Henry Ernest Dudeney/Puzzles and Curious Problems/224 - The Tank Puzzle/Solution

by : $224$

 * The Tank Puzzle
 * The area of the floor of a tank is $6$ square feet,
 * the water in it is $9$ inches deep.
 * $(1)$ How much does the water rise if a $1$ foot metal cube is put in it,
 * $(2)$ How much farther does it rise if another cube like it is put in by its side?

Solution

 * $(1): \quad 1 \tfrac 4 5 = 1.8$ inches


 * $(2): \quad 2 \tfrac 1 5 = 2.2$ more inches.

Proof
Let $d$ inches be the level that the water rises when the first cube is placed in it.

The volume of the water displaced by this cube equals $1$ square foot multiplied by $9$ inches.

This fills a volume $6$ square feet less the $1$ square foot the cube pokes up through the surface multiplied by $d$ inches.

That is:
 * $\paren {6 - 1} \times d = 1 \times 9$

and so $d = \dfrac 9 5 = 1 \tfrac 4 5 = 1.8$ inches.

The water is now $9 + 1 \tfrac 4 5 = 10 \tfrac 4 5$ inches deep.

Now let the second cube is placed in the tank.

We have seen from the first part that the water level will rise at least (in fact, more than) $1 \tfrac 4 5$ inches, while $1 \tfrac 1 5$ inches will take it over the top of both cubes.

Hence a different model is needed.

The total volume of the water in the tank is $6$ square feet times $9$ inches.

This is equal to the total volume of the water with two cubes in it, which are now completely submerged.

Let $d$ inches be the additional depth of water above the surface of those two cubes.

Thus the total volume of the water in the tank is $4$ square feet multiplied by $12$ inches added to $6$ square feet multiplied by $d$ inches.

That is:
 * $6 \times 9 = 4 \times 12 + 6 d$

which leads to $d = 1$.

So $d$ is $1$ inch above the top of the cubes, which is $4$ inches higher than its original depth.

Hence the water rose an addition $4 - 1 \tfrac 4 5$ inches, which is $2 \tfrac 1 5$ inches.