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Harish Dorai
 
 

A certain jar contains only "b" black marbles

by Harish Dorai 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
 
 

Re: GMATPrep - Practice Test 2 - Problem #7

by givemeanid 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(r-w) > (w-r)(w+r)
r-w > (w-r)(w+r)/b ----> We know b is positive. So, we can divide both sides without changing the inequality
r-w > k(w-r) ----> 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
 
 

A different way

by Anadi Fri Aug 10, 2007 9:29 am

Suppose total is T , T=r+b+w

r/(b+w) > w/(b+r)

r/(b+w+r-r) > w/(b+w+r-w)

r/(t-r) > w/(t-w)

Since t>r and t>w, we can cross multiply.

rt-rw > wt-rw

rt > wt

Since t > 0

r > w

So 1 is sufficient.

2 is obviously not sufficient.
Harish Dorai
 
 

by Harish Dorai Fri Aug 10, 2007 10:47 am

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

by givemeanid Fri Aug 10, 2007 10:50 am

Anadi, I like your solution. Good thinking.
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by StaceyKoprince Sat Aug 11, 2007 7:30 pm

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

Re: A different way

by cracker 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
 
 

A different way

by k0nc3pt10n 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
 
 

by Guest 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
 
 

by MBA Action 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)
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Re:

by RonPurewal Wed Feb 18, 2009 4:12 am

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.
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Re: A certain jar contains only "b" black marbles

by pddtruong Thu Dec 17, 2009 8:22 am

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.
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Re: A different way

by ishkaran Mon Sep 27, 2010 3:58 pm

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
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Re: A certain jar contains only "b" black marbles

by RonPurewal Mon Oct 04, 2010 8:46 am

good stuff
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Re: A certain jar contains only "b" black marbles

by sudeepkapoor Sat Oct 16, 2010 12:35 am

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.