## Inequalities

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### Inequalities

by akshatgupta87 » Tue Apr 12, 2011 12:40 pm
Q.) Is x<y
I) (1/x)<(1/y)
II) (x/y)<0
Someone pls explain.

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by clock60 » Tue Apr 12, 2011 1:03 pm
to me the answer is C here
is y>x
(1) (1/y)>(1/x)
(1/y)-(1/x)>0
(x-y)/(xy)>0
here two cases are possible
x>y and xy>0 or
x<y and xy<0
so if we can prove that xy<0 it cames that y>x
but 1 st is insuff
(2)x/y<0, from here we can`t estimate what is greater, so insuff, but we have one important point
x and y have opposite signs, is x is +ve then y is -ve and is x is -ve then y-+ve
both from 2 st we know that xy<0 and as xy<0 then y>x
so suff

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by Tani » Tue Apr 12, 2011 3:29 pm
Given: (1/y)>(1/x)

Multiply by x
a) if x is positive, you get x/y>1. Both must be positive and x must be greater than y
b) if x is negative you get x/y<1. that gives us two cases:
b1)If y is positive and x is negative, we have a negative number less than +1 and don't have any idea which is greater.
b2)If y is negative, we have a positive number and the absolute value of x must be less than the absolute value of y, but since both are negative that means the actual value of x is greater. Insufficient.

The second clue alone is insufficient since all it tells us is that x and y have the same sign.

Putting them together, if x and y are both positive (case a), we have shown x is greater than y.
Also, if x and y are both negative (case b2) we have also shown x is greater than y.

If the algebra isn't working for you, try picking numbers.
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by Ian Stewart » Tue Apr 12, 2011 5:35 pm
akshatgupta87 wrote:Q.) Is x<y
I) (1/x)<(1/y)
II) (x/y)<0
Someone pls explain.
Statement 1 will be true whenever x is negative and y is positive, in which case x < y. But Statement 1 is also true if x > y > 0 (take x = 3, y = 2, for example). So not sufficient.

Statement 2 tells us that one of x or y is negative, the other positive, but we don't know which. Not sufficient.

Together, we know from Statement 2 that one of x or y is negative, the other positive. Statement 1 can then only be true if x is negative and y is positive, in which case certainly x < y (negative numbers are always smaller than positive numbers), so together the two Statements are sufficient. C.
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