Percentages: Key Concepts, Formulas, and Practice Questions

Percentages: Key Concepts, Formulas, and Practice Questions

Percentage: Key Concepts, Formulas, and Applications

The concept of percentage is widely used in day-to-day life and plays a critical role in solving problems in competitive exams. It represents a part of a whole as a fraction of 100. Understanding percentage calculations is essential for success in exams, especially those involving quantitative aptitude.

1. What is a Percentage?

A percentage is a way of expressing a number as a fraction of 100. It is denoted by the symbol %.Percentage=PartWhole×100\text{Percentage} = \frac{\text{Part}}{\text{Whole}} \times 100Percentage=WholePart​×100

For example, if you have 25 out of 100 students who passed a test, the percentage of students who passed is:Percentage=25100×100=25%\text{Percentage} = \frac{25}{100} \times 100 = 25\%Percentage=10025​×100=25%

2. Key Percentage Formulas:

  • To calculate percentage: Percentage of a number=Value×Percentage100\text{Percentage of a number} = \frac{\text{Value} \times \text{Percentage}}{100}Percentage of a number=100Value×Percentage​ Example: Find 20% of 500: 500×20100=100\frac{500 \times 20}{100} = 100100500×20​=100
  • To calculate the total value from a percentage: Total=PartPercentage×100\text{Total} = \frac{\text{Part}}{\text{Percentage}} \times 100Total=PercentagePart​×100 Example: If 25% of a number is 50, find the total number: Total=5025×100=200\text{Total} = \frac{50}{25} \times 100 = 200Total=2550​×100=200
  • To find percentage increase or decrease: Percentage change=New Value−Old ValueOld Value×100\text{Percentage change} = \frac{\text{New Value} – \text{Old Value}}{\text{Old Value}} \times 100Percentage change=Old ValueNew Value−Old Value​×100 Example: If the price of a product increases from ₹100 to ₹120, the percentage increase is: 120−100100×100=20%\frac{120 – 100}{100} \times 100 = 20\%100120−100​×100=20%
  • To calculate discount percentage: Discount Percentage=Discount AmountOriginal Price×100\text{Discount Percentage} = \frac{\text{Discount Amount}}{\text{Original Price}} \times 100Discount Percentage=Original PriceDiscount Amount​×100

3. Applications of Percentage:

  1. Profit and Loss:
    • Profit Percentage: Profit Percentage=ProfitCost Price×100\text{Profit Percentage} = \frac{\text{Profit}}{\text{Cost Price}} \times 100Profit Percentage=Cost PriceProfit​×100
    • Loss Percentage: Loss Percentage=LossCost Price×100\text{Loss Percentage} = \frac{\text{Loss}}{\text{Cost Price}} \times 100Loss Percentage=Cost PriceLoss​×100
  2. Simple Interest:
    The formula for Simple Interest (SI) is: SI=P×R×T100\text{SI} = \frac{P \times R \times T}{100}SI=100P×R×T​ where:
    • PPP = Principal
    • RRR = Rate of Interest
    • TTT = Time Period
  3. Population Growth:
    If the population of a city increases by 5% annually, and the population is 1,00,000, the population after 1 year will be: 1,00,000×(1+5100)=1,00,000×1.05=1,05,0001,00,000 \times \left(1 + \frac{5}{100}\right) = 1,00,000 \times 1.05 = 1,05,0001,00,000×(1+1005​)=1,00,000×1.05=1,05,000

4. Percentage Problems in Competitive Exams:

Percentage-related problems are common in exams like SSC, CAT, GRE, and other competitive tests. These may include:

  • Profit/Loss: Calculating profit or loss percentage based on cost price or selling price.
  • Simple Interest: Questions based on calculating interest, total amount, or time period.
  • Discounts: Finding the effective price after a discount or multiple discounts.
  • Speed, Time, Distance: Problems involving speed, time, and distance with percentage increase/decrease.
  • Mixtures and Alligations: Finding the percentage composition of different substances in mixtures.

