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25 Feb 2019, 02:17
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:37) correct
27%
(02:16)
wrong
based on 56
sessions
History
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Time
Result
Not Attempted Yet
In the diagram above, the clock shows that it is five minutes after 9 o’clock. If the radius of the clock is 9 inches, what is the area of the sector created by the hands of the clock starting at 9 and moving clockwise?
A. 6π
B. 13.5π
C. 27π
D. 81π
E. 54π
Attachment:
2019-02-25_1316.png [ 42.62 KiB | Viewed 2021 times ]
OhsostudiousMJ
Joined: 28 Jan 2019
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Location: India
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WE:Manufacturing and Production (Manufacturing)
25 Feb 2019, 02:47
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Area of sector = (pi)r^2*C/360
C = 120 degrees (angle intended at center)
r = 9 inches (given)
Substituting values, Area = 27(pi)
IMO C
C = 120 degrees (angle intended at center)
r = 9 inches (given)
Substituting values, Area = 27(pi)
IMO C
Archit3110
Major Poster
Updated on: 25 Feb 2019, 08:35
Originally posted by Archit3110 on 25 Feb 2019, 04:49.
Last edited by Archit3110 on 25 Feb 2019, 08:35, edited 2 times in total.
Last edited by Archit3110 on 25 Feb 2019, 08:35, edited 2 times in total.
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Bunuel
angle made by clock
11/2 * min ; 11* 5/2 ; ~ 28 * 90 + 28 or say 117.5 * ; ~ 118 *
118/360 * pi * 9^2 ; ~ 27 pi
IMO C
