Printable worksheet for practicing decimal addition with missing numbers, ideal for 5th-grade math learners.

To solve the missing numbers in these decimal additions, we need to carefully align the decimal points and perform the addition or subtraction as required. Let's go through each problem step by step.

Problem 1:

\[
\begin{array}{r}
0.37 \\
+ 0.\_8 \\
\hline
0.1\_ \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
0.37 \\
+ 0.x8 \\
\hline
0.1y \\
\end{array}
\]
- The sum of the tenths place is 3 + x = 1 (with a carryover). This means x must be 7 (since 3 + 7 = 10, and we carry over 1).
- The sum of the hundredths place is 7 + 8 = 15. We write down 5 and carry over 1.
- The sum of the units place is 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 7 \) and \( y = 5 \).

\[
\boxed{7, 5}
\]

Problem 2:

\[
\begin{array}{r}
9.3 \\
+ 1.\_8 \\
\hline
73.\_9 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
9.30 \\
+ 1.x8 \\
\hline
73.y9 \\
\end{array}
\]
- The sum of the hundredths place is 0 + 8 = 8.
- The sum of the tenths place is 3 + x = y (with a carryover). This means x must be 4 (since 3 + 4 = 7).
- The sum of the units place is 9 + 1 + 1 (carryover) = 11. We write down 1 and carry over 1.
- The sum of the tens place is 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 4 \) and \( y = 7 \).

\[
\boxed{4, 7}
\]

Problem 3:

\[
\begin{array}{r}
3.382 \\
+ 3.\_6 \\
\hline
7.\_9 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
3.382 \\
+ 3.x60 \\
\hline
7.y90 \\
\end{array}
\]
- The sum of the thousandths place is 2 + 0 = 2.
- The sum of the hundredths place is 8 + 6 = 14. We write down 4 and carry over 1.
- The sum of the tenths place is 3 + x + 1 (carryover) = y. This means x must be 5 (since 3 + 5 + 1 = 9).
- The sum of the units place is 3 + 3 = 6.
- Therefore, the missing numbers are \( x = 5 \) and \( y = 6 \).

\[
\boxed{5, 6}
\]

Problem 4:

\[
\begin{array}{r}
\_86 \\
+ 2.\_7 \\
\hline
7.4\_ \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
x.86 \\
+ 2.y7 \\
\hline
7.4z \\
\end{array}
\]
- The sum of the hundredths place is 6 + 7 = 13. We write down 3 and carry over 1.
- The sum of the tenths place is 8 + y + 1 (carryover) = 4. This means y must be 5 (since 8 + 5 + 1 = 14, and we carry over 1).
- The sum of the units place is x + 2 + 1 (carryover) = 7. This means x must be 4 (since 4 + 2 + 1 = 7).
- Therefore, the missing numbers are \( x = 4 \), \( y = 5 \), and \( z = 3 \).

\[
\boxed{4, 5, 3}
\]

Problem 5:

\[
\begin{array}{r}
66.5 \\
+ 3.\_7 \\
\hline
9.\_21 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
66.50 \\
+ 3.x7 \\
\hline
9.y21 \\
\end{array}
\]
- The sum of the hundredths place is 0 + 7 = 7.
- The sum of the tenths place is 5 + x = y. This means x must be 6 (since 5 + 6 = 11, and we carry over 1).
- The sum of the units place is 6 + 3 + 1 (carryover) = 10. We write down 0 and carry over 1.
- The sum of the tens place is 6 + 0 + 1 (carryover) = 7.
- Therefore, the missing numbers are \( x = 6 \) and \( y = 1 \).

\[
\boxed{6, 1}
\]

Problem 6:

\[
\begin{array}{r}
\_62 \\
+ 6.\_7 \\
\hbox{-} \\
1.\_12 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
x.62 \\
+ 6.y7 \\
\hline
1.z12 \\
\end{array}
\]
- The sum of the hundredths place is 2 + 7 = 9.
- The sum of the tenths place is 6 + y = z. This means y must be 4 (since 6 + 4 = 10, and we carry over 1).
- The sum of the units place is x + 6 + 1 (carryover) = 1. This means x must be 4 (since 4 + 6 + 1 = 11, and we write down 1).
- Therefore, the missing numbers are \( x = 4 \), \( y = 4 \), and \( z = 0 \).

