Percentages — Explained

Detailed Explanation

Percentages form the bedrock of quantitative aptitude, offering a standardized way to express proportions and changes. Understanding them deeply is not just about formulas but about conceptual clarity and strategic application. This section delves into the core concepts, their applications, and Vyyuha's analytical insights.

1. Basic Percentage Definition and Conversion

At its core, a percentage is a fraction where the denominator is 100. It's a way to express a part of a whole. The symbol '%' means 'per hundred'.

Fraction to Percentage: — Multiply the fraction by 100. Example: 3/4 = (3/4) * 100% = 75%. This is a crucial skill for fraction to percentage conversion.
Decimal to Percentage: — Multiply the decimal by 100. Example: 0.65 = 0.65 * 100% = 65%. This connects directly to decimal calculations in CSAT.
Percentage to Fraction/Decimal: — Divide the percentage by 100. Example: 40% = 40/100 = 2/5 = 0.40.

Example 1: Basic Conversion

Problem: — Convert 7/8 to a percentage and 0.125 to a percentage.
Stepwise Solution:

1. For 7/8: (7/8) * 100% = (700/8)% = 87.5%. 2. For 0.125: 0.125 * 100% = 12.5%.

Alternate Approach (for 7/8): — Recognize common fractions. 1/8 = 12.5%, so 7/8 = 7 * 12.5% = 87.5%.
Final Answer: — 87.5% and 12.5%.

2. Percentage Increase and Decrease Calculations

These are fundamental for tracking changes over time or comparing values.

Percentage Increase: — ((New Value - Original Value) / Original Value) * 100%
Percentage Decrease: — ((Original Value - New Value) / Original Value) * 100%

Example 2: Price Increase

Problem: — A product's price increased from Rs. 80 to Rs. 100. Find the percentage increase.
Stepwise Solution:

1. Increase = New Value - Original Value = 100 - 80 = 20. 2. Percentage Increase = (20 / 80) * 100% = (1/4) * 100% = 25%.

Alternate Approach: — New value is 100, original is 80. Ratio is 100/80 = 5/4. This means the new value is 5/4 times the original. (5/4 - 1) * 100% = (1/4) * 100% = 25% increase.
Final Answer: — 25% increase.

Example 3: Population Decrease

Problem: — A town's population decreased from 12,000 to 10,800. What is the percentage decrease?
Stepwise Solution:

1. Decrease = Original Value - New Value = 12,000 - 10,800 = 1,200. 2. Percentage Decrease = (1,200 / 12,000) * 100% = (1/10) * 100% = 10%.

Alternate Approach: — New value is 10,800, original is 12,000. Ratio is 10,800/12,000 = 108/120 = 9/10. This means the new value is 9/10 times the original. (1 - 9/10) * 100% = (1/10) * 100% = 10% decrease.
Final Answer: — 10% decrease.

3. Successive Percentage Changes

When a quantity undergoes multiple percentage changes, the net change is not simply the sum of individual changes. The base for the second change is the result of the first change.

Formula: — If a quantity changes by x% and then by y%, the net percentage change is (x + y + (xy/100))%. Use positive for increase, negative for decrease.

Example 4: Successive Increase

Problem: — A salary is first increased by 10% and then by 20%. What is the overall percentage increase?
Stepwise Solution (Formula Method):

1. x = 10, y = 20. 2. Net Change = (10 + 20 + (10*20)/100)% = (30 + 200/100)% = (30 + 2)% = 32%.

Alternate Approach (Base Value Method):

1. Assume original salary = 100. 2. After 10% increase: 100 * (1 + 10/100) = 100 * 1.1 = 110. 3. After 20% increase on 110: 110 * (1 + 20/100) = 110 * 1.2 = 132. 4. Overall increase = (132 - 100) / 100 * 100% = 32%.

Final Answer: — 32% increase.

Example 5: Successive Increase and Decrease

Problem: — A product's price is increased by 20% and then decreased by 10%. What is the net percentage change?
Stepwise Solution (Formula Method):

1. x = 20, y = -10 (for decrease). 2. Net Change = (20 + (-10) + (20*(-10))/100)% = (10 - 200/100)% = (10 - 2)% = 8%.

