Percentage Problems.
Subtopic Navigator
Lesson Objectives
- Calculate percentage increase and decrease
- Solve multi-step percentage problems
- Understand and apply repeated percentage change
- Calculate compound interest using the formula $A = P(1 + r)^n$
- Solve reverse percentage problems (finding original amount)
- Apply percentage skills to real-world contexts (sales, population, finance)
Introduction to Percentage Problems
Percentages are used everywhere in daily life: discounts in shops, interest rates on loans and savings, population growth, inflation, and many other contexts. A percentage is a fraction out of 100. Solving percentage problems requires understanding how to find a percentage of a quantity, calculate percentage change, work with repeated percentage changes, and handle compound interest.
Key Percentage Formulas
Percentage of a quantity: $x\% \text{ of } Y = \frac{x}{100} \times Y$
Percentage change: $\frac{\text{new value} - \text{original value}}{\text{original value}} \times 100\%$
Compound interest: $A = P(1 + r)^n$ where $A$ = final amount, $P$ = principal, $r$ = interest rate per period (as decimal), $n$ = number of periods
Percentage of a quantity: $x\% \text{ of } Y = \frac{x}{100} \times Y$
Percentage change: $\frac{\text{new value} - \text{original value}}{\text{original value}} \times 100\%$
Compound interest: $A = P(1 + r)^n$ where $A$ = final amount, $P$ = principal, $r$ = interest rate per period (as decimal), $n$ = number of periods
Key Definitions:
• Percentage: A fraction with denominator 100, denoted by the symbol %.
• Percentage Increase: The amount added to a quantity expressed as a percentage of the original.
• Percentage Decrease: The amount subtracted from a quantity expressed as a percentage of the original.
• Multiplier: A decimal factor used to apply a percentage change (e.g., +20% = multiplier 1.20, -15% = multiplier 0.85).
• Compound Interest: Interest calculated on both the principal and previously earned interest.
• Principal: The initial amount of money invested or borrowed.
• Percentage: A fraction with denominator 100, denoted by the symbol %.
• Percentage Increase: The amount added to a quantity expressed as a percentage of the original.
• Percentage Decrease: The amount subtracted from a quantity expressed as a percentage of the original.
• Multiplier: A decimal factor used to apply a percentage change (e.g., +20% = multiplier 1.20, -15% = multiplier 0.85).
• Compound Interest: Interest calculated on both the principal and previously earned interest.
• Principal: The initial amount of money invested or borrowed.
Percentage Increase and Decrease
To increase a quantity by a percentage, multiply by $(1 + \frac{p}{100})$. To decrease by a percentage, multiply by $(1 - \frac{p}{100})$. The multiplier method is efficient for single and repeated changes.
Step-by-Step Method for Percentage Change:
1. Find the multiplier: for increase, multiplier = $1 + \frac{p}{100}$; for decrease, multiplier = $1 - \frac{p}{100}$.
2. Multiply the original quantity by the multiplier to find the new amount.
3. To find the percentage change, use: $\frac{\text{new} - \text{original}}{\text{original}} \times 100\%$.
1. Find the multiplier: for increase, multiplier = $1 + \frac{p}{100}$; for decrease, multiplier = $1 - \frac{p}{100}$.
2. Multiply the original quantity by the multiplier to find the new amount.
3. To find the percentage change, use: $\frac{\text{new} - \text{original}}{\text{original}} \times 100\%$.
Example 1: Percentage Increase
Problem: A laptop costs ₦50,000. The price increases by 15%. Find the new price.
Solution:
Multiplier = $1 + \frac{15}{100} = 1.15$
New price = ₦50,000 × 1.15 = ₦57,500
Answer: ₦57,500
Problem: A laptop costs ₦50,000. The price increases by 15%. Find the new price.
Solution:
Multiplier = $1 + \frac{15}{100} = 1.15$
New price = ₦50,000 × 1.15 = ₦57,500
Answer: ₦57,500
Example 2: Percentage Decrease
Problem: A smartphone priced at ₦80,000 is on sale with a 20% discount. Find the sale price.
