SOLUTION: The four oldest people in the Golden City have lived a total of 384 years put together. The difference in ages for the youngest and second oldest is 14. The second youngest is 3 ye

Question 173432: The four oldest people in the Golden City have lived a total of 384 years put together. The difference in ages for the youngest and 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. Find their ages and write from youngest to oldest.
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
let the ages from youngest to oldest be: a, b ,c and d respectively
:
a+b+c+d=384.....eq 1
b=a+3...........eq 2
c-a=14..........eq 3--->c=a+14 revised eq 3
d=((a+c)/2)+20..eq 4
:
take c's value from revise eq 3 and plug it into eq 4 so tht eq 4 is in terms of a. d=((a+(a+14))/2)+20--->d=(2a+14/2)+20...revised eq 4
:
now take value of d from revised eq 4 and the value of b from eq 2 and the value of c in revised eq 3 and plug them into eq 1
:
a+(a+3)+(a+14)+(2a+14)/2 +20=384:
:
3a+37+(2a+14)/2=384......multiply all terms by 2 to get rid of fraction.
:
6a+74+2a+14=768
:
8a=680
:
age of youngest
:
age of 2nd to the younges
:
age of 2nd to the oldest
:
age of oldest
:
that some healthy living there...lol

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