Let the price at 1993 = z
$$Therefore,\ price\ at\ 1994\ =\ \frac{\left(z\ +\ x2\right)}{100}$$
$$=z\ \frac{\left(1\ +\ x\right)}{100}$$
$$\Pr ice\ at\ 1995\ =\ \frac{\left(z\ +\ x2\right)}{100}+\ \frac{y\left(z\ +\ x2\right)}{100}$$
$$=\ z\left(\frac{\left(1+\ x\right)}{100} \ \frac{\left(1\ +\ y\right)}{100}\right)$$

Statement 1 = x + y = 300
This is NOT SUFFICIENT because, we will have different value of x and y to be equal to 300 from many solutions.

$$Statement\ 2=\ \frac{\left(1+\ x\right)}{100}\ \frac{\left(1\ +\ y\right)}{100}=\ 1.0035$$
From the equation
$$=\ z\left(\frac{\left(1+\ x\right)}{100}\ \frac{\left(1\ +\ y\right)}{100}\right)$$
= z (1.0035)
Price at 1995 = z1.0035
%increase from 1993 to 1994
$$=\ price\ at\ 1994\ -\ price\ at\ 1993\ \cdot\ \frac{100}{1}$$
$$\left(z1.0035\ -\ 2\cdot\ \frac{100}{1}\right)$$
= (z1.0035 - 5)%

Statement 2 is SUFFICIENT.
Option B is correct.