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

Target question: Is the point (r, s) in region R?

Given: Region R consists of all the points (x,y) such that 2x + 3y <6
So, what does Region R look like?
To find out, let's first graph the EQUATION, 2x + 3y = 6


Since Region R is described as an INEQUALITY, we can choose any point on the coordinate plane to test whether or not it is in Region R. An easy point to test is (0,0).
So, does x=0 and y=0 satisfy the inequality 2x + 3y <6? YES
2(0) + (3)(0) is less than or equal to 6.
So, the point (0,0) is in Region R. More importantly, EVERY POINT on the same side of the line will also be in Region R.


Statement 1: 3r + 2s = 6
The target question refers to the point (r, s)
In other words, the x-coordinate is r and the y-coordinate is s.
So, all of the points (r, s) that satisfy the above equation can be found on the line 3x + 2y = 6
In other words, statement 1 tells us that the point (r,s) lies somewhere on the red line below.

As you can see, some points are in Region R, and some points are not in Region R
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: r < 3 and s < 2
There are many points that satisfy this condition.
In fact, the point (r,s) can be ANYWHERE inside the red box shown below.

As you can see, some points are in Region R, and some points are not in Region R
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
When we combine the statements, we are saying that the point (r,s) is on the red line (2x + 3y = 6) AND inside the red box.


As you can see by the two blue points below, it's possible to have a point in Region R, and it's possible to have a point not in Region R


Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer =

Cheers,
Brent

Region R consists of all the points on or below the line \(y=-\frac{2}{3}x + 2\).

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

The figure above shows that some points on \(s=-\frac{3}{2}r + 3\) lie BELOW \(y=-\frac{2}{3}x + 2\), while others lie ABOVE \(y=-\frac{2}{3}x + 2\).
INSUFFICIENT.

Statement 2: \(r\leq{3}\) and \(s\leq{2}\)

Inside the green box are points such that \(r\leq{3}\) and \(s\leq{2}\).
Some of the points inside the green box lie BELOW \(y=-\frac{2}{3}x + 2\), while others lie ABOVE \(y=-\frac{2}{3}x + 2\).
INSUFFICIENT.

Statements combined:

Inside the green box are points on \(s=-\frac{3}{2}r + 3\) such that \(r\leq{3}\) and \(s\leq{2}\).
Some of these points lie BELOW \(y=-\frac{2}{3}x + 2\), while others lie ABOVE \(y=-\frac{2}{3}x + 2\).
INSUFFICIENT.

MATH REVOLUTION VIDEO SOLUTION:

Thanks Bunuel for the explanation! It's very clear. However, I have some questions about your criteria to choose the values:



How did you know that you had to use those values (\(r=0\) and \(s=0\))?



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,

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 !!

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 \(2r+3s\leq{6}\).

P.S. Edited the diagram in shrouded1's post: in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html#p794049

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.

(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).

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

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


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search.php?search_tags=all&selected_search_tags%5B%5D=62&selected_search_tags%5B%5D=216

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

There are 2 variables (x,y) and one equations (2x+3y<=6) and 2 more equations are given by the 2 conditions
This makes (D) a likely answer.
For condition 1, the answer to the question becomes 'yes' for (r,s)=(2,0), but 'no' for (0,3). This is therefore insufficient.
For condition 2, the answer to the question becomes 'yes' for (r,s)=(2,0), but 'no' for (2/3,2). This is also insufficient and the answer therefore becomes (E).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


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

-> When you modify the original condition and question, it becomes 2r+3s<=6? and there are 2 variables(r, s), which should match with the number equations. So, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is most likely to make C the answer. In 1) & 2), (r,s)=(2,0) yes, (r,s)=(2/3,2), which is no and therefore not sufficient. Thus, the answer is E.

-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

I tried the graph method and was right in plotting it out.

Could someone help me understand why
"(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"

From the graph, we see that any point with r<=3 and s<=2 will definitely lie under the blue line. Where am I going wrong here?

I understand that if we were to substitute r and s as 3 and 2 in the original equation,we would get the solution though. Just wanted to know what am I doing wrong here

This is a tricky question

1) Given an equation of region 2x +3y ≤ 6 i.e region R
2) Asking if point (r,s) lies in region R?

rephrasing the given:

3y ≤ 6 - 2x
y ≤ -2/3x + 2 ---> Slope = -2/3

y interecept = 2
x interecept = 3

Statement 1) This is an equation of a line. If we prove that this line passes within the region R than the point (r,s) on the line will fall under region R

3r + 2s = 6
2s = 6 - 3r
s = -3/2r + 3 ---> Slope = -3/2

y intercept = 3
x intercept = 2

Since the slope of statement 1 is greater than given for the region R, the line 3r+2s =6 will be steeper, therefore it will not completely fall under region R. Thus value of point (r,s) on line (3r+2s =6) can fall out side Region R -- (Not Sufficient)


Statement 2)

r ≤ 3 and s ≤ 2

In order for point (r,s) to be in the region of R, it should satisfy the equation (y ≤ -2/3x + 2)
case a: r=2 and s= 3, 3 ≤ -4/3 + 2 -- {NO}
case b: r=1 and s =0 , 0 ≤ -2/3 + 2 -- {YES}

Since we get a YES and NO for different values of r and s, therefore, not sufficient

Answer is E

what is definition of "region" ? this is what confused me.