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

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Senior Manager
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Is the point A(x, y) in 2rd Quadrant? 1. (x+1 y+1) in 2nd [#permalink]

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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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SVP
Joined: 30 Oct 2003
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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.

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Director
Joined: 03 Jul 2003
Posts: 651

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30 Apr 2004, 14:56
anandnk wrote:

Agree with Anand, the answer shoud be C

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CEO
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08 May 2004, 18:47

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Senior Manager
Joined: 02 Mar 2004
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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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Manager
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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.

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Re: DS-75   [#permalink] 11 May 2004, 05:49
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