BODMAS Rule-Sainik School Class 6 Math Study Material Notes pdf download free-Anand Classes
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What is BODMAS Rule ? |
BODMAS is an acronym and it stands for Bracket, Order, Division, Multiplication, Addition, and Subtraction.
In certain regions, PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction) is used, which is the synonym of BODMAS. Thus, the order of operations of BODMAS and PEMDAS is shown in the below figure.
According to the BODMAS rule, if an expression contains brackets ((), {}, []) we have first to solve or simplify the bracket followed by ‘order’ (that means powers and roots, etc.), then division, multiplication, addition and subtraction from left to right. Solving the problem in the wrong order will result in a wrong answer.
BODMAS Rule Full form
As we mentioned earlier, the full form of BODMAS is Brackets, Orders, Division, Multiplication, Addition, Subtraction. While applying the BODMAS rule we should follow the order of these operations.
B | Brackets | ( ), { }, [ ] |
O | Order of | Square roots, indices, exponents and powers |
D | Division | ÷, / |
M | Multiplication | ×, * |
A | Addition | + |
S | Subtraction | – |
Tips to Remember BODMAS Rule:
The rules to simplify the expression using BODMAS rule are as follows:
- First, simplify the brackets
- Solve the exponent or root terms
- Perform division or multiplication operation (from left to right)
- Perform addition or subtraction operation (from left to right)
This order must be followed to get accurate results.
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Here’s how you apply the BODMAS rule in mathematics:
- Brackets: First, perform any calculations within parentheses, brackets, or braces. Solve expressions inside the innermost set of brackets first. For example:
- (3 + 4) × 2 = 7 × 2 = 14
- Orders (Exponents and Roots): After dealing with the brackets, address exponents and roots (like square roots or cube roots).For example:
- 2^{3} = 2 × 2 × 2 = 8
- Division and Multiplication: From left to right, perform any division or multiplication operations. Both of these operations have the same level of precedence.For example:
- 6 ÷ 2 × 3 = 3 × 3 = 9
- Addition and Subtraction: Finally, from left to right, perform any addition or subtraction operations. Like division and multiplication, these have the same level of precedence.For example:
- 5 + 4 – 2 = 9 – 2 = 7
Using the BODMAS rule ensures that you perform calculations in the correct order, preventing errors in your math problems. This rule is particularly important in more complex mathematical expressions to maintain consistency and accuracy in your solutions.
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Example Question Using BODMAS Rule:
Take the following questions:
(3+5)-(5+2)
To solve this, the first bracket has to be solved which will give the following-
(8)-(5+2)
and then second bracket is solved, we get
(8)-(7)
Now, subtraction has to be solved:
8 – 7 = 1.
Lets take an another example to understand it better.
For example :-
84 × 3 – (3 ÷ 3) + 8 × 56 ÷ 7
According to BODMAS rule , in this question we should solve the the bracket first then division, multiplication, addition and then subtraction as;
84 × 3 – 1 + 8 × 56 ÷ 7
84 × 3 – 1 + 8 × 8
252 – 1 + 64
316 – 1
315
(Answer)
Simplification of Brackets
Simplification of terms inside the brackets can be done directly. That means we can perform the operations inside the bracket in the order of division, multiplication, addition and subtraction.
Note: The order of brackets to be simplified is (), {}, [].
Simplify:
14 + (8 – 2 × 3)
Solution:
14 + (8 – 2 × 3)
= 14 + (8 – 6)
= 14 + 2
= 16
Therefore, 14 + (8 – 2 × 3) = 16.
Simplify :
1800 ÷ [10{(12−6)+(24−12)}]
Solution:
1800 ÷ [10{(12−6)+(24−12)}]
Step 1: Simplify the terms inside {}.
Step 2: Simplify {} and operate with terms outside the bracket.
1800 ÷ [10{(12−6)+(24−12)}]
= 1800 ÷ [10{6+12}]
= 1800 ÷ [10{18}]
Step 3: Simplify the terms inside [ ].
