What is BODMAS?
Rule · Priority · Types of Brackets · Key Notes
Theory
BODMAS is a universally accepted mathematical rule that defines the correct order to perform arithmetic operations when an expression contains more than one operation. Without it, different people would get different answers for the same expression!
Full form: Brackets → Orders (Powers/Roots) → Division → Multiplication → Addition → Subtraction
Equal Priority: Division and Multiplication share the same priority — solve them left to right. Likewise, Addition and Subtraction share the same priority — solve left to right.
Also known as: PEMDAS (USA) · BIDMAS (UK) · BEDMAS (Australia) — all the same rule, different names!
| Letter | Stands For | Priority & Rule | Example |
|---|---|---|---|
| B | Brackets | Solve innermost bracket first — Vinculum → ( ) → { } → [ ] | (3 + 5) = 8 first |
| O | Orders | Powers (x²) and roots (√x) — second priority | 3² = 9 · √16 = 4 |
| D | Division | Equal priority with M — solve left to right | 12 ÷ 4 = 3 |
| M | Multiplication | Equal priority with D — solve left to right | 3 × 5 = 15 |
| A | Addition | Equal priority with S — solve left to right | 7 + 3 = 10 |
| S | Subtraction | Equal priority with A — solve left to right | 10 − 4 = 6 |
📌 Rule 1 — Bracket Order
Always solve the innermost bracket first. Vinculum (bar) → Round ( ) → Curly { } → Square [ ]
📌 Rule 2 — Left to Right
When D & M (or A & S) appear together, always go left to right. D does NOT always beat M.
📌 Rule 3 — "Of" = Multiply
"Of" is treated as multiplication (under O or M). Example: ½ of 20 = ½ × 20 = 10
📌 Rule 4 — Inside Brackets
Apply full BODMAS inside every bracket before working outside it. Don't skip inner steps.
Section 1 — Basic BODMAS
Simple expressions · No powers | Q 1 – 10
Easy ⭐
Q1
18 + 6 ÷ 2 − 3 × 2
Easy
— Solution —
D
18 + 6 ÷ 2 − 3 × 2 = 18 + 3 − 3 × 2
6 ÷ 2 = 3
M
18 + 3 − 3 × 2 = 18 + 3 − 6
3 × 2 = 6
A&S
18 + 3 − 6 = 21 − 6 = 15
Left to right: 18+3=21, then 21−6=15
✅ Answer = 15
Q2
3 × (4 + 8) ÷ 6 − 1
Easy
— Solution —
B
3 × (4 + 8) ÷ 6 − 1 = 3 × 12 ÷ 6 − 1
Bracket first: 4 + 8 = 12
M then D
3 × 12 ÷ 6 − 1 = 36 ÷ 6 − 1 = 36 ÷ 6 − 1 = 6 − 1
Left to right: multiply first (3×12=36), then divide (36÷6=6)
S
6 − 1 = 5
✅ Answer = 5
Q3
25 − 5 × 4 + 20 ÷ 5
Easy
— Solution —
D
25 − 5 × 4 + 20 ÷ 5 = 25 − 5 × 4 + 4
M
25 − 5 × 4 + 4 = 25 − 20 + 4
A&S
25 − 20 + 4 = 5 + 4 = 9
Left to right: 25−20=5, then 5+4=9
✅ Answer = 9
Q4
15 + (20 − 8) ÷ 4 × 3
Easy
— Solution —
B
15 + (20 − 8) ÷ 4 × 3 = 15 + 12 ÷ 4 × 3
D then M
15 + 12 ÷ 4 × 3 = 15 + 3 × 3 = 15 + 3 × 3 = 15 + 9
D first (12÷4=3), then M (3×3=9) — both left to right
A
15 + 9 = 24
✅ Answer = 24
Q5
7 + 7 × 7 − 7 ÷ 7
Easy
— Solution —
D
7 + 7 × 7 − 7 ÷ 7 = 7 + 7 × 7 − 1
M
7 + 7 × 7 − 1 = 7 + 49 − 1
A&S
7 + 49 − 1 = 56 − 1 = 55
✅ Answer = 55
💡 A classic trick question — all 7s! Many students rush and write 7 as the answer (wrong). Always follow BODMAS strictly.
