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OG 12 - DS 121

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OG 12 - DS 121

by getmore » Mon May 04, 2009 7:05 pm
Hi everyone... I'm hoping for some help with a logical and time efficient method of solving this one.

In the xy-plane, region R consists of all the points (x,y) such that 2x + 3y <= 6. Is the point (r,s) in the region R?

(1) 3r + 2s = 6
(2) r <= 3 and s <= 2

OA: E

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DS -121

by Bhattu » Fri May 08, 2009 5:01 am
Thanks for pointing my error - I will re do and post a graphical solution
Last edited by Bhattu on Sun May 10, 2009 2:13 pm, edited 1 time in total.

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question

by chinks » Sat May 09, 2009 3:10 am
I do not understand why y=-3 when x=0.
we have the equation 2x+3y<=6, when x=0, y<=3 and when y=0, x<=3. Thus the region R is for all x<3, y<2. The second equation satisfies this condition so i think the answer is B.

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Re: DS -121

by Svedankae » Sun May 10, 2009 1:04 pm
Bhattu wrote: So for x=0, y=-2 and x=3, y=0.
thats incorrect. when x=0, y=2.

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by iriijei.idwimd » Thu May 14, 2009 9:30 am
i strongly feel the answer is B.

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by abcdefg » Wed Jul 15, 2009 9:46 pm
The answer is E .

Does anyone have a quick way to getting an answer for this? the suggested solution from the back of the book was very long.

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Here it is

by niramay » Wed Jul 15, 2009 11:39 pm
I don know how to draw the graph here.
But by the problem statement, graph wud luk like:-
X axis having 3 as the intercept and Y axis having 2 as intercept and the region covered wud be towards the origin. Lets call this graph1.

By option 1 : X intercept is 2, Y intercept is 3 and the region is towards the origin(say graph2), thus covering both the areas, one inside the graph1 and outside as well. So cannot be determined.

By option 2: Check for point (2.5,1.5) the region is outside graph1, now check for point (0,0) and the region comes out to be inside graph1.
So cannot be determined.

Combining option 1 and 2: Draw X =3 and Y=2, combine with graph2 and still the graph lies both outside and inside graph1.

So answer is E


Doing it graphically took me 2 minutes.