Is 2x-3y<x^2?
1. 2x-3y = -2
2. x>2 and y>0
Thanks!
equalities
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- dhanda.arun
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IMO answer is D
because:-
1. 2x - 3y = -2
since x^2 is always positive except when x is imaginary, so this is sufficient to answer.
2. x>2 and y>0
for all postive x >2 => x^2 > 2x
and for y>0
2x -3y < 2x < x^2
so this is also sufficient
Please share the original answer
because:-
1. 2x - 3y = -2
since x^2 is always positive except when x is imaginary, so this is sufficient to answer.
2. x>2 and y>0
for all postive x >2 => x^2 > 2x
and for y>0
2x -3y < 2x < x^2
so this is also sufficient
Please share the original answer
IMO D.beater wrote:Is 2x-3y<x^2?
1. 2x-3y = -2
2. x>2 and y>0
Thanks!
x^2 will always be a positive, hence greater than -2. SUFFICIENT
Statement 2: 2x-3y < x^2 is same as -3y < x(x-2)
if y>0, the left side will always be negative and if x>2, right side will always be positive. SUFFICIENT
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I understand the logic of the answers above me but how can we graphically answer this question.
I am drawing it (x^2 being a parabola with center O), and 2x-3y=2 being a line, and x^2 is below the line for certain values of x.
Please help!
I am drawing it (x^2 being a parabola with center O), and 2x-3y=2 being a line, and x^2 is below the line for certain values of x.
Please help!
- cubicle_bound_misfit
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beater wrote:Is 2x-3y<x^2?
1. 2x-3y = -2
2. x>2 and y>0
Thanks!
this is how I understand
by using stmt 1. we are asked is -2<x^2 the answer is yes as x62 is always positive. SUFF
for 2 there can be two possibilities
x<y x>2 y>0
for putting integer values always, x^2 is greater.
or if x>y then minimum case x=2, y=0 where 2x-3y = x^2
for any other case when x>y and x>2 , y>0
2x-3y<x^2.
SUFF
hence D
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Statement 1. 2x-3y = -2.beater wrote:Is 2x-3y<x^2?
1. 2x-3y = -2
2. x>2 and y>0
Thanks!
assign this value in the inequality 2x-3y<x^2 it will becomes
-2 < x^2. No matter what ever the value of x, the x^2 will be positive. And this satisfy the inequality. Sufficient.
x>2 and y>0
if x = 3, y=1.... 3< 9 True.
if x = 3 and y = 1/3, 5<9 True.
if x=5/2 and y = 1/2, 3.5 < 6 (nearly) True
Sufficient.
Regards,
Farooq Farooqui.
London. UK
It is your Attitude, not your Aptitude, that determines your Altitude.
Farooq Farooqui.
London. UK
It is your Attitude, not your Aptitude, that determines your Altitude.