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19 May 2019, 01:55
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
15% (02:50) correct
85%
(02:51)
wrong
based on 46
sessions
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For some value 'd', an isosceles triangle has at least one angle of (2d-40) degrees and at least one angle of (d+20) degrees. For each of such triangles, let x denote the smallest angle in degrees. Then what is the sum of all possible x's?
A. 50
B. 56
C. 106
D. 126
E. 150
A. 50
B. 56
C. 106
D. 126
E. 150
19 May 2019, 04:07
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nick1816
Three cases
1) angle that is twice is 2d-40
So 2(2d-40)+d+20=180...5d=240.....d=48
So angles are 2d-40=2*48-40=56 and d+20=48+20=68..
Least angle or X is 56..
2) angle that is twice is d+20
So (2d-40)+2(d+20)=180...4d=180......d=45
So angles are 2d-40=2*45-40=50 and d+20=45+20=65.
Least angle or X is 50..
3) angle that is twice is 2d-40 and d+20
So (2d-40)=d+20......d=60.
So angles are 2d-40=2*60-40=80 and d+20=60+20=80, so third angle is 180-80-80=20
Least angle or X is 20.
Sum becomes 56+50+20=126
D
General Discussion
23 Oct 2020, 21:33
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I think everyone forgot the 3rd Case.....
Great Question!
Great Question!