= 1800 ÷ 180
= 10
Simplify :
1/2[{−2(1+2)}10]
Solution:
1/2[{−2(1+2)}10]
Step 1: Simplify the terms inside () followed by {}, then [].
Step 2: Operate terms with the terms outside the bracket.
1/2[{−2(1+2)}10]
= 1/2 [{-2(3)} 10]
= 1/2 [{-6} 10]
= 1/2 [-60]
= -30
BODMAS Rule without Brackets
The BODMAS rule can be applied to solve the mathematical expression without brackets too. Consider the following problems to verify.
Simplify:
17 – 24 ÷ 6 × 4 + 8
Solution:
17 – 24 ÷ 6 × 4 + 8
As per the BODMAS rule, we should perform the division first.
17 – 4 × 4 + 8
Let’s perform the multiplication.
17 – 16 + 8
Finally, addition and subtraction.
25 – 16 = 9
Simplify the expression:
1/7 of 49 + 125 ÷ 25 – 12
Solution:
1/7 of 49 + 125 ÷ 25 – 12
Here the term “of” refers to the operation of multiplication.
= (1/7) × 49 + 125 ÷ 25 – 12
= 7 + 125 ÷ 25 – 12
= 7 + 5 – 12
= 12 – 12
= 0
Solve :
8 + 9 ÷ 9 + 5 × 2 − 7.
Solution:
The problem given is 8 + 9 ÷ 9 + 5 × 2 − 7.
The division operation is performed first.
9 ÷ 9 = 1
So, the expression reduces to 8 + 1 + 5 × 2 − 7
The multiplication operation is taken next,
5 × 2 = 10
So, the expression reduces to 8 + 1 + 10 − 7
The addition operation is
8 + 1 + 10 = 19
The final answer is 19 – 7 = 12.
Simplify the expression
[25 – 3 (6 + 1)] ÷ 4 + 9
Solution:
The problem given is [25 – 3 (6 + 1)] ÷ 4 + 9.
The round bracket is (6 + 1) = 7.
The next bracket is 3 (7) = 21
Take [25 – 21] ÷ 4 + 9
(25 – 21) = 4
Then division operation is performed,
4 ÷ 4 = 1
Then 1 + 9 = 10
The final answer is 10.
Solve (1 / 4 + 1 / 8) of 64
Solution:
In the first step, consider (1 / 4 + 1 / 8) = (2 + 1) / 8 = 3 / 8
Take (3 / 8) of 64
Here the term “of” refers to the operation of multiplication.
(3 / 8) of 64 = (3 / 8) * 64
= 24
Simplify the given expression:
180 ÷ 15 {(12 – 6) – (14 – 12)}
Solution:
Initially, the first ( ) brackets are simplified,
180 ÷ 15 {(12 – 6) – (14 – 12)}
= 180 ÷ 15 (6 − 2) (solve round bracket)
= 180 ÷ 15 (4) (solve curly bracket)
= 12 (4) (divide 180 by 15 = 12)
= 12 × 4 (if no operator is mentioned behind any given bracket, multiplication operation can be performed)
= 48
The final answer is 48.
Simplify the following expression
3 + 2^{4} × (15 ÷ 3) using the BODMAS rule
Solution:
The expression given is 3 + 2^{4} × (15 ÷ 3).
The bracket is taken first.
(15 ÷ 3) = 5
Then 3 + 2^{4} × 5
The calculation is done in order 2^{4} = 2 × 2 × 2 × 2 = 16
16 × 5 = 80
The addition operation is performed next.
3 + 80 = 83
The final answer is 83.
Solve the expression using BODMAS rule
{50 – (2 + 3) + 15}
Solution:
Input Equation:
= {50 – (2 + 3) + 15}
= {50 – (5) + 15}
= {50 – 5 + 15}
= {45 + 15}
= {60}
= 60
Simplify the expression using the BODMAS rule
[18 – 2 (5 + 1)] ÷ 3 + 7.
