Order of Operation
While solving questions involving mathematical operations involving addition, subtraction, multiplication and division we follow a particular order while performing a calculation. To perform calculations we follow BODMAS.
BODMAS stands for
B = Bracket
O = of
D = Division
M= Multiplication
A = Addition
S= Subtraction
Let us see examples based on this
Example 1:
What is the value of N= {(4 +5)of4}/3 + 2 ?
Solution:
We will solve bracket first
{(9) x 4}/3 + 2
= 12 + 2
= 14
Example 2:
$ \displaystyle \begin{array}{l}What\text{ }is\text{ }the\text{ }value\text{ }of\text{ }N=\text{ }\{(12\div 4\times ~2\text{ }+5)\text{ }of4\}/2-1\text{ }?\\Solution:\,\\We\text{ }will\text{ }solve\text{ }terms\text{ }inside\text{ }bracket\text{ }first\\=\,\,\{(12\div 4\times ~2\text{ }+5)\text{ }of4\}/2-1\\=\{(3~\times 2\text{ }+5)\text{ }of4\}/2-1\text{ }\left( We\text{ }divided\text{ }12\text{ }and\text{ }4\text{ }first \right)\\=\left\{ \left( 6\text{ }+5 \right)\text{ }of4 \right\}/2-1\text{ }\left( We\text{ }multiplied\text{ }3\text{ }and\text{ }2 \right)\\=\,\,\left\{ \left( 11 \right)\text{ }of4 \right\}/2-1\text{ }\left( We\text{ }added\text{ }6\text{ }and\text{ }5 \right)\\=\left\{ 44 \right\}/2-1~\left( We\text{ }wrote\text{ }11\text{ }of\text{ }4 \right)\\=\text{ }22-1\left( We\text{ }divided\text{ }44\text{ }and\text{ }2 \right)\\=\,\,\,21\left( We\text{ }subtracted\text{ }1\text{ }from\text{ }22 \right)\end{array}$
Example 3:
What is the value of N = {(18 ÷ 3 × 2 + 1) of 8}/2 + 5?
Solution:
We will solve the terms inside the bracket first
{(18 ÷ 3 × 2 + 1) of 8}/2 + 5
= {6 × 2 + 1) of 8}/2 + 5 (We divided 18 and 3)
= {(12 + 1) of 8}/2 + 5 (We multiplied 6 and 2)
= {13 of 8}/ 2 + 5 (We added 12 and 1)
= (104)/2 + 5 (We wrote 13 of 8)
= 52 + 5 (We divide 104 and 2)
= 57 (We added 52 and 5)
$ \begin{array}{l}Example\,\,4:\,\,\\What\text{ }is\text{ }the\text{ }value\text{ }of\text{ }N=\text{ }\{(9\text{ }+\text{ }8\div ~2\times ~3)\text{ }of8\}/2\text{ }+\text{ }3\text{ }?\\Solution:\,\\We\text{ }will\text{ }solve\text{ }terms\text{ }inside\text{ }bracket\text{ }first\\\{(9\text{ +}8\div ~2~\times 3)\text{ }of8\}/2\text{ }+\text{ }3\\=\{(9\text{ }+\text{ }4\times ~3)\text{ }of8\}/2\text{ }+\text{ }3\left( We\text{ }divided\text{ }8\text{ }and\text{ }2\text{ }first \right)\\=\text{ }\left\{ \left( 9\text{ }+\text{ }12 \right)\text{ }of8 \right\}/2\text{ }+\text{ }3\left( We\text{ }multiplied\text{ }3\text{ }and\text{ }4 \right)\\=\text{ }\left\{ \left( 21 \right)\text{ }of8 \right\}/2\text{ }+\text{ }3\left( We\text{ }added\text{ }9\text{ }and\text{ }12 \right)\\=\text{ }\left\{ 168 \right\}/2\text{ }+\text{ }3\left( We\text{ }wrote\text{ }21\text{ }of\text{ }8 \right)\\=\text{ }84+\text{ }3\left( We\text{ }divided\text{ }168\text{ }and\text{ }2 \right)\\=\text{ }87\left( We\text{ }added\text{ }3\text{ }in\text{ }84\text{ } \right)\end{array}$
