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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 \(s=< 2\)

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
_________________

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 \(s=< 2\)

Though the solution provided by shrouded1 above is perfectly OK, it's doubtful that can be done in 2-3 minutes.

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.

Graphs are good for the soul ! When I was taught all this, we were always told to use graphs where possible (algebra is drab :p) ... Once you are used to it, questions like these should take < 30secs !!
_________________

Thanks Bunuel for the explanation! It's very clear. However, I have some questions about your criteria to choose the values:

Bunuel wrote:

(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;

How did you know that you had to use those values (\(r=0\) and \(s=0\))?

Bunuel wrote:

(1)+(2) We already have an example for YES answer in (1) which valid for combined statements: (...) 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.

The same. How did you know that you had to use the max possible value. How did you know that this value will give you a NO answer.

I think that's the key of the problem.

Thanks!

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

For (2) given that \(r\leq{3}\) and \(s\leq{2}\) and we should see whether \(2r+3s<{6}\) is true.

Now, as lower limits of \(r\) ans \(s\) are not given then it's really easy to get an YES answer if we choose small enough values for them, to simplify calculation let's try \(r=0\) and \(s=0\) --> \(2r+3s=0<{6}\), so the answer is YES;

For NO answer let's try max possible value of \(r\) ans \(s\): \(r=3\) and \(s=2\) then \(2r+3s=12>6\), so the answer is NO.

For (1)+(2) Yes answer is from (1). To get NO answer we should maximize \(2r+3s\) to see whetrher this expression can be more than \(6\): try max possible value of \(r\), which is \(r=3\), then from (1) \(s=-\frac{3}{2}<2\) --> \(2r+3s=6-\frac{9}{2}\leq{6}\), still YES answer, we need NO;

So now 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. Done.

Thanks Bunuel for the explanation! It's very clear. However, I have some questions about your criteria to choose the values:

Bunuel wrote:

(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;

How did you know that you had to use those values (\(r=0\) and \(s=0\))?

Bunuel wrote:

(1)+(2) We already have an example for YES answer in (1) which valid for combined statements: (...) 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.

The same. How did you know that you had to use the max possible value. How did you know that this value will give you a NO answer.

I think that's the key of the problem.

Thanks!
_________________

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

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04 Aug 2012, 05:13

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harshavmrg 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 \(s=< 2\)

I followed a very manual process of mapping the data and then finding out the answer E( which is said the OA)

I took 2.43 mins...Is there a easier way to solve such problems? Please help

Yes! Using a graphical/visual approach is much faster. Try to understand shrouded1's approach presented above. Just the link to the graph is not working/missing.

Once you have a good command of the basic necessary tools, it's a piece of cake. If you need more help, get in touch with me.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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07 Jul 2013, 22:50

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(1) Not sufficient. since we do not know the value of 2r+3s.

(2) If r and s are 0 and 0, it satisfies the inequality. however, if r=3 and s =2, it doesn't. So, insufficient.

We need to know the values of 2r+3s. We are given the value of 3r+2s. 2r+3s=3r+2s-r+s = 6 +(s-r). If s< r inequality is satisfied. Taking (1) +(2) together still doesn't give us the information of whether s <r.

In the xy-plane, region R consists of all the points (x,y) such that \(2x + 3y\leq{6}\). Is the point (r,5) in region R?

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

I took nearly 10 min to solve. Probably on the test i would have guessed and moved on. Please correct the highlighted error. IMO graphical approach will lead to a quicker solution.

This Questions isn't worth 10 Mins

and You are right about Graphical method to be superior in this question but please ensure that while plotting graphically the approximate correct locations are marked on the graph paper

Step 1: Plot the equation of line 2x+3y = 6 by graphing points (0,2) and (3,0) [The easiest points to be taken by substituting x and y zero and making two cases]

Step 2: Since the given relation is inequation with the sign < hence understand that the questions is asking whether point (r,s) lie below the line or not

Step 3: Draw the line given in Statement 1 by taking points (0,3) and (2,0) and conclude that this statement is NOT sufficient because some points on this line lie in the region R e.g.(2,0) and some are outside like point (0,3)

Step 4: Draw the two lines given in Statement 2 which are Vertical and Horizontal respectively and and conclude that this statement is NOT sufficient because some points on these two line lie in the region R and some are outside the region R

Step 5: Combine the two statements and see the graphs drawn together. You will realize that some of the enclosed region represented by graphs of Statement 1 and 2 together is outside the region R and most of the region is inside region R hence conclude INSUFFICIENT

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Bunuel Pardon me. Let me rephrase. Hypothetically if point (r,s) lie in the region R. The equation becomes 2r + 3s <=6. Hence at s=0 r<=3 and at r=0 s<=2. So my question is can i paraphrase ?

the question 2r+3s<=6? to whether r<=3 and s<=2?

