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# The area of the region bounded by the curves y = |x - 2|, x = 1, x =

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Manager
Joined: 11 Dec 2013
Posts: 114
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)
The area of the region bounded by the curves y = |x - 2|, x = 1, x =  [#permalink]

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Updated on: 01 Feb 2019, 03:23
3
00:00

Difficulty:

65% (hard)

Question Stats:

52% (02:59) correct 48% (02:16) wrong based on 28 sessions

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The area of the region bounded by the curves y = |x - 2|, x = 1, x = 5 and the x-axis is _________ sq. units

A) 1
B) 2
C) 3
D) 4
E) 5

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Originally posted by 4d on 28 Jan 2019, 22:38.
Last edited by Bunuel on 01 Feb 2019, 03:23, edited 1 time in total.
Renamed the topic and edited the question.
Director
Joined: 18 Jul 2018
Posts: 675
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: The area of the region bounded by the curves y = |x - 2|, x = 1, x =  [#permalink]

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28 Jan 2019, 23:53
When x = 1, y = |x-2| = 1
x = 5, y = 3
Plotting this on graph we get,
Base AB = 4.
Height BC = 2
Area = 1/2*AB*BC = 4

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Manager
Joined: 12 Apr 2011
Posts: 73
Location: United Arab Emirates
Concentration: Strategy, Marketing
Schools: CBS '21, Yale '21, INSEAD
GMAT 1: 670 Q50 V31
WE: Marketing (Telecommunications)
Re: The area of the region bounded by the curves y = |x - 2|, x = 1, x =  [#permalink]

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01 Feb 2019, 03:20
4d wrote:
The area of the region bounded by the curves $$y = \left | x-2 \right | , x = 1, x = 5$$ and the $$x-axis$$ is _________ sq. units
A) 1
B) 2
C) 3
D) 4
E) 5

As we can see the 4 lines in the rough image below.

Two triangles are formed by the four lines.

Area of ABC = 1/2 * b * h = 1/2 * AB * BC = 1/2 * 1 * 1 = 1/2
Area of CED = 1/2 * CD * ED = 1/2 * 3 * 3 = 9/2

Total Area = 1/2 + 9/2 = 10/2 = 5

Hence the correct answer is E.
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File comment: Rough Image

57C099A0-958F-49EE-89C1-D870CA48185F.jpeg [ 613.36 KiB | Viewed 210 times ]

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Re: The area of the region bounded by the curves y = |x - 2|, x = 1, x =   [#permalink] 01 Feb 2019, 03:20
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# The area of the region bounded by the curves y = |x - 2|, x = 1, x =

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