Q6
100 ÷ 4 ÷ 5 × 2
Easy
— Solution —
D (left→right)
100 ÷ 4 ÷ 5 × 2 = 25 ÷ 5 × 2
D again
25 ÷ 5 × 2 = 5 × 2
M
5 × 2 = 10
✅ Answer = 10
⚠️ Common mistake: doing 5 × 2 = 10 first, then 100 ÷ 4 ÷ 10 = 2.5. That is WRONG. D and M are equal — always go left to right!
Q7
80 ÷ (15 + 8 − 3) − 1
Easy
— Solution —
B
80 ÷ (15 + 8 − 3) − 1 = 80 ÷ 20 − 1
Inside bracket: 15+8=23, 23−3=20
D
80 ÷ 20 − 1 = 4 − 1
S
4 − 1 = 3
✅ Answer = 3
Q8
(8 + 4 − 2) × (17 − 12) × 10 − 89
Easy
— Solution —
B (both brackets)
(8+4−2) × (17−12) × 10 − 89 = 10 × 5 × 10 − 89
1st bracket: 8+4=12, 12−2=10. 2nd bracket: 17−12=5
M
10 × 5 × 10 − 89 = 50 × 10 − 89 = 500 − 89
S
500 − 89 = 411
✅ Answer = 411
Q9
3×7 + 4 − 6÷3 − 7 + 45÷5×4 + 49
Easy
— Solution —
D first (left→right)
3×7 + 4 − 6÷3 − 7 + 45÷5×4 + 49 = 3×7 + 4 − 2 − 7 + 9×4 + 49
M (left→right)
3×7 + 4 − 2 − 7 + 9×4 + 49 = 21 + 4 − 2 − 7 + 36 + 49
A&S
21+4−2−7+36+49 = 25−2−7+85 = 110−9 = 101
✅ Answer = 101
Q10
50 ÷ (15 + 8 − 3) × 2
Easy
— Solution —
B
50 ÷ (15+8−3) × 2 = 50 ÷ 20 × 2
D then M (left→right)
50 ÷ 20 × 2 = 2.5 × 2 = 5
✅ Answer = 5
Section 2 — Nested Brackets
Multiple brackets: ( ) { } [ ] · Q 11 – 20
Medium ⭐⭐
Q11
50 − [10 − {5 − (3 − 2)}]
Medium
— Solution —
( ) Innermost
50 − [10 − {5 − (3−2)}] = 50 − [10 − {5 − 1}]
{ } Curly
50 − [10 − {5−1}] = 50 − [10 − 4]
[ ] Square
50 − [10−4] = 50 − 6
S
50 − 6 = 44
✅ Answer = 44
Q12
2 × [3 + {4 × (5 − 1)}]
Medium
— Solution —
( )
2 × [3 + {4 × (5−1)}] = 2 × [3 + {4 × 4}]
{ } — M inside
2 × [3 + {4×4}] = 2 × [3 + 16]
[ ]
2 × [3+16] = 2 × 19
M
2 × 19 = 38
✅ Answer = 38
Q13
100 − [20 + {30 − (10 + 5)}]
Medium
— Solution —
( )
100−[20+{30−(10+5)}] = 100−[20+{30−15}]
{ }
100−[20+{15}] = 100−[20+15] = 100−[35] = 100−35
S
100 − 35 = 65
✅ Answer = 65
Q14
20 − [10 ÷ {5 × (3 − 1)}]
Medium
— Solution —
( )
20−[10÷{5×(3−1)}] = 20−[10÷{5×2}]
{ } — M inside
20−[10÷{5×2}] = 20−[10÷10]
[ ] — D inside
20−[10÷10] = 20−1
S
20 − 1 = 19
✅ Answer = 19
Q15
5 × [3 + {8 − (4 ÷ 2 + 1)}]
Medium
— Solution —
( ) — BODMAS inside
Inside ( ): 4÷2=2, then 2+1=3 → 5×[3+{8−3}]
