Check sibling questions

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

What is BODMAS? - Part 2

 

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

Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month


Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.