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The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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21 Apr 2010, 21:09
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The area bounded by the curves x + y = 1 and x  y = 1 is A. 3 B. 4 C. 2 D. 1 E. None
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Re: area bounded by the curves [#permalink]
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21 Apr 2010, 22:05
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anni wrote: The area bounded by the curves \(x + y = 1,\) \(x  y = 1\) is
A.3 B.4 C.2 D.1 E. None
please, help how to solve ? hi, these equations represent four different equations x+y=1 x+y=1 xy=1 xy=1 once you plot these on the graph, you will easily find the area under these equations. just for the record, the answer shall be 2.
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21 Apr 2010, 22:12
fivezero7 wrote: anni wrote: The area bounded by the curves \(x + y = 1,\) \(x  y = 1\) is
A.3 B.4 C.2 D.1 E. None
please, help how to solve ? hi, these equations represent four different equations x+y=1 x+y=1 xy=1 xy=1 once you plot these on the graph, you will easily find the area under these equations. just for the record, the answer shall be 2. thank you for the reply, is it required to plot these lines and then calculate? can we get directly i mean is there any formula to solve this problem? thanks



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21 Apr 2010, 22:20
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anni wrote: thank you for the reply, is it required to plot these lines and then calculate? can we get directly i mean is there any formula to solve this problem? thanks hi anni, there is no need to plot it, once you have mastered the art of visualization. the way i did it is as under. take x+y=1 it intersects the axes at (1,0) and (0,1) and makes a right angled triangle with the axes with each side (other than the hypotenuse) as 1 hence the area of this triangle is 0.5*1*1 = 0.5 sq. units. all other lines are symmetrical and form 3 more congruent triangles with the axes at different points. so the total area will be 0.5+0.5+0.5+0.5 = 2 sq units. hope i am clear
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Re: area bounded by the curves [#permalink]
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25 Feb 2012, 00:06
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fivezero7 wrote: anni wrote: thank you for the reply, is it required to plot these lines and then calculate? can we get directly i mean is there any formula to solve this problem? thanks hi anni, there is no need to plot it, once you have mastered the art of visualization. the way i did it is as under. take x+y=1 it intersects the axes at (1,0) and (0,1) and makes a right angled triangle with the axes with each side (other than the hypotenuse) as 1 hence the area of this triangle is 0.5*1*1 = 0.5 sq. units. all other lines are symmetrical and form 3 more congruent triangles with the axes at different points. so the total area will be 0.5+0.5+0.5+0.5 = 2 sq units. hope i am clear Very clear, and amazed to see how much was packed under that question. But, am curious if there's any other way to solve the problem?



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Re: area bounded by the curves [#permalink]
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25 Feb 2012, 01:24
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fortsill wrote: fivezero7 wrote: anni wrote: thank you for the reply, is it required to plot these lines and then calculate? can we get directly i mean is there any formula to solve this problem? thanks hi anni, there is no need to plot it, once you have mastered the art of visualization. the way i did it is as under. take x+y=1 it intersects the axes at (1,0) and (0,1) and makes a right angled triangle with the axes with each side (other than the hypotenuse) as 1 hence the area of this triangle is 0.5*1*1 = 0.5 sq. units. all other lines are symmetrical and form 3 more congruent triangles with the axes at different points. so the total area will be 0.5+0.5+0.5+0.5 = 2 sq units. hope i am clear Very clear, and amazed to see how much was packed under that question. But, am curious if there's any other way to solve the problem? The area bounded by the curves x + y = 1, x  y = 1 is A. 3 B. 4 C. 2 D. 1 E. None x+y=1 represents two lines: x+y=1 and x+y=1 > y=1x and y=1x. Find the x and y intercept of these lines to plot; xy=1 represents two lines: xy=1 and xy=1 > y=x1 and y=x+1. Find the x and y intercept of these lines to plot; Notice that these lines are mirror images of each other. Here is a square you get when you plot them: Attachment:
Area.gif [ 4.93 KiB  Viewed 36180 times ]
Notice that the diagonal of this square is equal to 2 (the difference between x intercepts). Area of a square is diagonal^2/2=2^2/2=2. Answer: C. Check similar questions to practice: m065absolutevalue108191.htmlgraphsmodulushelp86549.htmlm06q572817.htmlifequationenclosesacertainregion110053.htmlareaofregion126117.htmlHope it helps.
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Re: The area bounded by the curves x + y = 1, x  y = 1 is [#permalink]
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12 Apr 2012, 08:17
anni wrote: The area bounded by the curves x + y = 1, x  y = 1 is
A. 3 B. 4 C. 2 D. 1 E. None
please, help how to solve ? One more method: solve following equns: X+Y=1 Xy=1 Xy=1 X+y=1 Out put X= +_ 1 Y=+_1 Plot the values of X and Y on graph, you will see the square Now use pythagoras thm to find diagonal,which will be the side of that square. = Sqaure root 2 square it and ans will be 2



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Re: The area bounded by the curves x + y = 1, x  y = 1 is [#permalink]
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22 Sep 2012, 18:11
the question threw me away when it said "curves" and I started thinking about parabola, It should have said "lines" instead.
@Bunuel, Does gmat confuses us with this kind of language?
thanks!



