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I have been practising for a math competition and came across the following question:

  1. 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.

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    $\begingroup$ $100\times200\times$height$=1480000$. $\endgroup$
    – Rammus
    Jun 28, 2015 at 1:23
  • $\begingroup$ Two possible cases: Water does not rise above the inserted prism; (2) water does rise above the prism. $\endgroup$ Jun 28, 2015 at 1:31
  • $\begingroup$ @Rammus Oh. So that gives a height of 74cm and the height above equals 74 - 60 = 14. Is that right? $\endgroup$
    – anonymous
    Jun 28, 2015 at 1:34
  • $\begingroup$ The cube displaces $60\cdot80\cdot100$ cc. This would raise the water by $\frac{60\cdot80\cdot100}{100\cdot200}=24$ cm to $74$ cm. $\endgroup$
    – robjohn
    Jun 28, 2015 at 15:53

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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.

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