There is $\color{blue}1$ decimal place in $6.1$ and there are $\color{blue}2$ decimal places in $2.49.$

To subtract the numbers, they both need the same number of decimals. Therefore, we rewrite the first number so that it has $\color{blue}2$ decimal places: \[ 6.1=6.1{\color{red}0} \]

Now, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 6 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 1 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 2 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 9 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& & \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ \\ 5 } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & & \!\!\!\! \cancel 6 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel 1 \!\!\!\!& \!\!\!\! \cancel 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 2 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 9 \!\!\!\! \\ \hline & & \!\!\!\! 3 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 6 \!\!\!\!& \!\!\!\! 1 \!\!\!\! \end{array} \]

Therefore, \[ 6.1-2.49 = 3.61 \]

There are $\color{blue}3$ decimal places in $4.521$ and there are $\color{blue}2$ decimal places in $2.64.$

To subtract the numbers, they both need the same number of decimals. Therefore, we rewrite the second number so that it has $\color{blue}3$ decimal places: \[ 2.64{\color{red}0} \]

Now, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 4 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 5 \!\!\!\!& \!\!\!\! 2 \!\!\!\!& \!\!\!\! 1 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 2 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 6 \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& & & \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ \\ 3 } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 14 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\12 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ } \!\!\!\! \\ & & \!\!\!\! \cancel 4 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel 5 \!\!\!\!& \!\!\!\! \cancel 2 \!\!\!\!& \!\!\!\! 1 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 2 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 6 \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \hline & & \!\!\!\! 1 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 8 \!\!\!\!& \!\!\!\! 8 \!\!\!\!& \!\!\!\! 1 \!\!\!\! \end{array} \]

Therefore, \[ 4.521-2.64=1.881 \]

There is $\color{blue}1$ decimal place in $8.7$ and there are $\color{blue}3$ decimal places in $5.468.$

To subtract the numbers, they both need the same number of decimals. Therefore, we rewrite the first number so that it has $\color{blue}3$ decimal places: \[ 8.7=8.7{\color{red}00} \]

Now, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 8 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 7 \!\!\!\!& \!\!\!\! 0 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 5 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 6 \!\!\!\!& \!\!\!\! 8 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& & & \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 6 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 9 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & & \!\!\!\! 8 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel7 \!\!\!\!& \!\!\!\! \cancel 0 \!\!\!\!& \!\!\!\! \cancel 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 5 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 6 \!\!\!\!& \!\!\!\! 8 \!\!\!\! \\ \hline & & \!\!\!\! 3 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 2 \!\!\!\!& \!\!\!\! 3 \!\!\!\!& \!\!\!\! 2 \!\!\!\! \end{array} \]

Therefore, \[ 8.7-5.468 = 3.232 \]

There is $\color{blue}1$ decimal place in $64.2$ and there are $\color{blue}3$ decimal places in $33.148.$

To subtract the numbers, they both need the same number of decimals. Therefore, we rewrite the first number so that it has $\color{blue}3$ decimal places: \[ 64.2=64.2{\color{red}00} \]

Now, we line up the decimal points:

\[ \begin{array}{cccccccc} & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! . \!\!\!\!& \!\!\!\! 2 \!\!\!\!& \!\!\!\! 0 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & \!\!\!\! 3 \!\!\!\!& \!\!\!\! 3 \!\!\!\! & \!\!\!\! . \!\!\!\!& \!\!\!\! 1 \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 8 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& & & \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & \!\!\!\!\! \color{blue}\substack{\\ } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{\\ } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 1 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 9 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel2 \!\!\!\!& \!\!\!\!\cancel 0 \!\!\!\!& \!\!\!\! \cancel 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & \!\!\!\! 3 \!\!\!\!& \!\!\!\! 3 \!\!\!\! & \!\!\!\! . \!\!\!\!& \!\!\!\! 1 \!\!\!\!& \!\!\!\! 4 \!\!\!\!& \!\!\!\! 8 \!\!\!\! \\ \hline & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! . \!\!\!\!& \!\!\!\! 0 \!\!\!\!& \!\!\!\! 5 \!\!\!\!& \!\!\!\! 2 \!\!\!\! \end{array} \]

Therefore, \[ 64.2-33.148=31.052 \]

