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If equation x/2 + y/2 = 5 enclose a certain region
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30 Sep 2010, 04:10
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If equation \(\frac{x}{2} + \frac{y}{2} = 5\) encloses a certain region on the coordinate plane, what is the area of this region? A. 20 B. 50 C. 100 D. 200 E. 400 M2519 ME: well, since \(x + y = 10\) ; X can range from (10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400
I think I am making a silly mistake some where but I just can't figure it out.
Thanks
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Re: CMAT Club Test Question  m25
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30 Sep 2010, 04:22
Barkatis wrote: Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19) If equation \(\frac{x}{2} + \frac{y}{2} = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400 OA: 200 ME: well, since \(x + y = 10\) ; X can range from (10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400 I think I am making a silly mistake some where but I just can't figure it out. Thanks Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear. \(\frac{x}{2} + \frac{y}{2} = 5\) You will have 4 case: \(x<0\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=10x\); \(x<0\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10+x\); \(x\geq{0}\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=x10\); \(x\geq{0}\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10x\); So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\) > \(area=side^2=200\). Answer: D. Check similar problem at: graphsmodulushelp86549.html?hilit=horizontal#p649401 it might help to get this one better. Hope it helps.
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Re: CMAT Club Test Question  m25
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30 Sep 2010, 17:41
X/2 + Y/2 = 5 so when x = 0, y= 10 when y=0 , x=10
so the sides of the enclosed area touches (0,10),(10,0),(0,10) and (10,0). so its a square having the diagonal =20unit So the area of the region = (20/1.414)^2 = 200




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Re: CMAT Club Test Question  m25
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16 Nov 2010, 09:45
x+y=10
put x=0
you get y=10 .....y=+10
when you y=0
you get x = 10 ..... x=+10
plot this on the coordinate plane.
you will get a rhombus area of rhombus = 1/2 (d1 x d2)= 20X20/2= 200



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If equation x/2 + y/2 = 5 enclose a certain region
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15 Jan 2012, 10:26
Apex231 wrote: If equation x/2+y/2 = 5 encloses a certain region on the coordinate plane, what is the area of this region?
A 20 B 50 C 100 D 200 E 400 First of all to simplify the given expression a little bit let's multiply it be 2: \(\frac{x}{2}+\frac{y}{2}=5\) > \(x+y=10\). Now, find x and y intercepts of the region (xintercept is a value(s) of x for y=0 and similarly yintercept is a value(s) of y for x=0): \(y=0\) > \(x=10\) > \(x=10\) and \(x=10\); \(x=0\) > \(y=10\) > \(y=10\) and \(y=10\). So we have 4 points: (10, 0), (10, 0), (0, 10) and (10, 0). When you join them you'll get the region enclosed by \(x+y=10\): You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\) > \(area=side^2=200\). Answer: D. Similar questions: http://gmatclub.com/forum/m065absolut ... 08191.htmlhttp://gmatclub.com/forum/graphsmodulu ... 86549.htmlHope it's clear. Attachment:
Enclosed region.gif [ 2.04 KiB  Viewed 28978 times ]
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Re: Area of region
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15 Jan 2012, 10:58
I had solved till this point  So we have 4 points: (10, 0), (10, 0), (0, 10) and (10, 0).
But instead of joining these points i did this  4 * (10 * 10) = 400 , which is wrong of course.
So when we join these points, how x+y = 10 stays satisfied , what's the maths behind it?



