# BIDMAS or BODMAS Questions, Revision and Worksheets

GCSE 1 - 3KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

## 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:

• Brackets
• Indices (or as they can be called, Orders)
• Division
• Multiplication
• Subtraction
Level 1-3 GCSE KS3    ## 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,

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

Level 1-3 GCSE KS3    ## 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

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

Level 1-3 GCSE KS3    ## Note:

• For division and multiplication, work 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: 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.

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

Level 1-3 GCSE KS3    ## Example Questions

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}

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

$16\times 9= 144$

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,

$\dfrac{12}{4}=3$

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,

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

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.

## Worksheet and Example Questions

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

Level 1-3 GCSENewOfficial MME

Level 1-3 GCSE

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