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Re M2519
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16 Sep 2014, 01:23
Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D
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Re: M2519
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06 Nov 2014, 18:06
Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D For this equation x+y=10. These points also satisfy the equation ( 5 , 5) ( 5 , 5 ) ( 5 ,5) ( 5, 5) . If i use this then the area would be 100.



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07 Nov 2014, 04:40
parameswaranprasad wrote: Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D For this equation x+y=10. These points also satisfy the equation ( 5 , 5) ( 5 , 5 ) ( 5 ,5) ( 5, 5) . If i use this then the area would be 100. That's not correct. Though these points lie on the graph of x+y=10, they are not the vertices of a cube you find the area of.
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Re: M2519
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08 Nov 2014, 02:57
Bunuel wrote: parameswaranprasad wrote: Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D For this equation x+y=10. These points also satisfy the equation ( 5 , 5) ( 5 , 5 ) ( 5 ,5) ( 5, 5) . If i use this then the area would be 100. That's not correct. Though these points lie on the graph of x+y=10, they are not the vertices of a cube you find the area of. Sorry, i am not getting it. Why is that we are not suppose to calculate area of this region ? Is the question to calculate max area?



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Re: M2519
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09 Nov 2014, 06:04
parameswaranprasad wrote: Bunuel wrote: parameswaranprasad wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D For this equation x+y=10. These points also satisfy the equation ( 5 , 5) ( 5 , 5 ) ( 5 ,5) ( 5, 5) . If i use this then the area would be 100. That's not correct. Though these points lie on the graph of x+y=10, they are not the vertices of a cube you find the area of. Sorry, i am not getting it. Why is that we are not suppose to calculate area of this region ? Is the question to calculate max area? The figure you get when graphing x+y=10. is below: Points ( 5 , 5) ( 5 , 5 ) ( 5 ,5) ( 5, 5) ARE indeed on that graph but they are NOT the vertices of the square (figure), so you cannot say that the side is 10 and the area is 100. Similar questions to practice: theareaboundedbythecurvesxy1andxy1is93103.htmlifequationx2y25enclosesacertainregion126117.htmlinthexyplanetheareaoftheregionboundedbythe86549.htmlifequationx2y25encloseacertainregion101963.htmlwhatistheareaoftheregionenclosedbylinesyxxy150487.htmlnewsetofmixedquestions150204100.html#p1208441Hope this helps.
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Re: M2519
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01 Dec 2014, 21:03
Diagonal is 5+5 = 10.. area = 100 ????



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15 Dec 2014, 08:20
After drawing the graph, you can see that each diagonal is equal to 20 units on the coordinate plane. A must memorize formula for the GMAT is \(d=s*\sqrt{2}\) where D=diagonal s=side of square. By plugging in 20 for d in this formula we can find the distance of one side of the square. > 20=s*\(\sqrt{2}\)>\(\frac{20}{\sqrt{2}}=s\)
s=10*\(\sqrt{2}\). So the area equals \(s^2\) or 200.



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04 Nov 2015, 16:00
I think this is a highquality question and the explanation isn't clear enough, please elaborate. I have no idea where 20 comes from and how exactly 10 and 10 is calculated for x and y



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22 Dec 2015, 09:46
I think this is a poorquality question and I don't agree with the explanation. If we calculate the side of the square formed it comes out to be _/50 Hence, area= _/50*_/50=50



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A much simpler way to solve this is to realize that this should be a rhombus because each vertex is either \((\pm10,0)\) or \((0,\pm10)\), hence distance between vertices should be same (ie, either \(\sqrt{(\pm10)^2 + (0)^2}\) or \(\sqrt{(0)^2 + (\pm10)^2}\))
Hence simply use area of rhombus = \(\frac{1}{2}(d1*d2)\) = \(\frac{1}{2}(20 * 20)\) = 200



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Re: M2519
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25 Feb 2018, 07:29
Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D Hi Bunuel, I answered this question incorreclty because that formula looks like that of a circle with radius of 10. abs(x)+abs(y)=10 ...can this be transferred to x^2+y^2 =10^2(circle formula) or is it the case that I can only square the entire left side, which is obviously not a circle?



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25 Feb 2018, 07:52
Kontaxis wrote: Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D Hi Bunuel, I answered this question incorreclty because that formula looks like that of a circle with radius of 10. abs(x)+abs(y)=10 ...can this be transferred to x^2+y^2 =10^2(circle formula) or is it the case that I can only square the entire left side, which is obviously not a circle? \((a + b)^2=a^2+2ab+b^2\), so if you square \(x+y=10\) you'll get \(x^2 + 2xy+y^2 = 100\) not \(x^2+y^2 =100\).
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26 Feb 2018, 01:34
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?
A. 20 B. 50 C. 100 D. 200 E. 400 IMO IS D again i will go with plugin values i.e if we try to find most extreme case we know that 5 is possible if we take (10,0),(10,0),(0,10),(0,10). Rest other points will lie under this figure and if we plot it on graph we will know that its a square whose diagonal is 20. so side is diagonal/root 2 which is 20/root2 and area is square of side. so 400/2 which is 200 option D



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Great question Bunuel. I think it will help to note that the four smaller triangles are 454590 triangle, as the slope is 10/10 = 1 i.e slope of 1. thus processing the last part will be much easier. So we it can be simplified as 101010 √2, therefore 10 √2 * 10 √2 = 200.



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22 Jun 2018, 08:09
+1 for option D. Draw the graph and check. The figure will lie on all four quadrants. The area will be (1/2*10*10*4)=200. Option D it is !
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28 Jun 2018, 03:17
Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D When I saw this question, i immediately thought that 10 would be the radius of a circle. So I saw x^2 + y^2 = 100. Is this out of the scope of the GMAT?



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yimmyyam wrote: Bunuel wrote: Official Solution: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 First of all to simplify the given expression multiply it be 2: \(x+y=10\). 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 \(x=0\)): \(y=0\) gives \(x=10\), so \(x=10\) and \(x=10\) \(x=0\) gives \(y=10\), so \(y=10\) and \(y=10\) Thus 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. Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\). Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\). Answer: D When I saw this question, i immediately thought that 10 would be the radius of a circle. So I saw x^2 + y^2 = 100. Is this out of the scope of the GMAT? The solution shows that it's not a circle (not because the equation of a circle is our of scope but because it's NOT a circle) but because it's a square. We don't have x^2 +y^2 = 100, we have \(x+y=10\). I guess you squared \(x+y=10\) but if do so you'll get x^2 + 2xy + y^2 = 100, not \(x+y=10\). Hop it's clear.
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