BODMAS fractions: The \(BODMAS\) rule is the order of operations to be followed while solving mathematical problems. \(B\) stands for brackets, \(O\) is for order of powers or roots, \(D\) stands for division, and \(M\) stands for multiplication. \(A\) stands for addition, and \(S\) stands for subtraction. Expressions with numerous operators must be simplified in this order only, from left to right. The brackets are solved first, followed by powers or roots, division, or multiplication (whatever comes first from the left side of the formula), and finally subtraction or addition.

The BODMAS Simplification of Fractions is a solution for addressing fractions that include many operations such as addition \(\left( + \right),\) subtraction \(\left( – \right),\)multiplication \(\left( \times \right),\) division \(\left( \div \right),\) and brackets \(\left({} \right).\) Keep reading the article to know more about the BODMAS fractions, its full form along with some solved examples.

BODMAS Rule

BODMAS was designed to simplify and standardise the evaluation of mathematical statements and the handling of complex calculations.
To solve any arithmetic expression, use this rule: first, solve the terms in brackets, then simplify exponential terms and move on to division and multiplication operations, and last, move on to addition and subtraction. In this case, addition and subtraction are level \(2\) operations, but multiplication and division are level \(1\) operations because they must be solved first. The terms inside the brackets can be simplified immediately. This means we can do division, multiplication, addition, and subtraction in the sequence specified inside the bracket.
Example: Simplify: \(\left({2 – \frac{{13}}{8}} \right) \div \frac{{27}}{{56}} \times 1\frac{2}{7}.\)

BODMAS Full Form

To make numerical equations easier to understand, operations must be performed according to the \(BODMAS\) rule, which involves division, multiplication, addition, and subtraction to be performed in the order brackets.
For example, the expression \(\left({48 – 24} \right) \div 3 \times 4,\) as per \(BODMAS\) rule, is simplified by first removing the bracket that is \(48 – 24 = 24,\) next dividing \(24\) by \(3\) and then multiplying by \(4\) as shown below:
\(\left({48 – 24} \right) \div 3 \times 4 = \left({24} \right) \div 3 \times 4\)
\( = 8 \times 4 = 32.\)

Learn BODMAS Rule in detail here

Solution: The given expression is \(\left({2 – \frac{{13}}{8}}\right) \div \frac{{27}}{{56}} \times 1\frac{2}{7}\)
\( = \left({\frac{{16 – 13}}{8}} \right) \div \frac{{27}}{{56}} \times 1\frac{2}{7}\) (Removing bracket)
\(= \frac{3}{8} \div \frac{{27}}{{56}} \times \frac{9}{7}\) (Performing division)
\( = \frac{3}{8} \times \frac{{56}}{{27}} \times \frac{9}{7} = 1\) (performing multiplication)
Brackets within brackets are occasionally necessary for complex statements. Different types of brackets are utilised because having the same type of brackets within each other can be confusing. The following are the most used brackets:

Each bracket sign has a left part representing the beginning of the bracket and a right part representing the bracket’s conclusion. Parenthesis are employed in the innermost section of mathematical equations with more than one bracket, followed by braces, and square brackets cover these two.

Removal of brackets: We utilise the following steps to simplify expressions that use more than one bracket:

STEP I: Determine whether the provided phrase has a vinculum. Perform operations under a vinculum if one is present. Otherwise, proceed to the next step.
STEP II: Examine the innermost bracket and carry out any necessary procedures within it.
STEP III: Using the following rules, remove the innermost bracket:
Rule 1: If a plus sign precedes a bracket, remove it by stating the bracket’s terms as they are.
Rule 2: If a minus sign before a bracket, turn all positive signs inside it to negative and vice versa.
Rule 2: Multiplication is indicated when there is no sign between a number and a grouping symbol.
Rule 3: If a number appears before some brackets, the number is multiplied by the number outside the brackets.
STEP IV: Examine the next innermost bracket and conduct any necessary operations there. Using the criteria from step \({\text{III,}}\) remove the second innermost bracket. Carry on in this manner until all the brackets have been eliminated. Let us now use an example to demonstrate these stages.

