If y > 0, is y greater than x ?

[islands80]

If y > 0, is y greater than x ?

(1) 3x = 2y
(2) x + y = 5


OA A

[shantanu86]

IMO [spoiler][A][/spoiler]
Here is the explanation-

(1) 3x = 2y

since y is +ve, x must be +ve
y =x + 1/2*x
=> y = x + a positive number
=> y > x hence sufficient

(2) x + y = 5

x= 3, y=2 also satisfies this thus not sufficient

Hence [spoiler][A][/spoiler]

[aneesh.kg]

Let's solve this using Co-ordinate Geometry.
Since y > 0, we're concerned with only the Ist and the IInd quadrants.



Statement (1) is a straight line y = 3x/2



For the given line y is always greater than x. You can see this by comparing it with the line y = x.




SUFFICIENT.

Statement (2): Draw a line with x and y intercepts = 5


If we compare it with y = x again, we notice that a certain portion of this line is below y = x (made bold in figure) and a certain portion (made dotted in figure) is above it.

So, for the given statement y can be greater than x or smaller than it.
INSUFFICIENT.

[spoiler](A)[/spoiler] is the answer.

See another problem solved easily by Co-ordinate Geometry:
https://www.beatthegmat.com/if-b-1-and-2 ... tml#469327