In the rectangular coordinate system shown above, which quadrant, if

Question

https://gmatclub.com/forum/download/file.php?id=22555 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 Untitled.png

niyantg · Accepted Answer

My answer is Option E i.e. 4th Quadrant

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.

Bunuel · Answer

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  cannot 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.

Answer: E.

WoundedTiger · Answer

Untitled.png 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 Untitled1.png 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.

kshitij89 · Answer

Guys, I am still not clear with the solution. By any way can you simplify it further ?

EgmatQuantExpert · Answer

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:

https://e-gmat.com/blogs/wp-content/uploads/2015/04/Capture1-e1429615808374.jpg

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

kshitij89 · Answer

Hi Harsh,

Thanks for your reply.

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

EMPOWERgmatRichC · Answer

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.

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

EgmatQuantExpert · Answer

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

BrainLab · Answer

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)

JeffTargetTestPrep · Answer

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.

Answer: E

generis · Answer

quadrants.png Rewrite the inequality in slope-intercept form for the inequality's boundary line 2x – 3y ≤ –6 -3y ≤ -2x - 6 Because dividing by -3, switch the sign: y \geq \frac{2}{3} x + 2 Slope is positive. Positive slope always runs through Quadrants I and III Eliminate answers B and D Find the intercepts to graph the line and/or to assess signs Set x equal to 0 to find y-intercept of the inequality's boundary line y \geq \frac{2}{3} (0) + 2 y \geq 2 Set y equal to 0 to find x-intercept of boundary line 0 \geq \frac{2}{3} x + 2 -2 \geq \frac{2}{3} x (\frac{3}{2})(-2) \geq{x} x ≤ - 3 Intercepts of the inequality's boundary line are (0,2) and (-3,0) Assess I graphed the intercepts, connected the points and extended the line. See diagram. No point in Quadrant IV satisfies the inequality You don't have to graph. (I graph often. For me, it is quick. This problem took just under a minute.) Quadrants II has positive y-values and negative x-values. -- y-intercept is positive -- x-intercept is negative Those signs fit: The boundary line runs through Quadrant II. Eliminate Answer C At this point, we know the graphed inequality's boundary line runs through Quadrants I, II, and III. All straight lines pass through at least two, and at most three, quadrants. (Most straight lines pass through three quadrants. Never four.) By POE, Quadrant IV contains no points that satisfy the inequality. AND/OR Quadrant IV has no positive y-values and no negative x-values. -- y-intercept is positive. The x-intercept is negative. -- Quadrant IV's (x, y) values are exactly the opposite: (+x, -y) No point in Quadrant IV satisfies this inequality Answer E

gagan0303 · Answer

is there any article on graphical representation of inequality and how to find the shaded region available on club?

Bunuel · Answer

I think you might find the following useful: https://gmatclub.com/forum/graphic-appr ... 68037.html https://gmatclub.com/forum/quick-way-to ... 76255.html https://gmatclub.com/forum/solving-quad ... 70528.html Hope it helps.

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In the rectangular coordinate system shown above, which quadrant, if

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