Alternate Approach (Base Value Method):

1. Assume original price = 100. 2. After 20% increase: 100 * 1.20 = 120. 3. After 10% decrease on 120: 120 * (1 - 10/100) = 120 * 0.90 = 108. 4. Overall change = (108 - 100) / 100 * 100% = 8% increase.

Final Answer: — 8% increase.

4. Percentage of Percentage Problems

These involve calculating a percentage of an already existing percentage or a part of a part.

Example 6: Students in a Class

Problem: — In a class, 60% are boys. Of these boys, 20% play cricket. What percentage of the total class are boys who play cricket?
Stepwise Solution:

1. Let total class strength be 100. 2. Number of boys = 60% of 100 = 60. 3. Boys who play cricket = 20% of 60 = (20/100) * 60 = 12. 4. Percentage of total class = (12/100) * 100% = 12%.

Alternate Approach (Direct Multiplication):

1. Percentage of boys who play cricket = 20% of 60% = (20/100) * (60/100) = 0.20 * 0.60 = 0.12. 2. Convert to percentage: 0.12 * 100% = 12%.

Final Answer: — 12%.

5. Profit and Loss Percentage Applications

Profit and Loss are direct applications of percentage increase/decrease on Cost Price (CP) and Selling Price (SP). This is a core concept in Profit and Loss.

Profit % = ((SP - CP) / CP) * 100%
Loss % = ((CP - SP) / CP) * 100%

Example 7: Calculating Profit Percentage

Problem: — A shopkeeper buys an item for Rs. 400 and sells it for Rs. 500. Find the profit percentage.
Stepwise Solution:

1. CP = 400, SP = 500. 2. Profit = SP - CP = 500 - 400 = 100. 3. Profit % = (100 / 400) * 100% = (1/4) * 100% = 25%.

Alternate Approach: — SP is 500, CP is 400. SP/CP = 500/400 = 5/4. Profit is (5/4 - 1) * 100% = 25%.
Final Answer: — 25% profit.

Example 8: Calculating Selling Price with Loss

Problem: — An item is bought for Rs. 600. If it is sold at a loss of 15%, what is its selling price?
Stepwise Solution:

1. CP = 600, Loss % = 15%. 2. Loss amount = 15% of 600 = (15/100) * 600 = 90. 3. SP = CP - Loss = 600 - 90 = 510.

Alternate Approach (Multiplier Method):

1. If there's a 15% loss, the SP is (100 - 15)% = 85% of the CP. 2. SP = 0.85 * 600 = 510.

Final Answer: — Rs. 510.

6. Simple and Compound Interest Percentage Calculations

Interest calculations are direct applications of percentage concepts over time. This is a dedicated topic in Simple and Compound Interest.

Simple Interest (SI): — SI = (P * R * T) / 100, where P = Principal, R = Rate %, T = Time.
Compound Interest (CI): — A = P * (1 + R/100)^T, where A = Amount.

Example 9: Simple Interest

Problem: — Find the simple interest on Rs. 5,000 at 8% per annum for 3 years.
Stepwise Solution:

1. P = 5000, R = 8, T = 3. 2. SI = (5000 * 8 * 3) / 100 = 50 * 8 * 3 = 400 * 3 = 1200.

Alternate Approach (Yearly Calculation):

1. Interest for 1 year = 8% of 5000 = 400. 2. Interest for 3 years = 3 * 400 = 1200.

Final Answer: — Rs. 1,200.

Example 10: Compound Interest

Problem: — Calculate the amount if Rs. 10,000 is compounded annually at 10% per annum for 2 years.
Stepwise Solution:

1. P = 10000, R = 10, T = 2. 2. A = 10000 * (1 + 10/100)^2 = 10000 * (1.1)^2 = 10000 * 1.21 = 12100.

Alternate Approach (Successive Percentage Increase):

1. Year 1: 10% increase on 10000 = 1000. Amount = 11000. 2. Year 2: 10% increase on 11000 = 1100. Amount = 11000 + 1100 = 12100.

