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Graphical Solution

by knight247 » Tue Jan 20, 2015 5:31 am
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

I was able to solve this simply by trial and error. The answer is E

However, when I solve it graphically, I encounter a problem.

I'm able to prove statement (1) is insufficient because the graphs of 3r + 2s = 6 and 2x + 3y =< 6 don't overlap. (SEE ATTACHED GRAPH)

However, in statement (2) r=< 3 and s=< 2, don't the values of r and s coincide EXACTLY with the graph of 2x + 3y =< 6? I mean, if r=3 and s=2, don't r and s fall exactly ON the graph of 2x + 3y =< 6? Also if we select values of r and s that are less than 3 and 2 respectively, both r and s will lie on the LEFT of the graph of 2x + 3y =< 6.

So how is statement (2) insufficient?

Detailed explanations would be appreciated. Thanks in advance.

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Last edited by knight247 on Tue Jan 20, 2015 6:22 am, edited 1 time in total.
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by MartyMurray » Tue Jan 20, 2015 6:16 am
knight247 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

I was able to solve this simply by trial and error. The answer is E

However, when I solve it graphically, I encounter a problem.

I'm able to prove statement (1) is insufficient because the graphs of 3r + 2s = 6 and 2x + 3y =< 6 don't overlap. (SEE ATTACHED GRAPH)

However, in statement (2) r=< 3 and s=< 2, don't the values of r and s coincide EXACTLY with the graph of 2x + 3y =< 6. I mean, if r=3 and s=2, don't r and s fall exactly ON the graph of 2x + 3y =< 6? Also if we select values of r and s that are less than 3 and 2 respectively, both r and s will lie on the LEFT of the graph of 2x + 3y =< 6.

So how is statement (2) insufficient?

Detailed explanations would be appreciated. Thanks in advance.

Image
Looks as if you are doing it more like 2x <= 6 and 3y <= 6.

When you combine them, for the point (r,s), if x = 3 and y = 2, 2x + 3y = 12.That does not match at all.

Also, (3,2) does not lie within the region defined by 2x + 3y <= 6. The region is to the left of the line, and (3,2) is to the the right of the line. You can confirm this on your graph.

So with r <= 3 and s <= 2 you can generate points both to the left and to the right of the line.

Thus Statement 2 is insufficient.
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by GMATGuruNY » Tue Jan 20, 2015 7:37 am
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 & s≤2
Region R is composed of all the points on or below y=(-2/3)x + 2.

Statement 1: s = (-3/2)r + 3.
Image
The figure above shows that some points on s=(-3/2)r + 3 lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.

Statement 2: r≤3 and s≤2.
Image
Inside the green box are points such that r≤3 and s≤2.
Some of the points inside the green box lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.

Statements 1 and 2 combined:
Image
Inside the green box are points on s=(-3/2)r + 3 such that r≤3 and s≤2.
Some of these points lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.

The correct answer is E.
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by Brent@GMATPrepNow » Tue Jan 20, 2015 8:05 am
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
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
Image

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.
Image

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.
Image
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.
Image
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.
Image

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
Image

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

Answer = E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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