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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, region R consists of all the points (x, y) [#permalink]
07 Jul 2013, 21:50

2

This post received KUDOS

(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.

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.

Re: In the xy-plane, region R consists of all the points (x, y) [#permalink]
02 Jun 2015, 02:26

1

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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

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)
[#permalink]
05 Jul 2015, 17:35

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