Addition: Addition of 2-digit and 2-digit numbers with the regrouping of tens

To solve this problem, we'll break it down into manageable steps:

• Step 1: Identify the numbers involved in the addition. They are 11 and 29.
• Step 2: Add the units digits of both numbers. The numbers 11 and 29 have units digits of 1 and 9, respectively.

Let's add the units digits:

$1 + 9 = 10$

This results in a sum of 10. Place the 0 (which is the unit value) in the units place of the final answer, and carry over the 1 to the tens digit.

• Step 3: Add the tens digits of both numbers. The tens digits are 1 (from 11) and 2 (from 29). Don't forget to include the carry-over 1 from the previous step.

Let's add the tens digits, including the carry-over:

$1 + 2 + 1 = 4$

The result is 4, and there is no need to carry over this time.

Step 4: Combine the calculated digits. The tens digit is 4, and the units digit is 0 from our previous operation.

Thus, the final sum is $40$.

Therefore, the solution to the problem is $\mathbf{40}$.

In this problem, we need to add 66 and 24. Let's break this down step-by-step using the standard addition method:

• Step 1: First, we align the numbers by their place values:

$\begin{array}{c} \phantom{+}66 \\ +24 \\ \hline \end{array}$

• Step 2: Add the digits in the units place (rightmost column):

The units digits are 6 and 4. Adding these gives us $6 + 4 = 10$.

Write down 0 in the units place, and carry over 1 to the tens place.

• Step 3: Add the digits in the tens place (including the carry-over):

The tens digits are 6 and 2, along with the carried 1. Adding these numbers gives us $6 + 2 + 1 = 9$.

• Step 4: Combine the results:

Placing the digits together, we have a final result of 90.

Thus, the solution to the addition problem $66 + 24$ is $90$.

To solve the addition problem $12 + 58$, we will use a step-by-step column addition method:

• Step 1: Align the numbers vertically by their place value, ensuring that the ones place is directly over each other.
• Step 2: Add the digits in the ones place: $2$ (from $12$) + $8$ (from $58$) = $10$.
• Step 3: Since the result is $10$ (which is greater than $9$), regroup by writing down the $0$ in the ones place and carrying over the $1$ to the tens place.
• Step 4: Move to the tens place. Add the digits: $1$ (carrying over from the ones place) + $1$ (from $12$) + $5$ (from $58$) = $7$.

Therefore, the sum of the numbers is $70$.

The correct answer choice is:

$70$

To solve the addition problem $19 + 41$, let's perform the following steps:

• Step 1: Align the numbers according to place value, so we have:

$\begin{array}{c c} & 19 \\ + & 41 \\ \hline \end{array}$

• Step 2: Add the units digits (9 and 1). $9 + 1 = 10$. This result implies a 0 in the units place and a carry-over of 1 to the tens column.
• Step 3: Add the tens digits plus any carry-over. We have $1 + 4 + 1 (carry) = 6$.

Thus, combining these we get: $\begin{array}{c c} & 19 \\ + & 41 \\ \hline & 60 \\ \end{array}$

Therefore, the sum of $19 + 41$ is $\textbf{60}$.

To find the sum of $22$ and $38$, we'll perform the following steps:

• Step 1: Align the numbers vertically by place value:
  $\ \ \ 22$
  $+\, 38$
• Step 2: Add the digits in the ones place: $2 + 8 = 10$.
  Since the result $10$ is greater than $9$, we'll place $0$ in the ones place and carry $1$ to the tens place.
• Step 3: Add the digits in the tens place, including the carried over $1$: $2 + 3 + 1 = 6$.
  Place the $6$ in the tens place, making the sum $60$.

Therefore, the sum of $22$ and $38$ is $60$.

Among the provided answer choices, $\ 60$ is the correct answer.