Example: Simplify \(222 – \left[{\frac{1}{3}\left\{{42 + \left({56 – \overline {8 + 9} } \right)} \right\} + 108} \right]\)
Solution: \(222 – \left[{\frac{1}{3}\left\{{42 + \left({56 – \overline {8 + 9} } \right)} \right\} + 108} \right]\)
\( = 222 – \left[{\frac{1}{3}\left\{{42 + \left({56 – 17} \right)} \right\} + 108} \right]\) (Removing vinculum)
\( = 222 – \left[{\frac{1}{3}\left\{{42 + 39} \right\} + 108} \right]\) (Removing parentheses)
\( = 222 – \left[{\frac{{81}}{3} + 108} \right]\) (Removing braces)
\( = 222 – \left[{27 + 108} \right]\)
\(= 222 – 135\)
\( = 87\)

Simplification of Fractions using BODMAS Rule

The steps to simplify fractions using BODMAS rule are given below:

Fraction Operations

Addition and subtraction of fractions: To add or subtract like fractions, we add or subtract their numerators and retain the common denominator.
The steps for adding and subtracting unlike fractions are as follows:
STEP I: Get the fractions and their denominators.
STEP II: Calculate the denominators’ LCM.
STEP III: Convert each fraction to an equivalent fraction with the same denominator as the LCM produced in step \({\text{II}}{\text{.}}\)
STEP IV: Combine or subtract the like fractions from step \({\text{III}}{\text{.}}\)

Multiplication of Fractions

Multiplication of fractions: Product of fractions \( = \frac{{{\text{Product}}\,{\text{of}}\,{\text{their}}\,{\text{numerators}}}}{{{\text{Product}}\,{\text{of}}\,{\text{their}}\,{\text{denominators}}}}\)

Division of fractions: The division of a fraction \(\frac{a}{b}\) by a non-zero fraction fraction \(\frac{c}{d}\) is defined as the product of \(\frac{a}{b}\) with the multiplicative inverse or reciprocal of \(\frac{c}{d}.\)
i.e., \(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}\)

Simplification of Fractions using \(BODMAS\) Rule: When an expression in fractions contains brackets, other operation signs, we apply the \(BODMAS\) rule for simplification.
Example: Simplify: \(4\frac{1}{2} – \left\{{3\frac{1}{4} – \left({6\frac{1}{3} – 4\frac{1}{6}} \right)} \right\} + 7\frac{1}{2}.\) of \(1\frac{3}{5}.\)
Solution: The given expression is \(4\frac{1}{2} – \left\{{3\frac{1}{4} – \left({6\frac{1}{3} – 4\frac{1}{6}} \right)} \right\} + 7\frac{1}{2}.\) of \(1\frac{3}{5}\)
\( = \frac{9}{2} – \left\{{\frac{{13}}{4} – \left({\frac{{19}}{3} – \frac{{25}}{6}} \right)} \right\} + \frac{{15}}{2}\) of \(\frac{8}{5}\) (Changed to improper fractions)
\( = \frac{9}{2} – \left\{{\frac{{13}}{4} – \left({\frac{{38 – 25}}{6}} \right)} \right\} + \frac{{15}}{2} \times \frac{8}{5}\) (changed ‘of’ into multiplication symbol \(\times \))
\( = \frac{9}{2} – \left\{{\frac{{13}}{4} – \left({\frac{{13}}{6}} \right)} \right\} + \frac{{12}}{1}\)
\( = \frac{9}{2} – \left\{{\frac{{39 – 26}}{{12}}} \right\} + \frac{{12}}{1}\)
\(= \frac{9}{2} – \left\{{\frac{{13}}{{12}}} \right\} + \frac{{12}}{1}\)
\( = \frac{{54 – 13 + 144}}{{12}}\)
\(=\frac{{185}}{{12}}\)
\( = 15\frac{5}{{12}}.\)

Solved Examples – BODMAS Simplification of Fractions

Q.1. Simplify: \(\frac{1}{2} \times \frac{1}{7} \div \frac{2}{3}\) of \(1\frac{2}{7}.\)
Ans: The given expression is \(\frac{1}{2} \times \frac{1}{7} \div \frac{2}{3}\) of \(1\frac{2}{7}\)
\( = \frac{1}{2} \times \frac{1}{7} \div \frac{2}{3} \times \frac{9}{7}\) (Performing ‘of’)
\( = \frac{1}{2} \times \frac{1}{7} \div \frac{6}{7}\)
\( = \frac{1}{2} \times \frac{1}{7} \times \frac{7}{6}\) (Performing \( \div \))
\( = \frac{1}{{12}}\) (Performing \( \times \))

