Deserted long division 4

$\LARGE{ \begin{array}{rlll} \phantom{0}\ \boxed{\color{#EC7300}{1}} \ \boxed{\color{#EC7300}{2}} && \\[-3pt] \boxed{\color{#EC7300}{4}}\ \boxed{\color{#EC7300}{7}} \enclose{longdiv}{\boxed{5} \ \boxed{\color{#EC7300}{6}} \ \boxed{\color{#EC7300}{4}} } \\[-3pt] \underline{\boxed{\color{#EC7300}{4}} \ \boxed{\color{#EC7300}{7}} \, \phantom0 \phantom0} \\[-3pt] \boxed{\color{#EC7300}{9}} \ \boxed{\color{#EC7300}{4}} && \\[-2pt] \underline{\boxed9 \ \boxed{4}} && \\[-2pt] \end{array} }$

Sum= $\color{#EC7300}{1+2+6+4+7+6+4+4+7+9+4}=\huge\boxed{\color{#007fff}{48}}$
Hint:- By looking at $94$ we can easily identify the divisor as $47$ after which the question is really easy. :-)