Number: Operations (Bracket-BODMAS)

This post covers two information (i) Bracket, (ii) BODMAS

Bracket:

In BODMAS rule, we have seen that B stands for Bracket. So in sequence, first step is to open or break a bracket.

There are four types of brackets:

(i) Vinculum: This is represented by a bar on the top of an operation of numbers. Under Vinculum Bracket, numbers are solved first then other operations will be solved. This bracket as normally used at a time when others brackets have been used and there is no any option and we have to use bracket again.

(ii) Simple: These are represented by (  ). In the case of only one bracket, it is often used. if two or more than two brackets are used, this is found as the most inner bracket often.

(iii) Curly: These are represented by {  }. This is often used when two brackets are required to be used.

(iv) Square: These are represented by [  ]. This is often used when three brackets are required to be used.

While solving a mathematics, the most inner bracket is solved first.

BODMAS Rule:

If we describe BODMAS we find that it means

B            O           D              M                     A                      S

↓             ↓            ↓                ↓                      ↓                      ↓

Bracket   order   Division   Multiplication   Addition    Subtraction

Here Orders means Powers, Roots, of, and other like operations

The above rule actually describes the sequence in which operations are to be performed when we have a mathematical number equation having combination of many operations.

The sequence says

Step 1 (Bracket): First solve the bracket. If a bracket contains other brackets inside it we need to solve innermost bracket. as. As a bracket is solved, it is in actual removed and we get only solved number with the sign following the rules of multiplication operation.

Step 2 (order): We solve here the roots and power terms. We also solve 'of' in this step. Here (of) means x (multiply). If 'of' is immediately preceded or followed by any bracket then bracket must be solved first.

Step 3 (Division): Then every division operation is performed.

Step 4 (Multiplication): Then every multiplication operation is performed.

Step 5 (Addition): Then every addition operation is performed.

Step 6 (Subtraction): Then every subtraction operation is performed.

Example: 

Solution:

Example: 

Solution:

First we will solve for order (power) and of in the equation

Bracket operations are simultaneously solved in each step as required 

a=24 ÷2 ×(3- (-4) )- 2 × (2 ×25)+ 2/3  ×9 ÷3 

Then we will perform remaining brackets and then divide operations

a=12 × 7-2×50+2/3  ×3 

Then we will perform Multiply operations

a=84-100+2 

Then we will add and subtract

a=14 

Numerical for Exercise:

(a) 3 + 6 ÷ 2 × 2

(b) 4 × 3 - 24 ÷ 4 + 7

(c) 30 - (5 × 2³ - 15)

(d) (7 - √9) × (4² - 3 + 1)

Solutions:

(a) 3 + 6 ÷ 2 × 2

= 3 + 3 × 2
= 3 + 6
= 9

(b) 4 × 3 - 24 ÷ 4 + 7

First we do, Multiplication and division Operations

4 × 3 - 24 ÷ 4 + 7
= 12 - 24 ÷ 4 + 7
= 12 - 6 + 7
Then Addition and subtraction is done
12 - 6 + 7
= 19 - 6
= 13

(c) 30 - (5 × 2³ - 15)

Remember 2³ = 2 × 2 × 2 = 8

Start from the innermost Brackets, and then use "Orders" first:

30 - (5 × 2³ - 15) 

= 30 - (5 × 8 - 15)
Then we do multiply
30 - (5 × 8 - 15) 

= 30 - (40 - 15)

Then we subtract
30 - (40 - 15) 

= 30 - 25

Then we again subtract
30 - 25 

= 5

(d) (7 - √9) × (4² - 3 + 1)

First Bracket:

7 - √9
First we will do Order (in this case square root) first:
= 7 - √9 

= 7 - 3
Then we do Subtraction:
7 - 3 

= 4
Second Bracket:
4² - 3 + 1
Do Orders (the squaring) first:
4² - 3 + 1 = 16 - 3 + 1
Then Addition and Subtraction:
16 - 3 + 1 

= 17 - 3

= 14

So the problem will be solved by multiplying the two bracket contents

(7 - √9) × (4² - 3 + 1) 

= 4 × 14 

= 56

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