Top 10 Percentage Practice Questions:

  1. Find 30% of 750.
  2. A person bought a product for ₹2000 and sold it for ₹2400. What is the percentage profit?
  3. A student scored 80 marks out of 100 in an exam. What is the percentage of marks obtained?
  4. The price of a TV was reduced by 15%. If the original price was ₹30,000, what is the new price?
  5. In a class of 40 students, 25% are absent. How many students are present in the class?
  6. The price of a shirt is ₹600 after a 20% discount. What was the original price?
  7. A person invests ₹5000 at 8% simple interest per annum for 2 years. Find the amount of interest earned.
  8. The population of a town increases by 10% every year. If the current population is 50,000, what will be the population after 2 years?
  9. A shirt was sold for ₹800 after a 20% profit. What is the cost price of the shirt?
  10. A person’s salary increases by 10% in the first year and 15% in the second year. If the salary at the start was ₹50,000, what is the salary after 2 years?
Answer

1. Find 30% of 750.

  • Answer: 30% of 750=30×750100=22530\% \text{ of } 750 = \frac{30 \times 750}{100} = 22530% of 750=10030×750​=225

2. A person bought a product for ₹2000 and sold it for ₹2400. What is the percentage profit?

  • Answer: Profit=2400−2000=400\text{Profit} = 2400 – 2000 = 400Profit=2400−2000=400 Profit Percentage=4002000×100=20%\text{Profit Percentage} = \frac{400}{2000} \times 100 = 20\%Profit Percentage=2000400​×100=20%

3. A student scored 80 marks out of 100 in an exam. What is the percentage of marks obtained?

  • Answer: Percentage of Marks=80100×100=80%\text{Percentage of Marks} = \frac{80}{100} \times 100 = 80\%Percentage of Marks=10080​×100=80%

4. The price of a TV was reduced by 15%. If the original price was ₹30,000, what is the new price?

  • Answer: Discount=15100×30000=4500\text{Discount} = \frac{15}{100} \times 30000 = 4500Discount=10015​×30000=4500 New Price=30000−4500=25500\text{New Price} = 30000 – 4500 = 25500New Price=30000−4500=25500

5. In a class of 40 students, 25% are absent. How many students are present in the class?

  • Answer: Number of Absent Students=25100×40=10\text{Number of Absent Students} = \frac{25}{100} \times 40 = 10Number of Absent Students=10025​×40=10 Number of Present Students=40−10=30\text{Number of Present Students} = 40 – 10 = 30Number of Present Students=40−10=30

6. The price of a shirt is ₹600 after a 20% discount. What was the original price?

  • Answer:
    Let the original price be xxx. Discounted Price=x−20100×x=600\text{Discounted Price} = x – \frac{20}{100} \times x = 600Discounted Price=x−10020​×x=600 0.8x=600⇒x=6000.8=7500.8x = 600 \quad \Rightarrow \quad x = \frac{600}{0.8} = 7500.8x=600⇒x=0.8600​=750 So, the original price of the shirt is ₹750.

7. A person invests ₹5000 at 8% simple interest per annum for 2 years. Find the amount of interest earned.

  • Answer:
    The formula for simple interest is: SI=P×R×T100\text{SI} = \frac{P \times R \times T}{100}SI=100P×R×T​ Where P=5000P = 5000P=5000, R=8R = 8R=8, and T=2T = 2T=2. SI=5000×8×2100=800\text{SI} = \frac{5000 \times 8 \times 2}{100} = 800SI=1005000×8×2​=800 So, the interest earned is ₹800.

8. The population of a town increases by 10% every year. If the current population is 50,000, what will be the population after 2 years?

  • Answer:
    After 1 year: Population=50000×(1+10100)=50000×1.1=55000\text{Population} = 50000 \times \left(1 + \frac{10}{100}\right) = 50000 \times 1.1 = 55000Population=50000×(1+10010​)=50000×1.1=55000 After 2 years: Population=55000×1.1=60500\text{Population} = 55000 \times 1.1 = 60500Population=55000×1.1=60500 So, the population after 2 years will be 60,500.

9. A shirt was sold for ₹800 after a 20% profit. What is the cost price of the shirt?

  • Answer:
    Let the cost price be xxx. Selling Price=x+20100×x=800\text{Selling Price} = x + \frac{20}{100} \times x = 800Selling Price=x+10020​×x=800 1.2x=800⇒x=8001.2=666.671.2x = 800 \quad \Rightarrow \quad x = \frac{800}{1.2} = 666.671.2x=800⇒x=1.2800​=666.67 So, the cost price of the shirt is ₹666.67.

10. A person’s salary increases by 10% in the first year and 15% in the second year. If the salary at the start was ₹50,000, what is the salary after 2 years?

Answer:
After 1st year: New Salary=50000×(1+10100)=50000×1.1=55000\text{New Salary} = 50000 \times (1 + \frac{10}{100}) = 50000 \times 1.1 = 55000New Salary=50000×(1+10010​)=50000×1.1=55000 After 2nd year: New Salary=55000×(1+15100)=55000×1.15=63250\text{New Salary} = 55000 \times (1 + \frac{15}{100}) = 55000 \times 1.15 = 63250New Salary=55000×(1+10015​)=55000×1.15=63250 So, the salary after 2 years will be ₹63,250.