\[
\boxed{4, 4, 0}
\]

Problem 7:

\[
\begin{array}{r}
\_59 \\
+ 3.\_8 \\
\hline
6.\_2 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
x.59 \\
+ 3.y8 \\
\hline
6.z2 \\
\end{array}
\]
- The sum of the hundredths place is 9 + 8 = 17. We write down 7 and carry over 1.
- The sum of the tenths place is 5 + y + 1 (carryover) = z. This means y must be 5 (since 5 + 5 + 1 = 11, and we carry over 1).
- The sum of the units place is x + 3 + 1 (carryover) = 6. This means x must be 2 (since 2 + 3 + 1 = 6).
- Therefore, the missing numbers are \( x = 2 \), \( y = 5 \), and \( z = 1 \).

\[
\boxed{2, 5, 1}
\]

Problem 8:

\[
\begin{array}{r}
7.\_35 \\
+ 37.\_6 \\
\hline
\_9.31 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
7.x35 \\
+ 37.y60 \\
\hline
z9.31 \\
\end{array}
\]
- The sum of the thousandths place is 5 + 0 = 5.
- The sum of the hundredths place is 3 + 6 = 9.
- The sum of the tenths place is x + y = 3. This means x and y must be 6 and 7 (since 6 + 7 = 13, and we carry over 1).
- The sum of the units place is 7 + 7 + 1 (carryover) = 15. We write down 5 and carry over 1.
- The sum of the tens place is 0 + 3 + 1 (carryover) = 4.
- Therefore, the missing numbers are \( x = 6 \), \( y = 7 \), and \( z = 4 \).

\[
\boxed{6, 7, 4}
\]

Problem 9:

\[
\begin{array}{r}
77.32 \\
+ 71.\_60 \\
\hline
12.\_05 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
77.320 \\
+ 71.x60 \\
\hline
12.y05 \\
\end{array}
\]
- The sum of the thousandths place is 0 + 0 = 0.
- The sum of the hundredths place is 2 + 6 = 8.
- The sum of the tenths place is 3 + x = y. This means x must be 7 (since 3 + 7 = 10, and we carry over 1).
- The sum of the units place is 7 + 1 + 1 (carryover) = 9.
- The sum of the tens place is 7 + 7 + 1 (carryover) = 15. We write down 5 and carry over 1.
- The sum of the hundreds place is 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 7 \) and \( y = 9 \).

\[
\boxed{7, 9}
\]

Problem 10:

\[
\begin{array}{r}
\_53 \\
+ 7.\_9 \\
\hline
4.\_2 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
x.53 \\
+ 7.y9 \\
\hline
4.z2 \\
\end{array}
\]
- The sum of the hundredths place is 3 + 9 = 12. We write down 2 and carry over 1.
- The sum of the tenths place is 5 + y + 1 (carryover) = z. This means y must be 5 (since 5 + 5 + 1 = 11, and we carry over 1).
- The sum of the units place is x + 7 + 1 (carryover) = 4. This means x must be 6 (since 6 + 7 + 1 = 14, and we write down 4).
- Therefore, the missing numbers are \( x = 6 \), \( y = 5 \), and \( z = 1 \).

\[
\boxed{6, 5, 1}
\]

Problem 11:

\[
\begin{array}{r}
56.7 \\
+ 59.\_3 \\
\hline
12.\_83 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
56.70 \\
+ 59.x3 \\
\hline
12.y83 \\
\end{array}
\]
- The sum of the hundredths place is 0 + 3 = 3.
- The sum of the tenths place is 7 + x = y. This means x must be 1 (since 7 + 1 = 8).
- The sum of the units place is 6 + 9 = 15. We write down 5 and carry over 1.
- The sum of the tens place is 5 + 5 + 1 (carryover) = 11. We write down 1 and carry over 1.
- The sum of the hundreds place is 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 1 \) and \( y = 8 \).

\[
\boxed{1, 8}
\]

