Simplification, Approximation, Quadratic Equations

Simplification

Simplification is a fundamental topic in quantitative aptitude, focusing on the ability to perform arithmetic operations accurately and efficiently. It involves applying the order of operations and using basic mathematical rules to solve complex-looking expressions. Mastering simplification is crucial as it forms the basis for solving many other quantitative aptitude problems.

Order of Operations (BODMAS/PEMDAS)

The order of operations dictates the sequence in which mathematical operations should be performed to ensure a unique and correct answer. The acronyms BODMAS and PEMDAS are commonly used mnemonics for this.

• Brackets (or Parentheses)
• Orders (or Exponents, powers, square roots)
• Division and Multiplication (from left to right)
• Addition and Subtraction (from left to right)

Key Concepts and Techniques

To simplify expressions quickly and accurately, several key concepts and techniques are employed:

• Fractions: Understanding how to add, subtract, multiply, and divide fractions. This includes finding a common denominator for addition and subtraction, and inverting and multiplying for division.
• Decimals: Proficiency in decimal arithmetic, including multiplication and division. Multiplying by powers of 10 involves shifting the decimal point.
• Percentages: Converting percentages to fractions or decimals and vice versa. For example, 50% is 1/2 or 0.5, and 25% is 1/4 or 0.25.
• Square Roots and Cube Roots: Recognizing perfect squares and cubes to simplify these operations.
• Exponents: Applying rules of exponents, such as a^m * aⁿ = a^m+n and a^m / aⁿ = a^m-n.

Example:

Simplify the following expression: 18 + [16 - (6 + 2)] ÷ 4 × 3

1. Start with the innermost bracket: (6 + 2) = 8
2. Next bracket: [16 - 8] = 8
3. Now perform division: 8 ÷ 4 = 2
4. Next, perform multiplication: 2 × 3 = 6
5. Finally, perform addition: 18 + 6 = 24

Therefore, the simplified value is 24.

Practice Problems:

1. Calculate: 5/8 of 160 + 25% of 200 - 15 × 4

2. Evaluate: 36 ÷ 6 + (12 × 3) - 5

3. Find the value of: 75 + [20 - (5 + 3)] ÷ 2 × 4

Approximation

Approximation is a technique used to estimate the value of a complex calculation without performing the exact computation. This is particularly useful in multiple-choice questions where the options are spread out, allowing you to quickly select the closest answer. It involves rounding numbers to their nearest convenient values.

When to Use Approximation

Approximation is best used when:

• The question explicitly asks for an approximate value.
• The options provided are significantly different from each other.
• The calculation involves large numbers, decimals, or complex fractions that would be time-consuming to solve exactly.

Approximation Techniques

The core of approximation is rounding numbers. The common methods include:

• Rounding to the nearest integer: If the decimal part is 0.5 or greater, round up; otherwise, round down.
• Rounding to the nearest ten, hundred, or thousand: Identify the place value you are rounding to and look at the digit to its right.
• Rounding up or down strategically: Sometimes, rounding numbers up or down in a way that simplifies calculation, even if not to the nearest value, can be effective if the error introduced is manageable.

Example:

Approximate the value of: 39.87 × 49.92 + 101.03 ÷ 20.11

Round the numbers to the nearest convenient integer:

• 39.87 ≈ 40
• 49.92 ≈ 50
• 101.03 ≈ 100
• 20.11 ≈ 20

Now, perform the calculation with the rounded numbers:

(40 × 50) + (100 ÷ 20)

2000 + 5 = 2005

The approximate value is 2005. You would then look for the option closest to this value.

Important Considerations:

• Multiplication and Division: If you round numbers up, the product/quotient will likely be higher than the actual value. If you round down, it will likely be lower.
• Addition and Subtraction: The effect of rounding is generally less pronounced but still present.
• Combined Operations: In expressions with multiple operations, the cumulative effect of rounding can sometimes be significant. If the options are very close, exact calculation might be needed.

