**BIDMAS or BODMAS**

**BIDMAS** (sometimes **BODMAS**) is an acronym that helps us remember which order to perform operations in. We start from left to right. The letters stand for:

**B**rackets**I**ndices (or as they can be called,**O**rders)**D**ivision**M**ultiplication**A**ddition**S**ubtraction

**Using BIDMAS**

When performing calculations, always follow the **BIDMAS** order of operations.

**Example:** Work out the value of 3 \times(3^2 + 4) - 8

**Step 1:** The first letter of** BIDMAS** is **B**, meaning the first thing we should do is look to what’s inside the **brackets**. (If there are not brackets, move onto indices, then divide and so on..)

Here, we have two operations happening: a **power/index**, and an **addition**. The letter **I** comes before the letter **A** in **BIDMAS** which means we first work out the result of 3^2 and then add 4 to it.

(3^2 + 4) = (9 + 4) = 13

**Step 2:** We are left with a multiplication and a subtraction, so because **M** comes before **S**, we do the **multiplication first** and the **subtraction second**,

3 \times 13 - 8 = 39 - 8 = 31

**BIDMAS and Fractions**

For fractions, we work out what the values of the top (numerator) and bottom (denominator) are separately (using the rules of **BIDMAS**), and then lastly, we look at the fraction we have and see if it can be **simplified**.

**Example:** Simplify the fraction \dfrac{3 \times 4 - 5}{11 + (9 \div 3)}

**Step 1: **First, considering the numerator. There’s a **multiplication** and a **subtraction**, so we do the **multiplication first** and the **subtraction** **second**.

3 \times 4 - 5 = 12 - 5 = 7

**Step 2: **Now, the denominator. That contains a **division** inside **brackets**, so that will be the first bit of the calculation, and then the **addition** will be second.

11 + (9 \div 3) = 11 + (3) = 14

**Step 3: **Therefore, our fraction is \dfrac{7}{14}. Both top and bottom have a factor of 7, so the simplified answer is

\dfrac{7}{14} = \dfrac{1}{2}

**Note:**

- For
**division**and**multiplication**, calculate them out in the order that they appear (from left to right). - For
**addition**and**subtraction**, calculate them in the order that they appear (from left to right) when they are the only two operations left in the sum.

**Example 1: BIDMAS and Algebra**

Write the expression 4xy \times 9y - 13 \times xy^2 in its simplest form.

**[3 marks]**

**Step 1:** There are two multiplications in this expression, so it doesn’t matter which order we do them in providing we do them both before the subtraction. The first one becomes:

4xy \times 9y = 4 \times 9 \times x \times y \times y = 36xy^2

**Step 2:** The second multiplication becomes:

13 \times xy^2 = 13xy^2

**Step 3:** So, now we subtract the second from the first one, to get the expression in its simplest form.

36xy^2 - 13xy^2 = 23xy^2

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### Example Questions

**Question 1:** Calculate the value of (2 \times 3^3) \div (15 - 9).

**[2 marks]**

Mathematical operators must be carried out in the correct order. The acronym **BIDMAS** (or **BODMAS**) is a helpful way to remember this order.

There are two brackets (B) to first calculate,

(2 \times 3^3) and (15-9)

Inside the first bracket, there is a power or index number (I or O),

2 \times 3^3 = 2 \times 27

Carry out any divisions or multiplications (DM) then additions or subtractions (AM) inside the brackets,

(2 \times 27 = 54) and (15-9 = 6)

Complete the calculation,

\begin{aligned} 54 \div 6 &= 9 \\ (2 \times 3^3) \div (15 - 9) &= 9 \end{aligned}

**Question 2:** Calculate the value of 16 \times (12\div 4)^2 .

**[2 marks]**

The first operation to consider following BIDMAS is the calculation inside the brackets (B),

12 \div 4 = 3

As this does not simplify we can move onto the indices (I),

(3)^2 = 9

Again as this does not simplify, the last operation of the expression is multiplication (M), and we get

16\times 9= 144

**Question 3:** Calculate the value of \dfrac{(x^2+3)}{(10-6)} when x=-3.

**[3 marks]**

The first operation to consider following BIDMAS is the calculation inside the brackets (**B**) dealing with the numerator and denominator separately for the moment.

In the numerator, we have to first, substitute in the given value of x and apply the power (**I**), before the addition (**A**).

((-3)^2+3)=(9+3)=12

In the denominator, there are no indices nor any multiplications divisions to consider so we can move straight to the subtraction (**S**),

(10-6) = 4

The last operation of the expression is a division (**D**), so,

\dfrac{12}{4}=3

**Question 4:** Write the expression (y^2 + 5y^2) - 3y \times 7y in its simplest form.

**[3 marks]**

The first operation to consider following BIDMAS is the calculation inside the brackets (**B**),

y^2 + 5y^2 = 6y^2

There are no indices or divisions in this expression, but there is a multiplication (**M**),

3y \times 7y = 21y^2

The last operation of the expression is subtraction (**S**), and we get,

6y^2 - 21y^2 = -15y^2

**Question 5:** Write the expression \dfrac{42q^2 \times pq}{28p^3 \div (9p - 5p)} in its simplest form.

**[4 marks]**

The first operation to consider following BIDMAS is the calculation inside the brackets (**B**) dealing with the numerator and denominator separately for the moment.

In the numerator, there is only one operation in the form of multiplication (**M**), so

42q^2 \times pq = 42q^2 \times q \times p = 42q^{3}p

In the denominator, the first calculation is inside the brackets (**B**), which is a substitution (**S**),

9p - 5p = 4p

Then, the division (**D**) operator can be applied,

28p^3 \div 4p = 7p^2

So, we are left with a fraction with p on the top and bottom, as well as a factor of 7. Both of these cancel so,

\dfrac{42q^{3}p}{7p^2}=\dfrac{6q^3}{p}

As there are no more common factors, we can not simplify the expression any further.

### Worksheets and Exam Questions

#### (NEW) BIDMAS /BODMAS - Exam Style Questions - MME

Level 1-3 New Official MME### Drill Questions

#### BIDMAS and Prime Factors (Drill Questions)

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