# BODMAS Rule

Let us study the BODMAS rule.

##### Introduction

“BODMAS” is the rule for “order of operations” in mathematics.We know that +,- ,×,÷, powers, etc are different operations of mathematics. We have been performing these operations while solving.

But what if they all occur in the same problem? You may go for the”+” operation first, while others may go for a “-“,× or ÷ first. This may lead to different answers to the same problem!!
Thus, to approach mathematics in a proper and certain way rules are made.
It may be called as “BODMAS” rule. ##### Where,
B stands for BRACKETS.
This includes ¯, ( ), { } , [ ].

O stands for ORDER.
This includes powers and roots.

D stands for DIVISION.
which means we have to divide.

M stands for MULTIPLICATION.
which means we will multiply.

Thus we will add the numbers.

S stands for SUBTRACTION.
Hence we will subtract the numbers.

We will perform the operations from  LEFT TO RIGHT . In which we will first go for the “brackets” then “order” and then division, multiplication, addition and subtraction respectively.

##### Solving Brackets.

Note that, while solving brackets. We will solve ¯(bar) vinculum first, then ( ) small brackets, then { } curly brackets and then [ ] square brackets.

Question 1: Solve 2 × (1+4).
Solution: According to the BODMAS rule we will go for brackets first.
2 × (1 + 4)
= 2 ×
Therefore 2 × (1+4) = 10

Question 2: \frac { 2 }{ 3 }×[{(3× 2)+(3+6)}× 2]
Solution:
To solve.

\frac { 2 }{ 3 } × [ { ( 3 × 2 ) + ( 3 + 6 )}× 2]

Step 1 : Solve ( ) small brackets first.

\frac { 2 }{ 3 } × [ { ( 3 × 2 ) + ( 3 + 6 ) } ×  2 ]

\frac { 2 }{ 3 } × [ {
6 + 9 }× 2]

Step 2 : Solve {} curly brackets.

= \frac { 2 }{ 3 } × [ { 6  + 9} × 2]

= \frac { 2 }{ 3 } × [ 15 × 2]

Step 3 : Solve [] square brackets

= \frac { 2 }{ 3 } × [ 15 × 2]

=\frac { 2 }{ 3 } × 30

= 20

##### Exponents

Question 3: { 2 }^{ 2 }× (2+4)
Solution:
Step 1 : Solve ( ) small brackets first.

= { 2 }^{ 2 } × (2+4)
= { 2 }^{ 2 } × 6
Step 2 : Solve the exponent.
={ 2 }^{ 2 } × 6
=× 6
= 24

##### More questions

Question 4: 2×{ 3 }^{ 2 }×( 3+6) ÷ 9
Solution:
{ 3 }^{ 2 }×( 3+6) ÷ 9
= 2×{ 3 }^{ 2 }×( 3+6 ) ÷ 9  solve the brackets
= 2×{ 3 }^{ 2 }× 9 ÷ 9    solve the powers
=2× 9 × 9 ÷ 9   divide
= 2× 9 × 1 multiply
= 18

Question 5: 1 + { 2 }^{ 2 } × 3 – (6÷2)
Solution:
Here 1 + { 2 }^{ 2 } × 3 – (6÷2) solve the brackets
= 1 + { 2 }^{ 2 } × 3 – 3 solve the powers
= 1 + 4 × 3 – 3 multiply
= 1 + 12 – 3    add
= 13 – 3 subtract
=10

##### Some lengthy questions

Question 6: { ( { 3 }^{ 2 } + 7) × \frac { 2 }{ \sqrt { 64 } } } + 12 ÷ 2 – 20 + 5 + \frac { 1 }{ 2 } × 12
Solution:
{ ( { 3 }^{ 2 } + 7) × \frac { 2 }{ \sqrt { 64 } } } + 12 ÷ 2 – 20 + 5 + \frac { 1 }{ 2 } × 12

= {( 9 + 7 ) × \frac { 2 }{ \sqrt { 64 } } } + 12 ÷ 2 – 20 + 5 + \frac { 1 }{ 2 } × 12

= {16 × \frac { 2 }{ 8 } } + 12 ÷ 2 – 20 + 5 + \frac { 1 }{ 2 } × 12

= 4 + 12 ÷ 2 – 20 + 5 + \frac { 1 }{ 2 } × 12

= 4 + 6 – 20 + 5 + \frac { 1 }{ 2 } × 12

= 4 + 6 – 20 + 5 + 6

= 10 – 20 + 11

= 21 – 20

= 1

Question 7:( 21÷ 7 ÷ 3 ) + 12 × 4 ÷ 6 + ( \sqrt { 225 } × \sqrt { 36 }) – 20
Solution:
When there are so many “÷” signs, it may create confusion. To make it easier, just reciprocate the number with division sign.

( 21 ÷ 7 ÷ 3 ) + 12 × 4 ÷ 6 + ( \sqrt { 225 } × \sqrt { 36 }) – 20

=( 21 × \frac { 1 }{ 7 } × \frac { 1 }{ 3 } ) + 12 × 4 × \frac { 1 }{ 6 } + ( \sqrt { 225 } × \sqrt { 36 }) – 20

= 1 + 12 × 4 × \frac { 1 }{ 6 } + ( 15 × 6 ) – 20

=1 + 12 × 4 × \frac { 1 }{ 6 } + 90 – 20

=1 + 8 + 90 – 20

= 79

### 2 thoughts on “BODMAS Rule”

1. Nice and very informative post for Students, this bosmas idea will help to solve many math problem, thanks for sharing. By the way if you are looking for a job or a job change, then I recommend you to visit jobicon.in ,where you will find latest job news and solved question papers for any competitive exam.

1. Thanks for the appreciation.