Final Answer: — Rs. 12,100.

7. Population Growth/Decline Percentage Problems

These are essentially compound interest problems in disguise, where the 'rate' is the growth/decline percentage and 'time' is the number of years.

Example 11: Population Growth

Problem: — The population of a city is 20,000. If it increases by 5% annually, what will be the population after 2 years?
Stepwise Solution:

1. Initial Population = 20000, Growth Rate = 5%, Time = 2 years. 2. Population after 2 years = 20000 * (1 + 5/100)^2 = 20000 * (1.05)^2 = 20000 * 1.1025 = 22050.

Alternate Approach (Successive Percentage):

1. After 1 year: 20000 * 1.05 = 21000. 2. After 2 years: 21000 * 1.05 = 22050.

Final Answer: — 22,050.

Example 12: Population Decline

Problem: — A village population of 15,000 decreases by 10% in the first year and then increases by 5% in the second year. What is the population after 2 years?
Stepwise Solution:

1. Initial Population = 15000. 2. After 10% decrease: 15000 * (1 - 10/100) = 15000 * 0.90 = 13500. 3. After 5% increase on 13500: 13500 * (1 + 5/100) = 13500 * 1.05 = 14175.

Alternate Approach (Net Change Formula):

1. x = -10, y = 5. 2. Net Change = (-10 + 5 + (-10*5)/100)% = (-5 - 50/100)% = (-5 - 0.5)% = -5.5%. 3. Final Population = 15000 * (1 - 5.5/100) = 15000 * 0.945 = 14175.

Final Answer: — 14,175.

8. Discount and Markup Percentage Scenarios

These are common in commercial mathematics and involve applying percentages to Marked Price (MP), Cost Price (CP), and Selling Price (SP).

Discount % = (Discount Amount / Marked Price) * 100%
Markup % = ((MP - CP) / CP) * 100%

Example 13: Discount Calculation

Problem: — An item is marked at Rs. 800. A discount of 15% is offered. What is the selling price?
Stepwise Solution:

1. MP = 800, Discount % = 15%. 2. Discount Amount = 15% of 800 = (15/100) * 800 = 120. 3. SP = MP - Discount Amount = 800 - 120 = 680.

Alternate Approach (Multiplier Method):

1. SP = (100 - 15)% of MP = 85% of 800 = 0.85 * 800 = 680.

Final Answer: — Rs. 680.

Example 14: Markup and Discount

Problem: — A shopkeeper marks an item 20% above its cost price of Rs. 500. He then offers a 10% discount on the marked price. Find the selling price and profit percentage.
Stepwise Solution:

1. CP = 500. 2. Marked Price (MP) = CP + 20% of CP = 500 + (20/100)*500 = 500 + 100 = 600. 3. Discount = 10% of MP = (10/100)*600 = 60. 4. Selling Price (SP) = MP - Discount = 600 - 60 = 540. 5. Profit = SP - CP = 540 - 500 = 40. 6. Profit % = (40 / 500) * 100% = (4/50) * 100% = 8%.

Alternate Approach (Successive Percentage):

1. First change (markup) = +20%. Second change (discount) = -10%. 2. Net change on CP = (20 - 10 + (20 * -10)/100)% = (10 - 2)% = 8%. 3. SP = CP * (1 + 8/100) = 500 * 1.08 = 540. 4. Profit % = 8%.

Final Answer: — SP = Rs. 540, Profit % = 8%.

9. Percentage Error and Approximation Techniques

Percentage error measures the relative error in a measurement or calculation. Approximation techniques are vital for CSAT to save time.

Percentage Error = ((|Actual Value - Measured Value|) / Actual Value) * 100%

Example 15: Percentage Error

Problem: — A student measured a line as 10.5 cm, but its actual length was 10 cm. Find the percentage error.
Stepwise Solution:

1. Actual Value = 10, Measured Value = 10.5. 2. Error = |10 - 10.5| = 0.5. 3. Percentage Error = (0.5 / 10) * 100% = 0.05 * 100% = 5%.

