Data Sufficiency

If -2x > 3y, is x negative?
1) y > 0
2) 2x + 5y - 20 = 0

How to analyse statement 2 for data sufficiency?

Thanks

A is sufficient

5y-20 = -2x

1) -2x= 4y
y = 20, x = -40; good
2) -2x = 6y
y = -20 x = +ve

Not sufficient..

IMO A

shebinjs wrote:If -2x > 3y, is x negative?
1) y > 0
2) 2x + 5y - 20 = 0

How to analyse statement 2 for data sufficiency?

Thanks

Hi,

I think the answer is D
The given statement is -2x>3y --> (2x+3y)<0

Now statement 1 is simple to evaluate..Lets consider the second statement
2x+5y=20,
(2x+3y)+2y=20
Now, as seen above (2x+3y)<0, thus, 2y needs to be positive to satisfy the above equation. Thus, again y>0 and we can ascertain the nature of x.
Thus, statement 2 in a nutshell is again y>0

Answer is D

pankajks2010 wrote:

shebinjs wrote:If -2x > 3y, is x negative?
1) y > 0
2) 2x + 5y - 20 = 0

How to analyse statement 2 for data sufficiency?

Thanks

Hi,

I think the answer is D
The given statement is -2x>3y --> (2x+3y)<0

Now statement 1 is simple to evaluate..Lets consider the second statement
2x+5y=20,
(2x+3y)+2y=20
Now, as seen above (2x+3y)<0, thus, 2y needs to be positive to satisfy the above equation. Thus, again y>0 and we can ascertain the nature of x.
Thus, statement 2 in a nutshell is again y>0

Answer is D

2x+5y=20, now here different results are possible for different values of x and y,

for example consider y=-2, x become 15, and for y=2 x becomes 5, therefore 2 alone is not sufficient to answer the question..!! answer should be A

Wow... I am with pankaj's solution

I always knew that if I make 1 quant mistake in gmat... 2000 desi guys will sit on my head.. (LOL only sentense)

manpsingh87 wrote:

pankajks2010 wrote:

shebinjs wrote:If -2x > 3y, is x negative?
1) y > 0
2) 2x + 5y - 20 = 0

How to analyse statement 2 for data sufficiency?

Thanks

Hi,

I think the answer is D
The given statement is -2x>3y --> (2x+3y)<0

Now statement 1 is simple to evaluate..Lets consider the second statement
2x+5y=20,
(2x+3y)+2y=20
Now, as seen above (2x+3y)<0, thus, 2y needs to be positive to satisfy the above equation. Thus, again y>0 and we can ascertain the nature of x.
Thus, statement 2 in a nutshell is again y>0

Answer is D

2x+5y=20, now here different results are possible for different values of x and y,

for example consider y=-2, x become 15, and for y=2 x becomes 5, therefore 2 alone is not sufficient to answer the question..!! answer should be A

Hey Manp..Lets just test your assumptions on the given statement..ie; -2x>3y
1) y=-2 & x=15, this doesn't satisfies the condition
2) y=2 & x=5, neither this...

pankajks2010 wrote:

manpsingh87 wrote:

pankajks2010 wrote:

shebinjs wrote:If -2x > 3y, is x negative?
1) y > 0
2) 2x + 5y - 20 = 0

How to analyse statement 2 for data sufficiency?

Thanks

Hi,

I think the answer is D
The given statement is -2x>3y --> (2x+3y)<0

Now statement 1 is simple to evaluate..Lets consider the second statement
2x+5y=20,
(2x+3y)+2y=20
Now, as seen above (2x+3y)<0, thus, 2y needs to be positive to satisfy the above equation. Thus, again y>0 and we can ascertain the nature of x.
Thus, statement 2 in a nutshell is again y>0

Answer is D

2x+5y=20, now here different results are possible for different values of x and y,

for example consider y=-2, x become 15, and for y=2 x becomes 5, therefore 2 alone is not sufficient to answer the question..!! answer should be A

Hey Manp..Lets just test your assumptions on the given statement..ie; -2x>3y
1) y=-2 & x=15, this doesn't satisfies the condition
2) y=2 & x=5, neither this...

oops my bad i didn't check the initial condition, but then again that's the beauty of DS question, i should have paid more attention to the initial condition...!!!

Many thanks guys .... Answer is D and I got it ...

Solved the best way. D it is..

shebinjs wrote:If -2x > 3y, is x negative?
1) y > 0
2) 2x + 5y - 20 = 0

How to analyse statement 2 for data sufficiency?

Thanks

IMO : A
1. it's clear that 3y > 0, in order that -2x > 0 => x < 0
=> SUFFICIENT
2. 2x + 5y = 20
if y > 4, x <0
if y < 4, x > 0
=> INSUFFICIENT

OH, I forgot the condition
Thanks all