3rd Grade Common Core Math Worksheets - Free Printable

Here are the solutions to the problems on the worksheet, broken down step-by-step.

Part 1: Find what is missing in each number sentence

These are simple division problems. You can solve them by asking, "How many times does the second number fit into the first?" or by using multiplication facts (e.g., for $6 \div 2$, think $2 \times ? = 6$).

1. $6 \div 2 = 3$ (Because $2 \times 3 = 6$)
2. $10 \div 5 = 2$ (Because $5 \times 2 = 10$)
3. $28 \div 7 = 4$ (Because $7 \times 4 = 28$)
4. $18 \div 3 = 6$ (Because $3 \times 6 = 18$)
5. $16 \div 4 = 4$ (Because $4 \times 4 = 16$)
6. $0 \div 5 = 0$ (Zero divided by any number is always 0)
7. $4 \div 2 = 2$ (Because $2 \times 2 = 4$)
8. $20 \div 4 = 5$ (Because $4 \times 5 = 20$)
9. $8 \div 2 = 4$ (Because $2 \times 4 = 8$)
10. $15 \div 3 = 5$ (Because $3 \times 5 = 15$)
11. $32 \div 4 = 8$ (Because $4 \times 8 = 32$)
12. $45 \div 9 = 5$ (Because $9 \times 5 = 45$)

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Part 2: Balance the equation

For these, you need to make sure the left side equals the right side. First, solve the part of the equation that has all the numbers, then figure out what is missing on the other side.

13. $6 \times 2 = 24 \div 2$
* Right side: $24 \div 2 = 12$.
* Left side: We have a 6. To get 12, we must multiply by 2 ($6 \times 2 = 12$).
* Symbol: $\times$

14. $36 \div 6 = 2 \times 3$
* Left side: $36 \div 6 = 6$.
* Right side: We have $\_ \times 3$. To get 6, the missing number is 2 ($2 \times 3 = 6$).
* Number: 2

15. $45 \div 5 = 10 - 1$
* Left side: $45 \div 5 = 9$.
* Right side: We have $10 - \_$. To get 9, we subtract 1 from 10.
* Number: 1

16. $16 \div 4 = 4 \times 1$
* Left side: $16 \div 4 = 4$.
* Right side: We have $4 \times \_$. To keep it equal to 4, we multiply by 1.
* Number: 1

17. $40 \div 5 = 4 \times 2$
* Left side: $40 \div 5 = 8$.
* Right side: We have $\_ \times 2$. To get 8, the missing number is 4 ($4 \times 2 = 8$).
* Number: 4

18. $10 \div 5 = 2 - 1$
* Left side: $10 \div 5 = 2$.
* Right side: We have $2 \_ 1$. To get 2, we can subtract 1 ($2 - 1 = 1$... wait, $10\div5=2$. So right side needs to be 2. $2+0$? No. Let's re-read carefully. Ah, $10 \div 5 = 2$. The right side is $2 \_ 1$. If we use minus, $2-1=1$ (No). If we use plus, $2+1=3$ (No). If we use divide, $2/1=2$ (Yes). If we use multiply, $2*1=2$ (Yes). Usually, these use basic ops. Let's look at problem 13 again. It used division and multiplication. Let's look at 15. Subtraction.
* Let's re-evaluate #18. Left side is 2. Right side starts with 2. Ends with 1.
* $2 \div 1 = 2$.
* $2 \times 1 = 2$.
* Looking at the spacing, it looks like a single symbol slot. Both $\div$ and $\times$ work mathematically. However, usually, if the result is the same as the first number, dividing by 1 or multiplying by 1 are the standard identities. Let's look at the other problems. They mix operations. Let's assume $\div$ or $\times$. Given the previous problems involved division heavily, $\div$ is a strong candidate, but $\times$ is equally correct. Let's provide $\div$ as it balances the division on the left, but actually, looking at problem 21 ($15 \dots 3 = 45 \times 1$), multiplication is used there. Let's stick with the most common balance. Actually, let's look closer. Is it possible the answer is related to the operation on the left? No. Let's provide $\div$ (divide) as it is a very common way to show identity in these worksheets, though $\times$ is also correct. *Self-correction*: In elementary worksheets, $2 \div 1$ is less common than $2 \times 1$ or $2 + 0$. But 0 isn't an option usually unless specified. Let's look at the visual gap. It's small. Let's go with $\div$ because $10 \div 5$ is division. Wait, looking at #13, the operations were different sides ($6 \times 2$ vs $24 \div 2$). So the operation doesn't have to match. Let's provide $\div$ as the primary answer, but note that $\times$ works too. Actually, let's look at #20. $1 \times 7 = 7 \dots 1$. Left is 7. Right is $7 \dots 1$. To get 7, you do $7 + 0$ (no 0), $7 \times 1$, $7 \div 1$. The blank is likely $\times$ or $\div$.
* Let's re-solve #18 strictly. $10 \div 5 = 2$. Equation: $2 = 2 \text{ [op] } 1$. The only single-digit integer operations that result in 2 using 2 and 1 are $2 \times 1$ and $2 \div 1$. I will list $\div$ as it mirrors the left side's operation type often found in these specific "balance" drills, but $\times$ is perfectly valid. Let's pick $\div$.

19. $14 \times 3 = 28 + 14$
* Right side: $28 + 14 = 42$.
* Left side: $14 \times \_$. To get 42, we know $14 \times 3 = 42$.
* Number: 3

20. $1 \times 7 = 7 \div 1$ (or $7 \times 1$)
* Left side: $1 \times 7 = 7$.
* Right side: $7 \_ 1$. To get 7, we can divide by 1 or multiply by 1. Given the pattern of mixing operations, $\div$ is a good fit, but $\times$ is also correct. I will provide $\div$ to vary from the multiplication on the left, similar to #13.

21. $15 \times 3 = 45 \times 1$
* Right side: $45 \times 1 = 45$.
* Left side: $15 \_ 3$. To get 45, we multiply ($15 \times 3 = 45$).
* Symbol: $\times$

*(Note for 18 and 20: In many of these worksheets, if the number 1 is involved, both multiplication and division are technically correct answers. Dividing by 1 or Multiplying by 1 leaves the number unchanged.)*

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Part 3: Read and solve

Problem: Sandara spends $\$45$ on 9 raffle tickets. How much does each raffle ticket cost?

Step-by-step Solution:
1. We know the total amount spent is $\$45$.
2. We know the total number of tickets bought is 9.
3. To find the cost of *one* ticket, we need to divide the total money by the number of tickets.
4. Equation: $45 \div 9 = ?$
5. Calculation: $9 \times 5 = 45$, so $45 \div 9 = 5$.

Each ticket costs $\$5$.

Final Answer:
Part 1:
1. 3
2. 2
3. 4
4. 6
5. 4
6. 0
7. 2
8. 5
9. 4
10. 5
11. 8
12. 5

Part 2:
13. $\times$
14. 2
15. 1
16. 1
17. 4
18. $\div$ (or $\times$)
19. 3
20. $\div$ (or $\times$)
21. $\times$

Part 3:
Answer: $5