Question 119484
I call them A(dam) [youngest], B(ert) [second youngest], C(laus) [second oldest] and D(iddl) [oldest].
total of 384 years: A+B+C+D = 384 (1)
The difference in ages for the youngest and the second oldest is 14:
C-A = 14 -> C = A + 14 (2)
The second youngest is 3 years older than the youngest:
B = A+3 (3)
The oldest is 20 years older than the average of the second oldest and youngest:
D = (C+A)/2 + 20 (4)
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A+B+C+D = 384 
use (3): 384 = A+B+C+D = A+ A+3 +C+D = 2A+C+D+3
use (4): 384 = 2A+C+D+3 = 2A+C+ (C+A)/2 + 20 +3 = 2A+C +C/2 + A/2 +23
-23 (384-23=361), than multiply by 2: 722 = 4A+2C + C + A = 5A + 3C
use (2): 722 = 5A + 3*(A+14) = 5A+3A+42
-42: 680 = 8A
-> A = 680/8 = 85
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use (2):
C = A+14 = 85+14 = 99
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use (3):
B = A+3 = 85+3 = 88
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A+B+C = 85 + 88 + 99 = 272
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use (1):
272 + D = 384
-> D = 384-272 = 112
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The ages are 85, 88, 99, and 112.
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