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# In the rectangular coordinate system shown above, which quad

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03 Mar 2014, 01:21
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality $$2x - 3y\leq{- 6}$$ ?

(A) None
(B) I
(C) II
(D) III
(E) IV

Problem Solving
Question: 123
Category: Geometry Simple coordinate geometry
Page: 77
Difficulty: 600

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03 Mar 2014, 01:21
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SOLUTION

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality $$2x - 3y\leq{- 6}$$ ?

(A) None
(B) I
(C) II
(D) III
(E) IV

$${2x-3y}\leq{-6}$$ --> $$y\geq{\frac{2}{3}x+2}$$. Thi inequality represents ALL points, the area, above the line $$y={\frac{2}{3}x+2}$$. If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Else you can notice that if $$x$$ is positive, $$y$$ can not be negative to satisfy the inequality $$y\geq{\frac{2}{3}x+2}$$, so you can not have positive $$x$$, negative $$y$$. But IV quadrant consists of such $$(x,y)$$ points for which $$x$$ is positive and $$y$$ negative. Thus answer must be E.

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04 Mar 2014, 03:07
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6

To draw the line on the Coordinate system consider the Inequality : $$2x - 3y\leq{- 6}$$ as $$2x - 3y={- 6}$$

We get points (0,2) & (-3,0 )
So the line looks some what as in the attachment.

To find out which area is covered by the graph put the Cordinate (0,0) in the original question $$2x - 3y\leq{- 6}$$

We get: $$0 \leq {-6}$$. Which is False.

So (0,0) does not lie in the area covered by the graph, Therefore the equation covers the area above the line.

Thus 4th Quadrant does not contain any point that satisfies the inequality. ( Rest 3 Quadrants will have a few points that would satisfy the inequality)

Not experienced enough to comment on the difficulty level.
Attachments

Q.png [ 21.53 KiB | Viewed 9045 times ]

##### General Discussion
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08 Mar 2014, 11:41
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SOLUTION

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality $$2x - 3y\leq{- 6}$$ ?

(A) None
(B) I
(C) II
(D) III
(E) IV

$${2x-3y}\leq{-6}$$ --> $$y\geq{\frac{2}{3}x+2}$$. Thi inequality represents ALL points, the area, above the line $$y={\frac{2}{3}x+2}$$. If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Else you can notice that if $$x$$ is positive, $$y$$ can not be negative to satisfy the inequality $$y\geq{\frac{2}{3}x+2}$$, so you can not have positive $$x$$, negative $$y$$. But IV quadrant consists of such $$(x,y)$$ points for which $$x$$ is positive and $$y$$ negative. Thus answer must be E.

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09 Mar 2014, 03:28
1
1
Attachment:
The attachment Untitled.png is no longer available
In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality $$2x - 3y\leq{- 6}$$ ?

(A) None
(B) I
(C) II
(D) III
(E) IV

Sol: The inequality can be 2x=3y=-6 can be re-written in Y intercept form as y=2x/3 +2
Attachment:

Untitled1.png [ 11.93 KiB | Viewed 8873 times ]

Notice that the slop of the line is positive and hence it will definitely pass through Quad 1 and 3.
If the slope was negative then the line will definitely pass through Quad 2 and 4.

Now for x=0, y= 2 that means line will have to pass through Quad 2 as well.
Hence No point in Quad 4 will satisfy the given equation.
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18 Apr 2015, 11:53
Guys, I am still not clear with the solution. By any way can you simplify it further ?
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21 Apr 2015, 04:28
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kshitij89 wrote:
Guys, I am still not clear with the solution. By any way can you simplify it further ?

Hi kshitij89 - For every point lying on the line segment $$2x - 3y =-6$$, the x and y coordinates are such that subtracting 3 times the y coordinate from 2 times the x coordinate is equal to -6. Examples are (3,4), (6,6) etc.

For any other point not lying on this line segment, this difference of 2x and 3y is either less than -6 or greater than -6.

The question asks us to find the location of points for which $$2x - 3y <= -6$$ is not true i.e. points for which the difference of 2x and 3y is not less than or equal to 6. For finding such points, we need to first plot its equivalent line segment in the X-Y coordinate system.

