I have been practising for a math competition and came across the following question:
A fishtank with base $100\,\rm cm$by $200\,\rm cm$ and depth $100\,\rm cm$ contains water to a depth of $50\,\rm cm$. A solid metal rectangular prism with dimensions $80\,\rm cm$ by $100\,\rm cm$ by $60\,\rm cm$ is then submerged in the tank with an $80\,\rm cm$ by $100\,\rm cm$ face on the bottom.
The depth of water, in centimetres, above the prism is then $$(\text A)\,12\qquad(\text B)\,14\qquad(\text C)\,16\qquad(\text D)\,18\qquad(\text E)\,20$$
I first calculated the volume of the water in the tank: $$100 \cdot 200 \cdot 50 = 1 \ 000 \ 000 \ \text{cm}^3$$
Then, I calculated the volume of the solid metal prism: $$80 \cdot 100 \cdot 60 = 480 \ 000 \ \text{cm}^3 $$
When the prism is put in the tank, the total volume would be: $$1 \ 000 \ 000 + 480 \ 000 = 1 \ 480 \ 000 \ \text{cm}^3$$
I am not sure how to continue from this step.
It turns out that case $(2)$ mentioned by @martycohen happened.
(The part on why this is so is skipped.)
$$\text{Total volume of water} = 50 \times 100 \times 200.$$
$$\text{Volume of water at the }60\,\mathrm{cm}\text{ level with the block side by side} = 120 \times 100 \times 60.$$
$$\text{Volume of water above the }60\,\mathrm{cm}\text{ level}= h’ \times 100 \times 200,$$
where $h'$ is the height of water above the $60\,\rm cm$.
From the above three, we get $h’ = 26$.
Final depth of water: 14.