Learning Objectives
  • Calculate percentages of quantities and express one quantity as a percentage of another.
  • Calculate percentage increase and decrease.
  • Solve simple and compound interest problems.
  • Use reverse percentages to recover an original value.
Key Terms
Percentage
A fraction with denominator 100.
Percentage change
The change expressed as a percentage of the original amount.
Simple interest
Interest calculated only on the original principal.
Compound interest
Interest calculated on the current balance, including earlier interest.
Reverse percentage
Finding an original amount from a final amount after a percentage change.
Multiplier
A decimal factor used to apply a percentage change.
Basic Percentage Calculations

A percentage means “per hundred”. To find a percentage of a quantity, multiply by the percentage written as a fraction or decimal. To express one quantity as a percentage of another, divide the part by the original or whole quantity and multiply by 100.

\text{percentage}=\frac{\text{part}}{\text{whole}}\times100\%
Worked Example: Percentage Of A Quantity

Question: Find 17.5% of 640.

  1. Write 17.5% as 0.175.
  2. Calculate 0.175\times640=112.

Answer: 112.

Worked Example: One Quantity As A Percentage

Question: A student scores 54 out of 72. Express the score as a percentage.

  1. Calculate \frac{54}{72}\times100.
  2. \frac{54}{72}=0.75.

Answer: 75%.

Percentage Increase And Decrease

Percentage change is measured relative to the original value, not the final value. A multiplier is efficient: increasing by p% means multiply by 1+\frac{p}{100}, while decreasing by p% means multiply by 1-\frac{p}{100}.

\text{percentage change}=\frac{\text{change}}{\text{original}}\times100\%
Worked Example: Percentage Increase

Question: A price rises from $240 to $282. Find the percentage increase.

  1. Increase =282-240=42.
  2. Calculate \frac{42}{240}\times100=17.5.

Answer: 17.5%.

Repeated Percentage Change

Repeated changes act on a new amount each time, so percentages should not simply be added. Use the multiplier once for each period.

Worked Example: Depreciation

Question: A machine worth $18 000 depreciates by 12% each year. Find its value after 3 years.

  1. The yearly multiplier is 1-0.12=0.88.
  2. Calculate 18000(0.88)^3.
  3. This gives 12266.496.
  4. Round appropriately for money.

Answer: $12 266.50.

Simple Interest

Simple interest is based only on the original principal. If the annual rate is r% for n years, the interest is P\times\frac{r}{100}\times n. The total amount is principal plus interest.

I=\frac{Prn}{100}
Worked Example: Simple Interest

Question: $4500 is invested at 3.2% simple interest per year for 5 years. Find the total amount.

  1. Interest =4500\times0.032\times5=720.
  2. Total =4500+720=5220.

Answer: $5220.

Compound Interest

Compound interest uses a multiplier repeatedly. The formula is not supplied in the examination, so students should understand the repeated multiplication method.

A=P\left(1+\frac{r}{100}\right)^n
Worked Example: Compound Interest

Question: $8000 is invested at 4.5% compound interest per year for 3 years. Find the amount.

  1. Use multiplier 1.045.
  2. Calculate 8000(1.045)^3=9129.129.
  3. Round to two decimal places.

Answer: $9129.13.

Reverse Percentages

When a final amount is known, divide by the multiplier to find the original. Do not simply subtract the percentage from the final value because the percentage was calculated from the original.

Worked Example: Reverse Increase

Question: After a 15% increase, a salary is $46 000. Find the original salary.

  1. Final =1.15\times original.
  2. Original =46000\div1.15.

Answer: $40 000.

Worked Example: Profit And Cost Price

Question: An item is sold for $276 after a profit of 20% on cost price. Find the cost price.

  1. Selling price =1.20\times cost price.
  2. Cost price =276\div1.20.

Answer: $230.

