Order of Operations Worksheets (Grade 5, answers, examples) - Free Printable

To solve the given problems, we need to follow the Order of Operations, often remembered by the acronym PEMDAS:

- P: Parentheses
- E: Exponents
- M/D: Multiplication and Division (from left to right)
- A/S: Addition and Subtraction (from left to right)

Let's solve each problem step by step.

---

Problem 1: \( 5 \times (6 + 4) - 4^2 \)

1. Parentheses: Solve \( 6 + 4 \):
\[
6 + 4 = 10
\]
So the expression becomes:
\[
5 \times 10 - 4^2
\]

2. Exponents: Solve \( 4^2 \):
\[
4^2 = 16
\]
So the expression becomes:
\[
5 \times 10 - 16
\]

3. Multiplication: Solve \( 5 \times 10 \):
\[
5 \times 10 = 50
\]
So the expression becomes:
\[
50 - 16
\]

4. Subtraction: Solve \( 50 - 16 \):
\[
50 - 16 = 34
\]

Final Answer:
\[
\boxed{34}
\]

---

Problem 2: \( (7 - 2)^2 + (16 \div 8) \)

1. Parentheses: Solve \( 7 - 2 \) and \( 16 \div 8 \):
\[
7 - 2 = 5 \quad \text{and} \quad 16 \div 8 = 2
\]
So the expression becomes:
\[
5^2 + 2
\]

2. Exponents: Solve \( 5^2 \):
\[
5^2 = 25
\]
So the expression becomes:
\[
25 + 2
\]

3. Addition: Solve \( 25 + 2 \):
\[
25 + 2 = 27
\]

Final Answer:
\[
\boxed{27}
\]

---

Problem 3: \( (12 - 9)^2 + (8 \div 4) \)

1. Parentheses: Solve \( 12 - 9 \) and \( 8 \div 4 \):
\[
12 - 9 = 3 \quad \text{and} \quad 8 \div 4 = 2
\]
So the expression becomes:
\[
3^2 + 2
\]

2. Exponents: Solve \( 3^2 \):
\[
3^2 = 9
\]
So the expression becomes:
\[
9 + 2
\]

3. Addition: Solve \( 9 + 2 \):
\[
9 + 2 = 11
\]

Final Answer:
\[
\boxed{11}
\]

---

Problem 4: \( (20 - 4^2) \div (10 - 6) \)

1. Parentheses and Exponents: Solve \( 4^2 \), then \( 20 - 4^2 \), and \( 10 - 6 \):
\[
4^2 = 16 \quad \Rightarrow \quad 20 - 16 = 4
\]
\[
10 - 6 = 4
\]
So the expression becomes:
\[
4 \div 4
\]

2. Division: Solve \( 4 \div 4 \):
\[
4 \div 4 = 1
\]

Final Answer:
\[
\boxed{1}
\]

---

Problem 5: \( (64 - 2^2) \div (17 + 3) \)

1. Parentheses and Exponents: Solve \( 2^2 \), then \( 64 - 2^2 \), and \( 17 + 3 \):
\[
2^2 = 4 \quad \Rightarrow \quad 64 - 4 = 60
\]
\[
17 + 3 = 20
\]
So the expression becomes:
\[
60 \div 20
\]

2. Division: Solve \( 60 \div 20 \):
\[
60 \div 20 = 3
\]

Final Answer:
\[
\boxed{3}
\]

---

Problem 6: \( (32 - 2) \div 5 - 2^2 \)

1. Parentheses: Solve \( 32 - 2 \):
\[
32 - 2 = 30
\]
So the expression becomes:
\[
30 \div 5 - 2^2
\]

2. Exponents: Solve \( 2^2 \):
\[
2^2 = 4
\]
So the expression becomes:
\[
30 \div 5 - 4
\]

3. Division: Solve \( 30 \div 5 \):
\[
30 \div 5 = 6
\]
So the expression becomes:
\[
6 - 4
\]

4. Subtraction: Solve \( 6 - 4 \):
\[
6 - 4 = 2
\]

Final Answer:
\[
\boxed{2}
\]

---

Summary of Answers

1. \( 5 \times (6 + 4) - 4^2 = \boxed{34} \)
2. \( (7 - 2)^2 + (16 \div 8) = \boxed{27} \)
3. \( (12 - 9)^2 + (8 \div 4) = \boxed{11} \)
4. \( (20 - 4^2) \div (10 - 6) = \boxed{1} \)
5. \( (64 - 2^2) \div (17 + 3) = \boxed{3} \)
6. \( (32 - 2) \div 5 - 2^2 = \boxed{2} \)