Question 173432
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
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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
:
{{{highlight(a=85)}}}age of youngest
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{{{highlight(b=a+3=85+3=88)}}}age of 2nd to the younges
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{{{highlight(c=a+14=85+14=99)}}} age of 2nd to the oldest
:
{{{highlight(d=(a+c)/2+20=(85+99)/2+20=112)}}}age of oldest
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that some healthy living there...lol