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Is the point A(x, y) in 2rd Quadrant? 1. (x+1 y+1) in 2nd

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Is the point A(x, y) in 2rd Quadrant? 1. (x+1 y+1) in 2nd [#permalink] New post 30 Apr 2004, 09:17
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Is the point A(x, y) in 2rd Quadrant?

1. (x+1 y+1) in 2nd quadrant
2. (x-1, y-1) in 2nd quadrant
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 [#permalink] New post 30 Apr 2004, 12:35
A) (x+1) is -ve so x < -1 (y+1) is +ve so y could be -0.5 so we dont know from this as well
B) (x-1) is -ve and (y-1) is +ve to be in the 2nd Quadrant
mening y > 1 but x could be <0 or < 1 so we dont know

Combine both

(x+1) is -ve and (x-1) is -ve so x < -1
(y+1) is +ve and (y-1) is +ve so y > 1 and is +ve

so X and Y are in 2nd quadrant.
Answer should be C
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 [#permalink] New post 30 Apr 2004, 14:56
anandnk wrote:
Answer should be C


Agree with Anand, the answer shoud be C
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 [#permalink] New post 08 May 2004, 18:47
please post answers for your questions. thanks
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Re: DS-75 [#permalink] New post 09 May 2004, 03:11
hallelujah1234 wrote:
Is the point A(x, y) in 2rd Quadrant?

1. (x+1 y+1) in 2nd quadrant
2. (x-1, y-1) in 2nd quadrant


Is x < 0, y > 0?

1. x < 0, y > -1 Insufficient
2. x < 1, y > 0 insufficient

1 and 2: x < 0 and y > 0--sufficient

C it is
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Re: DS-75 [#permalink] New post 11 May 2004, 05:49
hallelujah1234 wrote:
Is the point A(x, y) in 2rd Quadrant?

1. (x+1 y+1) in 2nd quadrant
2. (x-1, y-1) in 2nd quadrant


If A and B are in the same quadrant, then a*A + (1-a)*B are in that quadrant(if and only if 1 > a > 0). Let a be 1/2 => 1/2(x+1,y+1)+1/2(x-1,y-1) = (x,y). => 1 and 2 together are sufficient.

1 alone is not (-1/2,1/2) is the example.

2 alone is also not sufficient, (-1/2,1/2) is the example again.
Re: DS-75   [#permalink] 11 May 2004, 05:49
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