Explanation

(b)
Concentration of mixture
$= \frac{conc\ of\ acid \times its\ vol - conc\ of\ base \times its\ vol}{total\ volume}$
$N = \frac{N_{1}V_{1} - N_{2}V_{2}}{V_{1} + V_{2}}$
$= \frac{50 \times 1 - 30 \times 1}{50 + 30}$
$= \frac{20}{80}$
$= \frac{1}{4} = 0.25$
Since, HCl is in excess, so mixture is acidic and mixture contains $\lbrack H^{+}\rbrack$ ion. Thus,
$\left\lbrack H^{+} \right\rbrack = \ 0.25$
$pH = - \log{\lbrack H^{+}\rbrack}$
$= \ - \log{\lbrack 0.25\rbrack}$
$\ \ \ \ \ \ \ = - \log\frac{2.5}{10}$
$\ \ \ \ \ \ \ = - \lbrack\log{2.5} - \log 10\rbrack$
$\ \ \ \ \ \ \ = - (0.3979 - 1) = 0.6021$