Clockwork Orange Juice

Assume that,

• the amount of liquid in one glass is $W$ ,
• the area of the base of the jug is $A$ ,
• the area of the opening at the bottom of the jug is $B$ , and
• the height of the jug is $h$ .

Initially, the jug contains $15W$ amount of liquid.

Because the jug is cylindrical, then the volume of liquid inside is given by,

$V = Ah$

$h = \frac{V}{A}$ (1)

By Toricelli's Law, the velocity of the liquid pouring out of the opening is given by,

$v = \sqrt{2gh}$

Thus,

$\frac{dV}{dt} = -B \sqrt{2gh}$ using eq (1)

$\frac{dV}{dt} = -B \sqrt{\frac{2gV}{A}}$

$V^{-1/2} dV = -B\sqrt{\frac{2g}{A}} dt$ (2)

Since it takes $12 seconds$ to fill one glass of juice, we could integrate eq (2),

$\int_{15W}^{14W}{V^{-1/2} dV} = \int_{0}^{12}{-B\sqrt{\frac{2g}{A}} dt}$

$2(\sqrt{14W} - \sqrt{15W}) = -12B\sqrt{\frac{2g}{A}}$

$\frac{\sqrt{15}-\sqrt{14}}{6} = B \sqrt{\frac{2g}{AW}}$ (3)

Let $T$ be the time it takes to fill the remaining glasses, we would integrate eq (2) again,

$\int_{14W}^{0}{V^{-1/2} dV} = \int_{0}^{T}{-B\sqrt{\frac{2g}{A}} dt}$

$-2\sqrt{14W} = -BT\sqrt{\frac{2g}{A}}$

$2\sqrt{14} = T(B\sqrt{\frac{2g}{AW}})$ using eq (3)

$2\sqrt{14} = T(\frac{\sqrt{15} - \sqrt{14}}{6})$

$\frac{12\sqrt{14}}{\sqrt{15} - \sqrt{14}} = T$

Thus,

$(a, b, c, d) = (12, 14, 15, 14)$

$a + b + c + d = 12 + 14 + 15 + 14 = \boxed{55}$

I used calculus method for finding it.. It took jst some time to think upon it :)

Use eqn of continuity and toricelli's eqn/