{ }
5×[3+{5}] = 5×[3+5] = 5×[8]
M
5 × 8 = 40
✅ Answer = 40
Q16
3 × {5 + (4 × 2 − 6) ÷ 2} − 8
Medium
— Solution —
( ) — BODMAS inside
Inside ( ): 4×2=8, 8−6=2 → 3×{5 + 2 ÷ 2} − 8
{ } — D then A
Inside { }: 2÷2=1, 5+1=6 → 3 × 6 − 8
M then S
3×6−8 = 18−8 = 10
✅ Answer = 10
Q17
60 ÷ [4 + {3 − (2 − 1)} × 2]
Medium
— Solution —
( )
60÷[4+{3−(2−1)}×2] = 60÷[4+{3−1}×2]
{ } then M
60÷[4+{2}×2] = 60÷[4+2×2] = 60÷[4+4] = 60÷[8]
D
60÷8 = 7.5
✅ Answer = 7.5
Q18
[20 − 5] − {10 − (3 × 2)}
Medium
— Solution —
( ) — M inside
[20−5] − {10−(3×2)} = [20−5] − {10−6}
[ ] and { }
[15] − {4} = 15 − 4
S
15 − 4 = 11
✅ Answer = 11
Q19
40 ÷ {2 + (6 × 3 − 8)}
Medium
— Solution —
( ) — M then S inside
Inside ( ): 6×3=18, 18−8=10 → 40÷{2+10}
{ }
40÷{12} = 40÷12
D
40÷12 = 10/3 ≈ 3.33
✅ Answer = 10⁄3
Q20
[3 × {5 + (½ × 8)} − 10] ÷ 7
Medium
— Solution —
( ) — M inside
Inside ( ): ½×8=4 → [3×{5+4}−10]÷7
{ }
[3×{9}−10]÷7 = [27−10]÷7 = [17]÷7
D
17÷7 = 17/7 ≈ 2.43
✅ Answer = 17⁄7
Section 3 — Orders: Powers & Roots
Exponents (² ³) · Square roots (√) · Q 21 – 30
Medium – Hard ⭐⭐⭐
Q21
3² + 4 × 2 − √25
Medium
— Solution —
O
3² + 4×2 − √25 = 9 + 4×2 − 5
3²=9 · √25=5
M
9 + 4×2 − 5 = 9 + 8 − 5
A&S
9 + 8 − 5 = 17 − 5 = 12
✅ Answer = 12
Q22
(3 + 5) × 2² − 4
Medium
— Solution —
B
(3+5) × 2² − 4 = 8 × 2² − 4
O
8 × 2² − 4 = 8 × 4 − 4
M then S
8×4 − 4 = 32 − 4 = 28
✅ Answer = 28
Q23
12 ÷ 4 × 2 + 3² − (9 + 4)
Medium
— Solution —
B
12÷4×2 + 3² − (9+4) = 12÷4×2 + 3² − 13
O
12÷4×2 + 3² − 13 = 12÷4×2 + 9 − 13
D&M left→right
12÷4×2 + 9 − 13 = 3×2 + 9 − 13 = 3×2+9−13 = 6+9−13
A&S
6+9−13 = 15−13 = 2
✅ Answer = 2
Q24
5² − 2 × (4 + 1) ÷ 5 + √9
Hard
— Solution —
B
5²−2×(4+1)÷5+√9 = 5²−2×5÷5+√9
O
5²−2×5÷5+√9 = 25−2×5÷5+3
M then D (left→right)
25−2×5÷5+3 = 25−10÷5+3 = 25−2+3
A&S
25−2+3 = 23+3 = 26
✅ Answer = 26
Q25
(3 + 2)³ − 2 × (4 + 1) ÷ 5
Hard
— Solution —
B (both brackets)
(3+2)³ − 2×(4+1)÷5 = 5³ − 2×5÷5
O
5³ − 2×5÷5 = 125 − 2×5÷5
M then D (left→right)
125−2×5÷5 = 125−10÷5 = 125−2 = 123
✅ Answer = 123
Q26
7 − 3 × (2² + 1) + 8
Medium
— Solution —
O inside bracket
