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# Coordinate plane

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Senior Manager
Joined: 04 Aug 2008
Posts: 372

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20 Oct 2008, 12:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

ds
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VP
Joined: 17 Jun 2008
Posts: 1373

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20 Oct 2008, 13:10
spiridon wrote:
ds

1) does not suffice since 2x+3y<=6

3r+2s=6 does not give the points same as given by above line !! again r,s may lie or may not ,(-2,6) is the point which satisfies this and does not above equation ,(0,3) is the point which satisfies the above equation

2) r<=3,s<=2 does not satisfy 2x+3y<=6 for some solutions INSUFFI

combine both we get :

(2/3,2 )which satifies 1) but not in the given R and

(2,0 ) satisfie above and lies in R
INSUFFI

IMO E
whats the OA?
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SVP
Joined: 17 Jun 2008
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20 Oct 2008, 21:00
Agree B. 2x + 3y <= 6 essentially gives the region to the left of line 2x + 3y = 6.
r<=3 and s <= 2 gives the subset of above region in first quadrant.

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Manager
Joined: 21 May 2008
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21 Oct 2008, 01:05
I agree with spriya. Plot the graph of 2x + 3y <= 6 first.

(1) Insuff. Maybe in or out of R.
(2) Insuff. Maybe in or out of R.

(1) and (2): Still exists a region that is questionable.

My ans is E.

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VP
Joined: 30 Jun 2008
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21 Oct 2008, 04:37
Given equation is 2x+3y <=6
put y=0 we get x<=3
put x=0 we get y<=2

if the point (r,s) has to lie in this area 2x+3y <=6 then r should be <=3 and s should be <=2

(1) I don't know how to proceed with this info

(2) it directly says what we are out to prove --- r<=3 and s<=2. Sufficient.

Answer must be B or D
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Re: Coordinate plane   [#permalink] 21 Oct 2008, 04:37
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# Coordinate plane

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