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A certain box contains only blue (b), green (g) and red(r)

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A certain box contains only blue (b), green (g) and red(r) [#permalink]

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A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1)\(\frac{b}{(g+r)}>\frac{r}{(g+b)}\)

(2) g > b
[Reveal] Spoiler: OA

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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 26 May 2013, 11:45
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A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1) b/(g+r) > r/(g+b)
Case b=10 g=1 r=1
\(\frac{10}{1+1}>\frac{1}{10+1}\) in this case the answer is blue
Case b=10 g=20 r=1
\(\frac{10}{20+1}>\frac{1}{10+20}\) in this case the answer is green
Not sufficient

(2) g > b
No info about r.
Not sufficient

(1+2)
\(b(g+b)>r(g+r)\)
\(bg+b^2>rg+r^2\)
\(b^2-r^2>rg-bg\)
\((b+r)(b-r)>g(r-b)\)
CASE if \(r>b\) we have \(r-b>0\) and \(b-r<0\)
\((+)(-)>(+)(+)\)
\((-)>(+)\) Not possible
CASE if \(b>r\) we have \(b-r>0\) and \(r-b<0\)
\((+)(+)>(+)(-)\) Possible
So we find out that \(b>r\) so \(g>b>r\)
The answer is G
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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 26 May 2013, 13:47
atalpanditgmat wrote:
A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1) b/(g+r) > r/(g+b)

(2) g > b

OA
[Reveal] Spoiler:
after discussion


Similar question to practice from GMAT PREP: a-certain-jar-contains-only-b-black-marbles-w-white-marbles-104924.html
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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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atalpanditgmat wrote:
A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1) b/(g+r) > r/(g+b)

(2) g > b



Using one of the Bunuel Method-

b/g+r +1 > r/g+b+1

= b+g+r/g+r > b+g+r/g+b

= g+b>g+r

=b>r ( not Suff as no clue of g)

2. Not Suff, as no clue of r

1+2 , Suff
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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 17 Mar 2015, 03:57
Zarrolou wrote:
A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1) b/(g+r) > r/(g+b)
Case b=10 g=1 r=1
\(\frac{10}{1+1}>\frac{1}{10+1}\) in this case the answer is blue
Case b=10 g=20 r=1
\(\frac{10}{20+1}>\frac{1}{10+20}\) in this case the answer is green
Not sufficient

(2) g > b
No info about r.
Not sufficient

(1+2)
\(b(g+b)>r(g+r)\)
\(bg+b^2>rg+r^2\)
\(b^2-r^2>rg-bg\)
\((b+r)(b-r)>g(r-b)\) --------------- (A)
CASE if \(r>b\) we have \(r-b>0\) and \(b-r<0\)
\((+)(-)>(+)(+)\)
\((-)>(+)\) Not possible
CASE if \(b>r\) we have \(b-r>0\) and \(r-b<0\)
\((+)(+)>(+)(-)\) Possible
So we find out that \(b>r\) so \(g>b>r\)
The answer is G
C


I reached till (A) but then had no clue and selected (E) as answer. How did could you reason beyond that?

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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 10 May 2015, 09:52
1. b/g+r > r/g+b, add 1 both sides we get

g+b+r/g+r > g+b+r/g+b ----> b>r ..not suff
2. g> b not suff
1 & 2 Suff.

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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 09 Dec 2015, 12:32
Let b, g and r be fraction of blue, green and red marbles in the box. so b+g+r=1

st1) b/(g+r) > r/(g+b)--> b/(1-b) > r/(1-r) --> (1-b)/b < (1-r)/r --> b>r. INSUF
St2) g > b INSUF

1+2--> g>b>r we know g is most likely to be picked. SUFF Ans:C
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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 17 Mar 2016, 18:37
atalpanditgmat wrote:
A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1)\(\frac{b}{(g+r)}>\frac{r}{(g+b)}\)

(2) g > b


my approach:
cross multiply:
bg+b^2 > rg + r^2 - subtract r^2 from both sides then subtract bg from both sides
b^2 - r^2 > rg - bg -> factor both sides
(b+r)(b-r) > g(r-b)

note that we work with marbles - these cannot be negative.
we can conclude that r-b is negative, and b>r.
otherwise if b-r is negative, then multiplying it by a positive number would get a negative number. and a positive number (g) when multiplied by a positive number (r-b) will yield a positive number, but in this case the equation given is not true.

from statement 1, we only know that b>r. alone is insufficient.

statement 2 alone is insufficient.

1+2
we are told that b>r, and g>b. so g>b>r. green marbles are the most likely to be picked.

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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 05 Aug 2016, 04:42
atalpanditgmat wrote:
A certain box contains only blue (b), green (g) and red(r) marbles. If one marble is to be picked out from the box at random, which color marble is most likely to be picked?

(1)\(\frac{b}{(g+r)}>\frac{r}{(g+b)}\)

(2) g > b


From stimulus we know that there are only Blue, Green and Red marbles.
So the probability of being picked will be directly proportional to the the number.
If Green has the highest number then green will have the highest probability
If Red has the highest number then red will have the highest probability
If Blue has the highest number then blue will have the highest probability

(1)\(\frac{b}{(g+r)}>\frac{r}{(g+b)}\)
This tells us that ratio of blue to other two colors is higher than ratio of red to other two colors
Meaning that Blue is more in number than Red.
But we don't know the number of green
INSUFFICIENT

(2) g > b[/quote]
This tells us that Green is more than blue
But we do not know anything about Red
INSUFFICIENT

COMBINE
GREEN > BLUE>RED

SUFFICIENT

ANSWER IS C
The probability of green is highest and red is lowest.
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Re: A certain box contains only blue (b), green (g) and red(r) [#permalink]

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New post 22 Oct 2017, 21:49
Another way to solve:

Let total t = b + g + r

Statement 1:
b/(g + r) > r/(g + b)
=> b/(t - b) > r/(t - r)
=> bt - br > rt - br
=> bt > rt
=> b > r

Not Suff since we don't know about g.

Statement 2:
g > b

Not Suff since we don't know about r.

Statement 1 and 2 together sufficient: g > b > r
=> g is most likely to be selected

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Re: A certain box contains only blue (b), green (g) and red(r)   [#permalink] 22 Oct 2017, 21:49
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