Let's see how we can plot the line. We need two points for plotting the line segment.We know that for all points on the X-axis, their Y coordinate is 0 and vice versa. So, putting x =0 in the equation of the line segment, we get the value of y = 2 and for y =0, we get the value of x = -3. Now, we have the x and the y - intercept for the line segment. Using this information, we can plot the line segment as shown below:

Now, we will need some test case to establish that on which side of the line the points do not satisfy the inequality $$2x - 3y <= -6$$. The best way is to test for the intersection point of X & Y axis i.e. (0, 0). If we put x =0 and y =0 in the inequality, we get $$0 <=-6$$ which is not true. So, we can say with certainty that the side of the line which contains ( 0, 0) does not satisfy the inequality.

This would mean that the area on the left side of the line segment $$2x - 3y =6$$ satisfy the inequality $$2x - 3y <= -6$$ and the area on the right side of the line segment do not satisfy the inequality. Since the question talks about such regions in terms of quadrants, we can observe that area on the left side of the line segment passes through Q- I, II & III only. So, we can say for sure that Q- IV does not contain any point that satisfy the inequality.

Hope this is clear. Let me know if you still have trouble in understanding of the solution.

Regards
Harsh
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21 Apr 2015, 10:52
Hi Harsh,

The explanation was crisp and clear but I am not sure if I will be able to solve similar questions with different nos.

Can you please provide examples of similar questions to further test my understanding ?

Regards
Kshitij
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21 Apr 2015, 19:59
Hi kshitij89,

How are your overall graphing "skills"? Are you comfortable with the basic concepts, formulas, drawing graphs, etc.? If so, then you'll probably handle the concept on Test Day just fine.

"Graphing", as a category, is relatively rare on the GMAT - you'll likely see just 1-2 graphing questions on Test Day and they will probably be considerably easier than this one.

Unless you've already mastered all of the 'big' categories (Algebra, Arithmetic, Formulas, Broader Geometry, Ratios, DS, etc.), then this nitpick category really isn't worth the extra time.

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21 Apr 2015, 22:43
kshitij89 wrote:
Hi Harsh,

The explanation was crisp and clear but I am not sure if I will be able to solve similar questions with different nos.

Can you please provide examples of similar questions to further test my understanding ?

Regards
Kshitij

Hi kshitij89 - what you need is more practice to get yourself comfortable with the graphical method of solving inequalities. To begin with, I would suggest you to plot lines on the X-Y coordinate system and find out points which lie on either side of the line. You may further extend this exercise to plotting of 2 lines in the X-Y coordinate system and finding points which satisfy corresponding inequalities of both the lines.

Once you get used to it, you will prefer using the graphical method for solving inequalities. Getting comfortable with the X-Y coordinate system will also strengthen your understanding of the Coordinate Geometry section

For your practice, you may refer the following posts which uses graphical method for solving questions on inequalities and coordinate geometry:

in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html
in-the-xy-plane-region-a-consists-of-all-the-points-x-y-154784.html?hilit=inequalities%20inequalities%20graph
point-x-y-is-a-point-within-the-triangle-what-is-the-139419.html
set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html

Hope it helps!

Regards
Harsh
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17 Jul 2015, 02:26
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1. Rewrite the equation (Y=mx+b) --> y=>2/3*x+2
2. Set x=0 and then y=0 --> (0, =>2); (-3<=, 0) so we have now two points the coordinate plane
3. Draw the line (see attachment) --Y you can see that Quadrant IV is not involved there (E)
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30 Dec 2015, 20:48
we have a positive slope and a 3 as y- coordinate when x=0.

well, regardless of what x is, line will never pass through IV, since the line must go up, and there is no way for the line to pass through IV, since it will imply that the slope is negative.
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06 Jul 2017, 17:33
Quote:

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality $$2x - 3y\leq{- 6}$$ ?

(A) None
(B) I
(C) II
(D) III
(E) IV

Let’s rewrite the inequality:

-3y ≤ -2x - 6

y ≥ (2/3)x + 2

We see that the graph of the inequality y ≥ (2/3)x + 2 consists of the line y = (2/3)x + 2, which is a positively sloped line with a y-intercept of 2 and the region above this line. Thus, the one quadrant that would not satisfy this inequality is quadrant IV.

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