Solution:
Multiplier = $1 - \frac{20}{100} = 0.80$
Sale price = ₦80,000 × 0.80 = ₦64,000
Answer: ₦64,000
Problem: A smartphone priced at ₦80,000 is on sale with a 20% discount. Find the sale price.
Solution:
Multiplier = $1 - \frac{20}{100} = 0.80$
Sale price = ₦80,000 × 0.80 = ₦64,000
Answer: ₦64,000
Example 3: Finding Percentage Change
Problem: The population of a town increased from 25,000 to 28,000. Find the percentage increase.
Solution:
Increase = 28,000 - 25,000 = 3,000
Percentage increase = $\frac{3,000}{25,000} \times 100\% = 0.12 \times 100\% = 12\%$
Answer: 12%
Problem: The population of a town increased from 25,000 to 28,000. Find the percentage increase.
Solution:
Increase = 28,000 - 25,000 = 3,000
Percentage increase = $\frac{3,000}{25,000} \times 100\% = 0.12 \times 100\% = 12\%$
Answer: 12%
Example 4: Finding Percentage Decrease
Problem: The price of a shirt dropped from ₦4,000 to ₦3,200. Find the percentage decrease.
Solution:
Decrease = 4,000 - 3,200 = 800
Percentage decrease = $\frac{800}{4,000} \times 100\% = 0.2 \times 100\% = 20\%$
Answer: 20%
Problem: The price of a shirt dropped from ₦4,000 to ₦3,200. Find the percentage decrease.
Solution:
Decrease = 4,000 - 3,200 = 800
Percentage decrease = $\frac{800}{4,000} \times 100\% = 0.2 \times 100\% = 20\%$
Answer: 20%
Watch Out!
Percentage increase and decrease are not reversible. A 20% increase followed by a 20% decrease does NOT return to the original value. For example, 100 → 120 (20% increase) → 96 (20% decrease of 120), not 100.
Percentage increase and decrease are not reversible. A 20% increase followed by a 20% decrease does NOT return to the original value. For example, 100 → 120 (20% increase) → 96 (20% decrease of 120), not 100.
Practice for Percentage Change
- Increase ₦30,000 by 25%.
- Decrease 150 kg by 30%.
- Find the percentage increase from 80 to 100.
- Find the percentage decrease from ₦12,000 to ₦9,000.
- A house worth ₦15,000,000 increases in value by 8%. Find the new value.
Repeated Percentage Change
When a quantity undergoes multiple percentage changes (increases or decreases), the overall multiplier is the product of the individual multipliers. This is efficient for multi-step problems.
Step-by-Step Method for Repeated Percentage Change:
1. Convert each percentage change into a multiplier.
2. Multiply all multipliers together to get the overall multiplier.
3. Multiply the original quantity by the overall multiplier.
4. Alternatively, apply multipliers one after another.
1. Convert each percentage change into a multiplier.
2. Multiply all multipliers together to get the overall multiplier.
3. Multiply the original quantity by the overall multiplier.
4. Alternatively, apply multipliers one after another.
Example 1: Two Percentage Increases
Problem: A salary of ₦100,000 increases by 10% in the first year and 15% in the second year. Find the final salary.
Solution:
Year 1 multiplier: $1.10$
Year 2 multiplier: $1.15$
Overall multiplier = $1.10 \times 1.15 = 1.265$
Final salary = ₦100,000 × 1.265 = ₦126,500
Answer: ₦126,500
Problem: A salary of ₦100,000 increases by 10% in the first year and 15% in the second year. Find the final salary.
Solution:
Year 1 multiplier: $1.10$
Year 2 multiplier: $1.15$
Overall multiplier = $1.10 \times 1.15 = 1.265$
Final salary = ₦100,000 × 1.265 = ₦126,500
Answer: ₦126,500
Example 2: Increase then Decrease
Problem: A price of ₦50,000 increases by 20%, then decreases by 10%. Find the final price.
