Simplification and Approximation

Welcome to the crucial topic of Simplification and Approximation, a cornerstone of the Quantitative Aptitude section for the SBI PO Preliminary Examination. This section tests your ability to quickly and accurately perform calculations involving numbers, fractions, percentages, and roots. Mastering these techniques will not only save you valuable time but also significantly boost your score. We will cover various types of problems, from basic arithmetic operations to more complex expressions involving powers and roots, along with approximation techniques.

1. Order of Operations (BODMAS/PEMDAS)

The fundamental rule governing the order of calculations in any mathematical expression is the BODMAS or PEMDAS rule. Understanding and applying this rule correctly is the first step towards accurate simplification.

PEMDAS is an alternative acronym where 'P' stands for Parentheses, 'E' for Exponents (similar to Orders), 'M' for Multiplication, 'D' for Division, 'A' for Addition, and 'S' for Subtraction. Division and Multiplication have the same precedence and are performed from left to right. Similarly, Addition and Subtraction also have the same precedence and are performed from left to right.

Example: Simplify $12 + (6 \times 5 - 3) \div 3^2$.

Step 1: Solve the expression inside the brackets: $6 \times 5 = 30$. Then $30 - 3 = 27$.

Step 2: Solve the exponent: $3^2 = 9$.

Step 3: Perform the division: $27 \div 9 = 3$.

Step 4: Perform the addition: $12 + 3 = 15$.

The simplified result is 15.

2. Simplification of Fractions and Decimals

Problems often involve fractions and decimals. It's essential to be comfortable converting between them and performing operations.

2.1. Fractions

Addition/Subtraction: To add or subtract fractions, they must have a common denominator. Find the Least Common Multiple (LCM) of the denominators.

Example: $\frac{2}{3} + \frac{1}{4}$. The LCM of 3 and 4 is 12. $\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$ $\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$ So, $\frac{8}{12} + \frac{3}{12} = \frac{8+3}{12} = \frac{11}{12}$.

Multiplication: Multiply the numerators together and the denominators together.

Example: $\frac{2}{5} \times \frac{3}{7} = \frac{2 \times 3}{5 \times 7} = \frac{6}{35}$.

Division: To divide by a fraction, multiply by its reciprocal.

Example: $\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times \frac{2}{1} = \frac{4}{3}$.

2.2. Decimals

Addition/Subtraction: Align the decimal points vertically and add or subtract as usual.

Example: $12.345 + 0.67 + 5$. ``` 12.345 0.670 5.000 ------- 18.015 ```

Multiplication: Multiply the numbers as if they were whole numbers. Then, count the total number of decimal places in the original numbers and place the decimal point in the result so that it has that many decimal places.

Example: $1.2 \times 0.34$. $12 \times 34 = 408$. $1.2$ has 1 decimal place, $0.34$ has 2 decimal places. Total = 3 decimal places. So, the result is $0.408$.

Division: To divide by a decimal, make the divisor a whole number by multiplying both the divisor and the dividend by a power of 10. Then perform the division.

Example: $4.5 \div 0.09$. Multiply both by 100: $(4.5 \times 100) \div (0.09 \times 100) = 450 \div 9 = 50$.

3. Simplification of Percentages

Percentages are a common feature in simplification problems. Converting percentages to fractions or decimals is key. Remember: $x\% = \frac{x}{100}$.

Example: Calculate $25\%$ of 200. $25\% \text{ of } 200 = \frac{25}{100} \times 200 = \frac{1}{4} \times 200 = 50$.

Example: Calculate $15\%$ of 80 + $30\%$ of 120. $15\% \text{ of } 80 = \frac{15}{100} \times 80 = \frac{3}{20} \times 80 = 3 \times 4 = 12$. $30\% \text{ of } 120 = \frac{30}{100} \times 120 = \frac{3}{10} \times 120 = 3 \times 12 = 36$. Total = $12 + 36 = 48$.

4. Simplification of Powers and Roots

These problems involve exponents and radicals. Key laws of exponents and properties of roots are crucial here.

4.1. Laws of Exponents

Let $a$ and $b$ be non-zero real numbers and $m$ and $n$ be integers.

• $a^m \times a^n = a^{m+n}$
• $a^m \div a^n = a^{m-n}$
• $(a^m)^n = a^{m \times n}$
• $a^0 = 1$
• $a^{-n} = \frac{1}{a^n}$
• $(ab)^n = a^n b^n$
• $(\frac{a}{b})^n = \frac{a^n}{b^n}$

Example: Simplify $3^4 \times 3^2 \div 3^3$. Using the laws: $3^{(4+2-3)} = 3^3 = 27$.

Example: Simplify $(5^2)^3 \div 5^4$. $(5^2)^3 = 5^{2 \times 3} = 5^6$. So, $5^6 \div 5^4 = 5^{6-4} = 5^2 = 25$.

