While dealing with the mathematical numbers and various mathematical operations, we come across various mathematical expressions for example - algebraic expression which looks very complicated and are not so easy to be solved and came across cbse syllabus many times. To solve such expressions, we need to follow certain rules to get the correct solution.
Solve the Following:
21 -( 12 - 6 ÷ 3)------- (1 )
In this given expression, we first consider the brackets,
We solve the bracket
( 12 - 6 ÷ 3),
Now in Bracket , we first take division i. e.
6 ÷ 3 =2,
= (12 -2)
= 10
Putting this value in eqn (1) we have
= 21- 2
= 19 Ans
In this way, by following proper procedure , order of operations, opening parenthesis systematically required for solution, any given numerical expression can be solved and if you want to know about Math Blog on Probability and Statistics then refer Internet,Read more maths topics of different grades such as Mathematical Reasoning in the next session here.
Whenever we are given some mathematical equations to solve, These equations are to be solved in a particular order in order to get the proper solution to the expressions. This Order of Operations includes the following sequence:
1 parenthesis, (called Braces)
2. Of
3. Division
4. Multiplication
5. Addition
6. Subtraction.
This sequence is remembered by the student as an acronym- BODMAS
First of all the braces are opened ,solved and then removed to make the equation in the simple form.(Learn more about mathematical expressions here),
After every step, the equation / expression takes a simpler form,
Let us consider some examples:
Solve the numerical expression 15 - [ 11 - 8 ÷ ( 17 + 3 * 2 - 19 )
Here we solve the brackets in the order ( ), followed by , then [ ]
= 15 - [ 11 - 8 ÷ (17 + 6 - 19) ]
we have 17 + 6 - 19 = 4
= 15 - [ 11 - 8 ÷ 4 ]
= 15 - [ 11- 2 ] As 8 ÷ 4 = 2
= 15 - 9
= 6 Ans.
Solve the given numerical expression:
11/2 of (2/3 - 3/5) + 1/2 ÷ 5/11-------(1)
Here , following the heirachy of the above simplification procedure, first we solve:
(2/3 - 3/5)
l.c.m. of 3 & 5 is 15, so by making the denominator =15 we get
= ( 2 * 5 - 3 * 3 ) / 15
= ( 10 - 9 )/15
= 1/15
Putting this value in ( 1) we get
= 11/2 of 1/15 + 1/2 ÷ 5/11------ (2)
In next step we solve "of "
11/2 of 1/15
= 11 /30
Putting this value 11/30 in (2) we get
= 11/30 + 1/2 ÷ 5/11 --------- (3)
In next step we solve " ÷ "
1/2 ÷ 5/11
= 1/2 * 11/5
= 11 / 10
Putting this value 11/ 10 in (3) we get
11 / 30 + 11 / 10
Now taking L.C.M. of 10 and 30 = 30,
we get
= ( 11 + 11* 3 ) /30
= ( 11+ 33 ) / 30
= 44 / 30
Now converting into simplest form , we divide Numerator & denominator 2, w get
= 22 / 15 Ans
1 parenthesis, (called Braces)
2. Of
3. Division
4. Multiplication
5. Addition
6. Subtraction.
This sequence is remembered by the student as an acronym- BODMAS
First of all the braces are opened ,solved and then removed to make the equation in the simple form.(Learn more about mathematical expressions here),
After every step, the equation / expression takes a simpler form,
Let us consider some examples:
Solve the numerical expression 15 - [ 11 - 8 ÷ ( 17 + 3 * 2 - 19 )
Here we solve the brackets in the order ( ), followed by , then [ ]
= 15 - [ 11 - 8 ÷ (17 + 6 - 19) ]
we have 17 + 6 - 19 = 4
= 15 - [ 11 - 8 ÷ 4 ]
= 15 - [ 11- 2 ] As 8 ÷ 4 = 2
= 15 - 9
= 6 Ans.
Solve the given numerical expression:
11/2 of (2/3 - 3/5) + 1/2 ÷ 5/11-------(1)
Here , following the heirachy of the above simplification procedure, first we solve:
(2/3 - 3/5)
l.c.m. of 3 & 5 is 15, so by making the denominator =15 we get
= ( 2 * 5 - 3 * 3 ) / 15
= ( 10 - 9 )/15
= 1/15
Putting this value in ( 1) we get
= 11/2 of 1/15 + 1/2 ÷ 5/11------ (2)
In next step we solve "of "
11/2 of 1/15
= 11 /30
Putting this value 11/30 in (2) we get
= 11/30 + 1/2 ÷ 5/11 --------- (3)
In next step we solve " ÷ "
1/2 ÷ 5/11
= 1/2 * 11/5
= 11 / 10
Putting this value 11/ 10 in (3) we get
11 / 30 + 11 / 10
Now taking L.C.M. of 10 and 30 = 30,
we get
= ( 11 + 11* 3 ) /30
= ( 11+ 33 ) / 30
= 44 / 30
Now converting into simplest form , we divide Numerator & denominator 2, w get
= 22 / 15 Ans
Solve the Following:
21 -( 12 - 6 ÷ 3)------- (1 )
In this given expression, we first consider the brackets,
We solve the bracket
( 12 - 6 ÷ 3),
Now in Bracket , we first take division i. e.
6 ÷ 3 =2,
= (12 -2)
= 10
Putting this value in eqn (1) we have
= 21- 2
= 19 Ans
In this way, by following proper procedure , order of operations, opening parenthesis systematically required for solution, any given numerical expression can be solved and if you want to know about Math Blog on Probability and Statistics then refer Internet,Read more maths topics of different grades such as Mathematical Reasoning in the next session here.
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