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by Buix0065 » Mon Jul 25, 2011 3:34 pm
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

[spoiler]OA: E the solve seems to be able to pick numbers that satisfy the equation 3r + 2s = 6, and points that do not, using this as evidence that we do not know if (r,s) is in region R or not. I'm having a hard time understanding the connection here, how would I know on a similar problem to approach it by choosing multiple sets of #'s to see if they all satisfy the equation?

Thanks![/spoiler]
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by edge » Mon Jul 25, 2011 5:38 pm
This wasn't a 2-minute-question for me.

Statement 1 is insufficient by itself because it tells us about 3r + 2s, not 2r + 3s (in the case of which it would have been sufficient).
Statement 2 is insufficient by itself because substituting max values for r and s in [2x + 3y] is greater than 6, but then you can also put in small values of r and s to satisfy the inequality.

Looking at the statements together, you find the range for r (2/3 <= r <= 3). Also, s = 3 - 3r/2. Plugging in values for r in the equation for s, you get two points: (2/3, 2) and (3, -3/2). Plugging these in the LHS of the original inequality, you will find that one of them DOESN'T satisfy it.

Therefore, both the statements together are not sufficient. E is the answer to this DS question.

This was a long way to go about it; does anyone have suggestions to cut down on the solve time?
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by dabral » Mon Jul 25, 2011 8:51 pm
This a tough question. Here is a video explanation:
https://www.gmatquantum.com/list-of-vide ... ds121.html

Dabral
Free Video Explanations: 2021 GMAT OFFICIAL GUIDE.
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by sandeep1306 » Mon Jul 25, 2011 10:40 pm
Please do not point to videos that dont open.Please provide a solution on the thread itself, if you wish to.
Thanks,
Sandeep
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by Anurag@Gurome » Mon Jul 25, 2011 11:01 pm
Buix0065 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

[spoiler]OA: E the solve seems to be able to pick numbers that satisfy the equation 3r + 2s = 6, and points that do not, using this as evidence that we do not know if (r,s) is in region R or not. I'm having a hard time understanding the connection here, how would I know on a similar problem to approach it by choosing multiple sets of #'s to see if they all satisfy the equation?

Thanks![/spoiler]
(1) 3r + 2s = 6 may or may not lie in region R. So, (1) is NOT SUFFICIENT to answer the question.

(2) If we take r = 3 and s = 2, then the point (3, 2) does not lie in region R.
r ≤ 3 and s ≤ 2 implies we can also take negative values for r and s. If r = -2, s = -3, then (-2, -3) lies in region R.
We don't get a unique answer, so (2) is NOT SUFFICIENT to answer the question.

Combining (1) and (2), if r = 2, s = 0 then (2, 0) lies in region R. But if r = 2/3 and s = 2 then (2/3, 2) lies above the line 2x + 3y = 6, which means (2/3, 2) does not lie in region R. Combining also doesn't give a unique answer.

The correct answer is E.

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by ajaykpat » Tue Aug 02, 2011 9:14 pm
Hi,
just one query on this que... is region R means area under the line 2X+3Y=6 or only points
which satisfy the line equation??

thanks,
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by gmatboost » Tue Aug 02, 2011 11:11 pm
The original question said
R consists of all the points (x,y) such that 2x + 3y <= 6
The <= means that we are interested in the area under the line and on the line itself.

I want to suggest another approach that does not depend on drawing any lines:

We want to know: Is it true that 2r + 3s <= 6?

Statement 1 tells us that 3r + 2s = 6. We can plug in points, as described above, or we can solve for r (or s) and plug that into the inequality to see what happens.

Using St. 1, r = (6 - 2s)/3. Plugging this into the inequality, we want to know:
Is it true that 2*(6 - 2s)/3 + 3s <= 6?
Is it true that (12 - 4s)/3 + 3s <= 6?
Is it true that (12 - 4s) + 9s <= 18?
Is it true that 12 + 5s <= 18?
Is it true that 5s <= 6?
Is it true that s <= 6/5?

Since we have no other information, Statement 1 is Insufficient.

As noted above, Statement 2 can be shown to be Insufficient pretty quickly by plugging in (r=3, s=2), and then (r=0, s=0) into the initial inequality.

Now when we combine the statements, we know from our work on St. 1 that the question we must answer is:
Is it true that s <= 6/5?

Statement 2 tells us that r <= 3 and s <= 2. So, s could be 2, or s could be 0. We do not know if s <= 6/5, so the answer is E.

If you prefer using coordinate geometry, that's totally fine, but I think it's worth remembering that you can also use algebra to solve many questions about two-variable inequalities.
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