To fill a rectangular tank with water, which has a length of $3 \mathrm{~m}$, a width of $50 \mathrm{~cm}$, and a height of $0.36 \mathrm{~m}$, a man uses a cylindrical bucket with a base diameter of $30 \mathrm{~cm}$ and a height of $48 \mathrm{~cm}$ to fetch water from a spring. Each time he goes to the spring, he fills the bucket to $4 / 5$ of its capacity and spills $10 \%$ of its content on the way. With the tank initially empty, how many trips to the spring will the man have to make so that the water in the tank reaches $3 / 4$ of its height?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-006.jpg?height=297&width=540&top_left_y=2433&top_left_x=1289)
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-007.jpg?height=457&width=308&top_left_y=260&top_left_x=1571)
In what follows, all volume measures are given in $\mathrm{cm}^{3}$.

The volume $V$ of the bucket is given by the usual formula for the volume of a cylinder, that is, $V=$ base area $\times$ height. The base of the bucket is a circle with a diameter of $30 \mathrm{~cm}$; its radius, then, measures $r=15 \mathrm{~cm}$ and its area is $\pi r^{2}=225 \pi \mathrm{cm}^{2}$. Therefore, $V=48 \times 225 \pi=10800 \pi$. For each trip, the volume of water that the man puts in the bucket is $4 / 5$ of $V$ and, of this volume, he loses $10 \%$. Therefore, $90 \%$ of $4 / 5$ of $V$ remains in the bucket, that is,

$$
\frac{9}{10} \times \frac{4}{5} V=\frac{18}{25} V=0.72 V=0.72 \times 10800 \pi=7776 \pi
$$

This amount $B=7776 \pi$ of water is what he effectively places in the tank with each trip. The volume of $3 / 4$ of the tank is $T=\frac{3}{4} \times 300 \times 36 \times 50=405000$. Therefore, the number of buckets needed to reach this volume is

$$
\frac{405000}{B}=\frac{405000}{7776 \pi}=\frac{625}{12 \pi}
$$

Using the approximation 3.14 for the number $\pi$, we get $\frac{625}{12 \pi} \approx \frac{625}{12 \times 3.14} \approx 16.587$.

Thus, the man will need 16 buckets, plus 0.587 of a bucket, and we conclude that he will have to make 17 trips.

Note: Above, we used an approximation for the value of $\pi$. It is important to understand what this means. As we know, $\pi$ is an irrational number and its decimal expansion is infinite and non-repeating. The approximate value of $\pi$, to 31 decimal places, is $\pi \approx 3.1415926535897932384626433832795$ (the symbol $\approx$ means "approximately"). Why, then, not use $\pi \approx 3.142$ or $\pi \approx 3.1416$ to solve our problem, instead of $\pi \approx 3.14$? To discuss this, let's look at an example.

Suppose you have a cylindrical bucket with a base radius of $1 \mathrm{~m}$ and a height of $1 \mathrm{~m}$, and a water tank with a volume of exactly $3.141 \mathrm{~m}^{3}$. The bucket must be filled from a source. How many trips to the source will be necessary to fill the tank, assuming that the volume of water from each bucket is completely transferred to the tank?

Using the approximation $\pi \approx 3.14$, we get $3.14 \mathrm{~m}^{3}$ for the volume of the bucket. Since $\frac{\text { volume of the tank }}{\text { volume of the bucket }} \approx \frac{3.141}{3.14}$ is greater than 1 (and, of course, less than 2), we conclude that two trips to the source will be necessary to fill the tank.

Now, let's use the approximation $\pi \approx 3.1416$. Here we calculate the volume of the bucket and get $3.1415 \mathrm{~m}^{3}$. Then, $\frac{\text { volume of the tank }}{\text { volume of the bucket }} \approx \frac{3.141}{3.1416}$ is less than 1, and we conclude, thus, that one trip to the source is enough to fill the tank, a result very different from the previous one!

It should be clear from this example that the initial choice of an approximation can strongly influence the final result. In this case, we say that the conditions of the problem are sensitive to the approximation. In our original problem of filling the tank, the initial data were not sensitive to the approximation used for $\pi$, which can be verified immediately by repeating the given solution with $\pi \approx 3.142$ or $\pi \approx 3.1416$. In both cases, the result is 17 trips.

In general, problems of this type proposed in elementary and high school textbooks are stated in a way that is not sensitive to the approximation. This partially justifies the use of " $\pi=3.14$ " as well as, for example, " $\sqrt{2}=1.41$ ".

(CURIOSITY: $\sqrt{2} \approx 1.4142135623730950488016887242097$ ).

We also note that a few decimal places make the calculations easier, especially when calculators are not used. It would be impractical, in practice, to work manually with the 31-decimal-place approximation we gave for $\pi$ at the beginning of this conversation.

The treatment of approximation problems is done through inequalities. Unfortunately, time and space do not allow us to address this topic in more detail at the moment, but we hope to have piqued your curiosity about the subject.