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

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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 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.
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 05 Jul 2013, 01:44
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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... [#permalink] New post 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?
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Re: In the xy-plane... [#permalink] New post 30 Sep 2013, 22:37
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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.
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Re: In the xy-plane... [#permalink] New post 30 Sep 2013, 23:13
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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.
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 30 Sep 2013, 23:19
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 06 Jul 2014, 15:45
I can't see why B is insufficient. Since r<= 3 means that x<=3 and s<=2 means y<=2 this covers the region under the line. which is f(x) <0 .
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 06 Jul 2014, 15:54
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amz14 wrote:
I can't see why B is insufficient. Since r<= 3 means that x<=3 and s<=2 means y<=2 this covers the region under the line. which is f(x) <0 .


Have you checked Graphic Approach HERE and Number Plugging approach HERE and HERE?
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 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.
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Re: In the xy-plane, region R consists of all the points (x, y) [#permalink] New post 16 Sep 2014, 09:17
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)

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




graph method is very clear.. thanks a lot :)
Re: In the xy-plane, region R consists of all the points (x, y)   [#permalink] 16 Sep 2014, 09:17
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