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M06#5 Absolute value [#permalink]
 23 Jan 2011, 20:37
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100% (01:48) correct
0% (00:00) wrong based on 2 sessions
If equation |X| + |Y|= 5 encloses a certain region on the coordinate plane, what is the area of that region? 5 10 25 50 100 The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .
The correct answer is D. How do we know that the area is square and not something else ? 
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Re: M06#5 Absolute value [#permalink]
24 Jan 2011, 03:27
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shrive555 wrote: If equation |X| + |Y|= 5 encloses a certain region on the coordinate plane, what is the area of that region? 5 10 25 50 100 The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .
The correct answer is D. How do we know that the area is square and not something else ? Below is the the region we get by joining X and Y intercepts which are (0, 5), (0, -5), (5, 0), (-5, 0): Attachment: 
MSP254019e3df2hii74417g00000g7c2817hhbe0a3b.gif [ 2.86 KiB | Viewed 982 times ]
Now, diagonals of the rectangle are equal (10 and 10), and also are perpendicular bisectors of each other (as the are on X and Y axises), so it must be a square. Area of a square equals to diagonal^2/2=10^2/2=50. Answer: D. Similar question: cmat-club-test-question-m25-101963.htmlgraphs-modulus-help-86549.htmlHope it helps.
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Re: M06#5 Absolute value [#permalink]
25 Jan 2011, 09:51
Thanks B one more question please whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B) For B we find the area through diagonal and for A we simply take one side and square it to get the area.
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Re: M06#5 Absolute value [#permalink]
25 Jan 2011, 10:01
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shrive555 wrote: Thanks B
one more question please
whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)
For B we find the area through diagonal and for A we simply take one side and square it to get the area. You can find the area of a square in either way: area_{square}=side^2=\frac{diagonal^2}{2}. The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees. |x+y|+|x-y|=4: Attachment: 
MSP86219e41c890f2h195700001928g4634f4e6h4h.gif [ 1.79 KiB | Viewed 946 times ]
|x|+|y| =5: Attachment: 
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif [ 2.74 KiB | Viewed 945 times ]
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Re: M06#5 Absolute value [#permalink]
26 Feb 2011, 16:40
Bunuel, why cant we assume other values for |x| + |y| = 5 it could also be |-3| + |8|...please advise. i may have a concept wrong
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Re: M06#5 Absolute value [#permalink]
26 Feb 2011, 17:42
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ajit257 wrote: Bunuel, why cant we assume other values for |x| + |y| = 5
it could also be |-3| + |8|...please advise. i may have a concept wrong We are not assuming the values for x and y. We are expanding |x| + |y| = 5 to get the figure. If x>0 and y>0 we'll have x+y=5; If x>0 and y<0 we'll have x-y=5; If x<0 and y>0 we'll have -x+y=5; If x<0 and y<0 we'll have -x-y=5; So we have 4 lines (4 linear equations which will give 4 segments) when drawn these lines give the square shown in my original post. But there is an easier way to get this figure: get X and Y intercepts (which are (0, 5), (0, -5), (5, 0), (-5, 0)) and join them. Follow the links in my previous posts for more.
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Re: M06#5 Absolute value [#permalink]
26 Feb 2011, 17:45
Bunuel...you're the best....kudos to you
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If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?
A) 5
B) 10
C) 25
D) 50
E) 100
I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:
e.g x=2 then y=3 or y=-3
x=1 y=4 or -4
I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?
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Manager
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Bunuel wrote: BN1989 wrote: If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?
A) 5
B) 10
C) 25
D) 50
E) 100
I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:
e.g x=2 then y=3 or y=-3
x=1 y=4 or -4
I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle? Merging similar topics. Please ask if anything remains unclear. Yes, why can't it be a circle?
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GMAT Club team member
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BN1989 wrote: Bunuel wrote: BN1989 wrote: If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?
A) 5
B) 10
C) 25
D) 50
E) 100
I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:
e.g x=2 then y=3 or y=-3
x=1 y=4 or -4
I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle? Merging similar topics. Please ask if anything remains unclear. Yes, why can't it be a circle? Well it cannot be a circle because it's a square. The reason why it's a square is given in previous posts (you can also see a diagram there). Next, in an xy plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: (x-a)^2+(y-b)^2=r^2. There is no resemblance whatsoever between this equation and the equation given in the question. Hope it's clear.
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
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