Practice Problems:

1. Approximate the value of: 74.9% of 199 + 50.1 × 14.8

2. Estimate the result of: 40.01 ÷ 19.98 + 99.89 × 4.99

3. Find the approximate value of: 30.12 × 5.98 + 120.05 ÷ 5.95

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They are characterized by having at least one term that involves a variable raised to the power of two. These equations are fundamental in algebra and appear frequently in quantitative aptitude tests, often in a comparative form.

Standard Form

A quadratic equation is typically written in the standard form:

ax² + bx + c = 0

Where:

• 'x' is the variable.
• 'a', 'b', and 'c' are coefficients (constants), with 'a' not equal to zero (a ≠ 0).
• 'ax²' is the quadratic term.
• 'bx' is the linear term.
• 'c' is the constant term.

Methods to Solve Quadratic Equations

There are several methods to find the roots (solutions) of a quadratic equation:

1. Factorization Method:

This method involves splitting the middle term ('bx') into two terms such that their product equals 'ac' and their sum equals 'b'.

Example: Solve x² + 5x + 6 = 0

Here, a=1, b=5, c=6. We need two numbers whose product is ac = 1 × 6 = 6 and whose sum is b = 5. These numbers are 2 and 3.

Split the middle term: x² + 2x + 3x + 6 = 0

Group terms: x(x + 2) + 3(x + 2) = 0

Factor out the common bracket: (x + 2)(x + 3) = 0

Set each factor to zero: x + 2 = 0 or x + 3 = 0

Solutions: x = -2 or x = -3

2. Quadratic Formula:

If factorization is difficult, the quadratic formula can be used. The roots of ax² + bx + c = 0 are given by:

x = [-b ± √(b² - 4ac)] / 2a

The term 'b² - 4ac' is called the discriminant (Δ). Its value determines the nature of the roots:

• If Δ > 0, there are two distinct real roots.
• If Δ = 0, there are two equal real roots (a single real root).
• If Δ < 0, there are two complex conjugate roots (not typically encountered in basic SBI Clerk exams).

Example: Solve 2x² - 7x + 3 = 0

Here, a=2, b=-7, c=3.

Using the formula: x = [ -(-7) ± √((-7)² - 4 × 2 × 3) ] / (2 × 2)

x = [ 7 ± √(49 - 24) ] / 4

x = [ 7 ± √25 ] / 4

x = [ 7 ± 5 ] / 4

Two solutions:

x₁ = (7 + 5) / 4 = 12 / 4 = 3

x₂ = (7 - 5) / 4 = 2 / 4 = 1/2

Solutions: x = 3 or x = 1/2

3. Completing the Square:

This method involves rewriting the equation in the form (x + k)² = m. While less common for direct solving in exams, understanding it helps grasp the derivation of the quadratic formula.

Comparing Roots of Two Quadratic Equations

Often, you will be given two quadratic equations and asked to compare their roots (values of x and y). You need to solve both equations and then compare the resulting values.

Example:

Equation 1: x² - 7x + 12 = 0

Equation 2: y² - 10y + 21 = 0

Solving Equation 1:

We need two numbers that multiply to 12 and add up to -7. These are -3 and -4.

So, (x - 3)(x - 4) = 0

Roots for x are: x = 3, x = 4

Solving Equation 2:

We need two numbers that multiply to 21 and add up to -10. These are -3 and -7.

So, (y - 3)(y - 7) = 0

Roots for y are: y = 3, y = 7

Comparing the roots:

x = 3, 4

y = 3, 7

We compare each value of x with each value of y:

• Is x > y? 3 > 3 (False), 3 > 7 (False), 4 > 3 (True), 4 > 7 (False)
• Is x < y? 3 < 3 (False), 3 < 7 (True), 4 < 3 (False), 4 < 7 (True)
• Is x = y? 3 = 3 (True)

Since x can be greater than, less than, or equal to y, the relationship cannot be established. The answer would be "Cannot be determined" or "x ≤ y".

Practice Problems:

1. If x² - 11x + 24 = 0 and y² - 13y + 36 = 0, compare x and y.

2. Given 2x² + 5x + 2 = 0 and 3y² - 7y + 2 = 0, compare x and y.

3. Solve for x: 3x² - 10x + 8 = 0