10 sums of BODMAS with answers
Answers
For example:
7 × (15 + 5)
= 7 × 20
= 140
2. Of → Then, solve Of part (Powers, Roots, etc.,) before Multiply, Divide, Add or Subtract.
For example:
6 + 3 of 7 - 5
= 6 + 3 × 7 - 5
= 6 + 21 - 5
= 27 - 5
= 22
3. Division/Multiplication → Then, calculate Multiply orDivide before Add or Subtract start from left to right.
For example:
20 + 21 ÷ 3 × 2
= 20 + 7 × 2
= 20 + 14
= 34
4. Addition/Subtraction → At last Add or Subtract start from left to right.
17 + (8 - 5) × 5
= 17 + 3 × 5
= 17 + 15
= 32
Worked-out problems for solving BODMAS rules - involving integers:
Simplify using BODMAS rule:
(a) 25 - 48 ÷ 6 + 12 × 2
Solution:
25 - 48 ÷ 6 + 12 × 2
= 25 - 8 + 12 × 2, (Simplifying ‘division’ 48 ÷ 6 = 8)
= 25 - 8 + 24, (Simplifying ‘multiplication’ 12 × 2 = 24)
= 17 + 24, (Simplifying ‘subtraction’ 25 - 8 = 17)
= 41, (Simplifying ‘addition’ 17 + 24 = 41)
Answer: 41
(b) 78 - [5 + 3 of (25 - 2 × 10)]
Solution:
78 - [5 + 3 of (25 - 2 × 10)]
= 78 - [5 + 3 of (25 - 20)], (Simplifying ‘multiplication’ 2 × 10 = 20)
= 78 - [5 + 3 of 5], (Simplifying ‘subtraction’ 25 - 20 = 5)
= 78 - [5 + 3 × 5], (Simplifying ‘of’)
= 78 - [5 + 15], (Simplifying ‘multiplication’ 3 × 5 = 15)
= 78 - 20, (Simplifying ‘addition’ 5 + 15 = 20)
= 58, (Simplifying ‘subtraction’ 78 - 20 = 58)
Answer: 58
(c) 52 - 4 of (17 - 12) + 4 × 7
Solution:
52 - 4 of (17 - 12) + 4 × 7
= 52 - 4 of 5 + 4 × 7, (Simplifying ‘parenthesis’ 17 - 12 = 5)
= 52 - 4 × 5 + 4 × 7, (Simplifying ‘of’)
= 52 - 20 + 4 × 7, (Simplifying ‘multiplication’ 4 × 5 = 20)
= 52 - 20 + 28, (Simplifying ‘multiplication’ 4 × 7 = 28)
= 32 + 28, (Simplifying ‘subtraction’ 52 - 20 = 32)
= 60, (Simplifying ‘addition’ 32 + 28 = 60)
Answer: 60
Given,
BODMAS
To find,
10 sums with answers
Solution,
1. Bracket → Solve inside the Brackets before Of, Multiply, Divide, Add or Subtract.
(a) 7 × (15 + 5)
= 7 × 20
= 140
(b) 6 x (54 ÷ 9)
= 6 x 6
= 36
2. Of → Then, solve Of part (Powers, Roots, etc.,) before Multiply, Divide, Add or Subtract.
(a) 6 + 3 of 7 - 5
= 6 + (3 × 7) - 5
= (6 + 21) - 5
= 27 - 5
= 22
(b) 3³ - 2 of 2²
= (27 - 2) x 4
= 25 x 4
= 100
3. Division/Multiplication → Then, calculate Multiply or Divide before Add or Subtract start from left to right.
(a) 20 + 21 ÷ 3 × 2
= 20 + 7 × 2
= 20 + 14
= 34
(b) 40 ÷ 8 x 2 +7
= (5 x 2) + 7
= 10 + 7
= 17
4. Addition/Subtraction → At last Add or Subtract start from left to right.
(a)17 + (8 - 5) × 5
= 17 + 3 × 5
= 17 + 15
= 32
Worked-out problems for solving BODMAS rules - involving integers:
Simplify using BODMAS rule:
(a) 25 - 48 ÷ 6 + 12 × 2
Solution:
25 - 48 ÷ 6 + 12 × 2
= 25 - 8 + 12 × 2, (Simplifying ‘division’ 48 ÷ 6 = 8)
= 25 - 8 + 24, (Simplifying ‘multiplication’ 12 × 2 = 24)
= 17 + 24, (Simplifying ‘subtraction’ 25 - 8 = 17)
= 41, (Simplifying ‘addition’ 17 + 24 = 41)
(b) 78 - [5 + 3 of (25 - 2 × 10)]
Solution:
78 - [5 + 3 of (25 - 2 × 10)]
= 78 - [5 + 3 of (25 - 20)], (Simplifying ‘multiplication’ 2 × 10 = 20)
= 78 - [5 + 3 of 5], (Simplifying ‘subtraction’ 25 - 20 = 5)
= 78 - [5 + 3 × 5], (Simplifying ‘of’)
= 78 - [5 + 15], (Simplifying ‘multiplication’ 3 × 5 = 15)
= 78 - 20, (Simplifying ‘addition’ 5 + 15 = 20)
= 58, (Simplifying ‘subtraction’ 78 - 20 = 58)
(c) 52 - 4 of (17 - 12) + 4 × 7
Solution:
52 - 4 of (17 - 12) + 4 × 7
= 52 - 4 of 5 + 4 × 7, (Simplifying ‘parenthesis’ 17 - 12 = 5)
= 52 - 4 × 5 + 4 × 7, (Simplifying ‘of’)
= 52 - 20 + 4 × 7, (Simplifying ‘multiplication’ 4 × 5 = 20)
= 52 - 20 + 28, (Simplifying ‘multiplication’ 4 × 7 = 28)
= 32 + 28, (Simplifying ‘subtraction’ 52 - 20 = 32)
= 60, (Simplifying ‘addition’ 32 + 28 = 60)
#SPJ3