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15 Feb 2022, 09:30
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
86% (01:46) correct
14%
(03:11)
wrong
based on 42
sessions
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Not Attempted Yet
Note: Not drawn to scale.
In the figure shown above, AB and DE are parallel. What is the value of y?
A. 110
B. 120
C. 135
D. 150
E. 160
Attachment:
2020-03-24_1744.png [ 18.89 KiB | Viewed 1626 times ]
24 Feb 2022, 07:21
Kudos
Bookmarks
Bunuel
Please refer to the attached fig. below:
Attachment:
Parallel lines GMAT.png [ 30.33 KiB | Viewed 1334 times ]
Draw line CE parallel to line AB and line DE bisecting angle \(y.\)
\(x+ \frac{y}{2} =180\) as these are supplementary angles.
\(\frac{y}{2} = 180-x\)
\(y =360-2x\)....(I)
We know sum of internal angles of a polygon \(= (n-2)180\), where \(n\) is the no. of sides.
For this \(5 \) sided fig. sum of internal angles = \( (5-2)180 = 540\)
Hence \(60+3x+360-2x= 540\)....using the value of angle \(y\) from (I)
\( x= 120\)
We know \(x+ \frac{y}{2} = 180 .....\) ( Supplementary angles)
\(120 + \frac{y}{2} = 180\)
\(y =120\)
Ans B
Hope it's clear.
