I would appreciate any help with the following:
Q1. Are x and y both positive?
(1) 2x- 2y = 1;
(2) (x/y) > 1.
Q2. In the xy-plane, does the line with the equation y = 3x + 2 contain the point (r,s)?
(1) (3r + 2 - s)(4r+9-s) =0;
(2) (4r-6-s)(3r + 2 - s) =0.
Q3. Is |x - y| > |x| - |y|?
(1) y < x;
(2) xy < 0.
Thanks,
3 Data Sufficiency Questions
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1. E
1. 2x - 2y = 1 -- x = 1, y = 1/2
x = -1/2, y = -1
in suff
2. x/y > 1, only tells us that x and y should be of same sign...
1. + 2. does not tell anything therefore, E
2. C, put (r,s ) in the equation of line 3x - y + 2 =0,
x = r and y = s,
1. 3r - s + 2 might be 0 -- insuff
2. again tells us that 3r - s + 2 might be 0 -- insuff
1. + 2. tells us that 3r- s + 2 = 0 hence suff
3. c
let me know if the answer is right..i will explain
1. 2x - 2y = 1 -- x = 1, y = 1/2
x = -1/2, y = -1
in suff
2. x/y > 1, only tells us that x and y should be of same sign...
1. + 2. does not tell anything therefore, E
2. C, put (r,s ) in the equation of line 3x - y + 2 =0,
x = r and y = s,
1. 3r - s + 2 might be 0 -- insuff
2. again tells us that 3r - s + 2 might be 0 -- insuff
1. + 2. tells us that 3r- s + 2 = 0 hence suff
3. c
let me know if the answer is right..i will explain
I actually got different answers. Like the post above though, I do not want to post all of my work only to find out I am wrong.
My differing answers are:
1)C
3)B
A brief overview of why I think C works for #1 is because in statement 1 you can move things around and get x=y+1/2
Plug that into statement two and you have (Y+1/2)/Y, which reduces to Y/2>0. This implies that Y is greater than 0.
Now if as you said in statement 1, both x and y need to be either both positive OR both negative. You know they are both positive.
Let us know the OA, thanks!
My differing answers are:
1)C
3)B
A brief overview of why I think C works for #1 is because in statement 1 you can move things around and get x=y+1/2
Plug that into statement two and you have (Y+1/2)/Y, which reduces to Y/2>0. This implies that Y is greater than 0.
Now if as you said in statement 1, both x and y need to be either both positive OR both negative. You know they are both positive.
Let us know the OA, thanks!
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|x-y| >= |x| - |y| (triangle inequality).zskhan wrote:I would appreciate any help with the following:
Q3. Is |x - y| > |x| - |y|?
(1) y < x;
(2) xy < 0.
|x-y| + |y| = |x| (extreme condition)
This can happen ---0===y===x--- or when ---x===y==0---
In other words, this is true when xy >= 0.
(2) is sufficient on its own. B
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2. y(y-x) < 0zskhan wrote:I would appreciate any help with the following:
Q1. Are x and y both positive?
(1) 2x- 2y = 1;
(2) (x/y) > 1.
Thanks,
1. x-y = 1/2
1. If y is positive, x is positive. Insuff
2. 0 < y < x or x < y < 0 (on its own)
Combining two gives us y is +ve. Sufficient.
C is the answer