Real Life Percentage Problems

Percentages are a fundamental concept in mathematics and are used widely in various real-life scenarios. Understanding percentages can help solve many practical problems involving discounts, interest rates, population changes, data interpretation, and more.

Here are some detailed examples of how percentages are used in real life:

1. Shopping Discounts

When shopping, discounts are often offered in percentages. For example, if a store advertises a 25% discount on all items, you need to calculate how much money you will save and what the final price will be.

Example Problem:

An item costs $80, and it is on sale for 25% off. How much do you pay?

Solution:

Discount amount:

80×0.25=20 dollars

Final price:

80−20=60 dollars

2. Interest Rates

Interest rates on savings accounts, loans, and credit cards are typically given in percentages.

Example Problem:

You have $1,000 in a savings account that earns 5% interest per year. How much interest will you earn in one year?

Solution:

Interest earned: 1000×0.05=50 dollars

3. Population Growth

Population growth rates are expressed as percentages. These rates help in understanding how quickly a population is increasing or decreasing.

Example Problem:

A town’s population increases by 2% per year. If the current population is 50,000, what will it be next year?

Solution:

Population increase:

50000×0.02=1000

Next year’s population: 50000+1000=51000

4. Grade Calculation

Teachers often calculate grades based on percentage scores.

Example Problem:

A student scores 85 out of 100 in a test. What is the percentage score?

Solution:

Percentage score: (85/100)×100=85%

5. Sales Tax

Sales tax is often added to the price of goods and services, expressed as a percentage.

Example Problem:

If you buy a product for $200 and the sales tax rate is 8%, how much is the total cost?

Solution:

Sales tax: 200×0.08=16 dollars

Total cost: 200+16=216 dollars

6. Tips in Restaurants

Tips at restaurants are generally calculated as a percentage of the total bill.

Example Problem:

Your restaurant bill is $45, and you want to leave a 15% tip. How much is the tip, and what is the total amount you pay?

Solution:

Tip amount: 45×0.15=6.75 dollars

Total amount: 45+6.75=51.75 dollars

7. Real Estate

Percentage changes are used to understand real estate price changes.

Example Problem:

A house was worth $300,000 last year, and its value increased by 10% this year. What is its current value?

Solution:

Increase in value: 300000×0.10=30000 dollars

Current value:

300000+30000=330000 dollars

8. Exam Performance

Percentages help compare performances across different subjects or exams.

Example Problem:

A student scored 75% in Math and 85% in Science. If each subject is out of 100 marks, how many marks did the student get in each?

Solution:

Math marks: 100×0.75=75

Science marks: 100×0.85=85

9. Medicine Dosages

Dosages are often calculated based on a percentage of a standard dose.

Example Problem:

A patient needs a dosage that is 20% less than the standard 500 mg. What is the adjusted dosage?

Solution:

Dosage reduction: 500×0.20=100 mg

Adjusted dosage: 500−100=400 mg

10. Nutrition Labels

Food packaging often shows nutritional content as a percentage of daily recommended values.

Example Problem:

A snack contains 8% of the recommended daily intake of iron. If the recommended daily intake is 18 mg, how much iron does the snack contain?

Solution:

Iron content: 18×0.08=1.44 mg

Example Problem: (11)

A shopkeeper bought 600 mangoes and 400 apples. He found 15% of mangoes and 8% of apples were rotten. Find the percentage of fruits in good condition?.

Solution:

Total number of fruits bought by shopkeeper = 600 + 400 = 1000

Number of rotten mangoes = 15% of 600 = 15/100 x 600 = 9000/100 =90

Number of rotten appels = 8% 0f 400 = 8/100 x 400 = 3200/100 = 32

Therefore, total number of rotten fruits = 90 +32 = 122

Number of friuts in good condition = 1000 – 122 = 878

percentage of fruits in good condition = (878/1000 x 100)%

= (87800/1000)% = 87.8%

Example Problem: (12)

John borrowed Rs 2000 at 2% per annum and Rs 1000 at 5% per annum. He cleared his debt after 2 years by giving Rs 2800 and a watch. What is the cost of the watch?

