Question

Bodmas Rules in Maths ​

Answer 1

BODMAS:-

$\: \: \: \: \: \: \: \:{\bf{\leadsto B \: denotes \: Brackets}}$

$\: \: \: \: \: \: \: \:{\bf{\leadsto O \: denotes \: Off}}$

$\: \: \: \: \: \: \: \:{\bf{\leadsto D \: denotes \: Division}}$

$\: \: \: \: \: \: \: \:{\bf{\leadsto M \: denotes \: Multiplication}}$

$\: \: \: \: \: \: \: \:{\bf{\leadsto A \: denotes \: Additional}}$

$\: \: \: \: \: \: \: \:{\bf{\leadsto S \: denotes \: Subtraction}}$

Let's see an example :-

★ Find the value of the following -

${\tt{\longrightarrow 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(2 + \dfrac{1}{2} \bigg)}}$

★ Let's solve this question...!

${\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(2 + \dfrac{1}{2} \bigg)}}$

${\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(2 + \dfrac{1}{2} \bigg)}}$

Let's take LCM first of those digits that are in small bracket..!

${\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(\dfrac{4+1}{2} \bigg)}}$

Now let's solve the bracket..!

${\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(\dfrac{5}{2} \bigg)}}$

Now let's open brackets..!

${\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \times \dfrac{5}{2}}}$

Remember we have to work according to BODMAS so now have to solve for division...!

${\sf{:\implies 2 \times 2 - \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{5}{4}}}$

${\sf{:\implies 4 - \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{5}{4}}}$

${\sf{:\implies 4 - \dfrac{1}{4} + \dfrac{5}{4}}}$

${\sf{:\implies 4 - \dfrac{5}{4}}}$

${\sf{:\implies \dfrac{20}{4}}}$

${\sf{:\implies 5}}$

${\small{\boxed{\bf{\bigstar{Hence, \: 5 \: is \: the \: value \: of \: the \: given \: question}}}}}$

Answer 2

Step-by-step explanation:

BODMAS:-

\: \: \: \: \: \: \: \:{\bf{\leadsto B \: denotes \: Brackets}}⇝BdenotesBrackets

\: \: \: \: \: \: \: \:{\bf{\leadsto O \: denotes \: Off}}⇝OdenotesOff

\: \: \: \: \: \: \: \:{\bf{\leadsto D \: denotes \: Division}}⇝DdenotesDivision

\: \: \: \: \: \: \: \:{\bf{\leadsto M \: denotes \: Multiplication}}⇝MdenotesMultiplication

\: \: \: \: \: \: \: \:{\bf{\leadsto A \: denotes \: Additional}}⇝AdenotesAdditional

\: \: \: \: \: \: \: \:{\bf{\leadsto S \: denotes \: Subtraction}}⇝SdenotesSubtraction

Let's see an example :-

★ Find the value of the following -

{\tt{\longrightarrow 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(2 + \dfrac{1}{2} \bigg)}}⟶2÷

2

1

−

2

1

÷2+

2

1

(2+

2

1

)

★ Let's solve this question...!

{\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(2 + \dfrac{1}{2} \bigg)}}:⟹2÷

2

1

−

2

1

÷2+

2

1

(2+

2

1

)

{\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(2 + \dfrac{1}{2} \bigg)}}:⟹2÷

2

1

−

2

1

÷2+

2

1

(2+

2

1

)

Let's take LCM first of those digits that are in small bracket..!

{\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(\dfrac{4+1}{2} \bigg)}}:⟹2÷

2

1

−

2

1

÷2+

2

1

(

2

4+1

)

Now let's solve the bracket..!

{\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \bigg(\dfrac{5}{2} \bigg)}}:⟹2÷

2

1

−

2

1

÷2+

2

1

(

2

5

)

Now let's open brackets..!

{\sf{:\implies 2 \div \dfrac{1}{2} - \dfrac{1}{2} \div 2 + \dfrac{1}{2} \times \dfrac{5}{2}}}:⟹2÷

2

1

−

2

1

÷2+

2

1

×

2

5

Remember we have to work according to BODMAS so now have to solve for division...!

{\sf{:\implies 2 \times 2 - \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{5}{4}}}:⟹2×2−

2

1

×

2

1

+

4

5

{\sf{:\implies 4 - \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{5}{4}}}:⟹4−

2

1

×

2

1

+

4

5

{\sf{:\implies 4 - \dfrac{1}{4} + \dfrac{5}{4}}}:⟹4−

4

1

+

4

5

{\sf{:\implies 4 - \dfrac{5}{4}}}:⟹4−

4

5

{\sf{:\implies \dfrac{20}{4}}}:⟹

4

20

{\sf{:\implies 5}}:⟹5

{\small{\boxed{\bf{\bigstar{Hence, \: 5 \: is \: the \: value \: of \: the \: given \: question}}}}}

★Hence,5isthevalueofthegivenanswer