# Using Place Value to Add Two Digit Numbers (Regrouping) — Let's Practice!

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### TranscriptUsing Place Value to Add Two Digit Numbers (Regrouping) — Let's Practice!

Razzi says get these items ready (...) Because today we're going to practice... Using Place Value to Add Two Digit Numbers (Regrouping) It's time to begin! Remember that there are many strategies to add, like using place value. Place value is the value of each digit depending on where it is in the number. Let's use base ten blocks to add and regroup with place value. What is twenty-five plus thirty-eight? You can add the tens rods together first, then the ones blocks. Then, if you have enough ones to make a ten, regroup and add it to the tens place. Did you also get sixty-three? Let's tackle the next problem! What is thirty-six plus thirty-four? You can add the tens rods together first, then the ones blocks. Then, if you have enough ones to make a ten, regroup and add it to the tens place. Did you also get seventy? Let's tackle the final problem! What is forty-six plus thirty-seven? You can add the tens rods together first, then the ones blocks. Then, if you have enough ones to make a ten, regroup and add it to the tens place. Did you also get eighty-three? Razzi had so much fun practicing with you today! See you next time!

## Using Place Value to Add Two Digit Numbers (Regrouping) — Let's Practice! exercise

Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Using Place Value to Add Two Digit Numbers (Regrouping) — Let's Practice!.
• ### Find the sum.

Hints

How many blue tens rods will there be at first? 2 + 3 = 5.

How many green ones blocks will there be at first? 6 + 5 = 11.

We have enough ones to make a ten, so we regroup and add them to the tens place on the left.

We add the tens rods together and the ones blocks together to calculate the sum.

Solution

• 2 is in the tens place of 26, so we draw 2 blue tens rods in the first row. 6 is in the ones place, so we draw 6 green ones blocks.
• 3 is in the tens place of 35, so we draw 3 blue tens rods in the second row. 5 is in the ones place, so we draw 5 green ones blocks.
• We have enough ones in the second row to make a ten, so we regroup and add them to the tens place on the left.
Finally, we add the tens rods together and the ones blocks together to calculate the sum:

• For the tens rods: 2 + 3 + 1 = 6. 6 is the number in the tens place of the sum.
• For the ones blocks: there is only 1 left. 1 is the number in the ones place of the sum.
The sum is 61.

• ### Find each sum.

Hints

For example, 29 + 27 has 2 + 2 = 4 tens rods.

29 + 27 has 9 + 7 = 16 ones blocks.

We have enough ones to make a ten, so we regroup and add them to the tens place on the left.

We add the tens rods together (2 + 2 + 1) and the ones blocks together (4 + 2) to calculate the sum.

Solution

To solve 29 + 27:

• 2 is in the tens place of 29, so we draw 2 blue tens rods in the first row. 9 is in the ones place, so we draw 9 green ones blocks.
• 2 is in the tens place of 27, so we draw 2 blue tens rods in the second row. 7 is in the ones place, so we draw 7 green ones blocks.
• We have enough ones in the second row to make a ten, so we regroup and add them to the tens place on the left.
Finally, we add the tens rods together and the ones blocks together to calculate the sum:

• For the tens rods: 2 + 2 + 1 = 5. 5 is the number in the tens place of the sum.
• For the ones blocks: 4 + 2 = 6. 6 is the number in the ones place of the sum.
The sum is 56.

* 23 + 17 = 40

* 45 + 28 = 73

• ### Find each sum.

Hints

For example, 18 + 49 has 1 + 4 = 5 tens rods.

For example, 18 + 49 has 8 + 9 = 17 ones blocks.

We have enough ones to make a ten, so we regroup and add them to the tens place on the left.

We add the tens rods together (1 + 4 + 1) and the ones blocks together (3 + 4) to calculate the sum.