Solution:
Increase multiplier: $1.20$
Decrease multiplier: $0.90$
Overall multiplier = $1.20 \times 0.90 = 1.08$
Final price = ₦50,000 × 1.08 = ₦54,000
Answer: ₦54,000
Problem: A price of ₦50,000 increases by 20%, then decreases by 10%. Find the final price.
Solution:
Increase multiplier: $1.20$
Decrease multiplier: $0.90$
Overall multiplier = $1.20 \times 0.90 = 1.08$
Final price = ₦50,000 × 1.08 = ₦54,000
Answer: ₦54,000
Example 3: Two Percentage Decreases
Problem: A car valued at ₦8,000,000 depreciates by 15% in the first year and 10% in the second year. Find its value after two years.
Solution:
Year 1 multiplier: $0.85$
Year 2 multiplier: $0.90$
Overall multiplier = $0.85 \times 0.90 = 0.765$
Value after 2 years = ₦8,000,000 × 0.765 = ₦6,120,000
Answer: ₦6,120,000
Problem: A car valued at ₦8,000,000 depreciates by 15% in the first year and 10% in the second year. Find its value after two years.
Solution:
Year 1 multiplier: $0.85$
Year 2 multiplier: $0.90$
Overall multiplier = $0.85 \times 0.90 = 0.765$
Value after 2 years = ₦8,000,000 × 0.765 = ₦6,120,000
Answer: ₦6,120,000
Example 4: Finding Overall Percentage Change
Problem: A quantity increases by 10% then decreases by 10%. Find the overall percentage change.
Solution:
Overall multiplier = $1.10 \times 0.90 = 0.99$
Overall change = $(0.99 - 1) \times 100\% = -1\%$ (a 1% decrease)
Answer: 1% decrease
Problem: A quantity increases by 10% then decreases by 10%. Find the overall percentage change.
Solution:
Overall multiplier = $1.10 \times 0.90 = 0.99$
Overall change = $(0.99 - 1) \times 100\% = -1\%$ (a 1% decrease)
Answer: 1% decrease
Watch Out!
A 10% increase followed by a 10% decrease results in a 1% decrease overall, not a return to the original. This is because the second percentage is applied to a different base amount.
A 10% increase followed by a 10% decrease results in a 1% decrease overall, not a return to the original. This is because the second percentage is applied to a different base amount.
Practice for Repeated Percentage Change
- A population increases by 5% each year for 3 years. Find the overall multiplier.
- A price increases by 30%, then decreases by 20%. Find the overall multiplier.
- An investment of ₦200,000 grows by 12% in year 1 and 8% in year 2. Find the final amount.
- A smartphone depreciates by 20% in year 1 and 15% in year 2. Find its value after 2 years if it started at ₦150,000.
- A quantity increases by 25%, then decreases by 20%. Find the overall percentage change.
Compound Interest
Compound interest is interest calculated on the initial principal and also on the accumulated interest from previous periods. The formula $A = P(1 + r)^n$ gives the final amount after $n$ periods at an interest rate of $r$ per period.
Compound Interest Formula
$A = P(1 + r)^n$
where:
$A$ = final amount (principal + interest)
$P$ = principal (initial amount)
$r$ = interest rate per period (as a decimal)
$n$ = number of periods
Interest earned = $A - P$
$A = P(1 + r)^n$
where:
$A$ = final amount (principal + interest)
$P$ = principal (initial amount)
$r$ = interest rate per period (as a decimal)
$n$ = number of periods
Interest earned = $A - P$
Step-by-Step Method for Compound Interest:
1. Identify $P$, $r$ (convert percentage to decimal), and $n$.
2. Calculate $(1 + r)^n$ using a calculator.
3. Multiply by $P$ to find $A$.
4. Subtract $P$ to find the interest earned.
1. Identify $P$, $r$ (convert percentage to decimal), and $n$.
2. Calculate $(1 + r)^n$ using a calculator.
3. Multiply by $P$ to find $A$.
4. Subtract $P$ to find the interest earned.
Example 1: Basic Compound Interest
Problem: ₦50,000 is invested at 8% compound interest per year for 3 years. Find the final amount.
