Inequality + function + coordinate

[MAAJ]

In the xy-plane, region R consist 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

Not satisfied with the answer from the OG, particularly on the second statement... it's easier to solve by picking numbers but can some1 try to solve it by graphing? I'm curious about the graphic of statement 2...

Correct Answer [spoiler](E)[/spoiler]

[anshumishra]

In the xy-plane, region R consist 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

Not satisfied with the answer from the OG, particularly on the second statement... it's easier to solve by picking numbers but can some1 try to solve it by graphing? I'm curious about the graphic of statement 2...

Correct Answer [spoiler](E)[/spoiler]

question : is 2r+3s <=6 ?

Statement 1:
3r+2s = 6
if r=2,s=0 => 2r+3s = 4 < 6
if r=0,s=3 => 2r+3s = 7 > 6 So, not sufficient

Statement 2:
r<=3 and s<=2
if r=2,s=0 => 2r+3s = 4 < 6
if r=2,s=1 => 2r+3s = 7 > 6 So, not sufficient

Combined 1 and 2 :
if r=2,s=0 => 2r+3s = 4 < 6
if s=2,r=2/3 => 2r+3s = 4/3+6 > 6 Still not sufficient.

Hence, E

[Rahul@gurome]

... it's easier to solve by picking numbers but can some1 try to solve it by graphing? I'm curious about the graphic of statement 2...

Here is the graphic approach.
If you are familiar with graphic approach you can solve these kind problem very fast. If you don't feel comfortable with it, don't go for it. Picking number is much easier for you.



Refer to the figure above.

The blue line is the line with equation 2x + 3y = 6.
Hence the region below the blue line, i.e. the side on which origin lies is the region satisfying 2x + 3y ≤ 6. Hence we need to see whether the points satisfying the statements are part of it or not.

Statement 1: 3r + 2s ≤ 6
The red line is the line with equation 3x + 2y = 6.
Hence the region below the blue line, i.e. the side on which origin lies is the region satisfying 3x + 2y ≤ 6. As 3x + 2y ≤ 6, (r, s) must lies within the region 3x + 2y ≤ 6.

From the figure we can see that some part of the coordinate plane is common with the region given in question stem and some part are not.

Not sufficient

Statement 1: r ≤ 3 and s ≤ 2
The green and the brown lines are the equation for x = 3 and y = 2 respectively. Hence the point (r, s) must lie below the brown line and left to the green line.

Again we see that some part of the coordinate plane is common with the region given in question stem and some part are not.

Not sufficient

1 & 2 Together: Now the point (r, s) must lie below the brown line and left to the green line and also below the red line. Now we have a small triangle which satisfies the above condition but lies above the blue line. And other part is common.

Not sufficient

The correct answer is E.

[GMATGuruNY]

In the xy-plane, region R consist 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

Not satisfied with the answer from the OG, particularly on the second statement... it's easier to solve by picking numbers but can some1 try to solve it by graphing? I'm curious about the graphic of statement 2...

Correct Answer [spoiler](E)[/spoiler]

While this looks like a geometry problem, it can be treated as an algebra problem. If (r,s) is in region R, then it must satisfy the conditions of the inequality. Thus, the question can be rewritten as:

Is 2r + 3s ≤ 6?

Statement 1: 3r + 2s ≤ 6
r=0 and s=3 works, because 3(0) + 2(3) ≤ 6. Is 2(0) + 3(3) ≤ 6? No.
r=2 and s=0 works, because 3(2) + 2(0) ≤ 6. Is 2(2) + 3(0) ≤ 6? Yes.
Since the answer can be both no and yes, insufficient.

Statement 2: r ≤ 3 and s ≤ 2
r=2 and s=0 works, because 2 ≤ 3 and 0 ≤ 2. Is 2(2) + 3(0) ≤ 6? Yes.
r=3 and s=2 works, because 3 ≤ 3 and 2 ≤ 2. Is 2(3) + 3(2) ≤ 6? No.
Since the answer can be both yes and no, insufficient.

Statements 1 and 2 together:
We saw above that r=2 and s=0 satisfy both statements. Is 2(2) + 3(0) ≤ 6? Yes.
To change the answer from yes to no, we need to maximize r and s. The largest combination that satisfies both statements is
r=2/3 and s=2. Is 2(2/3) + 3(2) ≤ 6? No.
Since the answer can be both yes and no, insufficient.

The correct answer is E.

[MAAJ]

So that little triangle above the blue line is the part that tell us that its insufficient. Guess I didn't have that much care when I was graphing

Thanks all!

[OneTwoThreeFour]

The best way to approach this problem is from a conceptual approach. Even by drawing graphs, I was having trouble doing this problem in under 2 minutes. (Plus it gets messy when you have multiple lines and try to determine the area that doesn't overlap and calculate the x intercepts too.) The first step is to realize that 2x+3y<=6 also equals to y <= 2 -2/3x. This means that the condition in the question is only satisfied when (r,s) is either a distinct point or has a slope of -2/3x and a y-intercept that is less than or equal to 2. Knowing this in mind, it is very easy to realize the first condition is insufficient and the second one is not either, since it gives a range of values, instead of a distinct point. Putting them together is also insufficient since the slope is not -2/3x.