Marinada '26

[lang=hr] Pljačkajući jednu stariju kuću, Harry i Marv pokušali su je poplaviti kao i obično te su ovoga puta otvorili jednu slavinu koja bi poplavila kuću za $3$ sata. Ipak, ovaj put nisu primijetili kako pukotine u starim zidovima propuštaju vodu. Na zidu nalazi se 8 identičnih pukotina raspoređenih jedna iznad druge po visini zida. Ako je zid visine $h$ onda se $k$-ta pukotina nalazi na visini $h * \frac{2^k-1}{2^k}$ za $k \in \{1, 2, 3, …, 8\}$. Kada visina vode dođe do neke pukotine kroz nju počne curiti voda. Kada bi kroz samo jednu pukotinu mogla cijelo vrijeme curiti voda, ona bi mogla isprazniti poplavljenu kuću za $30$ sati. Za koliko sati će se kuća poplaviti? Pretpostavi da je površina presjeka kuće konstantna po cijeloj visini, tako da je volumen vode proporcionalan njezinoj visini. (Oprez pri zaokruživanju u međukoracima, uvijek pišite više decimala no što je potrebno kako bi dobili dobro rješenje na dvije decimale!) [/lang] [lang=en] While robbing one older house, Harry and Marv tried flooding it like usual so they opened a faucet that would usually flood the house in $3$ hours. However, this time they didn't notice that the cracks in the old walls are draining water. On the wall there are 8 identical cracks vertically distributed along its height. If the wall is of height $h$ the height of the $k$th crack is $h * \frac{2^k-1}{2^k}$ for $k \in \{1, 2, 3, …, 8\}$. When the water level in the house reaches the height of some crack that crack will start draining water. If one crack could alone constantly drain water it would be able to drain a fully flooded house in $30$ hours. In how many hours will the house be fully flooded? Assume the area of the crossection of the house is constant, that is to say, that the volume of water is always proportional to its height. (Be careful if rounding partial results, always use more decimal places than necessary to get the final result rounded to two decimal places!) [/lang]