Answer
$R_{eq} =40\Omega$
Work Step by Step
let $R_1$ be the equivalent resistance of $10\Omega $ and $40\Omega $ which are in series
$R_1 =10\Omega +40\Omega =50 \Omega $
and let $R_2$ be the equivalent resistance of $55\Omega $ and $20\Omega $ which are also in series
$R_2=55\Omega +20\Omega =75\Omega $
now $R_1$ and $R_2$ are in parallel, let their equivalent is$R_3$
$\frac{1}{R_3} = \frac{1}{R_1}+ \frac{1}{R_2}$
$\frac{1}{R_3} = \frac{1}{50\Omega} + \frac{1}{75\Omega}$
$R_3 =30\Omega$
now $R_3$ and $10\Omega$ are in series
$R_{eq} =R_3 + 10\Omega$
$R_{eq} = 30\Omega+ 10\Omega$
$R_{eq} =40\Omega$
