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05 May 2020, 04:35
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (01:46) correct
37%
(01:46)
wrong
based on 569
sessions
History
Date
Time
Result
Not Attempted Yet
What is the area of the region enclosed by \(2|x| + 3|y| = 6\) ?
A. 3
B. 4
C. 6
D. 8
E. 12
Are You Up For the Challenge: 700 Level Questions
A. 3
B. 4
C. 6
D. 8
E. 12
Are You Up For the Challenge: 700 Level Questions
05 May 2020, 05:11
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Bookmarks
Bunuel
Whenever you have such questions, choose x and y as 0, and find the other coordinates y and x respectively..
1) Take x as 0..
\(2|x| + 3|y| = 6\).....\(2|0| + 3|y| = 6\)\(...... 3|y| = 6......|y|=2......y=2..y=-2\)..Coord (0,-2) and (0,2)
2) Take y as 0..
\(2|x| + 3|y| = 6\).....\(2|x| + 3|0| = 6\)\(...... 2|x| = 6......|x|=3......x=3...x=-3\)..Coord (3,0) and (-3,0)
Now you can draw the figure as shown
Area = \(2*(\frac{1}{2}*(3+3)*2)=2*6=12\)
OR
the figure is a rhombus with equal sides and diagonals as 4 and 6..Area = \(\frac{4*6}{2}=12\)
E
Attachments
Gmatclub area.png [ 11.61 KiB | Viewed 9363 times ]
General Discussion
13 May 2020, 06:15
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Bunuel
Plug \(x = 0\) ; \(y =\) + \(2\) & If \(y = 0 ; x =\) + \(3\) and plot the graph
Attachment:
Capture.PNG [ 50.79 KiB | Viewed 8883 times ]
Now, \(△AOD = △AOB = △BOC = △COD\) ; where all the triangles are right angles (Intersection of X and Y axis at O)
So, Area of △AOD is \(\frac{1}{2}*3*2= 3\)
So, Area of the enclosed region is \(4*3 = 12\), Answer must be (E)
