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Infographics: Simplification & Approximation

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Simplification & Approximation
Master these essential techniques for cracking quantitative aptitude questions in competitive 
examinations. This comprehensive guide covers fundamental rules, practical tips, and strategic 
approaches to tackle simplification and approximation problems with confidence and speed.
BODMAS Rule
Follow the correct 
sequence: Brackets, 
Orders, Division, 
Multiplication, Addition, 
Subtraction
Approximation 
Methods
Convert decimals to 
nearest values and 
simplify square roots 
efficiently
Modulus 
Operations
Understand absolute 
values: |x| is always 
positive or zero
BODMAS Rule: The Foundation
The BODMAS rule defines the correct 
sequence for performing mathematical 
operations. Always follow this order strictly:
Brackets - Solve (), {}, [] in order 1.
Orders - Powers, square roots, exponents 2.
Division & Multiplication - Left to right 3.
Addition & Subtraction - Left to right 4.
= 12 + 22 ÷ 11 × 6² - 10
= 12 + 22 ÷ 11 × 36 - 10
= 12 + 2 × 36 - 10
= 12 + 72 - 10
= 74
Example Problem
Solve: 12 + 22 ÷ 11 × (18 ÷ 3)² - 10
Solution:
Approximation Techniques
0 1
Convert Decimals
Round decimal numbers to 
their nearest whole values for 
quick calculations. Example: 
4433.764 becomes 4434
0 2
Simplify Percentages
Convert complex percentages 
to simpler values. Example: 
20.3% becomes 20%, 16.8% 
becomes 17%
0 3
Factor Square Roots
Break down numbers inside 
square roots into factors, 
extract pairs, and simplify 
systematically
Modulus of Real Numbers
Key Concept: The modulus |x| of any real number x is always positive or zero. If x g 0, 
then |x| = x. If x < 0, then |x| = -x.
Example: |8| = 8 and |-8| = 8
Quick Calculation Shortcuts
Multiplying by 99
For 64 × 99:
Subtract 1: 64 - 1 = 63
Find complement: 100 - 
64 = 36
Answer: 6336
Multiplying by 999
For 678 × 999:
Subtract 1: 678 - 1 = 677
Find complement: 1000 
- 678 = 322
Answer: 677322
Squaring Numbers
For 102²:
Add difference: 102 + 2 
= 104
Square difference: 2² = 
04
Answer: 10404
Practice Problems with Solutions
1
Denomination 
Problem
Question: A man has Rs. 
480 in one-rupee, five-
rupee, and ten-rupee 
notes equally. Total 
notes?
Solution: Let x = notes of 
each type
x + 5x + 10x = 480 ³ 16x 
= 480 ³ x = 30
Answer: Total notes = 3x 
= 90
2
Examination 
Rooms
Question: If 10 students 
move A³B, rooms equal. 
If 20 move B³A, A has 
double B's students. Find 
A?
Solution: x - 10 = y + 10 
and x + 20 = 2(y - 20)
Solving: x - y = 20 and x 
- 2y = -60
Answer: x = 100
3
Price Calculation
Question: 10 chairs = 4 
tables in price. 15 chairs 
+ 2 tables = Rs. 4000. 
Find 12 chairs + 3 tables?
Solution: Let chair = x, 
table = y
10x = 4y ³ y = 5x/2. 
Then 20x = 4000 ³ x = 
200, y = 500
Answer: 12x + 3y = Rs. 
3900
10-12
Questions Expected
Typical number of 
simplification questions in 
competitive exams
2-3
Minutes Per Question
Average time to solve with 
practice and shortcuts
100%
Scoring Potential
Master these basics for full 
marks in this section
Key Takeaways for Success
Practice BODMAS Daily
Solve 5-10 problems daily to build muscle 
memory for operation sequence. Speed 
comes with consistent practice.
Use Approximation Wisely
Don't waste time on exact calculations. 
Round smartly and save precious exam 
minutes.
Master Shortcuts
Learn multiplication tricks for 99, 999, 
and quick squaring methods to boost 
calculation speed.
Time Management
Allocate maximum 2-3 minutes per 
question. If stuck, move on and return 
later with fresh perspective.
"Success in quantitative aptitude isn't about complex mathematics4it's about applying 
simple rules correctly and consistently. Master the basics, practice shortcuts, and build 
speed through repetition."
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FAQs on Infographics: Simplification & Approximation

1. What is simplification in the context of the SSC CGL exam?
Ans. Simplification in the context of the SSC CGL exam refers to the process of reducing complex numerical expressions or calculations to simpler forms. This helps in making calculations quicker and more manageable, allowing candidates to solve problems efficiently within the time constraints of the exam.
2. How does approximation differ from simplification in mathematical problems?
Ans. Approximation involves estimating a value to make calculations easier, often by rounding numbers to the nearest whole number or a convenient figure. In contrast, simplification focuses on reducing expressions without changing their value. Both techniques are useful in the SSC CGL exam to save time during problem-solving.
3. Why are simplification and approximation important for SSC CGL aspirants?
Ans. Simplification and approximation are crucial for SSC CGL aspirants as they enhance speed and accuracy in solving quantitative aptitude questions. Mastering these techniques allows candidates to handle complex calculations quickly, which is essential for achieving high scores in the exam.
4. What types of questions typically involve simplification and approximation in SSC CGL?
Ans. Questions involving simplification and approximation in SSC CGL typically include arithmetic problems, percentage calculations, and numerical series. These questions often require candidates to quickly estimate or simplify calculations to arrive at the correct answer in a limited time.
5. Can you provide an example of a simplification problem?
Ans. An example of a simplification problem is: Calculate 48 ÷ (6 × 4) + 12. First, simplify the expression inside the brackets: 6 × 4 = 24. Then, simplify: 48 ÷ 24 = 2. Finally, add 12: 2 + 12 = 14. Thus, the answer is 14.
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