# 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.

A stands for ADDITION.
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”.

When it comes to division and multiplication, their operations are considered  to be of same level. That is, while going from left to right, if multiplication comes first, we will solve it first. If division comes first, we will solve division first.

Same goes for the addition and subtraction, Whichever comes first while going from left to right, we will perform that operation.

To summarise,

We perform the operations in following manner,

##### 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
= 9 × 9 ÷ 9 multiply
= 162  ÷ 9 divide
= 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

= -10 + 11

= 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

### 4 thoughts on “BODMAS Rule”

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