Question 119484: The four oldest people in Golden City have lived a total of 384 years put together. The difference in ages for the youngest and the second oldest is 14. The second youngest is 3 years older than the youngest. The oldest is 20 years older than the average of the second oldest and youngest. What are their ages?
The ages are ____, _____, _____, and _____ .
Found 2 solutions by MaxLiebling, Edwin McCravy:
Answer by MaxLiebling(13)   (Show Source):
You can put this solution on YOUR website! 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
=====================
use (3):
B = A+3 = 85+3 = 88
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A+B+C = 85 + 88 + 99 = 272
==========================
use (1):
272 + D = 384
-> D = 384-272 = 112
====================
The ages are 85, 88, 99, and 112.
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Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website! The four oldest people in Golden City have lived a total of 384 years put together. The difference in ages for the youngest and the second oldest is 14. The second youngest is 3 years older than the youngest. The oldest is 20 years older than the average of the second oldest and youngest. What are their ages?
The ages are ____, _____, _____, and _____ .
Suppose their ages, youngest to oldest are x, y, z, and w,
That is x < y < z < w
>>...have lived a total of 384 years put together...<<
So
x + y + z + w = 384
>>...The difference in ages for the youngest and the second
oldest is 14...<<
so
z - x = 14
>>...The second youngest is 3 years older than the youngest.,,<<
so
y = x + 3
>>...The oldest is 20 years older than the average of the second oldest and youngest...<<
The average of the second oldest, z, and the youngest x, is 
So the oldest, w, is  or
w =
So we have the system of four equations in four unknowns:
x + y + z + w = 384
z - x = 14
y = x + 3
w =
Clear the last one of fractions by multiplying through by 2:
x + y + z + w = 384
z - x = 14
y = x + 3
2w = x+z+40
Solve the 2nd for z, getting z = 14+x
Substitute 14+x for z in the 4th, getting 2w = x+(14+x)+40, or
2w = x+14+x+40 or 2w = 2x+54. Then divide thru by 2, getting
w = x+27.
Now substitute x+3 for y, 14+x for z and x+27 for w, all
in the 1st:
x + y + z + w = 384
x + (x+3) + (14+x) + (x+27) = 384
x+x+3+14+x+x+27 = 384
4x+44 = 384
4x = 340
x = 85
y = x+3, so y = 85 + 3 = 88
z = 14+x, so z = 85+14 = 99
w = x+27 = 85 + 27 = 112
Their ages are 85, 88, 99, and 112.
Edwin
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