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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]
26 Feb 2013, 23:44

I want to follow this posting.

this question test out ability to react to numbers, to plug in number. This process is done by Bruno already. og explanation is the same. this process of plugging takes me 2 minutes.

drawing or solving the equation is too time consuming and is not the correct the method.

Re: In the xy-plane... [#permalink]
27 Sep 2013, 10:26

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 for the explanation. I used the same incorrect approach to get to B and I missed that the equation defines a region and not a line.

If the equation was a line, then would "B" be the correct answer?

Re: In the xy-plane... [#permalink]
30 Sep 2013, 22:37

Expert's post

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, Just a quick clarification- Combining two statements, to get NO as answer, we can also try the first values from Stat.1 - If r=0 and s=3 then 2r+3s=9>6, so the answer is NO..

Re: In the xy-plane... [#permalink]
30 Sep 2013, 23:13

Expert's post

bagdbmba wrote:

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, Just a quick clarification- Combining two statements, to get NO as answer, we can also try the first values from Stat.1 - If r=0 and s=3 then 2r+3s=9>6, so the answer is NO..

Please correct me if I'm wrong.

When combining the values must satisfy both statements: r=0 and s=3 does not satisfy the second one. _________________

Re: In the xy-plane, region R consists of all the points (x, y) [#permalink]
08 Jul 2014, 12:58

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.

I made the same mistake as you and marked B. But the region enclosed by the second statement is a rectangular region where r < = 3 and s<= 2. Only some part of this rectangle lies below the original line 2x + 3y <= 6. So this option is not sufficient to say that all points (r,s) lie below the original line. Others have shown this via algebra.

In the xy-plane, region R consists of all the points (x, y) [#permalink]
05 Jul 2015, 17:35

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

I did it a little differently.

I solved for y to get \(y <=\frac{-2}{3}x + 2\)

(1) I solved for s, so \(s = \frac{-3}{2}r + 3\)

Both x-y and r-s lines are downward sloping and have different y-intercepts. We know that they deviate in the -x region if you draw or visualize a crude graph.

Plug in an arbitrary negative x-value for r. Say, -10. Then solve for s, which is 18.

Plug the x-value of r into the x-y equation. \(y <=\frac{-2}{3}*-10\)+ 2; y <= \(\frac{26}{3}\) or 8\(\frac{2}{3}\)

If (r,s) is within the x-y region, then the y-value (s) must be <= 8\(\frac{2}{3}\). 18 <= 8\(\frac{2}{3}\) does not hold.

Inconsistent answers so insufficient.

(2) Plug given r and s values into given 2r+3s≤6 equation. 12 ≤ 6 does not hold.

Insufficient.

Both (1) and (2) together neither confirms nor eliminates the large negative x-value concern that we have, and it also neither confirms nor eliminates the positive x-values. So both together are insufficient.

gmatclubot

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