Two oxides of metal contains 50% and 40% of metal $M$ respectively. If the formula of the first oxide is  $M{O}_2$. Then formula of the second oxide will be:

First oxide                     2nd oxide 
$M=50\%$                     $M=40\%$
$O=50\%$                    $O=60\%$
Formula $=M{O}_2$        formula = ?
Let the atomic weight of metal =x
$\% of m is M{O}_2=\frac{1\times x\times 100}{x+32}=50$

$x=32$

EF  of 2nd oxide.

$0=\frac{60}{16}=3.75 \frac{3.75}{1.25}=3$

$M=\frac{40}{32}=1.25 \frac{1.25}{1.25}=1$

Formula of 2nd oxide $=M{O}_3$