From the stimuli, -2x>3y gives x/y>-3/2
stem 1. y>0, if y is greater than 0, for x/y>-1.5 x may be +ve or -ve. NS
stem 2. 2x+5y-20=0
5y=20-2x
y=2/5*(10-x)
=0.4(10-x), x may have any value. NS
1+2, for y>0, x may be +ve or -Ve. hence NS
Is the ans E?
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Source: Beat The GMAT — Data Sufficiency |
I will go with D
Stmt 1 - y is +ve means 3y is +ve.. so, if -2x>3y that means x is -ve... SUFF
Stmt 2 - 2x+5y = 20..... (1)
q we have -2x>3y => 2x+3y < 0 .... (2)
Eq 1 can be written as 2x+3y+2y = 20.. from eq 2 we know 2x+3y is -ve. So, to satisfy eq 1 2y must be greater than 20. means y is +ve.. then same explanation as stmt 1... SUFF
OA??
Stmt 1 - y is +ve means 3y is +ve.. so, if -2x>3y that means x is -ve... SUFF
Stmt 2 - 2x+5y = 20..... (1)
q we have -2x>3y => 2x+3y < 0 .... (2)
Eq 1 can be written as 2x+3y+2y = 20.. from eq 2 we know 2x+3y is -ve. So, to satisfy eq 1 2y must be greater than 20. means y is +ve.. then same explanation as stmt 1... SUFF
OA??
The answer should be D.
-2x > 3y ----------------- (1)
=> 2x <3y> x <3> -2x = 5y – 20
Substituting in eq(1),
5y – 20 > 3y
2y > 20
y > 10
So from eqn (2), if y >10, then x must be negative.
Hence the answer is [D]
-2x > 3y ----------------- (1)
=> 2x <3y> x <3> -2x = 5y – 20
Substituting in eq(1),
5y – 20 > 3y
2y > 20
y > 10
So from eqn (2), if y >10, then x must be negative.
Hence the answer is [D]
The equation is simplified as
x < - ((3/2)y)
Now, we have -
Condition 1 -
gives y > 0, then x is always negative.
Condition 2 -
gives y > 10, than x is always negative.
Hence D
x < - ((3/2)y)
Now, we have -
Condition 1 -
gives y > 0, then x is always negative.
Condition 2 -
gives y > 10, than x is always negative.
Hence D
