Alternate Approach: — The measured value is 10.5, which is 0.5 more than 10. (0.5/10) * 100% = 5%.
Final Answer: — 5%.

Example 16: Approximation

Problem: — Approximately, what is 31.2% of 498?
Stepwise Solution:

1. Approximate 31.2% to 30% or 31%. 2. Approximate 498 to 500. 3. 30% of 500 = (30/100) * 500 = 150. 4. If we use 31% of 500 = (31/100) * 500 = 155. 5. More precise approximation: 31.2% of 498 is roughly 31% of 500. 31% of 500 = 155. Now, 31.2% is slightly more than 31%, and 498 is slightly less than 500. The two effects might partially cancel out. Let's calculate 31.2% of 498 = (31.2/100) * 498 = 0.312 * 498 = 155.376. So 155 is a good approximation.

Final Answer: — Approximately 155.

10. Data Interpretation Percentage Questions (Charts, Graphs, Tables)

Percentages are indispensable for analyzing and interpreting data presented in various formats. This is a crucial skill for data interpretation strategies.

Example 17: Bar Graph Analysis

Problem: — A bar graph shows sales of Product A as 200 units and Product B as 250 units. What percentage are sales of Product A compared to Product B?
Stepwise Solution:

1. Sales A = 200, Sales B = 250. 2. Percentage = (Sales A / Sales B) * 100% = (200 / 250) * 100% = (4/5) * 100% = 80%.

Alternate Approach: — 200 is to 250 as 4 is to 5. 4/5 as a percentage is 80%.
Final Answer: — 80%.

Example 18: Pie Chart Analysis

Problem: — A pie chart shows 'Expenditure on Education' as 18% of the total budget of Rs. 50,000. How much money is spent on education?
Stepwise Solution:

1. Total Budget = 50,000, Education % = 18%. 2. Amount on Education = 18% of 50,000 = (18/100) * 50,000 = 18 * 500 = 9,000.

Alternate Approach: — 10% of 50,000 = 5,000. 8% of 50,000 = 4,000. Total = 5,000 + 4,000 = 9,000.
Final Answer: — Rs. 9,000.

Example 19: Table Data Comparison

Problem: — A table shows the number of male employees as 300 and female employees as 200 in a company. What percentage of total employees are female?
Stepwise Solution:

1. Male = 300, Female = 200. 2. Total Employees = 300 + 200 = 500. 3. Percentage Female = (Female Employees / Total Employees) * 100% = (200 / 500) * 100% = (2/5) * 100% = 40%.

Alternate Approach: — Ratio of Female to Total is 200:500 or 2:5. Convert 2/5 to percentage: 40%.
Final Answer: — 40%.

Example 20: Combined Percentage Change in DI

Problem: — A company's revenue was Rs. 100 Cr in 2020. It increased by 15% in 2021 and then decreased by 5% in 2022. What was the revenue in 2022?
Stepwise Solution:

1. Initial Revenue = 100 Cr. 2. Revenue in 2021 = 100 * (1 + 15/100) = 100 * 1.15 = 115 Cr. 3. Revenue in 2022 = 115 * (1 - 5/100) = 115 * 0.95 = 109.25 Cr.

Alternate Approach (Successive Percentage Formula):

1. x = 15, y = -5. 2. Net Change = (15 - 5 + (15 * -5)/100)% = (10 - 75/100)% = (10 - 0.75)% = 9.25%. 3. Revenue in 2022 = 100 Cr * (1 + 9.25/100) = 100 * 1.0925 = 109.25 Cr.

Final Answer: — Rs. 109.25 Cr.

Vyyuha Analysis: Inter-topic Connections

Percentages are not an isolated topic. They are the 'glue' that connects various quantitative aptitude concepts. For instance, understanding percentages is critical for time and work percentage problems, where efficiency changes are often expressed in percentages. Similarly, average calculations often involve finding percentage deviations. A strong grasp here amplifies your ability across the entire CSAT arithmetic section .