Let the price of the smallest cup be $2a,$ and medium be $5a,$ and large be $b.$
Now, $2a \times 5a \times b = 800$
$\Rightarrow a^{2}b = 80 \quad \longrightarrow (1)$
And, $(2a+6) (5a+6)b = 3200 \quad \longrightarrow (2)$
Divide equations $(2)$ by equation $(1).$
$\Rightarrow \frac{(2a+6)(5a+6)b}{a^{2}b} = \frac{3200}{80}$
$\Rightarrow (2a+6)(5a+6) = 40 a^{2}$
$\Rightarrow 10a^{2} + 12a + 30a + 36 = 40a^{2}$
$\Rightarrow -30a^{2} + 42a + 36 = 0$
$\Rightarrow -5a^{2} + 7a + 6 = 0$
$\Rightarrow 5a^{2} - 7a - 6 = 0$
$\Rightarrow 5a^{2} - 10a + 3a - 6 = 0$
$\Rightarrow 5a(a – 2) – 3 (a - 2) = 0$
$\Rightarrow (a-2)(5a-3) = 0$
$\Rightarrow \boxed{a = 2}$
So,
Put the value of $a$ in equation $(1).$
$\Rightarrow a^{2}b = 80$
$\Rightarrow 2^{2}b = 80$
$\Rightarrow \boxed{b = 20}$
$\therefore$ The sum of the original prices of three different sizes $ = 2a + 5a + b = 4 + 10 + 20= 34.$
Correct Answer $:34$