Vertical Addition: Addition of 2-digit-and 1-digit numbers with the regrouping of both tens and ones

To solve this problem, we'll follow these steps:

• Step 1: Align the numbers by their place value.
• Step 2: Begin adding from the rightmost column (ones place).
• Step 3: Carry over any value over 9 to the next left column.
• Step 4: Continue with the next column (tens place), including any carried over value.

Now, let's perform the addition:

Vertical addition setup:
$\begin{aligned} &33 \\ + &~~8 \\ &\underline{\phantom{776}} \end{aligned}$

Step 1: Add the ones digits: $3 + 8 = 11$. We write 1 in the ones place and carry over 1 to the tens place.

Step 2: Add the tens digits (3 from 33 and 0 from 8, since it has no tens digit), including any carry: $3 + 1 = 4$.

This gives us:

$\begin{aligned} &33 \\ + &~~8 \\ &\underline{\phantom{776}} \\ &41 \end{aligned}$

Therefore, the solution to the problem is $\textbf{41}$.

To solve the vertical addition problem, we'll proceed through the following steps:

• Step 1: Align the numbers by place value: tens and ones.
  Given: $\begin{aligned} &43 \\ +& \\ &~9 \\ \end{aligned}$
• Step 2: Add the ones column:
  Add $3$ (from $43$) and $9$. The sum is $12$. Because $12$ is greater than $9$, we need to carry over:
• Write down the $2$ in the ones place.
• Carry over the $1$ to the tens column.
• Step 3: Add the tens column:
  Add the $4$ (from the tens place of $43$) and the carried over $1$, the sum is $5$.
• The tens column now becomes $5$.
• Step 4: Assemble the result:
  The result is the tens value $5$ and the ones value $2$, making the final answer $52$.

Therefore, the solution to the problem is $52$.

To solve this problem, we will use vertical addition to find the sum of 52 and 9. Let's follow the steps outlined:

• Step 1: Align the numbers.
  Write the numbers so that the ones digits align vertically:

$\begin{array}{cc} & 5 & 2 \\ + & & 9 \\ \hline \end{array}$

• Step 2: Add the ones digits.
  First, add the digits in the far right column (ones place): $2 + 9 = 11$.
• Step 3: Handle the carry-over.
  Since 11 is greater than 9, write the 1 in the ones place and carry over the 1 to the tens place:

$\text{Carry: } 1 \quad \quad$ $\begin{array}{cc} & 5 & 2 \\ + & & 9 \\ \hline & 6 & 1 \\ \end{array}$

• Step 4: Add the tens digits, including any carry-over.
  Next, add the digits in the tens place: $5 + 1 = 6$.
• Step 5: Write the final result.

Therefore, the sum of 52 and 9 is $61$.

The correct answer is choice $3$ with the final total being $61$.

To solve this problem, we'll follow these steps:

• Step 1: Add the ones digits: $7 + 8$.
• Step 2: Determine any carry-over from the addition.
• Step 3: Add the tens digits, incorporating any carry-over.

Now, let's work through each step:

Step 1: Add the ones digits:
$7 + 8 = 15$
Since the sum (15) is greater than 9, we write down the 5 in the ones place and carry over the 1 to the tens place.

Step 2: Add the tens digits with the carried-over 1:
$1 \text{ (carry)} + 7 = 8$

Therefore, the result of adding 77 and 8 is:
$\boxed{85}$

The correct choice from the given options is 85.

To solve this problem, we need to perform vertical addition of the numbers 69 and 5.

Step-by-step:

• Align the Numbers: Write 69 and 5 vertically, ensuring units are under units.
• Add the Units Place:
  • Add the units digits: $9 + 5 = 14$.
  • Regroup: Write down $4$ and carry over $1$ to the tens place.
• Add the Tens Place:
  • Add the digits in the tens place, including the carryover: $6 + 1 = 7$.
  • Since there is no other number in the tens column besides the initial carry, the result is $7$.

Therefore, $69 + 5 = 74$.

Conclusively, the answer to the problem is $74$.

To solve this problem, let's calculate the sum of the numbers 25 and 6 using vertical addition:

Step 1: Align the numbers in vertical format.
The tens and ones are aligned as follows:

• Tens place: $2$

• Ones place: $5$

• $+$ $6$ (ones place)

• $\underline{\phantom{52}}$

Step 2: Add the digits in the ones place.
$(5 + 6 = 11)$. Since the sum exceeds 9, write down $1$ in the ones place and carry over $1$ to the tens place.

Step 3: Add the digits in the tens place, considering the carryover.
$(2 + 1 = 3)$.

Thus, the complete sum is $31$.

Therefore, the sum of 25 and 6 is $\mathbf{31}$, making choice 2 the correct one.