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Re: Area of region
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15 Jan 2012, 11:07
Apex231 wrote: I had solved till this point  So we have 4 points: (10, 0), (10, 0), (0, 10) and (10, 0).
But instead of joining these points i did this  4 * (10 * 10) = 400 , which is wrong of course.
So when we join these points, how x+y = 10 stays satisfied , what's the maths behind it? Given: \(x+y=20\) You will have 4 case: \(x<0\) and \(y<0\) > \(xy=10\) > \(y=10x\); \(x<0\) and \(y\geq{0}\) > \(x+y=10\) > \(y=10+x\); \(x\geq{0}\) and \(y<0\) > \(xy=10\) > \(y=x10\); \(x\geq{0}\) and \(y\geq{0}\) > \(x+y=10\) > \(y=10x\); So we have equations of 4 lines. If you draw these four lines you'll get the figure shown in my previous post. Hope it's clear.
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If equation x/2+y/2 = 5 encloses a certain region
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10 Sep 2012, 12:25
CMcAboy wrote: Can someone help me with this question:
If equation x/2 + y/2 = 5 encloses a certain region on the coordinate plane, what is the area of this region?
A) 20 B) 50 C) 100 D) 200 E) 400 I believe this is the simplest & the quickest solution x/2 + y/2 = 5 Put x = 0 in the above equation we get y/2 = 5, which means y= 10,  10 Put y = 0 in the above equation we get y/2 = 5, which means x= 10,  10 If you see plot these four points you get a square with two equal diagonals of length 20 units Thus area = 1/2 * (Diagonal)^2 > 1/2 * 400 = 200 I hope this will help many.
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Re: If equation x/2+y/2 = 5 encloses a certain region
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05 Dec 2012, 23:30
(1) derive all equations x+y = 10 xy = 10 x+y=10 x+y=10 (2) Get your x and y intercepts (0,10), (10,0) (0,10),(10,0) (0,10),(10,0) (0,10),(10,0) (3) You will have a square with a diagonal of 20 (4) Calculate area = \((10 * \sqrt{2})\sqrt{^2}\) = 200 Answer : D
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If equation x/2+y/2 = 5 encloses a certain region
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15 Oct 2014, 19:17
Apex231 wrote: If equation x/2+y/2 = 5 encloses a certain region on the coordinate plane, what is the area of this region?
A. 20 B. 50 C. 100 D. 200 E. 400 Hello There, Equation of a straight line whose x and y intercepts are a and b resp. is (x/a) + (y/b) = 1 i.e., coordinates of two ends of the line are (a,0) and (0,b). Now, from the given question, x/2+y/2 = 5, reducing this to intercept form we get, x/10+y/10 = 1 Considering the equation without modulus, coordinates are (10,0) and (0,10). Since there is modulus, other two coordinates are (10,0) and (0,10). Now coordinates (10,0), (0,10), (10,0) and (0,10) form a square with diagonal length = 20. Here diagonal length can be obtained by calculating the distance between (10,0) and (10,0) or (0,10) and (0,10). In a square, Diagonal = Side * sqrt(2) Side = 10 * sqrt(2) Area = Side * Side = 200. Ans : D Hope this helps! Thanks!
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Re: If equation x/2 + y/2 = 5 enclose a certain region
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10 Jun 2015, 09:21
Bunuel wrote: Barkatis wrote: Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19) If equation \(\frac{x}{2} + \frac{y}{2} = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400 OA: 200 ME: well, since \(x + y = 10\) ; X can range from (10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400 I think I am making a silly mistake some where but I just can't figure it out. Thanks Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear. \(\frac{x}{2} + \frac{y}{2} = 5\) You will have 4 case: \(x<0\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=10x\); \(x<0\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10+x\); \(x\geq{0}\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=x10\); \(x\geq{0}\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10x\); So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\) > \(area=side^2=200\). Answer: D. Check similar problem at: graphsmodulushelp86549.html?hilit=horizontal#p649401 it might help to get this one better. Hope it helps. hey Bunuel, I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method. We can rewrite the question as below \(x^2/4 +y^2/4 = 5\) (since \(x = x^2\)) \(x^2 + y^2 = 20\) This is the equation is a circle having the centre at (0,0) (general form is \(x^2 + y^2= r^2\)) area =\(3.14 * R^2\) = \(3.14 * 20\) = 62.8 What am i assuming wrong here?? Thanks!



