If $a/b = 1/3$ , $a/c=1/9$ and $a+b+c=26$ find $a,b,c$
I solved it by making in the first ratio
$a=m$
$b=3m$
and in the second ratio
$a=n$
$c=9n$
since $a=m , a=n$ then $m=n$
so $c=9m$ (since m=n)
so we deduce that $a=m,b=3m,c=9m$ and we can solve it easily as equation
but my teacher said it is wrong and I do not know why, can you explain ?, or is it right and my teacher is wrong?
2026-03-29 03:19:09.1774754349
proportion and ratio solving by m,n question
27 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
I see nothing wrong with your calculation, though it is a bit inefficient. You don't need the $m,n$ for instance. You could just write $$\frac ab=\frac 13\implies b=3a\quad \&\quad \frac ac=\frac 19\implies c=9a$$
Then conclude as $$a+b+c=26\implies a+3a+9a=13a=26\implies a=2, b=6, c=18$$
But you get there in the end.