Am I assuming too much

Bunuel wrote:

Not sure that I understood your question. Anyway: region R is not some closed figure it's the the REGION below the line y=2-2/3*x.

Certainly you CANNOT rephrase "is \(2r+3s\leq{6}\)?" to "is \(r\leq{3}\) and \(s\leq{2}\)?". There is an example in my solution showing that.
_________________

This is how I solved the question and im arriving at the wrong answer. Please could someone correct me.

Given:- equation of the line is 2x+3y<6 Thus, y<2/3x+2 the y and x intercepts of the given line are y<2 and x <3 respectively.

Statement 2 gives us that r (x) < 2 and s (y) <3. Therefore i concluded statment 2 is suff to answer the question. Am i going wrong in the calculation?

Thanks.

First of all \(2x+3y\leq{6}\) can be rewritten as \(y\leq{2-\frac{2}{3}*x}\), not as \(y\leq{2+\frac{2}{3}*x}\). Next, it's not equation of a line, it gives the region which is below line \(y={2-\frac{2}{3}*x}\), so all your farther conclusions are wrong.
_________________

WE: General Management (Non-Profit and Government)

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

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25 Dec 2012, 03:47

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metallicafan wrote:

In the xy-plane, region R consists of all the points (x,y) such that \(2x + 3y\leq{6}\). Is the point (r,5) in region R?

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

I took nearly 10 min to solve. Probably on the test i would have guessed and moved on. Please correct the highlighted error. IMO graphical approach will lead to a quicker solution.

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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

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11 Sep 2016, 10:53

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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

Bunuel, Thanks for the solution ! Please correct me.

If (r,s) is inside the region R on the XY plane, as per equation 2r + 3s <=6 is true. That also means that r is bounded (when s=0) i.e. r<=3 s is also bounded (when r=0) is s<=2

Hence r<=3 and s<=2 should be sufficient. Am I missing something? Please clarify.

Bunuel, Thanks for the solution ! Please correct me.

If (r,s) is inside the region R on the XY plane, as per equation 2r + 3s <=6 is true. That also means that r is bounded (when s=0) i.e. r<=3 s is also bounded (when r=0) is s<=2

Hence r<=3 and s<=2 should be sufficient. Am I missing something? Please clarify.

Not sure that I understood your question. Anyway: region R is not some closed figure it's the the REGION below the line y=2-2/3*x.
_________________

Bunuel Pardon me. Let me rephrase. Hypothetically if point (r,s) lie in the region R. The equation becomes 2r + 3s <=6. Hence at s=0 r<=3 and at r=0 s<=2. So my question is can i paraphrase ?

the question 2r+3s<=6? to whether r<=3 and s<=2?

Am I assuming too much

Bunuel wrote:

Not sure that I understood your question. Anyway: region R is not some closed figure it's the the REGION below the line y=2-2/3*x.

Bunuel Pardon me. Let me rephrase. Hypothetically if point (r,s) lie in the region R. The equation becomes 2r + 3s <=6. Hence at s=0 r<=3 and at r=0 s<=2. So my question is can i paraphrase ?

the question 2r+3s<=6? to whether r<=3 and s<=2?

Am I assuming too much

Bunuel wrote:

Not sure that I understood your question. Anyway: region R is not some closed figure it's the the REGION below the line y=2-2/3*x.

Certainly you CAN NOT rephrase "is \(2r+3s\leq{6}\)?" to "is \(r\leq{3}\) and \(s\leq{2}\)?". There is an example in my solution showing that.

This is how I solved the question and im arriving at the wrong answer. Please could someone correct me.

Given:- equation of the line is 2x+3y<6 Thus, y<2/3x+2 the y and x intercepts of the given line are y<2 and x <3 respectively.

Statement 2 gives us that r (x) < 2 and s (y) <3. Therefore i concluded statment 2 is suff to answer the question. Am i going wrong in the calculation?

This is how I solved the question and im arriving at the wrong answer. Please could someone correct me.

Given:- equation of the line is 2x+3y<6 Thus, y<2/3x+2 the y and x intercepts of the given line are y<2 and x <3 respectively.

Statement 2 gives us that r (x) < 2 and s (y) <3. Therefore i concluded statment 2 is suff to answer the question. Am i going wrong in the calculation?

Thanks.

First of all \(2x+3y\leq{6}\) can be rewritten as \(y\leq{2-\frac{2}{3}*x}\), not as \(y\leq{2+\frac{2}{3}*x}\). Next, it's not equation of a line, it gives the region which is below line \(y={2-\frac{2}{3}*x}\), so all your farther conclusions are wrong.

Thanks . \(y\leq{2-\frac{2}{3}*x}\)- this was a typo.

But, Next, it's not equation of a line, it gives the region which is below line \(y={2-\frac{2}{3}*x}\), so all your farther conclusions are wrong., is where I was going wrong.