There are no decimal places in $7$ and there is $\color{blue}1$ decimal place in $1.7.$

To subtract the numbers, they both need the same number of decimals. Therefore, we rewrite the first number so that it has $\color{blue}1$ decimal place: \[ 7\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{0}_{\large\text{$\color{blue}1$ zero}}\!\!\! \]

Next, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 7 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 1 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 7 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& \end{array} \]

Then, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ \\ 6 } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & & \!\!\!\! \cancel7 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 1 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 7 \!\!\!\! \\ \hline & & \!\!\!\! 5 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 3 \!\!\!\! \end{array} \]

Therefore: \[ 7-1.7 = 5.3 \]

There are no decimal places in $5$ and there are $\color{blue}2$ decimal places in $0.84.$

To subtract the numbers, they both need the same number of decimals. Therefore, we rewrite the first number so that it has $\color{blue}2$ decimal places: \[ 5\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{00}_{\large\text{$\color{blue}2$ zeros}}\!\!\! \]

Now, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 5 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 0 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 0 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 8 \!\!\!\!& \!\!\!\! 4 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& & \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ \\ 4 } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 9 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & & \!\!\!\! \cancel5 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel0 \!\!\!\!& \!\!\!\! \cancel0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 0 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 8 \!\!\!\!& \!\!\!\! 4 \!\!\!\! \\ \hline & & \!\!\!\! 4 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 1 \!\!\!\!& \!\!\!\! 6 \!\!\!\! \end{array} \]

Therefore, \[ 5-0.84 = 4.16 \]

To find how much water was left in the container, we need to calculate the difference $ 12- 0.9.$

There are $\color{blue}0$ decimal places in $12$ and there is $\color{blue}1$ decimal place in $0.9.$

To subtract the numbers, both must have the same number of decimals. Hence, we rewrite the first number so that it has $\color{blue}1$ decimal place: \[ 12 = 12.{\color{red}0} \]

First, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 1 \!\!\!\!& \!\!\!\! 2 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & & \!\!\!\! 0 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 9 \!\!\!\! \\ \hline & & & & \!\!\!\! . \!\!\!\!& \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ \\ } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 1 } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & & \!\!\!\! 1 \!\!\!\!& \!\!\!\! \cancel{2} \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel{0} \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & & \!\!\!\! 0 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 9 \!\!\!\! \\ \hline & & \!\!\!\! 1 \!\!\!\!& \!\!\!\! 1 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 1 \!\!\!\! \end{array} \]

Therefore, $11.1$ liters of water was left in the container.

To find the length of the other piece, we need to calculate the difference $9- 4.38.$

There are $\color{blue}0$ decimal places in $9$ and there are $\color{blue}2$ decimal places in $4.38.$

To subtract the numbers, both must have the same number of decimals. Hence, we rewrite the first number so that it has $\color{blue}2$ decimal places: \[ 9 = 9.{\color{red}00} \]

First, we line up the decimal points:

\[ \begin{array}{cccccccc} & & \!\!\!\! 9 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 0 \!\!\!\!& \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 4 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 3 \!\!\!\!& \!\!\!\! 8 \!\!\!\! \\ \hline & & & \!\!\!\! . \!\!\!\!& & \end{array} \]

Next, we proceed by subtracting the two numbers just as we would with whole numbers:

\[ \begin{array}{cccccccc} & & \!\!\!\!\! \color{blue}\substack{ \\ 8 } \!\!\!\! & & \!\!\!\!\! \color{blue}\substack{ \\ 9 } \!\!\!\! & \!\!\!\!\! \color{blue}\substack{ \\ 10 } \!\!\!\! \\ & & \!\!\!\! \cancel{9} \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! \cancel{0} \!\!\!\!& \!\!\!\! \cancel{0} \!\!\!\! \\ \!\!\!\!-\!\!\!\! & & \!\!\!\! 4 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 3 \!\!\!\!& \!\!\!\! 8 \!\!\!\! \\ \hline & & \!\!\!\! 4 \!\!\!\!& \!\!\!\! . \!\!\!\!& \!\!\!\! 6 \!\!\!\!& \!\!\!\! 2 \!\!\!\! \end{array} \]

Hence, \[9- 4.38= 4.62.\] So, the other piece was $4.62$ feet long.