Solution:
$P = 50,000$, $r = 0.08$, $n = 3$
$A = 50,000(1 + 0.08)^3 = 50,000(1.08)^3$
$1.08^3 = 1.259712$
$A = 50,000 \times 1.259712 = 62,985.60$
Answer: ₦62,985.60
Problem: ₦50,000 is invested at 8% compound interest per year for 3 years. Find the final amount.
Solution:
$P = 50,000$, $r = 0.08$, $n = 3$
$A = 50,000(1 + 0.08)^3 = 50,000(1.08)^3$
$1.08^3 = 1.259712$
$A = 50,000 \times 1.259712 = 62,985.60$
Answer: ₦62,985.60
Example 2: Compound Interest with Half-Yearly Compounding
Problem: ₦100,000 is invested at 12% per annum compounded half-yearly for 2 years. Find the final amount.
Solution:
Half-yearly rate = $\frac{12\%}{2} = 6\% = 0.06$ per half-year
Number of periods = $2 \times 2 = 4$ half-years
$A = 100,000(1 + 0.06)^4 = 100,000(1.06)^4$
$1.06^4 = 1.26247696$
$A = 100,000 \times 1.26247696 = 126,247.70$
Answer: ₦126,247.70
Problem: ₦100,000 is invested at 12% per annum compounded half-yearly for 2 years. Find the final amount.
Solution:
Half-yearly rate = $\frac{12\%}{2} = 6\% = 0.06$ per half-year
Number of periods = $2 \times 2 = 4$ half-years
$A = 100,000(1 + 0.06)^4 = 100,000(1.06)^4$
$1.06^4 = 1.26247696$
$A = 100,000 \times 1.26247696 = 126,247.70$
Answer: ₦126,247.70
Example 3: Finding Interest Earned
Problem: ₦20,000 is invested at 5% compound interest per year for 4 years. Find the interest earned.
Solution:
$A = 20,000(1.05)^4 = 20,000 \times 1.21550625 = 24,310.125$
Interest = $24,310.125 - 20,000 = 4,310.125$
Answer: ₦4,310.13 (rounded to nearest kobo)
Problem: ₦20,000 is invested at 5% compound interest per year for 4 years. Find the interest earned.
Solution:
$A = 20,000(1.05)^4 = 20,000 \times 1.21550625 = 24,310.125$
Interest = $24,310.125 - 20,000 = 4,310.125$
Answer: ₦4,310.13 (rounded to nearest kobo)
Example 4: Finding Principal (Reverse)
Problem: After 3 years at 10% compound interest, an investment grows to ₦133,100. Find the principal.
Solution:
$A = P(1.10)^3$, $133,100 = P \times 1.331$
$P = \frac{133,100}{1.331} = 100,000$
Answer: ₦100,000
Problem: After 3 years at 10% compound interest, an investment grows to ₦133,100. Find the principal.
Solution:
$A = P(1.10)^3$, $133,100 = P \times 1.331$
$P = \frac{133,100}{1.331} = 100,000$
Answer: ₦100,000
Watch Out!
Compound interest is different from simple interest. Simple interest is calculated only on the principal, while compound interest is calculated on the principal plus accumulated interest. Always check the compounding frequency (yearly, half-yearly, quarterly, monthly).
Compound interest is different from simple interest. Simple interest is calculated only on the principal, while compound interest is calculated on the principal plus accumulated interest. Always check the compounding frequency (yearly, half-yearly, quarterly, monthly).
Comparison: Simple vs Compound Interest (₦100,000 at 10% for 3 years)
| Year | Simple Interest | Amount (Simple) | Compound Interest | Amount (Compound) |
|---|---|---|---|---|
| 1 | ₦100,000 | ₦110,000 | ₦100,000 | ₦110,000 |
| 2 | ₦100,000 | ₦120,000 | ₦110,000 | ₦121,000 |
| 3 | ₦100,000 | ₦130,000 | ₦121,000 | ₦133,100 |
Practice for Compound Interest
- Invest ₦30,000 at 6% compound interest per year for 2 years. Find the final amount.
