Simplification
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Solved Examples
Study MaterialSolved Examples – Simplification
Solved examples help students understand the practical application of simplification concepts in competitive examinations. These examples are designed from basic to advanced level and cover important questions frequently asked in SSC, Banking, Railway, CDS, NDA, CAT, UPSC, and placement aptitude tests.
Topics Covered in Solved Examples
- VBODMAS Rule
- Bracket Operations
- Fractions and Decimals
- Square and Cube Simplification
- Algebraic Identities
- Approximation Techniques
- Percentage and Fraction Conversions
- Absolute Value Problems
- Mixed Arithmetic Operations
- Mental Calculation Techniques
Example 1: Basic VBODMAS Rule
Question: Simplify:
18 + 6 × 4
Solution:
According to VBODMAS, multiplication is performed before addition.
6 × 4 = 24
18 + 24 = 42
Example 2: Bracket Simplification
Question: Simplify:
[12 + {8 − (3 × 2)}]
Solution:
First solve small bracket:
3 × 2 = 6
Now:
[12 + {8 − 6}]
= [12 + 2]
= 14
Example 3: Vinculum Operation
Question: Simplify:
8 + 3 + 2 × 4
Solution:
Solve the vinculum first:
3 + 2 = 5
Now:
8 + 5 × 4
= 8 + 20
= 28
Example 4: Fraction Addition
Question: Find:
1/2 + 1/3
Solution:
LCM of 2 and 3 = 6
1/2 = 3/6
1/3 = 2/6
Therefore:
3/6 + 2/6 = 5/6
Example 5: Decimal Addition
Question: Simplify:
12.5 + 7.25 + 0.75
Solution:
12.5 + 7.25 = 19.75
19.75 + 0.75 = 20.5
Example 6: Decimal Multiplication
Question: Find:
0.2 × 0.02 × 0.002
Solution:
Ignore decimal points initially:
2 × 2 × 2 = 8
Total decimal places:
1 + 2 + 3 = 6
Answer = 0.000008
Example 7: Square Shortcut
Question: Find:
297²
Solution:
297² = (300 − 3)²
Using identity:
(a − b)² = a² + b² − 2ab
= 300² + 3² − 2 × 300 × 3
= 90000 + 9 − 1800
= 88209
Example 8: Difference of Squares
Question: Simplify:
99 × 101
Solution:
99 × 101
= (100 − 1)(100 + 1)
Using:
(a − b)(a + b) = a² − b²
= 100² − 1²
= 10000 − 1
= 9999
Example 9: Cube Identity
Question: Simplify:
8³ − 2³
Solution:
Using identity:
a³ − b³ = (a − b)(a² + ab + b²)
= (8 − 2)(64 + 16 + 4)
= 6 × 84
= 504
Example 10: Absolute Value
Question: Simplify:
|−25| + |12|
Solution:
|−25| = 25
|12| = 12
Therefore:
25 + 12 = 37
Example 11: Percentage to Fraction Simplification
Question: Find:
50% of 240
Solution:
50% = 1/2
Therefore:
1/2 × 240 = 120
Example 12: Simplification with Division
Question: Simplify:
84 ÷ 7 + 5 × 3
Solution:
84 ÷ 7 = 12
5 × 3 = 15
12 + 15 = 27
Example 13: Decimal Division
Question: Find:
18.6 ÷ 3
Solution:
18.6 ÷ 3 = 6.2
Example 14: Multiplication by 25 Shortcut
Question: Find:
64 × 25
Solution:
Multiply by 100 and divide by 4:
64 × 100 = 6400
6400 ÷ 4 = 1600
Example 15: Approximation Technique
Question: Approximate:
19.98 × 5.02
Solution:
19.98 ≈ 20
5.02 ≈ 5
20 × 5 = 100
Example 16: Complex Expression
Question: Simplify:
24 ÷ 6 × 4 + 8 − 3
Solution:
24 ÷ 6 = 4
4 × 4 = 16
16 + 8 = 24
24 − 3 = 21
Example 17: Reciprocal Concept
Question: Find:
5 ÷ (2/3)
Solution:
Division by fraction becomes multiplication by reciprocal.
= 5 × 3/2
= 15/2
= 7.5
Example 18: Mixed Fraction Expression
Question: Simplify:
2/5 of 50 + 12
Solution:
2/5 of 50
= 2/5 × 50
= 20
20 + 12 = 32
Important Exam Tips
- Always follow VBODMAS sequence correctly.
- Practice multiplication tables, squares, and cubes regularly.
- Use algebraic identities for faster calculations.
- Apply approximation techniques in lengthy calculations.
- Be careful with negative signs and brackets.
- Simplify fractions before multiplication.
- Use mental calculation tricks whenever possible.
Practicing solved examples regularly improves speed, conceptual clarity, and accuracy in solving Simplification questions in competitive examinations.