A certain jar contains only "b" black marbles

[mschwrtz]

Yeah, that'll work. I wouldn't have thought to do it that way. How long did it take you?

[vicksikand]

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.

I used a similar technique:
r/(b+w) +1 > w/(b+r) +1
(r+b+w)/(b+w) > (r+b+w)/(b+r)
Take reciprocal
b/(r+b+w) + w/(r+b+w) < b/(r+b+w) + r/(r+b+w)
or P(w) < P(r)

2 is insufficient

[jnelson0612]

Nice discussion! Thank you all.

[rafael.odorizzi]

I used this argument to accept st1, is it correct ?

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

r / w > w / r, therefore, r > w

Is it possible to assume that ?

[jnelson0612]

I used this argument to accept st1, is it correct ?

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

r / w > w / r, therefore, r > w

Is it possible to assume that ?

I would feel very nervous dismissing the effect of b. If b were being multiplied to w and r, and you knew that all the numbers were positive integers (which we do know in this problem), you could safely disregard the b in the denominator. However, this seems a bit fast and loose to me. I'd prefer that you do the algebra here (which isn't complicated) to make sure that you are completely safe.

[krishnan.anju1987]

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.

I agree. Even I got the same answer. But I did not understand givemeanid's solution to this problem. He mentions

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

after this point, this is how I understand it.

-w+r/b is positive as w,r,b are all positive. Agreed
-since this is a positive number, consider this to be a positive number k. Agree
- Now, as r-w >(w-r)*k
since RHS needs to be positive, LHS must be positive. hence r>w. but if so, w-r on the right hand side will be negative. so LHS and RHS do not match anyway.

Do I make any sense or am I misunderstanding the scenario here?

[krishnan.anju1987]

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.

Loved this one since this is exactly how I solved this problem :) :)

[RonPurewal]

I used this argument to accept st1, is it correct ?

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

r / w > w / r, therefore, r > w

Is it possible to assume that ?

I would feel very nervous dismissing the effect of b. If b were being multiplied to w and r, and you knew that all the numbers were positive integers (which we do know in this problem), you could safely disregard the b in the denominator. However, this seems a bit fast and loose to me. I'd prefer that you do the algebra here (which isn't complicated) to make sure that you are completely safe.

actually that works, but maybe not for the reason this poster originally thought.

namely, if you have
r/(b + w) > w/(b + r)
then you can put that into words as
red / everything else > white / everything else
if this is true, then, since both sides are talking about the same group of marbles, we must have red > white.

[sachin.w]

I solved it the following way .
Not sure if the approach is right but I do get teh ans.

r / (b + w) > w / (b + r)
since all of them have to be +ve as they are real marbles
=>

r(b+w) >w(b+r) .... (if a/b>0 => ab>0)
=>
simplifying this we get r>w..

Hope this approach correct.

but adding 1 would be a universal approach as this may not work if we dont know the signs

[RonPurewal]

I solved it the following way .
Not sure if the approach is right but I do get teh ans.

r / (b + w) > w / (b + r)
since all of them have to be +ve as they are real marbles
=>

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

whoa, no, you can't do that -- you can't just take the denominators and magick them into numerators.

try it with actual numbers and you'll see the problem. what you're saying is that you could take, e.g., 1/2 > 1/3 (a true statement) and transform it, as if by magic, into 1(2) > 1(3) (a false statement).