Q.2. Simplify: \(\frac{1}{2} – \)(\(\frac{4}{5}\) of \(\frac{5}{6}\))\( + \frac{1}{4}\)
Ans: The given expression is \(\frac{1}{2} – \)(\(\frac{4}{5}\) of \(\frac{5}{6}\))\( + \frac{1}{4}\)
\( = \frac{1}{2} – \frac{4}{5} \times \frac{5}{6} + \frac{1}{4}\)
\( = \frac{1}{2} – \frac{2}{3} + \frac{1}{4}\)
\( = \frac{{6 – 8 + 3}}{{12}}\)
\( = \frac{{9 – 8}}{{12}}\)
\( = \frac{1}{{12}}.\)

Q.3. Simplify: \(\left({4\frac{2}{3} \times 5\frac{4}{7}} \right) \div 3\frac{1}{4}\)
Ans: The given expression is \(\left({4\frac{2}{3} \times 5\frac{4}{7}} \right) \div 3\frac{1}{4}\)
\( = \left({\frac{{14}}{3} \times \frac{{39}}{7}} \right) \div \frac{{13}}{4}\)
\( = 26 \div \frac{{13}}{4}\)
\(= 26 \times \frac{4}{{13}} = 8.\)

Q.4. Simplify: \(\frac{3}{{19}}\) of \(\left\{ {\frac{7}{9} + \left( {\frac{3}{4} \div \overline {\frac{1}{2} – \frac{1}{3}} } \right)} \right\}\)
Ans: The given expression is \(\frac{3}{{19}}\) of \(\left\{ {\frac{7}{9} + \left( {\frac{3}{4} \div \overline {\frac{1}{2} – \frac{1}{3}} } \right)} \right\}\)
\( = \frac{3}{{19}}\) of \(\left\{{\frac{7}{9} + \left({\frac{3}{4} \div \frac{{3 – 2}}{6}} \right)} \right\}\) (Remove vinculum)
\( = \frac{3}{{19}}\) of \(\left\{{\frac{7}{9} + \left({\frac{3}{4} \div \frac{1}{6}} \right)} \right\}\)
\( = \frac{3}{{19}}\) of \(\left\{{\frac{7}{9} + \frac{3}{4} \times \frac{6}{1}} \right\}\)(Remove common brackets)
\( = \frac{3}{{19}}\) of \(\left\{{\frac{7}{9} + \frac{9}{2}} \right\}\)
\( = \frac{3}{{19}}\) of \(\frac{{7 \times 2 + 9 \times 9}}{{18}}\) (Remove curly brackets)
\( = \frac{3}{{19}}\) of \(\frac{{95}}{{18}}\)
\( = \frac{3}{{19}} \times \frac{{95}}{{18}} = \frac{5}{6}.\)

Q.5. Simplify: \(\left({5\frac{1}{4} – 2\frac{1}{3}} \right) + \frac{1}{3}\) of \(\left({5\frac{1}{2} \div 2\frac{1}{2}} \right)\)
Ans: The given expression is \(\left({5\frac{1}{4} – 2\frac{1}{3}} \right) + \frac{1}{3}\) of \(\left({5\frac{1}{2} \div 2\frac{1}{2}} \right)\)
\( = \left({\frac{{21}}{4} – \frac{7}{3}} \right) + \frac{1}{3}\) of \(\left({\frac{{11}}{2} \div \frac{5}{2}}\right)\) (Convert mixed fraction to improper fraction)
\( = \left({\frac{{63 – 28}}{{12}}} \right) + \frac{1}{3}\) of \(\left( {\frac{{11}}{2} \times \frac{2}{5}} \right)\)
\( = \frac{{35}}{{12}} + \frac{1}{3}\) of \(\frac{{11}}{5}\)
\( = \frac{{35}}{{12}} + \frac{1}{3} \times \frac{{11}}{5}\)
\( = \frac{{35}}{{12}} + \frac{{11}}{{15}}\)
\( = \frac{{175 + 44}}{{60}} = \frac{{219}}{{60}}\)

