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Source: — Data Sufficiency |

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by [email protected] » Fri Aug 02, 2013 3:55 pm
Hi shibsriz,

Since you're curious about what this question is really asking you for, I'm going to give you some hints as to how to solve it. I'll post the full solution later, but I think it's better for you to try to do the work first.

This question is clearly a graphing question, but you'll notice that the line is NOT based on an equation, it's based on an inequality. This means that your graph will have a line AND part of the graph will be SHADED (to include all of the points that are LESS THAN OR EQUAL to what you've been told).

Start by graphing the equation 2x + 3y = 6
Next, shade in the part of the graph that is LESS than that.

Now, use the Facts to answer the question

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by Matt@VeritasPrep » Fri Aug 02, 2013 4:49 pm
If the region R is defined as 2x + 3y ≤ 6, then the point (x, y) is in the region if its coordinates satisfy the equation 2x + 3y ≤ 6.

For example, the point (1, 1) is in the region, as 2*1 + 3*1 ≤ 6. But the point (2, 4) is not in the region, as 2*2 + 3*4 ≥ 6.

So if 2r + 3s ≤ 6, (r,s) is in the region R. If not, it isn't.

I'll give an algebraic approach, as I usually do, and trust that you can plug in / test values on your own if you prefer that way (and it's certainly a good way!)

S1::
3r + 2s ≤ 6
r + (2/3)s ≤ 2
r ≤ 2 - (2/3)s

Now plug r ≤ 2 - (2/3)s into the equation for the region:

2r + 3s ≤ 2(2 - (2/3)s) + 3s

If the right hand side is less than 6, then 2r + 3s is also less than 6. Let's clean it up a bit. If 4 + (5/3)s ≤ 6, then (5/3)s ≤ 2, or s ≤ (6/5)

Now we can restate the original question as "Is s ≤ (6/5)"? We can't say, so this is INSUFFICIENT.

S2::
r ≤ 3 and s ≤2 so
2r + 3s ≤ 2*3 + 3*2
2r + 3s ≤ 12

Definitely INSUFFICIENT.

Combining the two statements, S1 allows us to state the question as "Is s ≤ 6/5"? S2 only tells us that s ≤ 2, so we still can't solve and both statements together are INSUFFICIENT.
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by [email protected] » Fri Aug 02, 2013 10:08 pm
[email protected] wrote:Hi shibsriz,

Since you're curious about what this question is really asking you for, I'm going to give you some hints as to how to solve it. I'll post the full solution later, but I think it's better for you to try to do the work first.

This question is clearly a graphing question, but you'll notice that the line is NOT based on an equation, it's based on an inequality. This means that your graph will have a line AND part of the graph will be SHADED (to include all of the points that are LESS THAN OR EQUAL to what you've been told).

Start by graphing the equation 2x + 3y = 6
Next, shade in the part of the graph that is LESS than that.

Now, use the Facts to answer the question

GMAT assassins aren't born, they're made,
Rich

I am confused :(
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by Matt@VeritasPrep » Sat Aug 03, 2013 6:49 am
Shibsriz, I think it's best to treat this question algebraically. That said, here's the coordinate geometry.

The basic idea here (geometrically speaking) is that in the coordinate a linear inequality (such as 2r + 3s ≤ 6) represents a REGION rather than a straight line.

For example, if 2r + 3s = 6, that equation represents a straight line. On that line we have all the coordinates (r, s) such that 2r + 3s = 6. Nothing unusual there, other than using (r,s) instead of (x,y).

But if 2r + 3s ≤ 6, we haven't got a line, we've got a region, as there are many more points that satisfy the equation (≤) than the equation (=). So any points (r,s) such that 2r + 3s ≤ 6 are in that region.

I've attached a graph of the region so you can get an idea. Any point in the blue region of the plane satisfies the equation 2x + 3y ≤ 6.

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by [email protected] » Mon Aug 05, 2013 12:58 am
Hi shibsriz,

Here's a video explanation that you'll find helpful:

Watch Rich CRUSH this DS question...

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by GMATGuruNY » Tue Aug 06, 2013 8:58 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
2x+3y ≤ 6
y ≤ (-2/3)x + 2.
Region R consists of all the points on or below the line y=(-2/3)x + 2.

Approach 1: DRAW

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.

Approach 2: TEST VALUES

An alternate way to combine the two statements is to treat this as MAX/MIN problem.
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
R MAXIMIZED:
In statement 2, the maximum possible value of r is 3.
If r=3 and s=0, both statements 1 and 2 are satisfied.
Check whether (3,0) is within the region defined by y ≤ (-2/3)x + 2:
0 = (-2/3)(3) + 2
0 ≤ 0.
YES.

S MAXIMIZED:
In statement 2, the maximum possible value of s is 2.
If s=2 and r=(2/3), both statements 1 and 2 are satisfied.
Check whether (2/3, 2) is within the region defined by y ≤ (-2/3)x + 2:
2 ≤ (-2/3)(2/3) + 2
2 ≤ 2/3
NO.

Since in the first case (r,s) is within the required region, but in the second case (r,s) is not within the required region, the two statements combined are INSUFFICIENT.
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