8th Edition Chapter Map
  • Percentages as fractions and decimals
  • A percentage of a quantity and one quantity as a percentage of another
  • Percentage change and percentage points
  • Reverse percentage, simple interest and compound interest
Percentage Multipliers

A multiplier provides one consistent method for percentage change. Increasing by r\% multiplies by 1+\frac{r}{100}; decreasing by r\% multiplies by 1-\frac{r}{100}. This method is essential for repeated change and reverse percentage.

Worked Example: Successive Changes

A price of 12 000 is increased by 15% and then reduced by 10%. The final price is 12000\times1.15\times0.90=12420. The net change is an increase of 420/12000\times100\%=3.5\%, not 5%.

Percentage Points Versus Percent Change

If an interest rate rises from 4% to 6%, the rise is 2 percentage points. Relative to the original rate, the percentage increase is \frac{6-4}{4}\times100\%=50\%. These statements describe different comparisons and should not be confused.

Reverse Percentage

When the final amount after a percentage change is known, divide by the multiplier. Do not simply add or subtract the stated percentage of the final value.

Worked Example: Original Price

After a 12% discount, a jacket costs 7 920. The multiplier is 0.88, so original price =7920\div0.88=9000.

Simple And Compound Interest

Simple interest is calculated only on the original principal. Compound interest is calculated on the current balance, so interest itself earns interest. For annual compounding, A=P\left(1+\frac{r}{100}\right)^n. The same multiplier model also handles depreciation by using a value below 1.

Worked Example: Comparing Interest

On 50 000 at 8% per year for 3 years:

  • Simple interest =50000\times0.08\times3=12000, total 62 000.
  • Compound total =50000(1.08)^3=62985.60.
Finding A Percentage In Context

Use consistent units before dividing. If 350 g is compared with 2.5 kg, convert 2.5 kg to 2500 g. Then 350/2500\times100\%=14\%. State whether the question asks for a percentage of the whole, a percentage change, or a percentage difference.

Extended Practice

A. A population falls from 48 000 to 43 200. Find the percentage decrease.

Decrease 4 800, so 4800/48000\times100\%=10\%.

B. After a 20% increase, a salary is 72 000. Find the original salary.

72000\div1.20=60000.

C. A machine depreciates by 18% each year. Its initial value is 250 000. Find its value after 2 years.

250000(0.82)^2=168100.

Examination Guidance
  • Identify the original quantity before calculating percentage change.
  • For repeated changes, use repeated multipliers rather than adding percentages.
  • Keep full calculator precision until the final line.
  • Reverse percentage questions require division by the multiplier.
Common Mistakes
  • Dividing percentage change by the final value instead of the original value.
  • Adding repeated percentage changes, e.g. saying two 10% increases equal exactly 20%.
  • Using 0.15 as the multiplier for a 15% increase; the multiplier is 1.15.
  • Subtracting 20% of the selling price to recover a cost price after a 20% markup.
Knowledge Check

1. Find 12% of 850.

Answer

102.

2. Express 42 as a percentage of 56.

Answer

75%.

3. A value falls from 320 to 272. Find the percentage decrease.

Answer

15%.

4. Find the amount after $6000 is invested for 4 years at 5% compound interest.

Answer

6000(1.05)^4=7293.04 dollars.

5. A jacket costs $68 after a 15% discount. Find its original price.

Answer

$80.

Mixed Review With Full Solutions

1. Express 84 as a percentage of 240.

84/240\times100\%=35\%.

2. A value increases by 25% and then decreases by 20%. Show that it returns to its original value.

The combined multiplier is 1.25\times0.80=1, so the final value equals the original.

3. A savings account grows to 132 651 after 3 years at 5% compound interest. Find the initial deposit.

P=132651/(1.05)^3=114600.

4. A school’s pass rate rises from 72% to 81%. State the rise in percentage points and the percentage increase.

Rise is 9 percentage points. Relative increase =9/72\times100\%=12.5\%.

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