📊 Percentages

Day 1 | ⏱️ Part of 4-Hour Quant Block | TCS Interview Prep

📌 Subtopics

• Basics of percentage calculation
• Percentage increase & decrease
• Successive percentage changes
• Percentage to fraction conversions (and vice-versa)
• Population-based problems
• Depreciation problems
• Percentage in Profit and Loss context

📖 Comprehensive Notes

1. Basics of Percentage Calculation

A percentage is a fraction with a denominator of $100$. It represents a part out of $100$ equal parts. $\text{Percentage} = \left(\frac{\text{Part}}{\text{Whole}}\right) \times 100$ To find $x\%$ of $y$: $\frac{x}{100} \times y$.

Example: Find $15\%$ of $240$. $\frac{15}{100} \times 240 = 0.15 \times 240 = 36$.

2. Percentage Increase & Decrease

• Percentage Increase: $\text{Increase \%} = \left(\frac{\text{Final Value} - \text{Initial Value}}{\text{Initial Value}}\right) \times 100$
  • Multiplier: To increase a number by $r\%$, multiply it by $\left(1 + \frac{r}{100}\right)$.
• Percentage Decrease: $\text{Decrease \%} = \left(\frac{\text{Initial Value} - \text{Final Value}}{\text{Initial Value}}\right) \times 100$
  • Multiplier: To decrease a number by $r\%$, multiply it by $\left(1 - \frac{r}{100}\right)$.

Example: A salary of $₹40,000$ is increased by $12\%$. Find the new salary. $\text{New Salary} = 40000 \times (1 + 0.12) = 40000 \times 1.12 = ₹44,800$.

3. Successive Percentage Changes

If a quantity is changed successively by $a\%$ and then by $b\%$, the net percentage change is: $\text{Net Change \%} = a + b + \frac{a \times b}{100}$

• Note: Use a negative sign for any percentage decrease.

Example: The price of an item is first increased by $20\%$ and then decreased by $10\%$. Find the net percentage change. $a = +20$, $b = -10$ $\text{Net Change \%} = 20 - 10 + \frac{20 \times (-10)}{100} = 10 - 2 = 8\%$ increase.

4. Percentage to Fraction Conversions (and vice-versa)

Memorizing common fraction-to-percentage equivalents saves critical time in exams:

5. Population-Based Problems

If the current population is $P$ and it increases at a rate of $R\%$ per annum:

• Population after $n$ years: $P_n = P \times \left(1 + \frac{R}{100}\right)^n$
• Population $n$ years ago: $P_{\text{past}} = \frac{P}{\left(1 + \frac{R}{100}\right)^n}$

Example: The population of a town is $50,000$ and grows at $10\%$ annually. What will it be in $2$ years? $P_2 = 50000 \times \left(1 + \frac{10}{100}\right)^2 = 50000 \times (1.1)^2 = 50000 \times 1.21 = 60,500$.

6. Depreciation Problems

If the value of a machine is $V$ and it depreciates at a rate of $R\%$ per annum:

• Value after $n$ years: $V_n = V \times \left(1 - \frac{R}{100}\right)^n$
• Value $n$ years ago: $V_{\text{past}} = \frac{V}{\left(1 - \frac{R}{100}\right)^n}$

Example: A laptop purchased for $₹60,000$ depreciates at $20\%$ per year. What is its value after $2$ years? $V_2 = 60000 \times (1 - 0.20)^2 = 60000 \times (0.8)^2 = 60000 \times 0.64 = ₹38,400$.

7. Percentage in Profit and Loss Context

Profit/loss calculations are directly based on percentages, where:

• Cost Price is always considered as the base ($100\%$).
• Selling Price at $x\%$ profit is $(100 + x)\%$ of Cost Price.
• Selling Price at $y\%$ loss is $(100 - y)\%$ of Cost Price.

🌐 Learning Resources

✅ Practice Links

📝 Key Formulas

Percentage = (Value / Total) × 100
Increase %  = (Increase / Original) × 100
Decrease %  = (Decrease / Original) × 100
Successive Change: a + b + (ab/100)

Tip: Memorize fraction-percentage equivalents: 1/2=50%, 1/3=33.33%, 1/4=25%, 1/5=20%, 1/6=16.67%, 1/8=12.5%

🔗 Related Topics

• Profit and Loss — Profit/loss percentages are a direct application
• Data Interpretation — DI questions heavily use percentage calculations
• Averages — Percentage change in averages
• Ratios and Proportion — Fraction-percentage conversions
• Allegations and Mixtures — Mixture problems use percentage concentration