Question

If $\frac{1}{3}A=\frac{1}{4}B=\frac{1}{5}C$. If A:B:C is $x:y:z$ then find $x+y+z$

Answer 1

Let $\frac{1}{3}A=\frac{1}{4}B=\frac{1}{5}C=k$

$\Rightarrow A:B:C=3k:4k:5k=3:4:5$

$\Rightarrow x+y+z=3+4+5=12$