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If equation x/2 + y/2 = 5 enclose a certain region
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10 Jun 2015, 09:50
arshu27 wrote: Bunuel wrote: If equation \(\frac{x}{2} + \frac{y}{2} = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400
OA: 200
I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method.
We can rewrite the question as below
\(x^2/4 +y^2/4 = 5\) (since \(x = x^2\))
\(x^2 + y^2 = 20\)
This is the equation is a circle having the centre at (0,0) (general form is \(x^2 + y^2= r^2\))
area =\(3.14 * R^2\) = \(3.14 * 20\) = 62.8
What am i assuming wrong here?? Thanks!
The part that I have highlighted above is WRONG which the first step in your solution x is NOT equal to x^2 for all values of x[/highlight] The Function "Modulus" only keeps the final sign Positive but that doesn't mean what you mentioned in the quoted Highlighted section. Alternatively you can solve this question in this way Step 1: Substitute y=0, \(\frac{x}{2} + \frac{0}{2} = 5\) i.e. \(\frac{x}{2} = 5\) i.e. \(x = 10\) i.e. \(x = +10\) So on the XY plane you get two Point (+10,0) and (10,0) Step 2:Substitute x=0, \(\frac{0}{2} + \frac{y}{2} = 5\) i.e. \(\frac{y}{2} = 5\) i.e. \(y = 10\) i.e. \(y = +10\) So on the XY plane you get two lines parallel to XAxis passing through Y=+10 and Y=10 So on the XY plane you get two Point (0, +10) and (0, 10) Join all the four points, It's a Square with Side \(10\sqrt{2}\) i.e. Area =\((10\sqrt{2})^2\) = 200
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If equation x/2 + y/2 = 5 enclose a certain region
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10 Jun 2015, 10:02
arshu27 wrote: Bunuel wrote: Barkatis wrote: Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19) If equation \(\frac{x}{2} + \frac{y}{2} = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400 OA: 200 ME: well, since \(x + y = 10\) ; X can range from (10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400 I think I am making a silly mistake some where but I just can't figure it out. Thanks Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear. \(\frac{x}{2} + \frac{y}{2} = 5\) You will have 4 case: \(x<0\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=10x\); \(x<0\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10+x\); \(x\geq{0}\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=x10\); \(x\geq{0}\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10x\); So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\) > \(area=side^2=200\). Answer: D. Check similar problem at: graphsmodulushelp86549.html?hilit=horizontal#p649401 it might help to get this one better. Hope it helps. hey Bunuel, I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method. We can rewrite the question as below \(x^2/4 +y^2/4 = 5\) (since \(x = x^2\)) \(x^2 + y^2 = 20\) This is the equation is a circle having the centre at (0,0) (general form is \(x^2 + y^2= r^2\)) area =\(3.14 * R^2\) = \(3.14 * 20\) = 62.8 What am i assuming wrong here?? Thanks! One More Clarification ( \(x is NOT equal to x^2\))Instead, \(x = \sqrt{(x^2)}\)
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Re: If equation x/2 + y/2 = 5 enclose a certain region
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12 Jun 2015, 14:43
why should I suppose that x or y equals + 10 & zeros ? what about + 5 as following : + 5 + 5 = 10 5++5 = 10 5+5= 10 +5++5=10 S0 we have Square with side of 10 length Its area is 100



