In the xy-plane, region R consists of all the points (x.y) : Quant Question Archive [LOCKED]
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# In the xy-plane, region R consists of all the points (x.y)

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Manager
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In the xy-plane, region R consists of all the points (x.y) [#permalink]

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26 Jun 2007, 18:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

(1) 3r + 2s = 6

(2) r=3 and s=2

OA - E
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Re: Is the point (r,s) in region R? [#permalink]

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26 Jun 2007, 21:34
humtum0 wrote:
In the xy-plane, region R consists of all the points (x.y) such that
2x + 3y = 6. Is the point (r,s) in region R?

(1) 3r + 2s = 6

(2) r=3 and s=2

OA - E

i prefer not releasing OA with the question. there is no incentive to solve the problem.
Director
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26 Jun 2007, 21:37
2x + 3y = 6 is the eqn of a line - so how does 'region' apply here - does it mean all the points on the given line ?

if that is the case, st2 - can be used to verify if 3,2 lie on the line ..

so what is it I am missing here ?
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Re: Is the point (r,s) in region R? [#permalink]

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26 Jun 2007, 22:15
humtum0 wrote:
In the xy-plane, region R consists of all the points (x.y) such that
2x + 3y = 6. Is the point (r,s) in region R?

(1) 3r + 2s = 6

(2) r=3 and s=2

why not B?

2x + 3y = 6 ==> y = -2/3x + 2 (I'm assuming that this is region R)
given statement (1) r and s can be 0,3 or 2,0 (respectively) or some other combination of whole and/or fractional #s.

once we graph y = -2/3x + 2 won't that tell us whether r=3 and s=2?
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27 Jun 2007, 01:31
Question is confusing. Is R a region or a line?
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Is the point (r,s) in region R? [#permalink]

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07 Jul 2007, 05:51
I feel the second statement r=3 and s=2 gives us the equations of two separate lines viz. r=3 and s=2. Had it been a point they would have reffered as (3,2)

Lets solve this
we have very first equation 2x+3y=6 which defines the region R

then the first statement 3r+2s=6. This line and the line mentioned above has a point of intersection. only this equation does not tell us whether (r,s) is in region R or not

Second statement is r=3 and s=2. These are the equations of lines parallel to x and y axis resp. Each of this line has a different pts of intersection with the original line. Hence second statement is also not sufficient to tell us whether the point is in region R

Both statements together : Draw all these 4 lines on graph. The region defined by 3r+2s=6, r=3 and s=2 may or may not be in the original region R. Hence both statements together are not sufficient. Option E
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Re: Is the point (r,s) in region R? [#permalink]

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07 Jul 2007, 06:48
kekara wrote:
I feel the second statement r=3 and s=2 gives us the equations of two separate lines viz. r=3 and s=2. Had it been a point they would have reffered as (3,2)

Lets solve this
we have very first equation 2x+3y=6 which defines the region R

then the first statement 3r+2s=6. This line and the line mentioned above has a point of intersection. only this equation does not tell us whether (r,s) is in region R or not

Second statement is r=3 and s=2. These are the equations of lines parallel to x and y axis resp. Each of this line has a different pts of intersection with the original line. Hence second statement is also not sufficient to tell us whether the point is in region R

Both statements together : Draw all these 4 lines on graph. The region defined by 3r+2s=6, r=3 and s=2 may or may not be in the original region R. Hence both statements together are not sufficient. Option E

Yeah but when statement 2 says that r=3 and s=2, that means we're looking at the POINT (3,2), not lines x=3 and y=2...

weird problem...still dont see how it's not B
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07 Jul 2007, 06:54
In the xy-plane, region R consists of all the points (x.y) such that
2x + 3y = 6. Is the point (r,s) in region R?

(1) 3r + 2s = 6

(2) r=3 and s=2

line equation y = ax+b

3y = 6-2x
y = -2/3*x + 2

point (r,s) = (3,2)

2 = -2/3*3 + 2
2 = -2+2
2 = 0 ---> false

this point is not on the line ! so I can't see how (E) is the OA ?

07 Jul 2007, 06:54
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