Question

Divide $Rs47$ among $A,B$ and $C$ so that $A$ may have $Rs10$ more then $B$ and $B$ has $Rs8$ more than $C$.

Answer 1

let $c=x$

so, $47=a+b+c$

$\Rightarrow 47=x+(x+8)+(x+8)+10$

because it is given that $c+8=b$ and $b+10=a$

$\Rightarrow 47=3x+26$

$\Rightarrow x=21$

$\Rightarrow x=7$

so, we get

$c=7$

$b=7+8=15$

$a=15+10=25$