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Re: The area bounded by the curves x + y = 1, x  y = 1 is [#permalink]
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05 Dec 2012, 21:27
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(1) Derive all equations x + y = 1 eq1: x + y = 1 eq2: x + y = 1 x  y = 1 eq3: x  y = 1 eq4: y  x = 1 (2) Plot your graph using x=0 and y=0. eq1: 0,1 and 1,0 eq2: 0,1 and 1,0 eq3: 01 and 1,0 eq4: 0,1 and 1,0 (3) You will recognize a region that is a square with a diagonal of 2 (4) Calculate the area. diagonal = side * \(\sqrt{2}\) side = \(\frac{2}{\sqrt{2}}\) side = \(\sqrt{2}\) Area = \(side^2\) = \(\sqrt{2}^2\) = \(2\) For detailed solutions for other similar problems. http://burnoutorbreathe.blogspot.com/2012/12/absolutevaluessolvingforareaof.html
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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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04 Aug 2014, 00:06
anni wrote: The area bounded by the curves x + y = 1 and x  y = 1 is
A. 3 B. 4 C. 2 D. 1 E. None The question seems confusing at first since it said the curves. But if you know how to deal with absolute values, you can come up with the solution pretty quickly. C
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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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04 Sep 2015, 05:11
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREwhy only (0,1) (1,0) should only pick? for x+y=1, if i chose (10,9) then it still become 1, can you please explain me what exactly the method of chosing number while finding area for this type of math



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The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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29 Dec 2015, 06:47
Hi Bunuel, just a question, I am getting confused with this. Can't we just find all the solutions for x and y by using just one equation? If we take the first equation: x + y = 1 and say that y = 0 then x = 1 or 1 and if x = 0 then y =1 and 1 Would that be incorrect? And if that's the case, what is the difference between this equation x + y = 1 and this one x+y=1 Thanks a lot
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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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anni wrote: The area bounded by the curves x + y = 1 and x  y = 1 is
A. 3 B. 4 C. 2 D. 1 E. None x+y=1 > y=x+1 > slope 1. x+y=1 > y=x1 > slope 1. we have 2 parallel lines. xy=1 > y=x1 => slope 1. xy=1 > y=x+1 > slope 1. > we have another 2 parallel lines. i simply drew the lines, and for the sake of getting the image, try x=1 then y=1 for all the equations. i got a square like shape, with the points of intersection at (1; 0); (0; 1); (1; 0); (0; 1) since the diagonal of the square is the x axis, and it has a length of 2, we can apply the 454590 triangle rule, and see that 2=x*sqrt(2), where x is the side of the square. the side of the square is sqrt(2). now, we need to find the area > sqrt(2) squared is equal to 2.



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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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23 Mar 2016, 19:03
anik19890 wrote: Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREwhy only (0,1) (1,0) should only pick? for x+y=1, if i chose (10,9) then it still become 1, can you please explain me what exactly the method of chosing number while finding area for this type of math you would find more points on THE LINE, but we are asked for the area of the figure when the 4 lines intersect. the 4 points of intersection are as shown in the figure in bunuel's post.



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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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06 Aug 2017, 08:56
Excellent question... Tested understanding of Inequalities, Geometry and visualization... Is it a Q 51 level question?
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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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19 Aug 2017, 07:56
I have a query with the wording of the question it says : The area bounded by the curves x + y = 1, x  y = 1 is A. 3 B. 4 C. 2 D. 1 E. None
Based on answers here it appears we consider them as four straight lines (not curves). Curves usually are written by x^2 + y^2 format; but since it said curves I thought it would be circular> so in case its a non  square in the equation;I assume it to be lines not curves?? Will keep in mind in case I come across such language/ format later



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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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30 Aug 2017, 08:49
Answer is C:An easy question instead <700 probably x+y=1 , gives 2 equations x+y = + 1 and x+y =1 xy =1 gives xy=1 and xy = 1 we know they will form an enclosed figure on x and y axis. so directly find the points = put x=0 and y=0 we get (0,1) (0,1) (1,0) (1,0) when u actually plot the distance between opposite points are = 2 on each diogonal which cut at 90 degrees at O , origin there fore A = 0.5 D1D2 = 0.5x2x2 = 2 which is C
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Re: The area bounded by the curves x + y = 1 andx  y = 1 is [#permalink]
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05 Feb 2018, 21:59
anni wrote: The area bounded by the curves x + y = 1 and x  y = 1 is
A. 3 B. 4 C. 2 D. 1 E. None The total Area = \(4*\frac{1}{2}*1*1\)
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