- Invest ₦50,000 at 8% per annum compounded half-yearly for 1.5 years. Find the final amount.
- Find the compound interest on ₦25,000 at 12% per annum for 3 years.
- After 2 years at 15% compound interest, an investment is worth ₦132,250. Find the principal.
- What is the difference between simple and compound interest on ₦10,000 at 10% for 2 years?
Reverse Percentage Problems
Reverse percentage problems involve finding the original quantity when you know the final amount after a percentage change. Divide by the multiplier to find the original.
Step-by-Step Method for Reverse Percentage:
1. Identify the multiplier that was applied to get the final amount.
2. Divide the final amount by the multiplier to find the original.
3. If multiple changes occurred, divide by the product of all multipliers.
1. Identify the multiplier that was applied to get the final amount.
2. Divide the final amount by the multiplier to find the original.
3. If multiple changes occurred, divide by the product of all multipliers.
Example 1: Finding Original Price After Increase
Problem: After a 25% increase, a price is ₦50,000. Find the original price.
Solution:
Multiplier = $1.25$
Original price = ₦50,000 ÷ 1.25 = ₦40,000
Answer: ₦40,000
Problem: After a 25% increase, a price is ₦50,000. Find the original price.
Solution:
Multiplier = $1.25$
Original price = ₦50,000 ÷ 1.25 = ₦40,000
Answer: ₦40,000
Example 2: Finding Original Price After Decrease
Problem: After a 15% discount, a shirt costs ₦34,000. Find the original price.
Solution:
Multiplier = $0.85$
Original price = ₦34,000 ÷ 0.85 = ₦40,000
Answer: ₦40,000
Problem: After a 15% discount, a shirt costs ₦34,000. Find the original price.
Solution:
Multiplier = $0.85$
Original price = ₦34,000 ÷ 0.85 = ₦40,000
Answer: ₦40,000
Example 3: Reverse with Two Changes
Problem: A price increased by 10%, then decreased by 10%, resulting in ₦99,000. Find the original price.
Solution:
Multiplier 1 = $1.10$, Multiplier 2 = $0.90$
Overall multiplier = $1.10 \times 0.90 = 0.99$
Original price = ₦99,000 ÷ 0.99 = ₦100,000
Answer: ₦100,000
Problem: A price increased by 10%, then decreased by 10%, resulting in ₦99,000. Find the original price.
Solution:
Multiplier 1 = $1.10$, Multiplier 2 = $0.90$
Overall multiplier = $1.10 \times 0.90 = 0.99$
Original price = ₦99,000 ÷ 0.99 = ₦100,000
Answer: ₦100,000
Example 4: Reverse Compound Interest (Finding P)
Problem: After 3 years at 8% compound interest, an investment is worth ₦125,971.20. Find the principal.
Solution:
$A = P(1.08)^3$, $1.08^3 = 1.259712$
$P = \frac{125,971.20}{1.259712} = 100,000$
Answer: ₦100,000
Problem: After 3 years at 8% compound interest, an investment is worth ₦125,971.20. Find the principal.
Solution:
$A = P(1.08)^3$, $1.08^3 = 1.259712$
$P = \frac{125,971.20}{1.259712} = 100,000$
Answer: ₦100,000
Watch Out!
When doing reverse percentage problems, remember that the final amount represents a certain percentage of the original. For a 20% increase, the final is 120% of original, so divide by 1.20, not 0.20.
When doing reverse percentage problems, remember that the final amount represents a certain percentage of the original. For a 20% increase, the final is 120% of original, so divide by 1.20, not 0.20.
Practice for Reverse Percentages
- After a 30% increase, a price is ₦65,000. Find the original.
- After a 25% discount, a laptop costs ₦45,000. Find the original price.
- A price increased by 20% then decreased by 10% to become ₦54,000. Find the original price.
- After 2 years at 12% compound interest, an investment is worth ₦125,440. Find the principal.
- After a 15% decrease, a quantity is 85. Find the original.
Methods & Techniques
Mastering percentage problems requires understanding multipliers and being able to apply them in different contexts.
