region R

[MK]

In the xy-plane, Region R consists of all the points (x,y) suc that 2x+3y = 6, Is the point (r,s) in the region R?
a) 3r+2s=6
b) r=3 ans s=2.

I beleive that the ans should be B. However it is E.
Can any one explain?

[Bhaskar]

First of all 2x+3y=6 is a straight line, it cannot represent a region. Something is wrong with the question.
2x+3y<6 could represent a region and in that case the answer would be E.

[Bhaskar]

Sorry I meant the answer would be B

[RonPurewal]

yeah, you should check your transcription again.

in fact, statement (2) must be sufficient, regardless of the equation that defines the region. if we are given a definitive set of coordinates - in this case (3, 2) - then that point is either "in" or "out", no matter what region we're considering. therefore, the answer to the question will be either "definitive yes" or "definitive no", so it's sufficient either way.

[jp.jprasanna]

The original poster has posted the Question wrong... here it is

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

1. Since 3r + 2s = 6 is a straight line we can get the coordinates

put r = 0 then (0,3) question becomes 9 > 6 . No
put s = 0 then (2 , 0) question becomes 4 < 6 Yes

hence NS

2. I randomly took 2 points

Say (2,2) so - 6 + 4 <= 6 No
Say (0,0) so 0 <= 6 Yes

hence NS

Combing both the statements (2,0) works hence its a Yes?
But the OA is E can you please advise how to approach these problems more effectively.

Cheers
Jp

[jnelson0612]

This is an ugly problem. Here's what I did:

1) First, take 2x + 3y <= 6, and convert it to the standard equation for a line, y=mx+b.
a) Subtract 2x from both sides
b) Divide both sides by 3
c) You obtain y <= -2/3x + 2

2) I then plotted that line on my graph. My y intercept is 2 and my x intercept is 3. I plotted out x=1, y=4/3 and x=2, y=2/3 by plugging them into the equation and just so I could see the line better. I shaded everything below this line--the line itself and the area below is Region R.

3) Start with statement 2. Not good enough, as you noted. I could have (3,2) which is outside the region or (0,0) which is inside the region. Insufficient.

4) Go to statement 1. Again, I can easily plug values that show this is insufficient; I can use (2,0) which is included in Region R or (0, 3) which is outside Region R. Insufficient.

5) I then combined both to try to disprove. I can easily use (3,0) which fits both statements and is on the line and thus included in Region R.

Now, I want something that fits both and is outside Region R. Let's play a bit. What if I say r=1? Then s must be 3/2 according to statement 1. If we look at our graph, (1, 3/2) is above the shaded line because the point at r=1 is (1, 4/3) and 3/2 is more than 4/3. Okay, I have proved insufficient. Whew!

[prashant.ranjan]

Thanks Jamie for the answer.
Just would like to add 1 more point. If we plot the graph of the Region R and the line1: 3r+2s = 6.
The question becomes really simple since there will be a potion below the line1: 3r+2s = 6 that does not come in Region R. Since a point has the probability of belonging to that region so answer comes to be (E).

Thanks
Prashant Ranjan

[tim]

thanks for sharing!

[Enrique17]

This is an ugly problem. Here's what I did:

1) First, take 2x + 3y <= 6, and convert it to the standard equation for a line, y=mx+b.
a) Subtract 2x from both sides
b) Divide both sides by 3
c) You obtain y <= -2/3x + 2

2) I then plotted that line on my graph. My y intercept is 2 and my x intercept is 3. I plotted out x=1, y=4/3 and x=2, y=2/3 by plugging them into the equation and just so I could see the line better. I shaded everything below this line--the line itself and the area below is Region R.

3) Start with statement 2. Not good enough, as you noted. I could have (3,2) which is outside the region or (0,0) which is inside the region. Insufficient.

4) Go to statement 1. Again, I can easily plug values that show this is insufficient; I can use (2,0) which is included in Region R or (0, 3) which is outside Region R. Insufficient.

5) I then combined both to try to disprove. I can easily use (3,0) which fits both statements and is on the line and thus included in Region R.

Now, I want something that fits both and is outside Region R. Let's play a bit. What if I say r=1? Then s must be 3/2 according to statement 1. If we look at our graph, (1, 3/2) is above the shaded line because the point at r=1 is (1, 4/3) and 3/2 is more than 4/3. Okay, I have proved insufficient. Whew!

Thanks for the response Jamie. However, I'm struggling to figure out how to conceivably go through this process within a reasonable time limit. I am tempted to make an educated guess for choice E by the following summary:

Original Information:
Region R of 2x + 3y <= 6 gives me slope of -2/3(X) and Y intercept of 2, with Region R below this line
X Y
0 2
3 0

Stmt (1):
Line Equation gives Slope of -3/2(X) and Y intercept of 3
X Y
0 3
2 0

Stmt (2):
Gives possible values of X=2 and Y=3, which could be both outside and within the limits by original information.

(1)(2) Together: I'd be tempted to make an educated guess of E, by the difference in slopes and Y intercepts and the additional information in Stmt 2, but still quite unsure about the attempt and the time already spent.

Do you have any suggestions on how to make an educated guess faster in such a problem?

Thanks,

[RonPurewal]

okay... so there are 3 pieces of basic knowledge regarding statement 1.

one:
2x + 3y ≤ 6 is the region on one side of a straight line, including the line itself**.

two:
3x + 2y = 6 is a different line.

three:
these lines do not have the same slope.

it's pretty clear that you (the previous poster—you who are worried about time / efficiency) are well aware of all three of these facts.

if you know all three of these facts, then you can INSTANTLY realize that statement 1 is insufficient... because the line 3x + 2y = 6 will contain points ON BOTH SIDES of the line 2x + 3y ≤ 6.
in other words, it will contain points inside the 'region' AND points outside the 'region'.
boom. not sufficient.

if you don't have to do the work, don't do the work!

[RonPurewal]

**footnote:
the 'including the line itself' part isn't particularly important.
in fact, for this exam it's pretty much not important at all. i've NEVER seen a gmat problem that depended on 'boundary conditions', as these things are called

[RonPurewal]

once you've done that ^^ then you can also dispatch statement 2 fairly quickly.

then you'll have plenty of idea to play around with the two statements together, even if you adhere to this silly notion that you have to try to solve every problem in 2 minutes or less. (did i mention that's a silly idea? that's a silly idea, by the way.)

[RonPurewal]

if you just look for extremes—a sensible plan for ANY problem involving inequalities—you won't have to look very far.

taking s = 2 (the extreme value of y from the second statement), plug that into statement 1 and get 3r + 4 = 6. so r = 2/3.
this point (2/3, 2) is NOT in the region.

now take r = 3 (the extreme value of x from the second statement). plug into statement 1 and get 9 + 2s = 6. so s = –3/2.
this point (3, –3/2) IS in the region.

done, it's E.