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 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
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 y D ear A unt S ally
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class