painted either red  [#permalink]

### Show Tags

21 Nov 2017, 12:23
Let R number of full read beads before merge
Let B number of full blue beads before merge

Let r number of completely red beads after merge
Let b number of completely blue beads after merge
Let p number of half-read && half-blue beads after merge

R = (2r + p)/2
B = (2b + p)/2

Question is, before merge, R > B?
=> $$(2r + p)/2 > (2b + p)/2$$ ?
=> r > b ?

Statement (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.

probability of picking a bead that is only red = $$r / (r + b + p)$$
probability of picking a bead that is at least half blue = probalility of picking full blue OR probablity of half-blue = $$(b/(r+b+p)) + (p/(r+b+p))$$

=> $$r / (r + b + p)$$ < $$(b/(r+b+p)) + (p/(r+b+p))$$
=> r < b + p => r can be less than or greater than b => Not suff

Statement (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.

probability of picking a bead that is only blue = $$b / (r + b + p)$$
probability of picking a bead that is at least half red = probalility of picking full red OR probablity of half-red = $$(r/(r+b+p)) + (p/(r+b+p))$$

=> $$b / (r + b + p)$$ > $$(r/(r+b+p)) + (p/(r+b+p))$$
=> b > r + p => b is definitely greater than r => Sufficient to say there were less Red beads than blue beads before merge

Answer (B)
Non-Human User
Joined: 09 Sep 2013
Posts: 13259
Re: There is a set of beads, each of which is painted either red  [#permalink]

### Show Tags

18 Dec 2018, 01:58
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There is a set of beads, each of which is painted either red   [#permalink] 18 Dec 2018, 01:58
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne