How do we solve questions involving mathematical operations involving addition, subtraction, multiplication and division? The answer to this question is that we follow a particular order to solve the problem. This forms the basis to solve any kind of question.

**The order is known as VBODMAS.**

VBODMAS: – It stands for (Vinculum- Bracket – Of –Division –Multiplication – Addition – Subtraction). This is an extended version of BODMAS.

V stands for vinculum or Bar (Bar which is used to show recurring numbers for e.g.

B stands for Bracket for e.g. () , {} ,[]

O stands for Of

D stands for Division [

M stands for Multiplication [X]

A stands for Addition [+]

S stands for subtraction [-]

**Let us see some examples based on this.**

**Example 1:**

What is the value of N=4+5-7?

**Solution: **

According to BODMAS, first addition should take place and then subtraction.

So, N =(4+5) -7 = 9-7 = 2

**Example 2**:

What is the value of N= {(4 +5) of4}/3 + 2?

**Solution:**

We will first solve the bracket, N = {(9) x 4}/3 + 2 = 12 + 2 = 14.

Let’s try out some more questions.

**EXERCISE: – **

**Question 1: **

### Answer and Explanation

*Question 2: *

*What is the value of N = {(18 ÷ 3 × 2 + 1) of 8}/2 + 5?*

### Answer and Explanation

*Solution: *

*We will solve the terms inside the bracket first*

*{(18 ÷ 3 × 2 + 1) of 8}/2 + 5*

*= {6 × 2 + 1) of 8}/2 + 5 ( divided 18 and 3)*

*= {(12 + 1) of 8}/2 + 5 (multiplied 6 and 2)*

*= {13 of 8}/ 2 + 5 ( added 12 and 1)*

*= (104)/2 + 5 ( wrote 13 of 8)*

*= 52 + 5 ( divide 104 and 2)*

*= 57 ( added 52 and 5)*

**Question 3: **

### Answer and Explanation

**Question 4:** Solve 2 × 6 + 3 – 4/2 – 5 + 20/5 × 3 + 50

### Answer and Explanation

**Solution:**We have 2 × 6 + 3 – 4/2 – 5 + 20/5 × 3 + 50

= 12+3 – 4/2 – 5+20/5 × 3+50

= 12+3-2-5+20/5 × 3+50

= 12+3-2-5+4 ×3+50

= 12+3-2-5+12+50

= 15-2-5+12+50 = 77 – 7 = 70

**Question 5:** Solve (3+3-5) × (15-5) × 10 – 99

### Answer and Explanation

**Solution:**

We have (3+3-5) × (15-5) × 10 – 99

= (6 – 5) × (15 – 5) × 10 – 99

= 1 × (15 – 5) × 10 – 99 = 10 × 10 – 99 = 1