Milky & Tricky

Consider my diagram, the volume of the two shaded regions are equal. So we have

$(10-h)(12)(10)=80h$

$1200-120h=80h$

$h=6$

So the desired answer is $(80)(6)=\color{#D61F06}\boxed{480}$

The total area of the two containers is 120+80=200(cm^2), and the total volume of milk is 1200 ml. Now we can calculate the height of milk in the pot: 1200/200 = 6 (cm) Hence, the total volume in the pot is 480ml.

Man I solve this in 5 minute

We have $1200ml$ milk ( $600 ml + 600 ml$ )

Let's count the container volume,

Assume $p$ = Length, $l$ = Width and $t$ = Heigth

$v = plt = 12 cm \times 10 cm \times 15 cm = 1800 ml$

Now find the Heigth of the milk,

$\frac{1800 ml}{1200 ml} = \frac{15 cm}{ t( Container )}$

t ( Container ) $= 10 cm$

Now make an equation,

Assume $n$ is the milk volume that have been moved

$t ( Container ) = t ( Pot )$

$\frac{v ( Container )}{Surface \space Area} = \frac{v ( pot)}{Surface \space Area}$

$\frac{1200 ml - n }{12 cm \times 10 cm} = \frac{n}{80 cm^2}$

$96000 - 80n = 120n$

$96000 = 200n$

$n = \boxed{480 ml}$

Good question, I love it

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CHEEERSSSSSSSS

Let $x$ be the height of milk in both containers

Then we have that $120x+80x = 1200$ , as the total volume is given to be $1200cm^3$ , and that the hight of milk in both containers is the same value $x$ . We can then work out the volume of milk in the pot:

$120x+80x=1200$

$200x=1200$

Therefore, $x=6$

Substituting into $80x$ , we have that the volume of milk in the pot is $480cm^3$

The volume of milk in the pot is equal to the volume of milk that was removed from the rectangular container. Consider my diagram. We have

$(10-h)(12)(10)=80h$

$120(10-h)=80h$

$1200-120h=80h$

$1200=200h$

$h=6$

So the desired area is $(80)(6)=$ $\boxed{480}$