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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 04 Oct 2010, 19:27
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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 \(s\leq{2}\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 25 Dec 2012, 03:01, edited 2 times in total.
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Re: In the xy-plane... [#permalink] New post 04 Oct 2010, 22:24
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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)

Image

(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
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Last edited by Bunuel on 05 Aug 2012, 06:29, edited 1 time in total.
Edited the diagram.
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Re: In the xy-plane... [#permalink] New post 05 Oct 2010, 02:37
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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 \(s=< 2\)


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.

Answer: E.

Hope it's clear.
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Re: In the xy-plane... [#permalink] New post 07 Oct 2010, 01:53
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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 !!
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 04 Aug 2012, 04:13
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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.
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 07 Jul 2013, 21:50
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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.

So answer is (E).
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 02 Jun 2015, 02:26
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gaurav0480 wrote:
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.


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

Answer: Option
[Reveal] Spoiler:
C

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Re: In the xy-plane... [#permalink] New post 06 Oct 2010, 23:15
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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... [#permalink] New post 07 Oct 2010, 01:34
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metallicafan wrote:
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.

Hope it's clear,
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Re: In the xy-plane... [#permalink] New post 25 Feb 2011, 11:59
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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.
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In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 18 Sep 2012, 05:27
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smartass666 wrote:
Bunuel wrote:
egiles 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


Merging similar topics.


Can you provide the solution for this question?


Graphic Approach: in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html#p794049
Number Plugging: in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html#p794138 and in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html#p795613

Hope it helps.
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 25 Dec 2012, 02:47
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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.
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 29 Jun 2015, 05:27
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Swaroopdev wrote:
Hi Bunuel,

Any link to similar problems like this ?

Thanks.


Please find the link below

search.php?search_tags=all&selected_search_tags%5B%5D=41&selected_search_tags%5B%5D=180


search.php?search_tags=all&selected_search_tags%5B%5D=41&selected_search_tags%5B%5D=222


search.php?search_tags=all&selected_search_tags%5B%5D=62&selected_search_tags%5B%5D=187

search.php?search_tags=all&selected_search_tags%5B%5D=62&selected_search_tags%5B%5D=216
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Re: In the xy-plane... [#permalink] New post 25 Feb 2011, 10:59
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.
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Re: In the xy-plane... [#permalink] New post 25 Feb 2011, 11:14
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gmat1220 wrote:
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.
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Re: In the xy-plane... [#permalink] New post 25 Feb 2011, 11:50
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.
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Re: In the xy-plane... [#permalink] New post 04 Aug 2012, 01:31
Bunuel wrote:
gmat1220 wrote:
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?

Thanks.
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Re: In the xy-plane... [#permalink] New post 04 Aug 2012, 01:52
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shreya717 wrote:

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.
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Re: In the xy-plane... [#permalink] New post 04 Aug 2012, 02:30
Bunuel wrote:
shreya717 wrote:

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.

Thnaks for the clarification.
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Re: In the xy-plane... [#permalink] New post 04 Aug 2012, 04:06
Bunuel 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 \(s=< 2\)


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.

Answer: E.

Hope it's clear.


Hi Bunuel,

I did not understand the the way u have proved taking points 1 and 2 in this problem....can u plz explain.... what i have always done is take the conditions which is satisfying both point 1 and 2 and then try to get a PASS and a fail in it to prove that the answer is OPTION is C or E....
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Re: In the xy-plane...   [#permalink] 04 Aug 2012, 04:06

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