7−3×(2²+1)+8 = 7−3×(4+1)+8 = 7−3×(5)+8
M
7−3×5+8 = 7−15+8
A&S
7−15+8 = −8+8 = 0
✅ Answer = 0
Q27
√64 + 3³ ÷ (2 + 7) × 3
Hard
— Solution —
B
√64+3³÷(2+7)×3 = √64+3³÷9×3
O
√64+3³÷9×3 = 8+27÷9×3
D then M
8+27÷9×3 = 8+3×3 = 8+3×3 = 8+9
A
8 + 9 = 17
✅ Answer = 17
Q28
[2³ + {3² − (1² + 2)}] × 4
Hard
— Solution —
O inside ( )
Inside ( ): 1²=1, 1+2=3 → [2³+{3²−3}]×4
O remaining
[2³+{3²−3}]×4 = [8+{9−3}]×4 = [8+{6}]×4 = [14]×4
M
14 × 4 = 56
✅ Answer = 56
Q29
2 + 4 ÷ (2² + 6) × 2
Hard
— Solution —
O inside ( )
2+4÷(2²+6)×2 = 2+4÷(4+6)×2 = 2+4÷(10)×2
D then M (left→right)
2+4÷10×2 = 2+0.4×2 = 2+0.4×2 = 2+0.8
A
2 + 0.8 = 2.8
✅ Answer = 2.8
Q30
3² − 2 × (4 ÷ 2 + 1) × 5
Hard
— Solution —
B — BODMAS inside ( )
Inside ( ): 4÷2=2, 2+1=3 → 3²−2×(3)×5
O
3²−2×3×5 = 9−2×3×5
M (left→right)
9−2×3×5 = 9−6×5 = 9−30 = −21
✅ Answer = −21
💡 Note the negative answer — perfectly valid! Make sure to keep track of signs through every step.
Section 4 — Fractions & Decimals
BODMAS with rational numbers | Q 31 – 40
Hard ⭐⭐⭐⭐
Q31
½ of 24 + 6 ÷ 3 − 2
Medium
— Solution —
"of" = M
½ of 24 + 6÷3−2 = ½×24+6÷3−2 = 12+6÷3−2
D
12+6÷3−2 = 12+2−2
A&S
12+2−2 = 14−2 = 12
✅ Answer = 12
Q32
¾ × 16 + ⅔ × 12 − 5
Hard
— Solution —
M (both, left→right)
¾×16 + ⅔×12 − 5 = 12 + 8 − 5
¾ × 16 = 12 · ⅔ × 12 = 8
A&S
12 + 8 − 5 = 20 − 5 = 15
✅ Answer = 15
Q33
0.5 × (4 + 2) − 0.6 ÷ 0.3 + 1
Hard
— Solution —
B
0.5×(4+2)−0.6÷0.3+1 = 0.5×6−0.6÷0.3+1
M then D (left→right)
0.5×6−0.6÷0.3+1 = 3−0.6÷0.3+1 = 3−2+1
A&S
3−2+1 = 1+1 = 2
✅ Answer = 2
Q34
2.4 + 1.6 × 5 ÷ (0.5 + 1.5)
Hard
— Solution —
B
2.4+1.6×5÷(0.5+1.5) = 2.4+1.6×5÷2
M then D (left→right)
2.4+1.6×5÷2 = 2.4+8÷2 = 2.4+4
A
2.4 + 4 = 6.4
✅ Answer = 6.4
Q35
(1/5 + 3/5) × 10 − 2²
Hard
— Solution —
B
(1/5+3/5)×10−2² = (4/5)×10−2²
O
(4/5)×10−2² = (4/5)×10−4
M then S
(4/5)×10−4 = 8−4 = 4
✅ Answer = 4
Q36
5 + [2 × {3.5 − (1.5 + 0.5)} + 4]
Hard
— Solution —
( )
5+[2×{3.5−(1.5+0.5)}+4] = 5+[2×{3.5−2}+4]
{ } — M inside
5+[2×{1.5}+4] = 5+[3+4]
[ ]
5+[7] = 12
✅ Answer = 12
Q37
(⅔ + ⅓) of 45 − 20 ÷ 4
Hard
— Solution —
B
(⅔+⅓) of 45−20÷4 = 1 of 45−20÷4
