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,

##### Let us look at some examples.

##### 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 ×

**5**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.

Step 1 : Solve ( ) small brackets first.

= { 2 }^{ 2 } ×

**(2+4)**

= { 2 }^{ 2 }

**× 6**

Step 2 : Solve the exponent.

={ 2 }^{ 2 } × 6

Step 2 : Solve the exponent.

**= 4 × 6**

= 24

**More questions**

**Question 4: 2×{ 3 }^{ 2 }**

**×( 3+6) ÷ 9**

**Solution:**

2×{ 3 }^{ 2 }×( 3+6) ÷ 9

= 2×{ 3 }^{ 2 }×

*÷ 9 solve the brackets*

**( 3+6 )**= 2×

**{ 3 }^{ 2 }**× 9 ÷ 9 solve the powers

=

**2×**

**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

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