SOLUTION: The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4√3

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Question 633036: The age of the father of two children is twice that of the elder one added to four times that of the younger one. If the geometric mean of the ages of the two children is 4√3 and their harmonic mean is 6, then what is the father's age?
Answer by mananth(16946) About Me  (Show Source):
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Let the childrens' age be x & y , x >y
Father's age 2x+4y
xy= 48
Harmonic mean
%282%2F%281%2Fx%2B1%2Fy%29%29

%282%2F%28%28x%2By%29%2Fxy%29%29%29=6

2xy%2F%28x%2By%29+=6

xy = 48
2%2A48%2F%28x%2By%29= 6
x%2By=+96%2F6
x+y = 16
xy = 48

x=48/y
48%2Fy+%2B+y+=+16
multiply equation by y
48%2By%5E2=16y
y%5E2-16y%2B48=0
y%5E2-8x-6x%2B48=0
y%28y-8%29-6%28y-8%29=0
(y-8)(y-6)=0
y= 8 OR 6
xy = 48
therefore x=8
The sons' ages are 8 & 6
Father's age 2x+4y -> 2*8+4*6
=40 years
m.ananth@hotmail.ca