Verification / Checking Strategy:
1. For percentage change: Check if the result is reasonable (e.g., 10% of 100 is 10, so 10% increase gives 110).
2. For compound interest: Verify that $(1 + r)^n$ is calculated correctly.
3. For reverse percentages: Multiply the original by the multiplier to get back to the final amount.
4. Use estimation: Round numbers to check if your answer is in the right ballpark.
1. For percentage change: Check if the result is reasonable (e.g., 10% of 100 is 10, so 10% increase gives 110).
2. For compound interest: Verify that $(1 + r)^n$ is calculated correctly.
3. For reverse percentages: Multiply the original by the multiplier to get back to the final amount.
4. Use estimation: Round numbers to check if your answer is in the right ballpark.
Example: Checking Reverse Percentage
Original problem: After 20% discount, price is ₦40,000. Original = ₦50,000.
Check:
20% of ₦50,000 = ₦10,000, discount gives ₦50,000 - ₦10,000 = ₦40,000 ✓
Original problem: After 20% discount, price is ₦40,000. Original = ₦50,000.
Check:
20% of ₦50,000 = ₦10,000, discount gives ₦50,000 - ₦10,000 = ₦40,000 ✓
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Using the wrong multiplier (e.g., 15% increase = 1.15, not 0.15) → Solution: Remember: increase = $1 + \frac{p}{100}$, decrease = $1 - \frac{p}{100}$.
• Pitfall 2: Forgetting that percentage changes are multiplicative, not additive → Solution: Use multipliers and multiply, not add percentages.
• Pitfall 3: Confusing compound interest with simple interest → Solution: Compound interest uses $(1+r)^n$; simple interest uses $P \times r \times n$.
• Pitfall 4: Reverse percentage errors (dividing by the wrong number) → Solution: If final = original × multiplier, then original = final ÷ multiplier.
• Pitfall 5: Misidentifying the base for percentage change → Solution: The base is always the original amount before the change.
• Pitfall 1: Using the wrong multiplier (e.g., 15% increase = 1.15, not 0.15) → Solution: Remember: increase = $1 + \frac{p}{100}$, decrease = $1 - \frac{p}{100}$.
• Pitfall 2: Forgetting that percentage changes are multiplicative, not additive → Solution: Use multipliers and multiply, not add percentages.
• Pitfall 3: Confusing compound interest with simple interest → Solution: Compound interest uses $(1+r)^n$; simple interest uses $P \times r \times n$.
• Pitfall 4: Reverse percentage errors (dividing by the wrong number) → Solution: If final = original × multiplier, then original = final ÷ multiplier.
• Pitfall 5: Misidentifying the base for percentage change → Solution: The base is always the original amount before the change.
Common Percentage Multipliers
| Percentage Change | Multiplier | Example (Original = 100) |
|---|---|---|
| Increase by 5% | 1.05 | 105 |
| Increase by 10% | 1.10 | 110 |
| Increase by 25% | 1.25 | 125 |
| Increase by 50% | 1.50 | 150 |
| Decrease by 5% | 0.95 | 95 |
| Decrease by 10% | 0.90 | 90 |
| Decrease by 25% | 0.75 | 75 |
| Decrease by 50% | 0.50 | 50 |
Technique Practice
- Verify that a 20% increase followed by a 25% decrease gives a 5% decrease overall.
- Check: If ₦10,000 grows to ₦12,100 at 10% compound interest for 2 years, is this correct?
- Identify the error: A student said a 10% increase followed by a 10% decrease returns to the original. Correct this.
- For a reverse percentage: Original = 100, final after 20% increase = 120. Check by dividing 120 by 1.20 = 100.
Real-World Applications
Percentage problems appear in finance, retail, population studies, and many other areas of daily life.
Application 1: Sales and Discounts
Scenario: A store offers a 20% discount on a ₦25,000 jacket, followed by an additional 10% off the sale price. Find the final price.
Problem: Apply successive discounts.
Solution:
First discount: ₦25,000 × 0.80 = ₦20,000
Second discount: ₦20,000 × 0.90 = ₦18,000
Answer: ₦18,000
Scenario: A store offers a 20% discount on a ₦25,000 jacket, followed by an additional 10% off the sale price. Find the final price.
