Suppose we want to calculate

    2 × 4 + 3 – (8 – 1)

 

How do we do it?

Do we multiply first?

Or Subtract?

 

We use BODMAS rule

BODMAS - Definitions

 

BODMAS stands for B racket, O rder, D ivision, M ultiplication, A ddition and S ubtraction

 

Let’s get back to our question

  2 × 4 + 3 – (8 – 1)

 

In BODMAS

  • We first solve B racket
  • Then O rder
  • Then D ivision & M ultiplication are at the same level, so we go from left to right and solve them.
  • A ddition and S ubtraction are at the same level, so we go from left to right and solve them

 

2 × 4 + 3 – (8 – 1)

2 × 4 + 3 – (8 – 1)

  = 2 × 4 + 3 – 7

  = 8 + 3 – 7

  = 11 – 7

  = 4

Note :

Here we first solve B racket – (8 – 1)

Then there is no O rder,

So, we solve D ivision & M ultiplication left to right – 2 × 4

And then, we solve A ddition and S ubtraction left to right – 8 + 3 and 11 – 7

Let’s take some more examples

 

15 ÷ 3 × 1 ÷ 5

15 ÷ 3 × 1 ÷ 5

  = 15 ÷ 3 × 1 ÷ 5

  = 15/3 × 1 ÷ 5

  = 5 × 1  ÷ 5

  = 5 ÷ 5

  = 5/5

  = 1

Note :

Here we have no B racket, or O rder

So, we solve D ivision & M ultiplication left to right – first 15 ÷ 3 , then 5 × 1 and then 5 ÷ 5

And then, A ddition and S ubtraction are not there

 

48 ÷ 2 (9 + 3)

48 ÷ 2 (9 + 3)

  = 48 ÷ 2 × (9 + 3)

  = 48 ÷ 2 × 12

  = 48/2 × 12

  = 24 × 12

  = 288

Note :

Here we first solve B racket – (9 + 3)

Then there is no O rder,

So, we solve D ivision & M ultiplication left to right – First 48 ÷ 2, then 24 ×  12

And then, A ddition and S ubtraction are not there

 

10² – 4 ÷ 2

10² – 4 ÷ 2

  = 10²  – 4 ÷ 2

  = 100 –  4 ÷  2

  = 100 – 4/2

  = 100 – 2

  = 98

Note :

Here, there is no B racket

Then, we solve O rder – 10 2 ,

Then, we solve D ivision & M ultiplication left to right – 4 ÷ 2

And then, A ddition and S ubtraction left to right – 100 – 2

 

14 ÷ 7  + 4 × 8 – 2 3

14 ÷ 7  + 4 × 8 – 2 3

  = 14 ÷ 7  + 4 × 8 – 8

  = 2 + 4 × 8 – 8

  = 2 + 32 – 8

  = 34 – 8

  = 26

Note :

Here, there is no B racket

Then, we solve O rder – 2 3 ,

Then, we solve D ivision & M ultiplication left to right – first 14 ÷ 7, then 4 × 8

And then, A ddition and S ubtraction left to right – first 7 + 32, then 39 – 8

 

(4 × 3 2 – 2 3 ÷ 4) + 10 – 5

(4 × 3 2 2 3 ÷ 4) + 10 – 5

  = (4 × 9 – 8 ÷ 4) + 10 – 5

  = (36 – 8 ÷ 4 ) + 10 – 5

  = (36 – 8/4) + 10 – 5

  = (36 – 2) + 10 – 5

  = (34) + 10 – 5

  = 34 + 10 – 5

  = 44 – 5

  = 39

Note :

Here, there is a B racket, so we solve it first

In bracket,

We first solve O rder – 3 2 , 2 3 ,


Then, we solve D ivision & M ultiplication left to right – first 4 × 9, then 8 ÷ 4


And then, A ddition and S ubtraction left to right – first 36 – 4 in bracket, then 32 + 10, 42 – 5

 

PEMDAS

BODMAS in some Regions is called PEMDAS

PEMDAS stands for P arenthesis, E xponents, M ultiplication, D ivision, A ddition, S ubtraction

PEMDAS.jpg

 

In PEMDAS

  • We first solve P arenthesis
  • Then E xponents
  • Then M ultiplication & D ivision are at the same level, so we go from left to right and solve them
  • A ddition and S ubtraction are at the same level, so we go from left to right and solve them

You can remember it by saying

  P lease  E xcuse  M D ear  A unt  S ally

  1. Chapter 1 Class 6 Knowing our Numbers
  2. Concept wise
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Transcript

Bracket - () or { } Order or Power - 25, 37 , √2 Division (÷) Multiplication (×) Addition (+) Subtraction (–) Parenthesis - () or { } Exponents - 25, 37 , √2 Multiplication (×) Division (÷) Addition (+) Subtraction (–)

About the Author

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .