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I don't mean cubes to pack so tightly that the last of a layer can't be added. My solution assumes it can:
The volume of water is 2*3*.4 = 2.4cc
Cube one reduces the bottom area to 5scm (square centimeters) so the depth is not 2.4/5 = .48cm
Cube two raises the depth to 2.4/4 = .6cm
Cube three raises it to 2.4/3 = .8cm
Cube four will bring the depth above the first layer. The bottom contains the first 2cc, above them will be the remaining .4cc over the full cross section. .4/6=.06667 so the full depth is 1.06667cm
Cube five leaves 1cc at the bottom. 1 + 1.4/6 = 1.23333cm
Cube six fills the bottom layer entirely so the water rises a full cm, to 1.4cm.
Cubes seven, eight and nine are just 1cm more than one, two and three. 1.48, 1.6, 1.8.
Cubes ten and eleven will overflow the tin, the water can't rise above 2cm.
Cube twelve forces out the rest of the water, so there is no water level to speak of.
Summary:
{.48, .6, .8, 1.06667, 1.2333, 1.24, 1.48, 1.6, 1.8, 2, 2, none} |
