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Is x + y > xy?
(1) x > 0 > y
(2) |y| = x
I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.
Thanks
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.
Thanks
Show more
1) x > 0 > y means x is +ve and y is -ve so xy will be -ve also from x > 0 > y we get x+(-y)>0 so x-y > xy ( +ve > -ve)
1 is sufficient
2) |y| = x
now x can be y or -y
if x=y then x+y is 2x or 2y and xy = x^2 or y^2 2x < x^2 or x+y < xy if x=-y then -y+y = 0 and xy is -ve x+y > xy
I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question.
Thanks
Show more
1)x>0>y => x+y>xy say x=1,y=-2 => x+y=-1 > -2 x+y<xy say x=0.5 y=-2 => x+y=-1.5 < -1 INSUFFI 2) |y|=x => x is +ve and and x=value of y where y is +ve or -ve say y=2 x+y =xy y=x=3 => x+y<xy => INSUFFI 1) and 2) now y<0 and x=|y| => y=-1 x=1 x+y>-1 y=-0.5 x=0.5 x+y>xy always since x+y=0 xy <0 IMO C
The two statements are essentially saying that x and y are equidistant from 0 on the number line and y is to the left and x is to the right.
Hence, x + y will always be zero and xy will be -ve.
However, if x and y are not equidistant then depending upon the absolute value of x or y (whether less than 1 or more than 1), the values of x+y and xy will be different (less or more).
Hence, C is the answer.
tarek99
Is x + y > xy?
(1) x > 0 > y
(2) |y| = x
I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question. Thanks
The two statements are essentially saying that x and y are equidistant from 0 on the number line and y is to the left and x is to the right.
Hence, x + y will always be zero and xy will be -ve.
However, if x and y are not equidistant then depending upon the absolute value of x or y (whether less than 1 or more than 1), the values of x+y and xy will be different (less or more).
Hence, C is the answer.
tarek99
Is x + y > xy?
(1) x > 0 > y
(2) |y| = x
I know that this can be simple when we experiment with numbers. However, I want to avoid experimenting because I can tell that there is a subtle message in the inequality Is x + y > xy. Would anyone please show how we could abstractly interpret it without doing much math? I know that it's possible, but I just couldn't figure it out in this question. Thanks
Show more
exactly! statement 2 tells us that x can ONLY be a zero and positive, while y can be either positive, a zero, or even negative. When you look at statement 1, when combining the 2 statements, it says that x and y aren't zero, while statement 2 says that both x and y have the same absolute value, which helps us to approach this inequality.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.