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

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02 Apr 2011, 07:27

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The initial set R is y<=2-2x/3

The set from (1) is line y=3-3x/2 As you could see the line has points in the set R and outside it

The set from (2) is y<=2, x<=3 As you can see the blue set has points lying into the red set R and outside of it

The intersection of both sets (1) and (2) together also does not give definiteness. As you could see from the picture, the green line goes both through blue set only and through red and blue sets.

SO, the answer is E.
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Re: In the xy-plane, region R consists of all the points (x, y) such [#permalink]

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02 Apr 2011, 13:52

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mirzohidjon 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

Can someone solve this problem graphically. I have trouble understanding the explanation given in OG. If it is explained graphically, it would be greatly appreciated. Thank you

Please see the attached image.

The turquoise line is the stem inequality 2x + 3y <= 6. Because the inequality says "<=", the region below the line will be valid region. Below means South-West side or the left side of the line.

(1) It is the equation of a line that is denoted by red line. Just look at the intersection of turquoise and red-line. It is at about (1.4,1.1). All red points above that intersection is outside the region defined by the turquoise line and all red points below that intersection is within the region. Thus, (r,s) may be within the region or outside the region. Not Sufficient.

(2) It is denoted by the blue lines. The point of intersection is (3,2). Thus, everything that is south-west from that point will be the region. South-west means below and left of the co-ordinate. We can clearly see that (0,0) is within the range that is defined by the turquoise line. And (3,2) is a point outside the region that is defined by the turquoise line. Thus (r,s) could be within the turquoise region or outside of it. Not Sufficient.

Combining both statements; We still have two regions: one within the turquoise, another outside. Outside the turquoise region: The red line between horizontal blue line and turquoise line. Within the turquoise region: The red line between vertical blue line and turquoise line. Not Sufficient.

Ans: "E"

Attachments

Inequality_Graphical.png [ 19.19 KiB | Viewed 3810 times ]

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

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02 Apr 2011, 22:02

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gmat1220 No. As you could see the set (1) is just a green line, not any area. The set (2) is blue infinite rectangle. Therefore the inersection of (1) and (2) can not be anything but the part of the line or separate points. The intersection of (1) and (2) is a part of green line, which goes through blue set, between the points of intersection of the green line and two blue lines: vertical and horizontal.
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If my post is useful for you not be ashamed to KUDO me! Let kudo each other!

Here is a video explanation to this problem: http://www.gmatquantum.com/list-of-vide ... ds121.html This is a really difficult problem, but I do recommend students to learn the pieces in this problem that are likely to be relevant to new GMAT questions.

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

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02 Aug 2011, 20:19

Solution for the changed problem:

Using statement (1), if 3r + 2s=6, then at (2,0) the point (r,s) lies within region R but at (0,3) it does not. Insufficient.

Using statement (2), if r<=3 and s<=2, then the point (0,2) lies within the region R but the point (3,2) does not. Insufficient.

Combining statements (1) and (2), the point (2,0) lies within region R and satisfies both the statements too. However, the point (1,3/2) does not lie within the region R even though it satisfies both the other statements. Insufficient.

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

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09 Nov 2011, 19:59

how are you supposed to do this question in 1-2 minutes? or can we give ourselves more time for difficult questions like these? if one anticipates scoring highly on the quant section does it mean every question is going to be hard? (and hence every question will be this difficult, and thus can't take more than avg for every question?)

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

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14 Nov 2011, 08:55

This question is rather difficult and I know that I would struggle to do it under 2 minutes. However, if you can get sketch the graph relatively quickly and see what they are asking by sketching the 2 conditions, it is doable.

OG's explanation is not as intuitive and I would have a difficult time using that method to solve this in 2 minutes.

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

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?

Q: is \(2r+3s\leq{6}\)?

(1) \(3r + 2s = 6\) --> very easy to see that this statement is not sufficient: If \(r=2\) and \(s=0\) then \(2r+3s=4<{6}\), so the answer is YES; If \(r=0\) and \(s=3\) then \(2r+3s=9>6\), so the answer is NO. Not sufficient.

(2) \(r\leq{3}\) and \(s\leq{2}\) --> also very easy to see that this statement is not sufficient: If \(r=0\) and \(s=0\) then \(2r+3s=0<{6}\), so the answer is YES; If \(r=3\) and \(s=2\) then \(2r+3s=12>6\), so the answer is NO. Not sufficient.

(1)+(2) We already have an example for YES answer in (1) which valid for combined statements: If \(r=2<3\) and \(s=0<2\) then \(2r+3s=4<{6}\), so the answer is YES; To get NO answer try max possible value of \(s\), which is \(s=2\), then from (1) \(r=\frac{2}{3}<3\) --> \(2r+3s=\frac{4}{3}+6>6\), so the answer is NO. Not sufficient.