Stmt 2)

2x + 5y = 20

y = (20 - 2x)/5

We know that -2x > 3y

-2x > (60 - 6x) / 5

-10x > 60 - 6x

-4x > 60

x < -15 (remember, sign changes when you multiply or divide by a negative number)

If x < -15, it must be negative.

Thus, sufficient to answer the question

himanshi wrote:Hi,

I tried to search this question but couldn't find any link.

Q . If -2x > 3y ,is x negative ?

1) y > 0
2) 2x + 5y -20 =0

The AO is : D

But I marked A. I don't understand why 2) is sufficient alone... ?

Thanks

If -2x > 3y, then x<0 when y>0. Thus, we can rephrase the question as:

Is y>0?

Statement 1:
Tell us that y>0. Sufficient.

Statement 2:
2x+5y-20 = 0.
5y-20 = -2x.
Substituting 5y-20 for -2x in -2x>3y, we get:
5y-20>3y
2y>20
y>10. Sufficient.

The correct answer is D.

Oh ..That was Simple...
I just calculated halfway and left....

Thanks Guys

Hi,

I might be going a bit off topic, but what if Y = 0 in this case? How do you justify the answers?


I will be glad if you can clear this doubt.

Thanks

gdk800 wrote:Hi,

I might be going a bit off topic, but what if Y = 0 in this case? How do you justify the answers?


I will be glad if you can clear this doubt.

Thanks

If y=0, then -2x > 0

x < 0, so the question stem itself is enough to answer the question.

himanshi wrote: Q . If -2x > 3y ,is x negative ?

1) y > 0
2) 2x + 5y -20 =0

I think Important thing to learn over here is that equation with alligators { < , > } are as important as equal to { = }

I always had the perception that only equation with equal to { = } form a equation, but here in this problem we can see that -2x > 3y Helped us in finding a specific value,

Or i must we generally don't get the specific value , We can find the range in which the answer will lie,
example if x < -1 , { Now we know that x will always be -ve } This may not be very Helpful in problem solving questions But in DS questions, It is as important a information as x = 2 ,

I know, This is a general thing but i thought its good i to rewind because i was not able to figure out why answer is D.