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Re: If equation x/2 + y/2 = 5 enclose a certain region
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12 Jun 2015, 23:02
hatemnag wrote: why should I suppose that x or y equals + 10 & zeros ? what about + 5 as following : + 5 + 5 = 10 5++5 = 10 5+5= 10 +5++5=10 S0 we have Square with side of 10 length Its area is 100 Hi Hatemnag, The given equation is basically representing FOUR linear equations which are representing 4 lines on the plane
One Linear equation when x is +ve and y is +ve i.e. X+Y = 10 Second Linear equation when x is +ve and y is ve i.e. XY = 10 Third Linear equation when x is ve and y is +ve i.e. X+Y = 10 Forth Linear equation when x is ve and y is ve i.e. XY = 10
So you need to plot these equation and then take the area of Quadrilateral formed
Also, Please Note that Four Vertices of Quadrilateral are obtained where two lines Intersect, and The intersections of the lines are obtained at points (10,0), (10,0), (0,10) and (0,10)Whereas, what you have done is taking any FOUR RANDOM POINTS on those four lines as per your convenience and then have assumed that these points form the SquareI hope this clears your doubt!
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Re: If equation x/2 + y/2 = 5 enclose a certain region
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20 Jun 2015, 01:15
Bunuel wrote: Barkatis wrote: Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19) If equation \(\frac{x}{2} + \frac{y}{2} = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400 OA: 200 ME: well, since \(x + y = 10\) ; X can range from (10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400 I think I am making a silly mistake some where but I just can't figure it out. Thanks Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear. \(\frac{x}{2} + \frac{y}{2} = 5\) You will have 4 case: \(x<0\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=10x\); \(x<0\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10+x\); \(x\geq{0}\) and \(y<0\) > \(\frac{x}{2}\frac{y}{2}=5\) > \(y=x10\); \(x\geq{0}\) and \(y\geq{0}\) > \(\frac{x}{2}+\frac{y}{2}=5\) > \(y=10x\); So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\) > \(area=side^2=200\). Answer: D. Check similar problem at: graphsmodulushelp86549.html?hilit=horizontal#p649401 it might help to get this one better. Hope it helps. Sorry, i dont know what i am missing, how do i get the diagonal to be 20?...from the square i got, i have all the sides equal to 20, hence the area=400.



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Re: If equation x/2 + y/2 = 5 enclose a certain region
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20 Jun 2015, 01:53
jayanthjanardhan wrote: Sorry, i dont know what i am missing, how do i get the diagonal to be 20?...from the square i got, i have all the sides equal to 20, hence the area=400.
Hi Jayanthjanardan, The given equation is basically representing FOUR linear equations which are representing 4 lines on the plane One Linear equation when x is +ve and y is +ve i.e. X+Y = 10Second Linear equation when x is +ve and y is ve i.e. XY = 10Third Linear equation when x is ve and y is +ve i.e. X+Y = 10Forth Linear equation when x is ve and y is ve i.e. XY = 10NOTE: PLEASE PLOT THE LINES TO UNDERSTAND THE FIGURE (REFER THE FIGURE) and see that Diagonal of Square is 10So you need to plot these equation and then take the area of Quadrilateral formed Also, Please Note that Four Vertices of Quadrilateral are obtained where two lines Intersect, and The intersections of the lines are obtained at points (10,0), (10,0), (0,10) and (0,10) Whereas, what you have done is taking any FOUR RANDOM POINTS on those four lines as per your convenience and then have assumed that these points form the Square I hope this clears your doubt!
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Re: If equation x/2 + y/2 = 5 enclose a certain region
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20 Jun 2015, 03:01
Hi GMATInight, thanks a lot for the expalnantion. I get the logic now. However, kindly refer to this link below. inthexyplanetheareaoftheregionboundedbythe8654940.html#p1540033Its a similar problem, but the diagram we end getting is a square and not a rhombus...what am i missing here?



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If equation x/2 + y/2 = 5 enclose a certain region
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20 Jun 2015, 03:10
jayanthjanardhan wrote: Hi GMATInight, thanks a lot for the expalnantion. I get the logic now. However, kindly refer to this link below. inthexyplanetheareaoftheregionboundedbythe8654940.html#p1540033Its a similar problem, but the diagram we end getting is a square and not a rhombus...what am i missing here? Even that is a square but never forget that a Square is a specific type of Rhombus onlyI hope, You can understand that the Product of the slopes of the adjacent sides is 1 in that fugure which proves the angle between the adjacent sides as 90 degree a Square is a "Rhombus with all angles 90 degrees". So calling it a Rhombus won;t be wrong either but you are right about the figure being a Square.
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