"of" = M
1 × 45−20÷4 = 45−20÷4
D then S
45−20÷4 = 45−5 = 40
✅ Answer = 40
Q38
0.25 × (12 + 8) ÷ 0.5 + 3²
Hard
— Solution —
B
0.25×(12+8)÷0.5+3² = 0.25×20÷0.5+3²
O
0.25×20÷0.5+3² = 0.25×20÷0.5+9
M then D (left→right)
0.25×20÷0.5+9 = 5÷0.5+9 = 10+9 = 19
✅ Answer = 19
Q39
1.5² + (3.6 ÷ 0.9) − 2
Hard
— Solution —
B
1.5²+(3.6÷0.9)−2 = 1.5²+4−2
O
1.5²+4−2 = 2.25+4−2
A&S
2.25+4−2 = 6.25−2 = 4.25
✅ Answer = 4.25
Q40
3 × {5 + (4 × 2 − 6) ÷ 2} − 8
Hard
— Solution —
( ) — M then S inside
Inside ( ): 4×2=8, 8−6=2 → 3×{5 + 2 ÷ 2} − 8
{ } — D then A inside
Inside { }: 2÷2=1, 5+1=6 → 3 × 6 − 8
M then S
3×6−8 = 18−8 = 10
✅ Answer = 10
Common Mistakes to Avoid
Where Class 8 students go wrong most often
Must Read
❌Mistake 1 — Division ALWAYS before Multiplication
Wrong: In
Correct: Left to right —
D and M have equal priority — always work left to right!
8 ÷ 4 × 2, doing 4×2=8 first → 8÷8 = 1 ✗Correct: Left to right —
8÷4=2, then 2×2 = 4 ✓D and M have equal priority — always work left to right!
❌Mistake 2 — Adding before Multiplying
Wrong: In
Correct: Multiply first → 3×4=12, then 5+12 = 17 ✓
M always comes before A — no exceptions!
5 + 3 × 4, doing 5+3=8 first → 8×4 = 32 ✗Correct: Multiply first → 3×4=12, then 5+12 = 17 ✓
M always comes before A — no exceptions!
❌Mistake 3 — Wrong bracket order
Wrong: Solving
Correct: Always — Vinculum →
Innermost bracket must always be solved first!
[ ] before ( ) in nested expressions ✗Correct: Always — Vinculum →
( ) → { } → [ ] ✓Innermost bracket must always be solved first!
❌Mistake 4 — Forgetting negative answers
In
Don't panic at negatives — they are correct if BODMAS is followed!
9 − 2×5×2: 9 − 20 = −11 (negative is valid!) ✓Don't panic at negatives — they are correct if BODMAS is followed!
❌Mistake 5 — Distributing power across bracket
Wrong:
Correct: Solve bracket first → (3+2)=5, then 5² = 25 ✓
A power applies to the entire bracket, not each term!
(3+2)² = 3²+2² = 9+4 = 13 ✗Correct: Solve bracket first → (3+2)=5, then 5² = 25 ✓
A power applies to the entire bracket, not each term!