Solution

• 18 + 49 = 67
• 15 + 17 = 32
• 27 + 23 = 50
• 42 + 38 = 80
To solve 18 + 49:

• 1 is in the tens place of 18, so we draw 1 blue tens rod in the first row. 8 is in the ones place, so we draw 8 green ones blocks.
• 4 is in the tens place of 49, so we draw 4 blue tens rods in the second row. 9 is in the ones place, so we draw 9 green ones blocks.
• We have enough ones in the second row to make a ten, so we regroup and add them to the tens place on the left.
Finally, we add the tens rods together and the ones blocks together to calculate the sum:

• For the tens rods: 1 + 4 + 1 = 6. 6 is the number in the tens place of the sum.
• For the ones blocks: 3 + 4 = 7. 7 is the number in the ones place of the sum.
The sum is 67.

• ### Solve the addition problem.

Hints

In order to find how many total guests Razzi needs to invite, we add the number of friends and the number of family members.

Razzi has 16 friends and 55 family members.

Therefore, we add 16 + 55.

16 + 55 has 6 tens rods and 11 ones blocks.

We have enough ones to make a ten, so we regroup and add them to the tens place on the left.

Solution

In order to find how many total guests Razzi needs to invite, we add the number of friends and the number of family members. Razzi has 16 friends and 55 family members. Therefore, we add 16 + 55.

• 16 + 55 has 6 tens rods and 11 ones blocks.
• We have enough ones to make a ten, so we regroup and add them to the tens place on the left.
Finally, we add the tens rods together and the ones blocks together to calculate the sum:

• For the tens rods: 1 + 5 + 1 = 7. 7 is the number in the tens place of the sum.
• For the ones blocks: there is only 1 left. 1 is the number in the ones place of the sum.
The sum is 71.

Razzi needs to invite 71 guests to his party.

• ### Complete the addition problem.

Hints

How many blue tens rods will there be at first? 2 + 1 = 3.

How many green ones blocks will there be at first? 9 + 5 = 14.

We have enough ones to make a ten, so we regroup and add them to the tens place on the left.

We add the tens rods together and the ones blocks together to calculate the sum.

Solution

• 2 is in the tens place of 29, so we draw 2 blue tens rods in the first row. 9 is in the ones place, so we draw 9 green ones blocks.
• 1 is in the tens place of 15, so we draw 1 blue tens rod in the second row. 5 is in the ones place, so we draw 5 green ones blocks.
• We have enough ones in the second row to make a ten, so we regroup and add them to the tens place on the left.
Finally, we add the tens rods together and the ones blocks together to calculate the sum:

• For the tens rods: 2 + 1 + 1 = 4. 4 is the number in the tens place of the sum.
• For the ones blocks: there are 4 left. 4 is the number in the ones place of the sum.
The sum is 44.

• ### Complete each addition problem.

Hints

For 79 + ___ = 93:

To get a 3 in the ones place of the sum, we need to find a number that when added to the 9 in the ones place of 79 makes 13.

9 + 4 = 13. Therefore, 4 is in the ones place of our answer.

To solve 81 + 29, we follow the same process from the video.

How many tens rods will there be? How many ones blocks will there be?

Solution

To solve 79 + ___ = 93:

• We need to find the number that we add to 79 to get 93. (see image)
• To get a 3 in the ones place of the sum, we need to find a number that when added to the 9 in the ones place of 79 makes 13.
• 9 + 4 = 13. Therefore, 4 is in the ones place of our answer.
• 79 + 14 = 93. The answer is 14.

To solve ___ + 23 = 81:

• We need to find the number that we add to 23 to get 81.
• To get a 1 in the ones place of the sum, we need to find a number that when added to the 3 in the ones place of 23 makes 11.
• 8 + 3 = 11. Therefore, 8 is in the ones place of our answer.
• 58 + 23 = 81. The answer is 58.

To solve 81 + 29 = ___:

• We are finding the sum, so we follow the same process from the video.
• We have 10 tens rods and 10 ones blocks.
• We have enough ones to make a ten, so we regroup and add them to the tens place on the left. There are now 0 ones blocks on the right.
• Finally, we add the tens rods together and the ones blocks together to calculate the sum.
• 81 + 29 = 110. The answer is 110.