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XY plane and R point

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by user123321 » Tue Nov 22, 2011 2:40 pm
karthikpandian19 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 & s<=2
IMO E
1) two lines are clearly non parallel lines(first line being a region to be exact). hence we cannot comment whether (r,s) is in region R or not.
2) if r,s is -ve then it will be in the region R
if r = 3 and s = 2 then it will be not in the region R
hence insufficient.

using both you can have r=2,s=3 outside the region R
and r=2,s=0 inside the region R
hence both are insufficient

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by pemdas » Tue Nov 22, 2011 4:17 pm
interesting q.
slope of function 2x+3y<=6 may be derived from y<=(2/3)x+2
slope of function 3r+2s=6 in st(1) could be derived from s=(3/2)r+2. Since two lines are not parallel (slopes are different) there should be some cross point (at least one point) which may be (r,s). However this is only one point and other points may not be in the region required. Not Sufficient.
r<=3 & s<=2 in st(2) suggests if we supply our data into y<=(2/3)x+2 and equate x with r=3, then y equated with s=4. However we are given s<=2 and again our slopes will be different here. The point (r,s) may or may not be in the region required.
Combined st(1&2) two sets of properties of linear functions with varying slopes will be Not Sufficient

e
karthikpandian19 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 & s<=2
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by GMATGuruNY » Tue Nov 22, 2011 6:12 pm
karthikpandian19 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 & s<=2
Region R is composed of all the points on or below y=(-2/3)x + 2.

Statement 1: s = (-3/2)r + 3.
Image
The figure above shows that some points on s=(-3/2)r + 3 lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.

Statement 2: r≤3 and s≤2.
Image
Inside the green box are points such that r≤3 and s≤2.
Some of the points inside the green box lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.

Statements 1 and 2 combined:
Image
Inside the green box are points on s=(-3/2)r + 3 such that r≤3 and s≤2.
Some of these points lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.

The correct answer is E.
Last edited by GMATGuruNY on Fri Apr 05, 2013 1:49 am, edited 2 times in total.
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by karthikpandian19 » Tue Nov 22, 2011 10:43 pm
I can now understand this with pictorial rep.

Thank you very much
GMATGuruNY wrote:
karthikpandian19 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 & s<=2
Region R comprises all the points on or below y=(-2/3) + 2.

Statement 1: s = (-3/2)r + 3.
Image
The figure above shows that some points on s=(-3/2)r + 3 lie BELOW y=(-2/3)r + 2, while others lie ABOVE y=(-2/3)r + 2.
INSUFFICIENT.

Statement 2: r≤3 and s≤2.
Image
Inside the green box are points such that r≤3 and s≤2.
Some of the points inside the green box lie BELOW y=(-2/3)r + 2, while others lie ABOVE y=(-2/3)r + 2.
INSUFFICIENT.

Statements 1 and 2 combined:
Image
Inside the green box are points on s=(-3/2)r + 3.
Some of these points lie BELOW y=(-2/3)r + 2, while others lie ABOVE y=(-2/3)r + 2.
INSUFFICIENT.

The correct answer is E.
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by [email protected] » Fri Mar 16, 2012 6:29 am
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 & s<=2


The thing is could we solve the combined 1 and 2 statements without the diagrams as in the exam you cannot have the time of drawing the diagrams...
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by [email protected] » Mon Apr 30, 2012 2:09 am
I only have one question here, GMATguru... At the face of it when you solve this question and you come to C i.e Are both the statements sufficed or no??? then this is the point that I found...

There are two points that can be sufficed are (2,0) and (0,3).

These are the two points that is going with both the equations 2x + 3y = 6 and r<= 3 and s <= 2

And therefore (0,3) will not hold and (2,0) is the only point of (r,s).


Are there any more points that suffice both the equations.

Basically I searched for the points that go in both the equations and then does that point suffice for the main equation given in the main question...


Help needed guyzzzz....
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by Anurag@Gurome » Mon Apr 30, 2012 2:52 am
(1) 3r + 2s = 6 may or may not lie in region R. So, (1) is NOT SUFFICIENT to answer the question.

(2) If we take r = 3 and s = 2, then the point (3, 2) does not lie in region R.
r ≤ 3 and s ≤ 2 implies we can also take negative values for r and s. If r = -2, s = -3, then (-2, -3) lies in region R.
We don't get a unique answer, so (2) is NOT SUFFICIENT to answer the question.

Combining (1) and (2), if r = 2, s = 0 then (2, 0) lies in region R. But if r = 2/3 and s = 2 then (2/3, 2) lies above the line 2x + 3y = 6, which means (2/3, 2) does not lie in region R. Combining also doesn't give a unique answer.

The correct answer is E.

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by killer1387 » Mon Apr 30, 2012 3:34 am
[email protected] wrote:I only have one question here, GMATguru... At the face of it when you solve this question and you come to C i.e Are both the statements sufficed or no??? then this is the point that I found...

There are two points that can be sufficed are (2,0) and (0,3).

These are the two points that is going with both the equations 2x + 3y = 6 and r<= 3 and s <= 2

And therefore (0,3) will not hold and (2,0) is the only point of (r,s).


Are there any more points that suffice both the equations.

Basically I searched for the points that go in both the equations and then does that point suffice for the main equation given in the main question...


Help needed guyzzzz....
there are infinite number of points (as nowhere its mentioned the points are integral). Also graphically you can solve this question very easily and under less time.

Hope this helps..!!
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by ronnie1985 » Mon Apr 30, 2012 11:07 am
Use the graphical method to solve the problem

The equation of the straight in intercept form is x/a+y/b = 1 where a and b are the x and y intercepts respectively of the give straight line...

Given that x/3+y/2 <=1 Means all the point in the lower half of the line in xy plane satisfy the condition. Now check for the given conditions.

S1 r/2+s/3 = 1 crosses the line and hence some of its points are above the boundary defined in the question statement and some below. Not sufficient.

S2 r<=3 and s<=2. The point (3,2) is outside whereas (0,0) is inside the given boundary.

Not Sufficient

Comb.

Check in the coordinate plane for points outside the boundary. (2/3, 2) lies outside it.

Hence (E)
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