Explanation

(a) All the resistances are in parallel order, so voltage across them will be equal.
$\therefore\ \ \ \ \ 60I = (15 + 5)I_{1}$
$\Longrightarrow \ \ 60I = 20I_{1}$
$\Longrightarrow \ \ \ I_{1} = 3I$
$Again\ \ \ \ (15 + 5)I_{1} = 10\left( 1 - I - I_{1} \right)$
$\Longrightarrow \ \ \ 2I_{1} = 1 - I - I_{1}$
$\Longrightarrow \ \ \ \ \ 2(3I) = 1 - I - 3I$
$\Longrightarrow \ \ \ \ \ 6I + 4I = 1$
$\Longrightarrow \ \ \ \ 10I = 1\ $
$\therefore\ \ \ \ \ \ \ \ I = \frac{1}{10} = 0.1A$