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19 Aug 2018, 09:53
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:39) correct
24%
(02:14)
wrong
based on 358
sessions
History
Date
Time
Result
Not Attempted Yet
If \(d = 2c\) and \(e = \frac{1}{2}a\), what is x in terms of a, b, and c?
(A) \(\frac{3}{2}a + b + 3c - 540\)
(B) \(\frac{3}{2}a + b + 3c\)
(C) \(720 - \frac{3}{2}a - b - 3c\)
(D) \(720 - \frac{1}{2}a - b - 2c\)
(E) \(540 - \frac{1}{2}a - b - \frac{3}{2}c\)
Attachment:
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19 Aug 2018, 10:02
Kudos
Bookmarks
Bunuel
sum of internal angles of a polygon = (n-2)*180
n= 6
sum of angles = 720
now,
720=a+b+c+d+e+(180-x)
x=a+b+c+d+e-540
there is only one option with a -540 (a)
still
substituting the values
x=\(\frac{3}{2}a + b + 3c - 540\)
19 Aug 2018, 10:10
Kudos
Bookmarks
Bunuel
Given, \(d = 2c\) and \(e = \frac{1}{2}a\)
Sum of the interior angles of a polygon=(n-2)*180=(6-2)*180=720
We have, a+b+c+d+e+180-x=720
Or, a+b+c+d+e-x=720-180=540
Or, x=a+b+c+d+e-540.
Substituting the value of d and e,
\(x=a+b+c+2c+\frac{a}{2}-540\)
Or, \(x=\frac{3a}{2}+b+3c-540\)
Ans. (A)