Solution:
Input Equation can be rewritten:
= [18 – 2 * (5 + 1)] / 3 + 7
= [18 – 2 * (6)] / 3 + 7
= [18 – 2 * 6] / 3 + 7
= [18 – 12] / 3 + 7
= [6] / 3 + 7
= (6 / 3) + 7
= 2 + 7
= 9
Solved Problems integers and decimals
Problems based on the BODMAS rule involving integers and decimals values are given below.
Find the value of
2[2+2{39-2(17+2)}]
Solution :
Given, 2[2+2{39-2(17+2)}]
By using BODMAS Rule,
First, simplify the value inside the brackets ()
= 2[2+2{39-2(19)}]
= 2[2+2{39-38}]
subtract the value inside the brackets {}
= 2[2+2{1}]
= 2[2+2]
Add the value inside the brackets [ ]
=2[4]
Finally, multiply both values, which gives 8.
Therefore, the value of 2[2+2{39-2(17+2)}] is 8
Find the value of
4.5 – [2 + 0.5 of (2.1 – 1.3 x 1.01)]
Solution:
Given, 4.5 – [2 + 0.5 of (2.1 – 1.3 x 1.01)]
By using BODMAS Rule,
First simplify the value inside the brackets ()
= 4.5 – [2 + 0.5 of (2.1 – 1.3 x 1.01)] “Multiply 1.3 x 1.01”
= 4.5 – [2 + 0.5 of 0.787] “Subtract 1.313 from 2.1”
Note : Here the term “of” refers to the operation of multiplication.
Simplify the value inside the brackets [ ]
= 4.5 – [2 + 0.3935] “Multiply 0.5 x 0.787”
= 4.5 – 2.3935 “Add two values inside the brackets [ ]”
Finally, Subtract the values
= 2.1065
Therefore, the value of 4.5 – [2 + 0.5 of (2.1 – 1.3 x 1.01)] is 2.1065.
Evaluate:
2 + 4 ÷ (22 + 6) × 2.
Solution:
We have
2 + 4 ÷ (22 + 6) × 2
First, we solve the parentheses, and we get
= 2 + 4 ÷ 28 × 2
Now, perform the division 4/28
= 2 + 1/7 × 2
Now, we do the multiplication 1/7 × 2
= 2 + 2/7
= (14 + 2)/7
= 16/7 = 2 ^{2}/_{7}.
Evaluate:
{15 × 32 ÷ 2 × 5} ÷ 75
Solution:
Now, {15 × 32 ÷ 2 × 5} ÷ 75 = {15 × (32 ÷ 2) × 5} ÷ 75
= {15 × 16 × 5} ÷ 75
= 16
Evaluate:
5 × (2 × 3^{4}) ÷ 6 + 7 – 8
Solution:
5 × (2 × 3^{4}) ÷ 6 + 7 – 8
= 5 × (2 × 81) ÷ 6 + 7 – 8
= 5 × 162 ÷ 6 + 7 – 8
= 5 × 27 + 7 – 8
= 135 + 7 – 8
= 142 – 8 = 134.
Evaluate:
2 of 5 ÷ 5 + 3
Solution:
2 of 5 ÷ 5 + 3 = (2 × 5) ÷ 5 + 3
Note : Here the term “of” refers to the operation of multiplication.
= 10 ÷ 5 + 3
= 2 + 3 = 5.
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Types of BracketsSometimes we need to use more than one type of bracket. The brackets which are used are –
We should start solving from the innermost bracket. Generally, vinculum is used as innermost bracket, then parentheses, then braces, then square brackets. |
Evaluate: [15% of 150] + 23 ÷ 115
Solution:
[15/100 × 150] + 23 ÷ 115
= 22.5 + 23 ÷ 115
= 22.5 + ⅕ = 22.7
Simplify:
12 + 6 × 27 ÷ 3 + 2 – 16 ÷ 8 × 2
Solution:
12 + 6 × 27 ÷ 3 + 2 – 16 ÷ 8 × 2
= 12 + 6 × 9 + 2 – 2 × 2
= 12 + 54 + 2 – 4
= 68 – 4 = 64.
