Question

Divide 243 into 3 parts such that half of the first part, one-third of the second part and one -fouth of the third part are all equal

Answer 1

let first part be x
second be y
third be z

According to condition,
x/2=y/3=z/4

by theorem on equal ratios
(x+y+z)/(2+3+4)=k

Also, x/2=y/3=z/4=k

therefore,
(x+y+z)=9k
but
x+y+z= 243

therefore
9k=243
hence
k=27

x=54
y=81
z=108

Answer 2

let first part be x
second by y
third by z
A/Q
x/2=t/3=z/4
by theorem on equal ration
(x+t+z)/(2+3+4)=k
also,x/2= y/3=z/4=k
therefore
(x+y+z=243
therefor
9k=243
k= 27
hence
x= 34
y= 81
z=108