Let the percentage loss he had at the end of first be $x\%.$ Then at the end of the second year his gain is $5x\%.$

Now, $-x+5x+\frac{(-x) \times (5x)}{100} = 35 \quad [\because \color{Blue}{\text{Successive percentage}} ]$

$\Rightarrow 4x–\frac{x^{2}}{20} = 35$

$\Rightarrow 80x-x^{2} = 700$

$\Rightarrow x^{2}-80x+700 = 0$

$\Rightarrow x^{2}-70x-10x+700 = 0$

$\Rightarrow x(x-70)-10(x-70) = 0$

$\Rightarrow (x-70)(x-10) = 0$

$\Rightarrow x=10 \; \text{(or)}\; x=70\;\color{Red}{\text{(rejected)}}$

$\Rightarrow \boxed{x=10}$

$\therefore$ The percentage of loss in the first year is $10\%.$

Correct Answer $:\text{B}$

$\textbf{PS:}$ Let the successive increase in percentages be $a\%$ and $b\%$. Then, the total increase will be $\left(a+b+\frac{ab}{100}\right)\%.$

- If there's an increase and a decrease, in that case, the decrease will be considered a negative value.
- In the case of discounts, the value of discount percentages will be considered negative.