BIDMAS | Questions and Worksheets | MME

# BIDMAS Questions, Revision and Worksheets

Level 1 Level 2 Level 3

## BIDMAS or BODMAS

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

• Brackets
• Indices (or as they can be called, pOwers)
• Division
• Multiplication
• Subtraction

### BIDMAS and Fractions:

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

Basically, do the fraction bit last!

### Example 1: Using BIDMAS

What is the value of $3 \times\textcolor{red}{(3^2 + 4)} - 8$?

The first letter of BIDMAS is B, meaning the first thing we should do is look to what’s inside the brackets. In there 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.

$3^2 + 4 = 9 + 4 = \textcolor{red}{13}$

Then, our calculation becomes

$3 \times \textcolor{red}{13} - 8$

Here, there is a multiplication and a subtraction, so because M comes before S, we do the multiplication first and the subtraction second. We get

$3 \times \textcolor{red}{13} - 8 = 39 - 8 = 31$

### Example 2: BIDMAS and Fractions

Simplify the fraction $\dfrac{3 \times 4 - 5}{11 + (9 \div 3)}$.

First, we’ll sort out the numerator. There’s a multiplication and an addition, so we do the multiplication first and the addition second.

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

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 the second.

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

Therefore, our fraction is $\frac{7}{14}$. Both top and bottom have a factor of 7, so the simplified answer is

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

### Example 3: BIDMAS and Algebra

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

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:

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

The second multiplication becomes:

$13 \times xy^2 = 13xy^2$

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

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

### 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.

(2×33)÷(15−9)

1. There are two brackets (B) to first calculate, $(2 \times 3^3)$ and $(15-9)$

(2×33)

2. Inside the first bracket, there is a power or index number (I or O), $2 \times 3^3 = 2 \times 27$

3. Carry out any divisions or multiplications (DM) then additions or subtractions (AM) inside the brackets, $(2 \times 27 = 54)$ and $(15-9 = 6)$

4. Complete the calculation, $54 \div 6 = 9$

(2×33)÷(15−9)=9

$(2 \times 3^3) \div (15 - 9) = 9$

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

$9 \div 4 = \dfrac{9}{4}$

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

$(9 \div 4)^2 = \dfrac{81}{16}$

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

$16\times \dfrac{81}{16}= 81$

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.

Level 1-3

Level 1-3

GCSE MATHS

GCSE MATHS

GCSE MATHS

### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.