BODMAS fractions: Summary

In this article, we learnt about \(BODMAS\) Full Form, \(BODMAS\) Rule, Simplification of fractions using \(BODMAS\) rule, solved examples on \(BODMAS\) simplification of fractions, and FAQs on \(BODMAS\) simplification of fractions.
The learning outcome of this article is, the \(BODMAS\) rule is an acronym that helps students recall the correct order in which to solve math problems. The BODMAS rule states that division, multiplication, addition, and subtraction to be performed in the order brackets. We also learnt to use the BODMAS rule to simplify fractions.

Frequently Asked Questions (FAQs) – BODMAS Simplification of Fractions

The most commonly asked questions about BODMAS Simplification of Fractions are answered here:

Q.1. How do you simplify fractions in BODMAS?
Ans: When an expression in fractions contains brackets or other operation signs, we apply the \(BODMAS\) rule for simplification.
To make numerical equations easier to understand, operations must be performed according to the \(BODMAS\) rule, which involves division, multiplication, addition, and subtraction to be performed in the order brackets.

Q.2. How do you simplify fractions with different denominators?
Ans: Addition and subtraction of fractions with different denominators:
The steps for adding and subtracting unlike fractions are as follows:
STEP I: Get the fractions and their denominators.
STEP II: Calculate the denominators’ LCM.
STEP III: Convert each fraction to an equivalent fraction with the same denominator as the LCM produced in step \({\text{II}}{\text{.}}\)
STEP IV: Combine or subtract the like fractions from step \({\text{III}}{\text{.}}\)
Example: Add \(\frac{1}{5} + \frac{3}{{10}}.\)
Solution: The given expression is \(\frac{1}{5} + \frac{3}{{10}}.\)
The LCM of denominators \(5\) and \(10\) is \(10.\)
Now, \(\frac{1}{5} \times \frac{2}{2} + \frac{3}{{10}} = \frac{{2 + 3}}{{10}} = \frac{5}{{10}} = \frac{1}{2}.\)

Q.3. What is the simplification of fractions?
Ans: If the numerator and denominator have no common factors other than \(1,\) the fraction is simplified. If there are common elements in the numerator and denominator of a fraction, we can simplify the fraction by deleting the common factors.

Q.4. How does the BODMAS Simplification of Fractions rule work?
Ans: \(BODMAS\) was designed to simplify and standardise the evaluation of mathematical statements and the handling of complex calculations.
To solve any arithmetic expression, use this rule: first, solve the terms in brackets, then simplify exponential terms and move on to division and multiplication operations, and last, move on to addition and subtraction. In this case, addition and subtraction are level \(2\) operations, but multiplication and division are level \(1\) operations because they must be solved first. The terms inside the brackets can be simplified immediately. This means we can do division, multiplication, addition, and subtraction in the sequence specified inside the bracket.

Q.5. How do you simplify fractions with brackets?
Ans: Each bracket sign has a left part representing the beginning of the bracket and a right part representing the bracket’s conclusion. Parenthesis are employed in the innermost section of mathematical equations with more than one bracket, followed by braces, and square brackets cover these two.
Removal of brackets: We utilise the following steps to simplify expressions that use more than one bracket:
STEP I: Determine whether the provided phrase has a vinculum. Perform operations under a vinculum if one is present. Otherwise, proceed to the next step.
STEP II: Examine the innermost bracket and carry out any necessary procedures within it.
STEP III: Using the following rules, remove the innermost bracket:
Rule 1: If a plus sign precedes a bracket, remove it by stating the bracket’s terms as they are.
Rule 2: If a minus sign before a bracket, turn all positive signs inside it to negative and vice versa.
Rule 3: Multiplication is indicated when there is no sign between a number and a grouping symbol.
Rule 4: If a number appears before some brackets, the number is multiplied by the number outside the brackets.
STEP IV: Examine the next innermost bracket and conduct any necessary operations there. Using the criteria from step\({\text{III,}}\) remove the second innermost bracket. Carry on in this manner until all the brackets have been eliminated. Let us now use an example to demonstrate these stages.