Evaluate:
0.07 × 0.28 ÷ 0.02 + 0.48 – 2.48 ÷ 0.04
Solution:
0.07 × 0.28 ÷ 0.02 + 0.48 – 2.48 ÷ 0.04
= 0.07 × 14 + 0.48 – 62
= 0.98 + 0.48 – 62
= 1.46 – 62 = – 60.54.
Frequently Asked Questions (FAQs) related to studying for the Sainik School Class 6 Math entrance exam
To help you prepare for Sainik School Class 6 Math, it’s important to use appropriate study materials. Here are some frequently asked questions (FAQ) related to studying for the Sainik School Class 6 Math entrance exam:
1. What is the syllabus for the Sainik School Class 6 Math entrance exam?
- The syllabus for the entrance exam may vary slightly from one Sainik School to another. However, it generally covers topics from the standard Class 6 mathematics curriculum, including arithmetic, geometry, algebra, and basic mathematical concepts.
2. Where can I find official information about the exam pattern and syllabus?
- You can find official information about the exam pattern and syllabus on the official website of the specific Sainik School you’re applying to. Each school may have its own admission criteria.
3. Which textbooks should I use for Class 6 Math preparation?
- You should primarily use the NCERT Class 6 Math textbook. It covers the fundamental concepts and is widely accepted in Indian schools. Additionally, consider supplementary Math textbooks that are designed for competitive exams.
4. Are there any online resources for Sainik School Class 6 Math preparation?
- Yes, you can find online resources such as video tutorials, practice questions, and mock tests on educational websites. Websites like ANAND CLASSES offer free Math materials that can be helpful for your preparation.
5. Where can I get sample papers and previous year’s question papers?
- You can find sample papers and previous year’s question papers at online bookstore of ANAND CLASSES that sell competitive exam preparation materials. Additionally, ANAND CLASSES website offer downloadable PDFs of these papers for free or at a minimal cost.
6. Should I consider enrolling in coaching classes for Sainik School Math preparation?
- Enrolling in coaching classes is a personal choice. While they can provide structured guidance and additional practice, they are not mandatory. You can achieve success through self-study and the use of appropriate study materials.
7. How should I manage my study time effectively for Class 6 Math preparation?
- Create a study schedule that allocates specific time for Math preparation daily. Focus on understanding concepts, practicing regularly, and taking regular breaks to avoid burnout. Consistency is key.
8. Is there any specific advice for tackling the Math section of the Sainik School entrance exam?
- Pay close attention to the BODMAS rule (Order of Operations) and practice solving a variety of math problems. Make sure you’re familiar with the types of questions that are commonly asked in the entrance exam.
Remember to check the specific requirements and guidelines provided by the Sainik School you are applying to, as these may vary from school to school.
To prepare effectively for the Sainik School Class 6 Math entrance exam, you should consider the following general sources:
Sainik School Official Website: Visit the official website of the Sainik School you are applying to. They often provide information about the exam pattern, syllabus, and sample question papers.
NCERT Textbooks: The National Council of Educational Research and Training (NCERT) textbooks are widely used in Indian schools and are a valuable resource for exam preparation. Ensure you have the NCERT Math textbook for Class 6.
Solved Sample Papers and Previous Year Question Papers: You can find solved sample papers and previous year question papers for Sainik School entrance exams at bookstores or online at ANAND CLASSES website. These papers can give you an idea of the exam pattern and types of questions asked.
Math Study Guides: The Math study guides and reference books are publish under publication department of ANAND CLASSES and are designed to help students prepare for entrance exams. Look for books specifically tailored to the Sainik School entrance exam.
- Online Resources: www.anandclasses.co.in
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