Simplification and Approximation

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.

Approximation is the process of finding a value or solution that is close to the exact one, often used when obtaining the precise value is difficult or impossible. Together, these techniques enable clearer analysis, easier calculations, and practical problem-solving in real-world situations.

How to Simplify Fractions?

To solve the above question, follow the steps below:

Step 1: Convert each mixed number to an improper fraction:,

• 2 \frac{1}{3} = \frac{7}{3}
• 3 \frac{1}{2} = \frac{7}{2}
• 4 \frac{1}{4} = \frac{17}{4}

Step 2: Find the Common Denominator

The denominators are 3, 2, and 4. Their Least Common Denominator (LCD) is 12.

Convert all fractions:

• \frac{7}{3} = \frac{28}{12}
• \frac{7}{2} = \frac{42}{12}
• \frac{17}{4} = \frac{51}{12}

Step 3: Add Fractions

\frac{28}{12} + \frac{42}{12} + \frac{51}{12} = \frac{121}{12}

Step 4: Convert Back to a Mixed Number

\frac{121}{12} = 10 \frac{1}{12}

In the article below we will learn some methods, rules, tricks, to simplify the questions related to simplification and appr

Simplification Rules

Several simplification rules can be applied to solve quantitative aptitude questions efficiently. Some of the most commonly used simplification rules are:

Tips and Tricks to Solve Simplification Questions

• When attempting to solve questions that involve, simplification, it is important to remember the VBODMAS rule. 
• Keep in mind the following formulas to solve the questions accurately:
• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² – 2ab
• a² – b² = (a+b) (a-b)
• a³ + b³ = (a+b) (a² – ab + b²)
• (a+b)³ = a³ + b³ + 3ab (a+b)
• (a-b)³ = a³ – b³ – 3ab (a-b)
• Always put on a timer while solving questions so that you can solve questions on time.

Rules to Solve the Mathematical Expressions by Approximation:

Rule 1:

To solve complex problems, take the closest value of the number given in the expression. for example, 77.8 is round off to 78;

33.02 is round off to 33 etc.

Example 1:  19% of (399.88/20 × 400) + 30 =?

          20/100 × (400/20 × 400) + 30 =?

          1/5 × (8000) + 30 = 1600 + 30 = 1630

Rule 2:

To solve problems with large numbers involved in multiplication, we can consider the approximate value of the large numbers by increasing or decreasing the round off values making the computation easy. for example, 239 × 111 is approximated to 240 × 110.

Example 2: 192 × 397 + 560 × 5/7 + 729.80 =?

         192 × 397 + 560 × 5/7 + 730 =?

         190 × 400 + 400 + 730 = 76000 + 1130 = 77130

Rule 3:

To solve problems with large numbers involved in the division, we can consider the approximate value of the large numbers by increasing or decreasing the round off values making the computation easy. for example, 6198.36/38.69 is approximated as 6200/40.

Example 3: 862.5/18.64 =?

        860/20 = 43

Example 1:

Practice Problem: Simplify the expression 15/5 ​+ 3(4 − 2)

Solution:

We can simplify by first evaluating the expression inside the parentheses:

15​/5 + 3(4 − 2) = 15/5​ + 3(2)

Then, we can calculate each part:

15/5 ​= 3

and

3(2) = 6

Now, we add the two results together:

3 + 6 = 9

Therefore, the expression 15/5 ​+ 3(4 − 2) simplifies to 9.

Example 2:

Practice Problem: Approximate the value of 1.28÷0.41.28 \div 0.41.28÷0.4 using rounding to two decimal places.

Solution:

• Round 1.28 to two decimal places.
• Round 0.4 to two decimal places.
• Perform the division with the rounded values.

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Practice Quiz on Simplification and Approximation.