Problem: Apply successive discounts.
Solution:
First discount: ₦25,000 × 0.80 = ₦20,000
Second discount: ₦20,000 × 0.90 = ₦18,000
Answer: ₦18,000
Application 2: Population Growth
Scenario: A city has a population of 500,000 and grows at 3% per year. What is the population after 5 years?
Problem: Compound growth.
Solution:
$A = 500,000(1.03)^5 = 500,000 \times 1.159274 = 579,637$
Answer: Approximately 579,637 people
Scenario: A city has a population of 500,000 and grows at 3% per year. What is the population after 5 years?
Problem: Compound growth.
Solution:
$A = 500,000(1.03)^5 = 500,000 \times 1.159274 = 579,637$
Answer: Approximately 579,637 people
Application 3: Inflation and Purchasing Power
Scenario: Inflation is 4% per year. An item costs ₦10,000 now. What will it cost in 3 years?
Problem: Compound increase.
Solution:
$A = 10,000(1.04)^3 = 10,000 \times 1.124864 = 11,248.64$
Answer: Approximately ₦11,249
Scenario: Inflation is 4% per year. An item costs ₦10,000 now. What will it cost in 3 years?
Problem: Compound increase.
Solution:
$A = 10,000(1.04)^3 = 10,000 \times 1.124864 = 11,248.64$
Answer: Approximately ₦11,249
Application 4: Depreciation (Value Loss)
Scenario: A car loses 15% of its value each year. If it was bought for ₦4,000,000, what is its value after 4 years?
Problem: Compound decay.
Solution:
$A = 4,000,000(0.85)^4 = 4,000,000 \times 0.52200625 = 2,088,025$
Answer: Approximately ₦2,088,025
Scenario: A car loses 15% of its value each year. If it was bought for ₦4,000,000, what is its value after 4 years?
Problem: Compound decay.
Solution:
$A = 4,000,000(0.85)^4 = 4,000,000 \times 0.52200625 = 2,088,025$
Answer: Approximately ₦2,088,025
Application 5: Tax Calculation
Scenario: Income tax is 20% on earnings above ₦300,000. If someone earns ₦500,000, how much tax do they pay?
Problem: Percentage of a portion.
Solution:
Taxable income = ₦500,000 - ₦300,000 = ₦200,000
Tax = 20% of ₦200,000 = ₦40,000
Answer: ₦40,000
Scenario: Income tax is 20% on earnings above ₦300,000. If someone earns ₦500,000, how much tax do they pay?
Problem: Percentage of a portion.
Solution:
Taxable income = ₦500,000 - ₦300,000 = ₦200,000
Tax = 20% of ₦200,000 = ₦40,000
Answer: ₦40,000
Cross-Curricular Connections
- Finance: Interest rates, loans, mortgages, investments, inflation
- Economics: GDP growth, unemployment rates, price indices
- Biology: Population growth, bacterial reproduction, decay rates
- Chemistry: Concentration percentages, reaction yields
- Business: Profit margins, markups, discounts, VAT/GST
Cumulative Practice Exercises
Try these problems on your own. Show all working steps. Use the verification strategies to check your answers.
- Increase 240 by 35%.
- Decrease ₦80,000 by 12%.
- Find the percentage increase from 150 to 180.
- Find the percentage decrease from ₦5,000 to ₦3,750.
- A price increases by 15%, then decreases by 15%. Find the overall multiplier.
- Invest ₦40,000 at 7% compound interest per year for 3 years. Find the final amount.
- Invest ₦60,000 at 10% per annum compounded half-yearly for 2 years. Find the final amount.
- After a 12% increase, a salary is ₦56,000. Find the original salary.
- After a 25% discount, a TV costs ₦225,000. Find the original price.
- A population of 120,000 decreases by 8% each year. Find the population after 4 years.
- A car depreciates by 20% in the first year and 15% in the second year. If it cost ₦3,000,000, find its value after 2 years.