Solution: Given Principle amount P = Rs 2000,

Rate of Interest R = 2% per annum and

Time T = 2 years

Given Principal amount P = Rs 2000

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation we get SI = (2000 x 2 x 2)/100 = Rs 80

Principal Amount P =Rs 1000

Rate of Interest R = 5% per annum and

Time T = 2 years

We know that simple interest SI = (P x R x T)/100.

Substituting these values in above equation we get SI = (1000 x 5 x 2)/100 = Rs 100

Principal Amount P =Rs 1000

The amount that he will have to return = 2000 +1000 + 80 + 100 = 3180

Amount repaid = 2800

Value of the watch = 3180 – 2800 = 380

Example Problem: (13)

David deposited Rs 2000 in abank which pays 6% simple interest. She withdraw Rs 700 at the end of first year. What will be her balance after 3 years?

Rate of Interest R = 6% per annum and Time T = 2 years

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation we get

SI = (2000 x 6 x 1)/100 = Rs 120

Amount after 1 year

Principal amount + Interest = 2000 + 120 = 2120

after 1 year, amount withdrawn = 700

Principle left = 2120 – 700 = 1420

Rate of Interest R = 6% per annum and

Time T = 2 years

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation

we get SI = (1420 x 6 x 2)/100 = Interest after 2 years = Rs 170.4

Total amount after 3 years = 1420 + 170.4 = 1590.4

Principle amount P = 60000,

Rate of Interest R = 15% per annum and

Time T = 3 years

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation

we get SI = (60000 x 9 x 2)/100 = Rs 10800

Amount gained by david = 12000 – 10800

= 1200

= Rs 1200

Example Problem: (14)

David lent Rs 15000 from his friend he charged 15% per annum on Rs 12500 and 18% on the rest. How much interest does he earn in 3 years?

Solution:

Given Principle amount P = Rs 12500,

Rate of Interest R = 15% per annum and Time T = 3 years

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation

we get SI = (12500 x 15 x 3)/100 = 5625

Rest of amount lent = 15000 – 12500 = 2500

Rate of Interest R = 18% per annum and Time = 3 years

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation

we get SI = (2500 x 18 x 3)/100 = 1350

Total interest earned = 5625 + 1350 = 6975

Interest after 2 years = Rs 170.4

Total amount after 3 years = 1420 + 170.4 = 1590.4

Example Problem: (15)

John took a loan of Rs 8000 from money lender, who charged interest at rate of 18% per annum. After 2 years david paid him Rs 10400 and a watch to clear the debt. What is the price of the watch?

Solution: Given Principle amount P = 8000,

Rate of Interest R = 18% per annum and

Time T = 2 years

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation we get

SI = (8000 x 18 x 2)/100 = Rs 2880

Total amount payable by david after 2 years = 8000 + 2880 = 10880

Amount paid = 10400

Value of watch = 10880 – 10400 = 480

= Rs 480

Rate of Interest R = 9% per annum and

Time T = 2 years

Example Problem: (16)

David deposited Rs 20000 as a fixed deposit in bank at 10% per annum. If 30% is deducted as income tax on the interest earned. Find his annual income?

Solution:

Given Principle amount P = 20000,

Rate of Interest R = 10% per annum and

Time T = 1 year

We know that simple interest SI = (P x R x T)/100

Substituting these values in above equation we get

SI = (20000 x 10 x 1)/100 = Rs 2000

Amount deduct as income tax = 30% of 2000

= (30 x 2000)/100

= 600 Rs

Annual interest after tax deduction = Rs 2000 – Rs 600 =

Rs 1400

Understanding percentages is crucial for making informed decisions in everyday life, from financial planning and shopping to understanding statistical data and academic performance.