4.2. Properties of Roots

The square root of a number $x$ is a number $y$ such that $y^2 = x$. We denote it as $\sqrt{x}$.

• $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$
• $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$
• $(\sqrt{a})^2 = a$
• $\sqrt{a^2} = |a|$ (absolute value of a)
• Cube root: $\sqrt[3]{x}$ is a number $y$ such that $y^3 = x$.

Example: Simplify $\sqrt{144 \times 16}$. $\sqrt{144 \times 16} = \sqrt{144} \times \sqrt{16} = 12 \times 4 = 48$.

Example: Simplify $\sqrt{\frac{81}{25}}$. $\sqrt{\frac{81}{25}} = \frac{\sqrt{81}}{\sqrt{25}} = \frac{9}{5}$.

Example: Calculate $\sqrt{64} + \sqrt{100} - \sqrt{36}$. $\sqrt{64} = 8$ (since $8^2 = 64$) $\sqrt{100} = 10$ (since $10^2 = 100$) $\sqrt{36} = 6$ (since $6^2 = 36$) So, $8 + 10 - 6 = 18 - 6 = 12$.

5. Approximation Techniques

In this type of problem, you are asked to find an approximate value of a complex expression. The key is to round the numbers in the expression to the nearest easy-to-calculate values (like multiples of 10, 100, or simple fractions/percentages) before performing the calculations.

General Strategy:

1. Identify the numbers in the expression.
2. Round each number to its nearest easy-to-handle value.
3. Perform the calculations with the rounded numbers.
4. The result obtained is the approximate value.

Rounding Rules:

• If the digit after the decimal place is 5 or greater, round up.
• If the digit after the decimal place is less than 5, round down (keep the digit as it is).
• For whole numbers, round to the nearest 10, 100, 1000, etc. For example, 48 rounds to 50, 152 rounds to 150.

Example: Find the approximate value of $49.87\%$ of $803 + 19.91\%$ of $1010$.

Step 1: Round the numbers. $49.87\% \approx 50\%$ $803 \approx 800$ $19.91\% \approx 20\%$ $1010 \approx 1000$

Step 2: Substitute the rounded values into the expression. Approximate value $\approx (50\% \text{ of } 800) + (20\% \text{ of } 1000)$.

Step 3: Calculate. $50\% \text{ of } 800 = \frac{50}{100} \times 800 = \frac{1}{2} \times 800 = 400$. $20\% \text{ of } 1000 = \frac{20}{100} \times 1000 = \frac{1}{5} \times 1000 = 200$.

Step 4: Add the results. Approximate value $\approx 400 + 200 = 600$.

Example: Find the approximate value of $24.8 \times 39.7 + 10.2 \times 59.6$.

Step 1: Round the numbers. $24.8 \approx 25$ $39.7 \approx 40$ $10.2 \approx 10$ $59.6 \approx 60$

Step 2: Substitute and calculate. Approximate value $\approx (25 \times 40) + (10 \times 60)$. $25 \times 40 = 1000$. $10 \times 60 = 600$.

Step 3: Add the results. Approximate value $\approx 1000 + 600 = 1600$.

6. Common Pitfalls and Tips

Pitfall 1: Incorrect order of operations. Always remember BODMAS/PEMDAS. Division and multiplication must be done before addition and subtraction.

Tip: Use parentheses to explicitly group operations if you are unsure. For example, $(6 \times 5) - (3 \div 3^2)$.

Pitfall 2: Errors in fraction or decimal conversion. Misplacing the decimal point or incorrect common denominator can lead to wrong answers.

Tip: Double-check your conversions. For percentages, remember $x\% = x/100$. For decimals, move the decimal point two places to the left.

Pitfall 3: Over-approximation or under-approximation. Rounding numbers too drastically or not enough can lead to an answer far from the correct one.

Tip: For approximation, aim to round to numbers that are easy to compute mentally or with simple calculations. For example, rounding 19.91 to 20 is usually better than rounding to 19.9.

Pitfall 4: Calculation errors. Simple multiplication or addition mistakes can be costly.

Tip: Practice mental math and quick calculation techniques. For approximations, the goal is speed, so focus on estimation.

Pitfall 5: Confusing simplification and approximation. Simplification requires an exact answer, while approximation requires a close estimate.

Tip: Read the question carefully. If it asks for an "approximate value" or "nearly equal to," use approximation techniques. If it asks for the "exact value" or to "simplify," perform precise calculations.

Mastering simplification and approximation is about consistent practice and understanding the underlying mathematical rules. By applying BODMAS, knowing your fractions, decimals, percentages, and practicing approximation techniques, you can tackle these questions efficiently and accurately in the SBI PO exam.