- Find the compound interest on ₦15,000 at 9% per annum for 2 years.
- After 3 years at 6% compound interest, an investment is worth ₦119,101.60. Find the principal.
- Error analysis: A student said a 10% increase followed by a 10% increase is the same as a 20% increase. Is this correct? Explain.
- A shop offers a 20% discount, then an additional 10% off the discounted price. Is this the same as a single 30% discount? Calculate and compare.
Answers to Cumulative Exercises
- Problem: Increase 240 by 35%.
Answer: 240 × 1.35 = 324 - Problem: Decrease ₦80,000 by 12%.
Answer: 80,000 × 0.88 = ₦70,400 - Problem: Percentage increase from 150 to 180.
Answer: (30 ÷ 150) × 100% = 20% - Problem: Percentage decrease from ₦5,000 to ₦3,750.
Answer: (1,250 ÷ 5,000) × 100% = 25% - Problem: Increase 15% then decrease 15%.
Answer: 1.15 × 0.85 = 0.9775 (2.25% decrease overall) - Problem: ₦40,000 at 7% compound for 3 years.
Answer: 40,000 × 1.07³ = 40,000 × 1.225043 = ₦49,001.72 - Problem: ₦60,000 at 10% p.a. half-yearly for 2 years.
Answer: Rate per half-year = 5% = 0.05, n = 4, 60,000 × 1.05⁴ = 60,000 × 1.21550625 = ₦72,930.38 - Problem: After 12% increase, salary is ₦56,000.
Answer: 56,000 ÷ 1.12 = ₦50,000 - Problem: After 25% discount, TV costs ₦225,000.
Answer: 225,000 ÷ 0.75 = ₦300,000 - Problem: Population 120,000 decreases 8% per year for 4 years.
Answer: 120,000 × 0.92⁴ = 120,000 × 0.71639296 = 85,967.16 (about 85,967) - Problem: Car ₦3,000,000 depreciates 20% then 15%.
Answer: 3,000,000 × 0.80 × 0.85 = 3,000,000 × 0.68 = ₦2,040,000 - Problem: Compound interest on ₦15,000 at 9% for 2 years.
Answer: A = 15,000 × 1.09² = 15,000 × 1.1881 = 17,821.50, Interest = 17,821.50 - 15,000 = ₦2,821.50 - Problem: After 3 years at 6%, worth ₦119,101.60. Find principal.
Answer: 119,101.60 ÷ 1.06³ = 119,101.60 ÷ 1.191016 = ₦100,000 - Problem: Error analysis: 10% + 10% = 20% increase?
Answer: Incorrect. 10% then 10% gives multiplier 1.10 × 1.10 = 1.21, which is a 21% increase, not 20%. - Problem: 20% discount then 10% off vs single 30% discount.
Answer: Two discounts: 1 × 0.80 × 0.90 = 0.72 (28% total discount). Single 30% discount = 0.70 (30% discount). They are not the same. Two discounts give a lower total discount (28% vs 30%).
Conclusion & Summary
Percentage problems are essential for understanding real-world situations involving changes, growth, and decay. The multiplier method simplifies calculations for single and repeated percentage changes. Compound interest is a powerful application of repeated percentage increase, and reverse percentages allow us to find original amounts.
Key Takeaways:
1. Multipliers: Increase: $1 + \frac{p}{100}$, Decrease: $1 - \frac{p}{100}$.
2. Percentage change: $\frac{\text{change}}{\text{original}} \times 100\%$.
3. Repeated changes: Multiply individual multipliers.
4. Compound interest: $A = P(1 + r)^n$ — interest on interest.
5. Reverse percentages: Original = Final ÷ multiplier.
6. Applications: Discounts, population growth, depreciation, inflation.
Keep practising with real-world problems. Understanding percentages is a vital life skill!
Video Resource
Watch this video for more examples of compound interest and repeated percentage change.
A Digital Learning & Technology Solutions Hub Ltd. Package (RC123456).
Share your ANSWER in the Chat. Indicate TITLE e.g Linear Equation 1. .....2. e.t.c
