Simplification- Basic Formulas With Practice Questions & Answers

8 mins read

Students can better understand the underlying principles and relationships by simplifying mathematical expressions, equations, and concepts. This process not only enhances problem-solving skills but also promotes critical thinking and analytical reasoning.

Basics Of Simplification

Simplification in mathematics involves reducing complex expressions to their simplest form by combining like terms or eliminating unnecessary elements. Simplifying equations and expressions makes them easier to work with, understand, and manipulate.

BODMAS Rule in Simplification Explained

Let us study the role of BODMAS in simplification:

Order Of Operations

BODMAS, which stands for Brackets, Orders, Division, Multiplication, Addition and Subtraction, is a fundamental mathematical rule to ensure accurate calculations. The order of operations dictates the sequence in which mathematical operations should be performed.

First, calculations within brackets are solved, followed by any exponents or roots. Then, division and multiplication are carried out from left to right before finally adding and subtracting similarly from left to right.

Vinculum In Mathematics

The vinculum, a horizontal line in mathematical expressions, is crucial in simplification. It often indicates the grouping of numbers or terms within an expression, indicating which operations should be performed first.

Click here to enhance and upskill your quantitative aptitude related to simplification right away!

Simplification & Approximation Relation

Simplification involves reducing a mathematical expression to its simplest form, while approximation focuses on finding close or near values to the actual result. When simplifying, exact values are sought after, whereas in approximation, approximate values are acceptable.

When doing math, simplification gives exact answers without rounding. Approximation is better for quick and efficient calculations when decimals work fine. In real-world applications like data analysis or statistics, approximation is often preferred over exact simplification due to time constraints and practicality.

Key Terms in Simplification

Let us study some of the key terms used in simplification:

Examples Of Simplification Techniques

Let us study some of the examples of simplification:

Fractions: Simplifying fractions means making them simpler by dividing the top and bottom numbers by their biggest shared number.

Exponents: When simplifying expressions with exponents, apply the rule that states when you divide two terms with the same base, subtract the exponents.

Algebraic Expressions: In algebra, simplification often involves combining like terms and performing operations following the order of operations.

Simplification Questions With Solved Answers

For your practice, we have selected some questions with detailed answers. Practice them and get ready for any placement interview or competitive exams:

Question 1. Find the value of (3 * 4) – (8 + 12).

(a) 9 

(b) 9.25 

(c) –9.25 

(d) None of these

Solution b) 9.25 

Explanation: Let us apply the BODMAS rule to solve this question,

(3 * 4) – (8 + 12) = –37 – (–4). [Note here that the ‘–’ sign between –37 and –4 is the operation defined above.]

= 37/4 = 9.25

(Define the following functions for the questions (2 - 4):

(a) (a M b) = a – b

(b) (a D b) = a + b

(c) (a H b) = ab

(d) (a P b) = a/b)

Question 2. What is the value of (3M4H2D4P8M2)? Define the following functions: 

(a) 6.5   

(b) 6   

(c) –6.5   

(d) None of these

Solution: c) -6.5 

Explanation: 3– 4 x 2 + 4/8 – 2 = 3 – 8 + 0.5 – 2 = – 6.5

Question 3. Which of the following functions will represent a² – b²? 

(a) (a M b) H (a D b)               

(b) (a H b) M (a P b)          

(c) (a D b)/(a M b)      

(d) None of these 

Solution: a) (a M b) H (a D b) 

Explanation: Solve using the options

option a = (a – b) (a + b) = a2 – b2.

Question 4. Which of the following represents a²? 

(a) (a M b) H (a D b) + b²              

(b) (a H b) M (a P b) + b²                                                         

(c) (a M b)/(a P b)              

(d) Both (a) and (c)

Solution: a) (a M b) H (a D b) + b² 

Explanation: Solve using the options

Option a = (a2 – b2 ) + b2 = a2

Question 5. If 381A is divisible by 9, find the value of the smallest natural number A.

(a) 5   

(b) 4 

(c) 7 

(d) 6

Solution: d) 6 

Explanation: For the given number 381A to be divisible by 9, its sum 3 + 8 + 1 + A should be divisible by 9. Choosing from among the giving options, A should be 6. Option (d) is correct.

Question 6. What will be the remainder obtained when (96 + 1) will be divided by 8?

(a) 0 

(b) 3 

(c) 7 

(d) 2

Solution: d) 2 

Explanation: 96, when divided by 8, would give a remainder of 1. Hence, the required answer would be 2.

Question 7. Determine the ratio between the LCM and HCF of 5, 15 and 20 is

(a) 8:1 

(b) 14:3  

(c) 12:2 

(d) 12:1

Solution: d) 12:1

Explanation: Given 5, 15 and 20. The LCM of these numbers is 60. 

Likewise, 5 is the HCF of 5, 15 and 20. 

Therefore, the ratio is 60:5 = 12:1

Question 8. Find the LCM of 5/2, 8/9, 11/14.

(a) 280 

(b) 360 

(c) 420 

(d) None of these

Solution: d) None of these 

Here's a step-by-step explanation:

LCM of 5/2, 8/9 and 11/14 would be given by: (LCM of numerators)/(HCF of denominators) = 440/1 = 44.

Question 9. A five-digit number is taken. The sum of the first four digits (excluding the number at the unit digit) equals the sum of all five digits. Which of the following will not necessarily divide this number?

(a) 10 

(b) 2

(c) 4 

(d) 5

Solution: c) 4

Explanation: The essence of this question is in the fact that the last digit of the number is 0. 

Naturally, the number is necessarily divisible by 2,5 and 10. Only 4 does not necessarily divide it.

Question 10. Find the number of divisors of 1420.

(a) 14 

(b) 15 

(c) 13 

(d) 12

Solution: d) 12

Explanation: 1420 = 142 x 10 = 22 x 711 x 51. 

The total factors of the number can be calculated by adding 1 to each exponent in the prime factorization and then multiplying them together as (2+1) x (1+1) x (1+1) = 3 x 2 x 2 = 12. Option (d) is correct. 

Conclusion

The exploration of simplification techniques has provided a comprehensive understanding of the fundamental principles of solving mathematical problems efficiently. From grasping the basics to mastering advanced strategies, the journey through BODMAS rules, vinculum applications, and approximation relations has equipped individuals with the tools to simplify complex calculations effectively.

To further enhance mathematical proficiency, individuals are encouraged to practice regularly, seek challenging problems for application, and remain diligent in honing their simplification skills.

Time For A Short Quiz

Frequently Asked Questions (FAQs)

1. What are the key principles of simplification in mathematics?

Simplification in mathematics involves applying basic rules like BODMAS, understanding the vinculum, and using approximation techniques to solve complex problems efficiently.

2. How can one effectively master the art of simplification in mathematics?

To master simplification, one should practice regularly, understand key terms, use strategies like BODMAS, learn various techniques through examples, and focus on precision while approximating values.

3. Can you explain the significance of the BODMAS rule in simplification?

The BODMAS rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction) is crucial for maintaining the correct order of operations when simplifying mathematical expressions to ensure accurate results.

4. What role do approximation techniques play in simplification?

Approximation techniques help simplify complex calculations by replacing numbers with easier values while ensuring the result is close enough to the actual value for practical purposes.

5. How do strategies like breaking down problems aid in effective simplification?

Breaking down problems into smaller steps or components allows individuals to handle each part systematically, making it easier to simplify complex expressions and arrive at accurate solutions efficiently.

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