Problem 12:

\[
\begin{array}{r}
97.71 \\
+ 6.\_06 \\
\hline
\_2.46 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
97.71 \\
+ 6.x06 \\
\hline
z2.46 \\
\end{array}
\]
- The sum of the thousandths place is 1 + 6 = 7.
- The sum of the hundredths place is 7 + 0 = 7.
- The sum of the tenths place is 7 + x = 4. This means x must be 7 (since 7 + 7 = 14, and we carry over 1).
- The sum of the units place is 7 + 0 + 1 (carryover) = 8.
- The sum of the tens place is 9 + 6 + 1 (carryover) = 16. We write down 6 and carry over 1.
- The sum of the hundreds place is 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 7 \) and \( z = 1 \).

\[
\boxed{7, 1}
\]

Problem 13:

\[
\begin{array}{r}
\_24 \\
+ 2.\_6 \\
+ 5.\_7 \\
\hline
11.7 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
x.24 \\
+ 2.y6 \\
+ 5.z7 \\
\hline
11.70 \\
\end{array}
\]
- The sum of the hundredths place is 4 + 6 + 7 = 17. We write down 7 and carry over 1.
- The sum of the tenths place is 2 + y + z + 1 (carryover) = 7. This means y + z must be 4 (since 2 + 4 + 1 = 7).
- The sum of the units place is x + 2 + 5 + 1 (carryover) = 11. This means x must be 3 (since 3 + 2 + 5 + 1 = 11).
- Therefore, the missing numbers are \( x = 3 \), \( y = 1 \), and \( z = 3 \).

\[
\boxed{3, 1, 3}
\]

Problem 14:

\[
\begin{array}{r}
43.3 \\
+ 2.\_6 \\
+ 62.\_1 \\
\hline
\_3.3 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
43.30 \\
+ 2.x6 \\
+ 62.y1 \\
\hline
z3.3 \\
\end{array}
\]
- The sum of the hundredths place is 0 + 6 + 1 = 7.
- The sum of the tenths place is 3 + x + y = 3. This means x + y must be 0 (since 3 + 0 = 3).
- The sum of the units place is 3 + 2 + 2 + 1 (carryover) = 8.
- The sum of the tens place is 4 + 0 + 6 + 1 (carryover) = 11. We write down 1 and carry over 1.
- The sum of the hundreds place is 0 + 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 0 \), \( y = 0 \), and \( z = 1 \).

\[
\boxed{0, 0, 1}
\]

Problem 15:

\[
\begin{array}{r}
78.42 \\
+ 17.29 \\
+ 24.\_18 \\
\hline
\_18 \\
\end{array}
\]
- Align the decimals:
\[
\begin{array}{r}
78.42 \\
+ 17.29 \\
+ 24.x18 \\
\hline
z18 \\
\end{array}
\]
- The sum of the thousandths place is 2 + 9 + 8 = 19. We write down 9 and carry over 1.
- The sum of the hundredths place is 4 + 2 + 1 + 1 (carryover) = 8.
- The sum of the tenths place is 8 + 7 + x = 1. This means x must be 5 (since 8 + 7 + 5 = 20, and we carry over 2).
- The sum of the units place is 8 + 7 + 4 + 2 (carryover) = 21. We write down 1 and carry over 2.
- The sum of the tens place is 7 + 1 + 2 + 2 (carryover) = 12. We write down 2 and carry over 1.
- The sum of the hundreds place is 0 + 0 + 0 + 1 (carryover) = 1.
- Therefore, the missing numbers are \( x = 5 \) and \( z = 1 \).

\[
\boxed{5, 1}
\]

The final answers are:
\[
\boxed{7, 5}, \boxed{4, 7}, \boxed{5, 6}, \boxed{4, 5, 3}, \boxed{6, 1}, \boxed{4, 4, 0}, \boxed{2, 5, 1}, \boxed{6, 7, 4}, \boxed{7, 9}, \boxed{6, 5, 1}, \boxed{1, 8}, \boxed{7, 1}, \boxed{3, 1, 3}, \boxed{0, 0, 1}, \boxed{5, 1}
\]