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In the xy-plane, region R consists of all the points (x, y)

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In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 31 Mar 2005, 06:25
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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

Can soebody explain the question plz?
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 [#permalink] New post 31 Mar 2005, 06:47
I think they consider the line 2x + 3y = 6 as a region and they ask you if the point (r,s) is on this line ?

I would answer B because with statement 2 we can obviously answer as we have the details of point r,s

In statement 1 we have the equation of the line where r,s is located and if you draw it you can see that it crosses the line 2x + 3y = 6 once !
so we can not be sure at 100% that r,s is not the intersection point between the 2 lines...so it's not sufficient. Maybe it's the intersection, or maybe not.
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 [#permalink] New post 31 Mar 2005, 07:14
"B".....this is defining a line and calling it as a region. For r,s to be on that region, it shud satisfy this eqn i.e. 2r+3s = 6.....state 1 gives you 3r+2s = 6....notsuff

state 2 ...for r = 3 and s=2 doesn't satisfy the eqn...so ans is NO...suff

Last edited by banerjeea_98 on 31 Mar 2005, 07:17, edited 1 time in total.
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 [#permalink] New post 31 Mar 2005, 07:23
surya, I have edited my post...made a mistake in reading the line eqn....state 1 doesn't tell u whether 2r+3s = 6 , so insuff......if we were to think that r,s do fall in that region, then we will have 2r+3s = 6.....in other words we have 2 eqn......2r+3s = 6 and 3r+2s = 6 (from state 1)

2r+3s = 3r+2s

r = s....but we don't know if r = s...so we can't ans...this state 1 is insuff...
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 [#permalink] New post 31 Mar 2005, 07:42
well the OA I have is E but I am not sure if that is correct.
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 [#permalink] New post 31 Mar 2005, 08:10
For interested people, more discussions can be found here:

http://www.gmatclub.com/phpbb/viewtopic ... 7933#87933

(I noticed that I asked for the source in there. I wonder if I've seen the same question in my test ...)
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 [#permalink] New post 31 Mar 2005, 22:29
how do you define a region with a line without stating which other axes form the boundaries for the region? Is there even such a thing ?
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Re: DS Regin on line? [#permalink] New post 31 Mar 2005, 23:32
saurya_s 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

agree with E.

from the question, 2x + 3y = 6........................................................i
y= -(2/3)x+2.
so slope =-2/3 and intercept = 3

from i, 3r+2s=6
let us suppose, x=r and y=s
3x+2y=6 .................................................................................ii

solving eq (i) and (ii), x=y=6/5. this is the only value with which it can be said that point (r,s) is in region R. but all other points with different values for r and s are not in region R.

from ii, r and s with values 3 and 2 could have any equations,which may or maynot lie on region R. if any line equation with r and s values 3 and 2 crosses the original equation 2x + 3y = 6, then it can be said that region R consists points r and s otherwise not.

So the answer is E. explain later if any.
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Re: DS Regin on line? [#permalink] New post 31 Mar 2005, 23:38
MA wrote:
saurya_s 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

agree with E.

from the question, 2x + 3y = 6........................................................i
y= -(2/3)x+2.
so slope =-2/3 and intercept = 3

from i, 3r+2s=6
let us suppose, x=r and y=s
3x+2y=6 .................................................................................ii

solving eq (i) and (ii), x=y=6/5. this is the only value with which it can be said that point (r,s) is in region R. but all other points with different values for r and s are not in region R.

from ii, r and s with values 3 and 2 could have any equations,which may or maynot lie on region R. if any line equation with r and s values 3 and 2 crosses the original equation 2x + 3y = 6, then it can be said that region R consists points r and s otherwise not.

So the answer is E. explain later if any.


I think we can safely assume r is x, and s is y. This is given by the question:
Is the point
(r, s) in region R ? . Normally the notation for a x-y pair is in this order (x,y).
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 [#permalink] New post 01 Apr 2005, 05:29
can somebody shed light on it. I am just not getting MA's point?
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 [#permalink] New post 01 Apr 2005, 07:06
MA,

we are not talking abt a line eqn passing thro (r,s), we r talking abt whether or not the point (r,s) lies in the region, and not any line that passes thro r,s lies in the region.

I think a line can define a region, however Infinitesimal it may be.
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 [#permalink] New post 01 Apr 2005, 08:36
A line is a set of points, just as a triangle, or a circle is.
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 [#permalink] New post 01 Apr 2005, 21:20
saurya_s wrote:
can somebody shed light on it. I am just not getting MA's point? S


Surya, probably you are not following explanation for statement ii. I think statement i is pretty straight. let me try it again.

from ii, with (r=3, s=2) we can form many equations that may or may not lie on region R. Why it is difficult to understand here is that we might messed up with r,s values and equations. lets not forget to understand r and s are x and y coordinates of the same xy-plane.

if we know only one of the following two position of the line equation for r,s (3,2) values, then we can answer the question:

1. any line equation that is parallel to the original equation. (note here: parallel lines have same slope)
2. any line equation that crosses/touches the original equation. (note here: for simplicity perpendicular lines have same slope but in opposite sign)

if we can form both line equations, we can not answer the question.
But with the above values for r and s, we can have both types of line equations as under:

1. the equation of a prependicular line:
3s-2r = 0
s = (2/3)r ............................................................i
slope = 2/3

2. the equation of a parallel line:
2r+3s = 12
s = (-2/3)r ..........................................................ii
slope = -2/3

[Note that we have the original line equation: 2x + 3y = 6 ..........iii,
y= -(2/3)x+2, slope = -2/3]

therefore. we cannot answer the question. therefore it is E.
pls do also correct me, if any.
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 [#permalink] New post 02 Apr 2005, 03:15
In my opinion it's still B
We have the equation of the line, 2x + 3y = 6
The region is defined by this line.

Then we have the coordinates of a point, given by statement 2, so we can figure out if the point is on the line or not...

Am I missing something ? :oops:
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 [#permalink] New post 02 Apr 2005, 07:35
I agree with u Antmavel....not sure why we keep talking abt a line passing thro (r,s), while the stem asks if the point (r,s) lies in the given line eqn, two different concepts.
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 [#permalink] New post 02 Apr 2005, 10:56
There is no mention that (r,s) and (x,y) are in the same plane i.e.,
co-planar.

So I dont think we can substitute r for x and s for y. That is a trap.
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 [#permalink] New post 04 Apr 2005, 08:08
Answer is definitely B. We have the region, defined by the line. We have the exact point. We can be certain that we know whether this point is in this line.
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 [#permalink] New post 04 Apr 2005, 08:31
saurya_s wrote:
gmat2me2 wrote:
There is no mention that (r,s) and (x,y) are in the same plane i.e.,
co-planar.

So I dont think we can substitute r for x and s for y. That is a trap.

Yes, you have a valid point.
S


GMAT DS instructions clearly states that all coordinates are are to be assumed in one plane, unless otherwise noted.
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 [#permalink] New post 05 Apr 2005, 08:27
banerjeea_98 wrote:
saurya_s wrote:
gmat2me2 wrote:
There is no mention that (r,s) and (x,y) are in the same plane i.e.,
co-planar.

So I dont think we can substitute r for x and s for y. That is a trap.

Yes, you have a valid point.
S


GMAT DS instructions clearly states that all coordinates are are to be assumed in one plane, unless otherwise noted.


Thanks for bringing that. I never noticed it yet.
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  [#permalink] 05 Apr 2005, 08:27
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