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 Cubes in a tin (Posted on 2006-05-15)
There is a rectangular tin 2cm by 3cm at the base, 2cm deep and open at the top.
It is filled with water to a depth of .4cm.
Twelve 1cm steel cubes are placed in the tin one by one. The first six form a single layer at the bottom. The next six form a second layer.

To what height does the water level rise after the placement of each cube?
Assume the cubes fit together tightly, leaving no spaces.

 See The Solution Submitted by Jer Rating: 2.6667 (6 votes)

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 solution | Comment 2 of 23 |

The first cube leaves a base area of 5 cm^2, so the depth will be .4 * 6/5 = .48 cm.

The second cube leaves a base area of 4 cm^2, so the depth will be .4 * 6/4 = .6 cm.

The third cube leaves a base area of 3 cm^2, so the depth will be .4 * 6/3 = .8 cm.

The fourth cube leaves a base area of 2 cm^2 up to a level of 1 cm, and a clear 6 cm^2 above that.  There are .4 * 6 = 2.4 cm^3 of water. The portion below the 1-cm level is 2 cm^3, leaving .4 cm^3 to occupy the full 6-cm^2 higher base. Dividing, .4 / 6 = 1/15, so the top of the water is 1 1/15 cm above the original base. (1.066666... cm)

With 5 cubes in place, there is room for only 1 cm^2 within 1 cm of the base, leaving 1.4 cm^3 for the free-and-clear next level, so the water level extends another 1.4 / 6 = 7/30 above the 1-cm mark, making the level 1 7/30 cm. (1.233333... cm)

The 6th cube would completely fill the bottom layer. The new water level would be at 1.4 cm.

The rest would follow, 1 cm higher than before: 1.48 cm, 1.6 cm, 1.8 cm, 2 1/15 cm, 2 7/30 cm, 2.4 cm.

 Posted by Charlie on 2006-05-15 09:58:35

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