BODMAS math worksheet with arithmetic expressions and a cartoon mouse.

You're working on a BODMAS worksheet — that stands for:

> Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction.

We must follow this order to evaluate expressions correctly.

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## ✔ PART 1: Solve the 16 expressions

Let’s solve each one step by step using BODMAS.

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1. 100 – (20 × 3)

→ Brackets first: 20 × 3 = 60
→ Then: 100 – 60 = 40

✔ Answer: 40

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2. (35 – 15) + (27 – 7)

→ Brackets: 35 – 15 = 20, 27 – 7 = 20
→ Then: 20 + 20 = 40

✔ Answer: 40

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3. 15 + (6 × 6)

→ Brackets: 6 × 6 = 36
→ Then: 15 + 36 = 51

✔ Answer: 51

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4. (4 + 5) × (3 + 6)

→ Brackets: 4+5=9, 3+6=9
→ Then: 9 × 9 = 81

✔ Answer: 81

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5. (5 + 5) × (5 – 2)

→ Brackets: 5+5=10, 5–2=3
→ Then: 10 × 3 = 30

✔ Answer: 30

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6. 50 – (6 × 6)

→ Brackets: 6×6=36
→ Then: 50 – 36 = 14

✔ Answer: 14

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7. (4 + 8) × (3 – 2)

→ Brackets: 4+8=12, 3–2=1
→ Then: 12 × 1 = 12

✔ Answer: 12

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8. (9 – 3) + (6 × 6)

→ Brackets: 9–3=6, 6×6=36
→ Then: 6 + 36 = 42

✔ Answer: 42

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9. (5 × 7) – (2 × 5)

→ Brackets: 5×7=35, 2×5=10
→ Then: 35 – 10 = 25

✔ Answer: 25

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10. 56 – (4 × 7)

→ Brackets: 4×7=28
→ Then: 56 – 28 = 28

✔ Answer: 28

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11. 78 – (10 × 7)

→ Brackets: 10×7=70
→ Then: 78 – 70 = 8

✔ Answer: 8

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12. (7 × 7) + (4 × 8)

→ Brackets: 7×7=49, 4×8=32
→ Then: 49 + 32 = 81

✔ Answer: 81

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13. (45 – 23) + (5 × 8)

→ Brackets: 45–23=22, 5×8=40
→ Then: 22 + 40 = 62

✔ Answer: 62

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14. 38 – (5 × 7)

→ Brackets: 5×7=35
→ Then: 38 – 35 = 3

✔ Answer: 3

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15. (100 – 45) + (7 × 7)

→ Brackets: 100–45=55, 7×7=49
→ Then: 55 + 49 = 104

✔ Answer: 104

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16. 45 – (9 × 4)

→ Brackets: 9×4=36
→ Then: 45 – 36 = 9

✔ Answer: 9

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## 🧩 PART 2: Tricky Problems — Insert brackets to get different answers

We are to insert brackets in different places in these two expressions and find how many different sums/answers we can get.

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Problem 1: 4 + 4 × 5 – 3

Without brackets, BODMAS says: multiplication first → 4×5=20, then 4+20–3 = 21

But we can insert brackets to change the order.

Let’s list all possible meaningful bracket placements (we won’t count redundant ones like `(4)`).

#### Option A: (4 + 4) × 5 – 3
→ 8 × 5 = 40 → 40 – 3 = 37

#### Option B: 4 + (4 × 5) – 3 → same as no brackets → 21

#### Option C: 4 + 4 × (5 – 3)
→ 5–3=2 → 4×2=8 → 4+8= 12

#### Option D: (4 + 4 × 5) – 3
→ Inside: 4×5=20 → 4+20=24 → 24–3= 21 → same as original

#### Option E: 4 + (4 × 5 – 3)
→ 4×5=20 → 20–3=17 → 4+17= 21 → again same

#### Option F: (4 + 4) × (5 – 3)
→ 8 × 2 = 16

#### Option G: 4 + 4 × 5 – 3 → already did → 21

#### Option H: ((4 + 4) × 5) – 3 → same as A → 37

#### Option I: 4 + (4 × (5 – 3)) → same as C → 12

So unique results so far:
- 21 (original)
- 37 (A)
- 12 (C)
- 16 (F)

✔ Different answers: 4

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Problem 2: 8 + 5 × 1 + 3 – 6

Without brackets: multiplication first → 5×1=5 → then left to right:
8 + 5 = 13 → 13 + 3 = 16 → 16 – 6 = 10

Now try different bracket placements.

#### Option A: (8 + 5) × 1 + 3 – 6
→ 13 × 1 = 13 → 13 + 3 = 16 → 16 – 6 = 10 → same

#### Option B: 8 + (5 × 1) + 3 – 6 → same as original → 10

#### Option C: 8 + 5 × (1 + 3) – 6
→ 1+3=4 → 5×4=20 → 8+20=28 → 28–6= 22

#### Option D: (8 + 5 × 1) + 3 – 6
→ 5×1=5 → 8+5=13 → 13+3=16 → 16–6= 10

#### Option E: 8 + 5 × 1 + (3 – 6)
→ 3–6= -3 → 5×1=5 → 8+5=13 → 13 + (-3) = 10

#### Option F: (8 + 5 × 1 + 3) – 6
→ 5×1=5 → 8+5+3=16 → 16–6= 10

#### Option G: 8 + (5 × 1 + 3) – 6
→ 5×1=5 → 5+3=8 → 8+8=16 → 16–6= 10

#### Option H: (8 + 5) × (1 + 3) – 6
→ 13 × 4 = 52 → 52 – 6 = 46

#### Option I: 8 + 5 × (1 + 3 – 6)
→ 1+3–6= -2 → 5×(-2)= -10 → 8 + (-10) = -2

#### Option J: (8 + 5 × 1 + 3 – 6) → same as original → 10

#### Option K: ((8 + 5) × 1) + 3 – 6 → same as A → 10

So unique results:
- 10 (original)
- 22 (C)
- 46 (H)
- -2 (I)

✔ Different answers: 4

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## ✔ FINAL ANSWERS:

Part 1: Expressions 1–16

1. 40
2. 40
3. 51
4. 81
5. 30
6. 14
7. 12
8. 42
9. 25
10. 28
11. 8
12. 81
13. 62
14. 3
15. 104
16. 9

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Part 2: Tricky Problems

1. 4 + 4 × 5 – 3 → Can make 4 different answers: 21, 37, 12, 16

2. 8 + 5 × 1 + 3 – 6 → Can make 4 different answers: 10, 22, 46, -2

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✔ Total different answers for tricky problems: 4 each → 8 total unique values across both.

But the question asks: *“how many different sums and answers can you find for these two sequences”* — likely meaning per sequence.

So:

> For each tricky problem, you can find 4 different answers.

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🎉 All done! You’ve mastered BODMAS!