Math worksheet practicing order of operations with integers using BIDMAS.

Let's solve each of the problems step by step using the BIDMAS rule (Brackets, Indices, Division/Multiplication, Addition/Subtraction). BIDMAS is the order of operations used in mathematics to evaluate expressions correctly.

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Problem 17)

$$
(-16 + 12)^3 + (-8)^2
$$

Step 1: Solve inside the brackets:
$$
-16 + 12 = -4
$$

Step 2: Apply exponents:
$$
(-4)^3 = -64 \quad \text{(because } (-4)^3 = -4 \times -4 \times -4 = -64\text{)}
$$
$$
(-8)^2 = 64 \quad \text{(because negative squared becomes positive)}
$$

Step 3: Add:
$$
-64 + 64 = 0
$$

✔ Answer: 0

---

Problem 18)

$$
(-20) \div 5 + (-25) \div (-5)
$$

Step 1: Do division from left to right:
$$
(-20) \div 5 = -4
$$
$$
(-25) \div (-5) = 5 \quad \text{(negative ÷ negative = positive)}
$$

Step 2: Add:
$$
-4 + 5 = 1
$$

✔ Answer: 1

---

Problem 19)

$$
8 \times (-2) + 2 \times (-7)
$$

Step 1: Multiplication first:
$$
8 \times (-2) = -16
$$
$$
2 \times (-7) = -14
$$

Step 2: Add:
$$
-16 + (-14) = -30
$$

✔ Answer: -30

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Problem p)

$$
10^2 \div (5 + 5)^2
$$

Step 1: Brackets:
$$
5 + 5 = 10
$$

Step 2: Indices:
$$
10^2 = 100
$$
$$
(10)^2 = 100
$$

Step 3: Division:
$$
100 \div 100 = 1
$$

✔ Answer: 1

---

Problem r)

$$
3 + 7^2 - (2 + 4)
$$

Step 1: Brackets:
$$
2 + 4 = 6
$$

Step 2: Indices:
$$
7^2 = 49
$$

Step 3: Now expression is:
$$
3 + 49 - 6
$$

Step 4: Left to right:
$$
3 + 49 = 52
$$
$$
52 - 6 = 46
$$

✔ Answer: 46

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Problem t)

$$
(1 + 1)^4 + 3^0 - (3 - 1)^3
$$

Step 1: Brackets:
$$
1 + 1 = 2
$$
$$
3 - 1 = 2
$$

Step 2: Indices:
$$
2^4 = 16
$$
$$
3^0 = 1 \quad \text{(Any non-zero number to power 0 is 1)}
$$
$$
2^3 = 8
$$

Step 3: Expression becomes:
$$
16 + 1 - 8 = 9
$$

✔ Answer: 9

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Problem c)

$$
\frac{48 \div 4^2}{30 \div 10}
$$

Step 1: Evaluate numerator and denominator separately.

Numerator: $48 \div 4^2$

- First, $4^2 = 16$
- Then $48 \div 16 = 3$

Denominator: $30 \div 10 = 3$

Step 2: Divide:
$$
\frac{3}{3} = 1
$$

✔ Answer: 1

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Problem e)

$$
\frac{7^2 + (15 - 10)^0}{\text{?}}
$$

Wait — it seems like the denominator is missing. But based on the image, only the numerator is visible:

$$
7^2 + (15 - 10)^0
$$

But since this is a fraction, we need both numerator and denominator. However, let’s assume the full expression is:

$$
\frac{7^2 + (15 - 10)^0}{\text{something}} \quad \text{but it's not fully shown}
$$

But looking at your image, it might be incomplete. Let's assume it's meant to be:

$$
\frac{7^2 + (15 - 10)^0}{\text{missing part}}
$$

But since the denominator isn't visible, I’ll compute just the numerator:

Step 1: $7^2 = 49$
Step 2: $15 - 10 = 5$, then $5^0 = 1$
Step 3: $49 + 1 = 50$

So if the denominator were, say, 10, it would be 5, but since it's missing, we can’t finish.

⚠️ Note: The problem seems cut off. Please check if the full expression is available.

But assuming the question was just the numerator:
✔ Numerator: 50

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Bonus Problem (Last one):

$$
\frac{-40 + 3 \times (3 + 7)}{5 \times (2^1 - 1^2)}
$$

Step 1: Simplify inside brackets:

Numerator:
- $3 + 7 = 10$
- $3 \times 10 = 30$
- $-40 + 30 = -10$

Denominator:
- $2^1 = 2$
- $1^2 = 1$
- $2 - 1 = 1$
- $5 \times 1 = 5$

Now:
$$
\frac{-10}{5} = -2
$$

✔ Answer: -2

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 17) | 0 |
| 18) | 1 |
| 19) | -30 |
| p) | 1 |
| r) | 46 |
| t) | 9 |
| c) | 1 |
| Bonus | -2 |

Let me know if you want explanations for any specific step or need help with the missing denominator in problem e!