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Re: In the xy-plane... [#permalink]
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)

(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 _________________

Re: In the xy-plane... [#permalink]
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.

Re: In the xy-plane... [#permalink]
06 Oct 2010, 23:15

1

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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! _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Re: In the xy-plane... [#permalink]
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.

Re: In the xy-plane... [#permalink]
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 !! _________________

Re: In the xy-plane... [#permalink]
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.

Re: In the xy-plane... [#permalink]
25 Feb 2011, 11:14

Expert's post

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. _________________

Re: In the xy-plane... [#permalink]
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.

Re: In the xy-plane... [#permalink]
25 Feb 2011, 11:59

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Expert's post

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 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. _________________

Re: In the xy-plane... [#permalink]
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?

Re: In the xy-plane... [#permalink]
04 Aug 2012, 01:52

Expert's post

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. _________________

Re: In the xy-plane... [#permalink]
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.

Re: In the xy-plane... [#permalink]
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.... _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Re: In the xy-plane, region R consists of all the points (x, y) [#permalink]
04 Aug 2012, 04:13

2

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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]
05 Aug 2012, 06:26

Expert's post

shirisha091 wrote:

How did you get 2r + 3s ≤ 6 in the beginning of this problem?

Since we are told that region R consists of ALL the points (x, y) such that \(2x+3y\leq{6}\), then in order point (r,s) to be in region R the same must hold true for that point too, hence it must be true that \(2r+3s\leq{6}\).

Re: In the xy-plane, region R... [#permalink]
02 Sep 2012, 12:41

2x +3y <= 6 3y <= 6 - 2x y <= 2 - 2x/3

Statement (1): 3r + 2s = 6 2s = 6 - 3r s = 3 - 2r

Since this line is not parralel to the edge of the region (when 2x+3y=6), it will at some point be above, and at some point be below it. So we don't have enough information.

Statement (2): Let's calculate point (3, 2): 2*3+3*2=12>6, so not inside region R But since we can go as low as we want with this point, it will eventually be in region R.

Combining (1) + (2): Let's see, where region R intersects with the line of statement (1). y=2-2x/3 s=3-2r --> y=3-2x 2-2x/3=3-2x 4x/3 = 1 x = 3/4 y=3-2*(3/4)=1.5

Intersection at (0.75, 1.5). So our line from statement (1) will intersect with region R within the region of statement (2). So there will be points on that line that also fullfill statement (2), but are not in region R. And there will be points that are in region R. --> not sufficient

Re: In the xy-plane, region R consists of all the points (x, y) [#permalink]
17 Dec 2012, 11:10

look, the question is about the point (r,5) not the point (r,s). if the question was about the point (r,s), I think the (2) is sufficient, the correct answer is B However, if the question is about the point (r,5), (1) and (2) together are sufficient but none of them alone is sufficient. The corrent answer is C

Re: In the xy-plane, region R consists of all the points (x, y) [#permalink]
25 Dec 2012, 02:47

1

This post received KUDOS

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.

gmatclubot

Re: In the xy-plane, region R consists of all the points (x, y)
[#permalink]
25 Dec 2012, 02:47

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