Xavier Aptitude Test 2022 Paper

We have
$3.1416141614161416......$
$=3+0.1416+0.00001416+0.000000001416.....$
$3+\frac{1416}{{10}^4}+\frac{1416}{{10}^8}+\frac{1416}{{10}^{12}}+.....$
Now excluding 3 we get a series with infinite Geometric progression such that first term is $\frac{1416}{{10}^4}$ and common ratio is $\frac{1}{{10}^4}$
Therefore we get sum as $\frac{\mathit{a}}{1-\mathit{r}}$
$\frac{\frac{1416}{{10}^4}}{1-\frac{1}{{10}^4}}$
we get sum as $\frac{1416}{{10}^4-1}$
$=\frac{1416}{9999}$
Now adding 3 we get value as :
$\frac{31413}{9999}$
Now taking 3 common we get ratio of $\mathrm{A}:\mathrm{B}$ as
$=\frac{10471}{3333}$
So $\mathrm{A}-\mathrm{B}$ will be $=7138$