Let price of smallest cup be $2x$ and medium be $5x$ and large be $y$
Now by condition 1
we get $2x × 5x × y = 800$
we get $ x^2 y =80 $
...(1)
Now as per second condition :
$(2x + 6) × (5x + 6) y = 3200$
...(2)
Now dividing (1) and (2)
We get :
${(2x + 6) × (\x + 6)} / x^2 = 40 $
$10x^2 + 42x +36 = 40x^2$
$30x^2 - 42x -36 =0 $
$5x^2 -7x -6 =0 $
We get $x = 2$
So, $2x = 4$ and$5x = 10 $
Now substituting in (1) we get $y =20 $
Now therefore sum $= 4+10+20 =34$