Live MCQ Quiz — 20 Questions
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Quick Revision — All 40 Answers
Glance before exams · Full answer reference
Exam Prep
| # | Expression | Key Operations | Answer |
|---|---|---|---|
| SECTION 1 — BASIC | |||
| 1 | 18 + 6÷2 − 3×2 | D→M→A&S | 15 |
| 2 | 3×(4+8)÷6 − 1 | B→M→D→S | 5 |
| 3 | 25 − 5×4 + 20÷5 | D→M→A&S | 9 |
| 4 | 15 + (20−8)÷4×3 | B→D→M→A | 24 |
| 5 | 7 + 7×7 − 7÷7 | D→M→A&S | 55 |
| 6 | 100÷4÷5×2 | D→D→M (left→right) | 10 |
| 7 | 80÷(15+8−3) − 1 | B→D→S | 3 |
| 8 | (8+4−2)×(17−12)×10−89 | B→M→S | 411 |
| 9 | 3×7+4−6÷3−7+45÷5×4+49 | D→M→A&S | 101 |
| 10 | 50÷(15+8−3)×2 | B→D→M | 5 |
| SECTION 2 — NESTED BRACKETS | |||
| 11 | 50−[10−{5−(3−2)}] | ( )→{ }→[ ] | 44 |
| 12 | 2×[3+{4×(5−1)}] | ( )→{ }→[ ]→M | 38 |
| 13 | 100−[20+{30−(10+5)}] | ( )→{ }→[ ] | 65 |
| 14 | 20−[10÷{5×(3−1)}] | ( )→{ }→[ ] | 19 |
| 15 | 5×[3+{8−(4÷2+1)}] | BODMAS inside each | 40 |
| 16 | 3×{5+(4×2−6)÷2}−8 | ( )→{ }→M→S | 10 |
| 17 | 60÷[4+{3−(2−1)}×2] | ( )→{ }→M→[ ]→D | 7.5 |
| 18 | [20−5]−{10−(3×2)} | ( )M→[ ]→{ } | 11 |
| 19 | 40÷{2+(6×3−8)} | ( )→{ }→D | 10/3 |
| 20 | [3×{5+(½×8)}−10]÷7 | ( )→{ }→[ ]→D | 17/7 |
| SECTION 3 — POWERS & ROOTS | |||
| 21 | 3² + 4×2 − √25 | O→M→A&S | 12 |
| 22 | (3+5)×2² − 4 | B→O→M→S | 28 |
| 23 | 12÷4×2 + 3² − (9+4) | B→O→D&M→A&S | 2 |
| 24 | 5²−2×(4+1)÷5+√9 | B→O→M→D→A&S | 26 |
| 25 | (3+2)³−2×(4+1)÷5 | B→O→M→D | 123 |
| 26 | 7−3×(2²+1)+8 | O inside→M→A&S | 0 |
| 27 | √64 + 3³÷(2+7)×3 | B→O→D→M→A | 17 |
| 28 | [2³+{3²−(1²+2)}]×4 | All brackets+O→M | 56 |
| 29 | 2+4÷(2²+6)×2 | O inside B→D→M→A | 2.8 |
| 30 | 3²−2×(4÷2+1)×5 | B→O→M | −21 |
| SECTION 4 — FRACTIONS & DECIMALS | |||
| 31 | ½ of 24 + 6÷3 − 2 | "of"=M→D→A&S | 12 |
| 32 | ¾×16 + ⅔×12 − 5 | M→A→S | 15 |
| 33 | 0.5×(4+2)−0.6÷0.3+1 | B→M→D→A&S | 2 |
| 34 | 2.4+1.6×5÷(0.5+1.5) | B→M→D→A | 6.4 |
| 35 | (1/5+3/5)×10 − 2² | B→O→M→S | 4 |
| 36 | 5+[2×{3.5−(1.5+0.5)}+4] | ( )→{ }M→[ ]→A | 12 |
| 37 | (⅔+⅓) of 45 − 20÷4 | B→"of"→D→S | 40 |
| 38 | 0.25×(12+8)÷0.5 + 3² | B→O→M→D→A | 19 |
| 39 | 1.5²+(3.6÷0.9)−2 | B→O→A&S | 4.25 |
| 40 | 3×{5+(4×2−6)÷2}−8 | ( )→{ }→M→S | 10 |
// BODMAS MASTERY TIPS
Rule 1: Start with the INNERMOST bracket — always.
Rule 2: D and M are equal priority — left to right only.
Rule 3: A and S are equal priority — left to right only.
Rule 4: "of" = multiplication (solve like O/M).
Rule 5: Power applies to entire bracket — NEVER distribute.
Rule 1: Start with the INNERMOST bracket — always.
Rule 2: D and M are equal priority — left to right only.
Rule 3: A and S are equal priority — left to right only.
Rule 4: "of" = multiplication (solve like O/M).
Rule 5: Power applies to entire bracket — NEVER distribute.
