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

Question

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

Zarrolou · Accepted Answer

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
C

HarveyS · Answer

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

Bunuel · Answer

Similar question to practice from GMAT PREP: a-certain-jar-contains-only-b-black-marbles-w-white-marbles-104924.html

b2bt · Answer

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

awanish1r · Answer

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.

NoHalfMeasures · Answer

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

mvictor · Answer

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.

LogicGuru1 · Answer

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 
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.

gotonishu · Answer

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

TheMBmonster · Answer

Start with stmt 2.

g>b

That allows for 3 options for the red marble

(1) r>g>b

(2) g>r> b

(3) g>b>r

stmt is Not suff.

Stmt 1

b(g+b) > r(g+r)

bg>r^2 > rg +r^2

we can conclude that b>r, but cannot infer anything about g 
 
Both stmts together 
 
From our analysis of statement 2, we can conclude that it's case 3. Suff.

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

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