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Difficulty:
95%
(hard)
Question Stats:
44% (02:05) correct
56%
(02:12)
wrong
based on 286
sessions
History
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Let S be the set of all points (x, y) in the xy plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then what is the area of the region represented by S?
A. 1/2
B. 1
C. 2
D. 3
E. 4
_________________
A. 1/2
B. 1
C. 2
D. 3
E. 4
_________________
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
Given that S be the set of all points (x, y) in the xy plane such that |x| + |y| ≤ 2 and |x| ≥ 1 and we need to find what is the area of the region represented by S
Plotting |x| + |y| ≤ 2
We will have four case for values of x and y each being ≥ 0 and ≤ 0 respectively (in four quadrants)
Mod x + Mod y LTEQ 2.jpg [ 32.92 KiB | Viewed 7362 times ]
Plotting |x| ≥ 1
We will have two cases x≥0 and x<0
Mod x + Mod y LTEQ 2 intersection with mod x GTEQ 1.jpg [ 21.96 KiB | Viewed 7175 times ]
Point of intersection of both the graphs
There will be four points of intersections
So, common area between the two graphs will be given by the two shaded triangles shown below:
Common area.jpg [ 23.42 KiB | Viewed 7032 times ]
Area of each triangle = \(\frac{1}{2}\) * Base * Height = \(\frac{1}{2}\) * 2 * 1 = 1
=> Total area of two triangles= 2*1 = 2 units
So, Answer will be C
Hope it helps!
Watch the following video to learn Basics of Absolute Values
_________________
Given that S be the set of all points (x, y) in the xy plane such that |x| + |y| ≤ 2 and |x| ≥ 1 and we need to find what is the area of the region represented by S
Plotting |x| + |y| ≤ 2
We will have four case for values of x and y each being ≥ 0 and ≤ 0 respectively (in four quadrants)
- 1. x≥0, y≥0 => |x| = x and |y| = y => x + y ≤ 2 (Quadrant 1). We will plot the line x + y = 2 and will consider the area below the line as we have ≤ sign
- 2. x≤0, y≥0 => |x| = -x and |y| = y => -x + y ≤ 2 (Quadrant 2). We will plot the line -x + y = 2 and will consider the area below the line as we have ≤ sign
3. x≤0, y≤0 => |x| = -x and |y| = -y => -x - y ≤ 2 (Quadrant 3) => x + y ≥ -2. We will plot the line x + y = -2 and will consider the area above the line as we have ≥ sign
4. x≥0, y≤0 => |x| = x and |y| = -y => x - y ≤ 2 (Quadrant 4) => y ≥ x - 2. We will plot the line x - y = 2 and will consider the area above the line as we have ≥ sign
Attachment:
Mod x + Mod y LTEQ 2.jpg [ 32.92 KiB | Viewed 7362 times ]
Plotting |x| ≥ 1
We will have two cases x≥0 and x<0
- 1. x≥0 => |x| = x => x ≥ 1 and Intersection of x≥0 and x ≥ 1 equals x ≥ 1 => Area will be to the right of x = 1
2. x<0 => |x| = -x => -x ≥ 1 => x ≤ - 1 and Intersection of x<0 and x ≤ - 1 equals x ≤ - 1 => Area will be to the left of x = -
Attachment:
Mod x + Mod y LTEQ 2 intersection with mod x GTEQ 1.jpg [ 21.96 KiB | Viewed 7175 times ]
Point of intersection of both the graphs
There will be four points of intersections
- 1. x=1 and x+y=2 will intersect at (1,1)
2. x=-1 and -x+y=2 will intersect at (-1,1)
3. x=-1 and -x-y=2 will intersect at (-1,-1)
4. x=1 and x-y=2 will intersect at (1,-1)
So, common area between the two graphs will be given by the two shaded triangles shown below:
Attachment:
Common area.jpg [ 23.42 KiB | Viewed 7032 times ]
Area of each triangle = \(\frac{1}{2}\) * Base * Height = \(\frac{1}{2}\) * 2 * 1 = 1
=> Total area of two triangles= 2*1 = 2 units
So, Answer will be C
Hope it helps!
Watch the following video to learn Basics of Absolute Values
_________________
How to Solve :
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
General Discussion
MARVEL13
Joined: 08 Mar 2020
Last visit: 13 Oct 2020
Posts: 32
Location: India
Schools: Ivey '21 HEC Montreal '21
WE:Corporate Finance (Non-Profit and Government)
Schools: Ivey '21 HEC Montreal '21
Posts: 32
for the condition this to be true lxl +lyl <=2, the area will be a square with vertices (-2,0)(2,0)(0,-2)(0,2)
Now we need to subtract the the area which does not satisfy the condition lxl >=1
Area of Set will be the area of Non Overlapping Part
2 x (1/2) x base x height
2 x (1/2) x 1 x 2 = 2
Now we need to subtract the the area which does not satisfy the condition lxl >=1
Area of Set will be the area of Non Overlapping Part
2 x (1/2) x base x height
2 x (1/2) x 1 x 2 = 2
Attachments
Area.PNG [ 11.53 KiB | Viewed 11626 times ]
TheMongoose
Joined: 21 Sep 2012
Last visit: 15 Jun 2020
Posts: 8
Location: United States
Concentration: Operations, General Management
GPA: 3.5
Posts: 8
Possible values are (1,1);(1,-1);(-1,-1);(-1,1);(-2,0)and(2,0)
So there are two triangle each with an area of 1 so 1+1 = 2
So there are two triangle each with an area of 1 so 1+1 = 2
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