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M06#5 Absolute value

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M06#5 Absolute value [#permalink] New post 23 Jan 2011, 19:37
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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

[Reveal] Spoiler:
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 ? :roll:
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Re: M06#5 Absolute value [#permalink] New post 24 Jan 2011, 02:27
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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

[Reveal] Spoiler:
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 ? :roll:


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
MSP254019e3df2hii74417g00000g7c2817hhbe0a3b.gif [ 2.86 KiB | Viewed 1794 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.html
graphs-modulus-help-86549.html

Hope it helps.
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Re: M06#5 Absolute value [#permalink] New post 25 Jan 2011, 08:51
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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] New post 25 Jan 2011, 09:01
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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
MSP86219e41c890f2h195700001928g4634f4e6h4h.gif [ 1.79 KiB | Viewed 1756 times ]

|x|+|y| =5:
Attachment:
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif [ 2.74 KiB | Viewed 1755 times ]

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Re: M06#5 Absolute value [#permalink] New post 26 Feb 2011, 16:42
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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] New post 26 Feb 2011, 15:40
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] New post 26 Feb 2011, 16:45
Bunuel...you're the best....kudos to you
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m06#5 [#permalink] New post 22 Apr 2012, 05:26
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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Re: m06#5 [#permalink] New post 22 Apr 2012, 05:31
Expert's post
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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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., 11 New DS set.


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Re: m06#5 [#permalink] New post 22 Apr 2012, 05:44
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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Re: m06#5 [#permalink] New post 22 Apr 2012, 05:52
Expert's post
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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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., 11 New DS set.


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Re: M06#5 Absolute value [#permalink] New post 14 Apr 2014, 21:04
Hi Bunuel,

I understood your concept of equations and plotting points. But, i have a small doubt.

Why can't the graph in the second question (|x|+|y| =5 )be a horizontal square than a square turned by 45 degrees in the first question (|x+y|+|x-y|=4)

My question is there any reason why in the second question we have plotted a graph which is turned by 45 degrees rather than a horizontal square.

Help is appreciated.

Thanks in advance :)

Bunuel wrote:
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

|x|+|y| =5:
Attachment:
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif
Re: M06#5 Absolute value   [#permalink] 14 Apr 2014, 21:04
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