Author 
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Harish Dorai

Post subject: A certain jar contains only "b" black marbles Posted: Thu Aug 09, 2007 12:45 pm 


A certain jar contains only "b" black marbles, "w" white marbles and "r" red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen is red greater than the probability that the marble chosen will be white?
1) r/(b+w) > w/(b+r)
2) b  w > r





givemeanid

Post subject: Re: GMATPrep  Practice Test 2  Problem #7 Posted: Thu Aug 09, 2007 2:04 pm 


Harish Dorai wrote: A certain jar contains only "b" black marbles, "w" white marbles and "r" red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen is red greater than the probability that the marble chosen will be white?
1) r/(b+w) > w/(b+r)
2) b  w > r
The question asks whether r/(b+w+r) > w/(b+w+r) or in other words is r > w?
1. r(b+r) > w(b+w)
br + r^2 > bw + w^2
br  bw > w^2  r^2
b(rw) > (wr)(w+r)
rw > (wr)(w+r)/b > We know b is positive. So, we can divide both sides without changing the inequality
rw > k(wr) > Where k > 0 as b,r and w are all positive
This is true only when r > w.
If r < w, left side is ve and right side is +ve and the inequality doesn't hold.
SUFFICIENT.
2. b  w > r
b > w + r
This doesn't tell us anything about relationship between w and r.
INSUFFICIENT.
Answer is A.





Anadi

Post subject: A different way Posted: Fri Aug 10, 2007 9:29 am 


Suppose total is T , T=r+b+w
r/(b+w) > w/(b+r)
r/(b+w+rr) > w/(b+w+rw)
r/(tr) > w/(tw)
Since t>r and t>w, we can cross multiply.
rtrw > wtrw
rt > wt
Since t > 0
r > w
So 1 is sufficient.
2 is obviously not sufficient.





Harish Dorai

Post subject: Posted: Fri Aug 10, 2007 10:47 am 


You guys are brilliant! The explanation makes perfect sense and the answer is (A).





givemeanid

Post subject: Posted: Fri Aug 10, 2007 10:50 am 


Anadi, I like your solution. Good thinking.





StaceyKoprince

Post subject: Posted: Sat Aug 11, 2007 7:30 pm 


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Posts: 7754 Location: San Francisco

You guys are all doing a great job here  you don't even need me! :)
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cracker

Post subject: Re: A different way Posted: Thu Oct 23, 2008 2:13 pm 


adding +ve no r to lhs and w to rhs
proper frac behave properly
therefore
we have
2r/(b+r+w) > 2 w/(b +r+w)
divinding by +ve no 2
r/(b+r+w)[prob of red bll] > w/(b +r+w)[prob of white ball]
2 in insuffincent
hence ,A





k0nc3pt10n

Post subject: A different way Posted: Sat Oct 25, 2008 10:06 pm 


I like Anadi's method better than mine, but...
1) r/(b+w) > w/(b+r)
2)r/(b+w) + (b+w)/(b+w) > w/(b+r) + (b+r)/(b+r) > I added 1 to both sides inthe form of (b+w)/(b+w) & (b+r)/(b+r)
3)(r+b+w)/(b+w) > (r+b+w)/(b+r) > Next I cross multiply
4)(b+r)(r+b+w) > (r+b+w)(b+w) > divide out the (r+b+w)
5)(b+r) > (b+w) > It's pretty obvious now, but you can take away the b's if you want
6)r>w
Also I don't agree with crackers' math. Can anyone explain it better?





Guest

Post subject: Posted: Wed Oct 29, 2008 3:33 pm 


A certain jar contains only "b" black marbles, "w" white marbles and "r" red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen is red greater than the probability that the marble chosen will be white?
P(red)=r/(b+w+r)
P(white)=w/(b+w+r)
P(red) > P (White) if r > w
1) r/(b+w) > w/(b+r)
gives
r/(b+w+r) > w/(b+w+r) [ this follows from: if A/B > C/D then A/(A+B) > C/(C+D) ]
hence r > w
Sufficient
2) b  w > r
Not Sufficient
I believe this also helps to explain cracker's method





MBA Action

Post subject: Posted: Fri Jan 30, 2009 10:23 am 


Ok, I have another approach
Can we say "safely" that since the total is fixed then
if: r/(rest(r)) > w/(rest(w)) as given in statement 1 then definitely r>w?
(rest(r) means rest of the marbles without r)





RonPurewal

Post subject: Re: Posted: Wed Feb 18, 2009 4:12 am 


ManhattanGMAT Staff 

Posts: 13427

MBA Action wrote: Ok, I have another approach
Can we say "safely" that since the total is fixed then if: r/(rest(r)) > w/(rest(w)) as given in statement 1 then definitely r>w? (rest(r) means rest of the marbles without r) yes, you can. if you have a fixed total, then, as the quantity of something increases, the quantity of everything else will decrease accordingly. therefore, the higher the quantity of the "something", the higher the ratio of that "something" to "everything else" will be. nice observation.
_________________ Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions ã Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi
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pddtruong

Post subject: Re: A certain jar contains only "b" black marbles Posted: Thu Dec 17, 2009 8:22 am 


Forum Guests 

Posts: 2

1) a) r(b+r) > w(b+w) b) r(b+r) + rw > w (b+w) + rw [Add same number to both sides] c) r[(b+r)+w] > w[(b+w)+ r] [factor out] d) divide both sides: r/(b+w+r) > w/(b+w+r). Sufficient.
2) Ins.





ishkaran

Post subject: Re: A different way Posted: Mon Sep 27, 2010 3:58 pm 


Students 

Posts: 1

k0nc3pt10n wrote: I like Anadi's method better than mine, but...
1) r/(b+w) > w/(b+r) 2)r/(b+w) + (b+w)/(b+w) > w/(b+r) + (b+r)/(b+r) > I added 1 to both sides inthe form of (b+w)/(b+w) & (b+r)/(b+r) 3)(r+b+w)/(b+w) > (r+b+w)/(b+r) > Next I cross multiply 4)(b+r)(r+b+w) > (r+b+w)(b+w) > divide out the (r+b+w) 5)(b+r) > (b+w) > It's pretty obvious now, but you can take away the b's if you want 6)r>w
Also I don't agree with crackers' math. Can anyone explain it better? Found this approach the best. Thanks





RonPurewal

Post subject: Re: A certain jar contains only "b" black marbles Posted: Mon Oct 04, 2010 8:46 am 


ManhattanGMAT Staff 

Posts: 13427

good stuff
_________________ Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions ã Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi
Un bon vêtement, c'est un passeport pour le bonheur. – Yves SaintLaurent





sudeepkapoor

Post subject: Re: A certain jar contains only "b" black marbles Posted: Sat Oct 16, 2010 12:35 am 


Students 

Posts: 1

Taking statement (1)
r/(b+w) > w/(b+r) :
taking reciprocal ,
(b+w)/r < (b+r)/w
[take the example of 1/2 and 1/3 ; 1/2 > 1/3 but if one takes the reciprocal , 2<3 ]
now, add 1 to both sides, (b+w)/r +1 < (b+r)/w +1 [inequality holds good when a positive constant is added]
This implies , (b+w+r)/r < (b+r+w)/w
Again take the reciprocal and the sign changes
r/(b+w+r) > w/(b+r+w)
also we know that :
P(red)=r/(b+w+r) P(white)=w/(b+w+r)
therefore P(red) > P(white)
Therefore statement 1 is sufficient
Statement (2) does not give any relation between red and white marbles and is obviously not sufficient ;
Answer is A.





