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

(1) \(3r + 2s = 6\) (2) \(r\leq{3}\) and \(s\leq{2}\)

Target question: Is the point (r, s) in region R?

Given: Region R consists of all the points (x,y) such that 2x + 3y <6 So, what does Region R look like? To find out, let's first graph the EQUATION, 2x + 3y = 6

Since Region R is described as an INEQUALITY, we can choose any point on the coordinate plane to test whether or not it is in Region R. An easy point to test is (0,0). So, does x=0 and y=0 satisfy the inequality 2x + 3y <6? YES 2(0) + (3)(0) is less than or equal to 6. So, the point (0,0) is in Region R. More importantly, EVERY POINT on the same side of the line will also be in Region R.

Statement 1: 3r + 2s = 6 The target question refers to the point (r, s) In other words, the x-coordinate is r and the y-coordinate is s. So, all of the points (r, s) that satisfy the above equation can be found on the line 3x + 2y = 6 In other words, statement 1 tells us that the point (r,s) lies somewhere on the red line below. As you can see, some points are in Region R, and some points are not in Region R Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: r < 3 and s < 2 There are many points that satisfy this condition. In fact, the point (r,s) can be ANYWHERE inside the red box shown below. As you can see, some points are in Region R, and some points are not in Region R Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined When we combine the statements, we are saying that the point (r,s) is on the red line (2x + 3y = 6) AND inside the red box.

As you can see by the two blue points below, it's possible to have a point in Region R, and it's possible to have a point not in Region R

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

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

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11 Sep 2016, 09:53

1

This post received KUDOS

I tried the graph method and was right in plotting it out.

Could someone help me understand why "(2) : r<=3 & s<=2. Again easy to see from the graph even with that constraint, the point (r,s) may lie above or below the line in question"

From the graph, we see that any point with r<=3 and s<=2 will definitely lie under the blue line. Where am I going wrong here?

I understand that if we were to substitute r and s as 3 and 2 in the original equation,we would get the solution though. Just wanted to know what am I doing wrong here

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

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21 Nov 2016, 12:48

rajarams wrote:

I tried the graph method and was right in plotting it out.

Could someone help me understand why "(2) : r<=3 & s<=2. Again easy to see from the graph even with that constraint, the point (r,s) may lie above or below the line in question"

From the graph, we see that any point with r<=3 and s<=2 will definitely lie under the blue line. Where am I going wrong here?

I understand that if we were to substitute r and s as 3 and 2 in the original equation,we would get the solution though. Just wanted to know what am I doing wrong here

I'm having the same difficulty, in understanding why (2) is insufficient. I don't understand why any point where r<=3 and s<=2 will be above the line in question :/

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

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21 Nov 2016, 12:51

shrouded1 wrote:

metallicafan 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 \(s=< 2\)

Ok, the easiest way to solve this is to visualize the graph with the lines plotted on it. 2x+3y<=6, is the region below the line with X-intercept 3 and Y-intercept 2. We know it is that region, because (0,0) lies below the line and it satisfies the inequality. And all points on one side of the line satisfy the same sign of inequality. (BLUE LINE)

(1) : The line 3r+2s=6 (PURPLE LINE) represents the second line shown in the figure. It can be above or below the other line. So insufficient. (2) : r<=3 & s<=2. Again easy to see from the graph even with that constraint, the point (r,s) may lie above or below the line in question

(1+2) : the two conditions together, only take a section of the 3r+2s=6 line as a solution, but again even with r<=3, s<=2, its not sufficient to keep solutions below the 2x + 3y =< 6 line

Answer is (e)

In questions like these, once you are comfortable with graphs, you can solve in less than 30 seconds fairly easily. Let me know if the method isn't clear

Shrouded1, I'm unable to see why your below statement holds good: (2) : r<=3 & s<=2. Again easy to see from the graph even with that constraint, the point (r,s) may lie above or below the line in question

Can you please provide an example for a point to show that (2) is insufficient?
_________________

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

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29 Mar 2017, 17:52

Bunuel wrote:

So I'd say the best way for this question would be to try boundary values.

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.

Answer: E.

Very good answer, very clear. To anyone else having issue with this problem, it may be because you are not understanding the question itself (like me). The question gives you a formula and asks if (r,s) can fit inside the formula. Basically, the question is asking you if you can replace (x,y) with (r,s) without any problems.

The first statement gives a new formula so that you can create your own (r,s). You are free to create any (r,s) as long as it fits this statement 1 formula. Does every (r,s) from statement 1 fit inside the main question formula? Not always. If you set (r,s) to (0,3) you can fit inside the statement 1 formula, but it does not fit into the original question formula. However, if you set (r,s) to (2,0) then it will fit in both the statement 1 formula and the original question. Since it (r,s) created by statement 1 formula can either fit or not fit in the main question formula, then statement 1 is insufficient.

The second statement is similar to the first statement, except you don't have to fit it in the formula. You just have to check for all r<=3, and s<=3. You can actually use the same points from statement 1 which are (2,0) and (0,3) to prove that this statement 2 is insufficient.

What about both? Well since (2,0) and (0,3) fit in both statements, you've already checked both. It is possible for both statements together to either fit or not fit in the main question. Thus both statements even combined are not sufficient, so answer is E. Hard question if you do not work with formulas often!

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

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27 Aug 2017, 09:27

This is a tricky question

1) Given an equation of region 2x +3y ≤ 6 i.e region R 2) Asking if point (r,s) lies in region R?

rephrasing the given:

3y ≤ 6 - 2x y ≤ -2/3x + 2 ---> Slope = -2/3

y interecept = 2 x interecept = 3

Statement 1) This is an equation of a line. If we prove that this line passes within the region R than the point (r,s) on the line will fall under region R

Since the slope of statement 1 is greater than given for the region R, the line 3r+2s =6 will be steeper, therefore it will not completely fall under region R. Thus value of point (r,s) on line (3r+2s =6) can fall out side Region R -- (Not Sufficient)

Statement 2)

r ≤ 3 and s ≤ 2

In order for point (r,s) to be in the region of R, it should satisfy the equation (y ≤ -2/3x + 2) case a: r=2 and s= 3, 3 ≤ -4/3 + 2 -- {NO} case b: r=1 and s =0 , 0 ≤ -2/3 + 2 -- {YES}

Since we get a YES and NO for different values of r and s, therefore, not sufficient