Let a = 1 b = 2 for simplicity

(-a ,b ) = -1 ,2
(-b ,a ) = -2 ,1



From stmt 1 we have

a) x + ve y + ve ----1
b) x - ve y - ve


So (-x,y) can lie two quadrants. Hence Insufficient


From stmt 2 we have

a) a + ve x + ve -----2
b) a - ve x - ve

Again two quadrants and nothing known about y.

Combining eliminate cases b of both we can x + ve and y + ve .
Thus -x,y will lie in the second quadrant

The probablity question ..i got the exact same one in my gmat prep ..dont know how the answer is A ...

Heres an explanation for the probability problem

Consider St 1

If r/(w+b) > w/(r+b) is true, then r>w must also be true

Heres how I proved it...

Assume r > w

So, r+b > w+b OR 1/(w+b) > 1/(r+b)

Multiplying the two equations...
r.[1/(w+b)] > w.[1/(r+b)]

Hence 1 is SUFF.

St 2 leads does not give a relation between r and w hence INSUFF.

So answer is A

Must confess though, it took me over 5 mins to get this right!

thanks i had no clue whatsoever

sujaysolanki wrote:Let a = 1 b = 2 for simplicity

(-a ,b ) = -1 ,2
(-b ,a ) = -2 ,1



From stmt 1 we have

a) x + ve y + ve ----1
b) x - ve y - ve


So (-x,y) can lie two quadrants. Hence Insufficient


From stmt 2 we have

a) a + ve x + ve -----2
b) a - ve x - ve

Again two quadrants and nothing known about y.

Combining eliminate cases b of both we can x + ve and y + ve .
Thus -x,y will lie in the second quadrant

Can you please expalin why elminate case (b)?

Hi

I don't understand what you're multiplying here:
Multiplying the two equations...
r.[1/(w+b)] > w.[1/(r+b)]

Isn't this what was originally given in stmt 1?

xcise_science wrote:Hi

I don't understand what you're multiplying here:
Multiplying the two equations...
r.[1/(w+b)] > w.[1/(r+b)]

Isn't this what was originally given in stmt 1?

I was just trying to prove that if r/(w+b) > w/(r+b), then r > w by going backwards. i.e. assuming r > w and consequently deriving [r/(w+b) > w/(r+b)]

Having done all this, what I realised was...

If a, b, c, d, .... are ratios of individual elements in a mixture and if a > b then ratio of a to the rest (i.e. b+c+d+...) is greater than b to the rest (a+c+d+.....)

Please explain:

"Combining eliminate cases b of both we can x + ve and y + ve .
Thus -x,y will lie in the second quadrant"

I didn’t pick numbers, but here’s what I did…..I think its similar to the prior explanation.

stmt 1 says x(+) & y(+) OR x(-) & y(-) ns
stmt 2 says nothing about y and x(+) & a (+) OR x(-) & y(-) ns
both: x is (+) in both, therefore x(+) (and so is y, from stmt 1) and you know what quadrant the point falls in. so you can answer yes or no to the question.

Not that I get the point of ANY gmat question, but I thought this one was a very pointless question.
In the xy coordinate, if the x is (-), wouldn’t (x,y) be in the 2nd quadrant anyway if the y is (+)?
But I guess we had to determine if x was first + or -, before the negative sign was added to it.

For a = 1, b = 2

(-a,b) -> (-1,2) ......IInd quadrant

(-b,a) -> (-2,1).......IInd quadrant

So,(-a,b) and (-b,a) are in the same quadrant

For a = 1, b = -2

(-a,b) -> (-1,-2)........IVth quadrant

(-b,a) -> (2,1)..........Ist quadrant

(-a,b) and (-b,a) are not in the same quadrant.

For a= -1, b = 2

(-a,b) -> (1,2)............Ist quadrant

(-b,a) -> (-2,-1)..........IVth quadrant

Again,(-a,b) and (-b,a) are not in the same quadrant

For a = -1,b = -2

(-a,b) -> (1,-2)............IVth quadrant

(-b,a) -> (2,-1)............IVth quadrant

Again,(-a,b) and (-b,a) are in the same quadrant

We can conclude that only when the a and b have the same sign

=>(-a,b) and (-b,a) are in the same quadrant

None is sufficient alone.

Combining,its sufficient to indicate that its sufficient.

Hence (C)