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Number and Algebra - Percentages and percentage change

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A percentage is a fraction where the denominator is always 100. It is denoted by the symbol %\%. For example, x%=x100x\% = \frac{x}{100}.

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To find a percentage of a quantity, multiply the quantity by the percentage expressed as a decimal or fraction: Value=p100×Total\text{Value} = \frac{p}{100} \times \text{Total}.

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Percentage change represents the difference between a final value and an initial value as a percentage of the initial value. A positive result indicates a percentage increase, and a negative result indicates a percentage decrease.

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The Multiplier Method is used for efficient calculations. To increase a value by r%r\%, multiply by (1+r100)(1 + \frac{r}{100}). To decrease a value by r%r\%, multiply by (1−r100)(1 - \frac{r}{100}).

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Reverse percentages are used to find the original value before a percentage change occurred. This is done by dividing the new value by the multiplier: Original=NewMultiplier\text{Original} = \frac{\text{New}}{\text{Multiplier}}.

📐Formulae

Percentage Change=New Value−Original ValueOriginal Value×100%\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%

New Value=Original Value×(1±r100)\text{New Value} = \text{Original Value} \times (1 \pm \frac{r}{100})

Original Value=New Value1±r100\text{Original Value} = \frac{\text{New Value}}{1 \pm \frac{r}{100}}

Simple Interest=P×r×t100\text{Simple Interest} = \frac{P \times r \times t}{100}

💡Examples

Problem 1:

A car was purchased for Rs 25000. After one year, its value depreciated by 15%15\%. Calculate the value of the car after one year.

Solution:

Multiplier=1−15100=0.85\text{Multiplier} = 1 - \frac{15}{100} = 0.85 New Value=25000×0.85=21250\text{New Value} = 25000 \times 0.85 = 21250

Explanation:

To find the value after depreciation, we use a multiplier for a 15%15\% decrease, which is 0.850.85. Multiplying the original price by this multiplier gives the new value of Rs 21250.

Problem 2:

The price of a gold ring increased from Rs 1200 to Rs 1350. Calculate the percentage increase.

Solution:

Percentage Change=1350−12001200×100%\text{Percentage Change} = \frac{1350 - 1200}{1200} \times 100\% Percentage Change=1501200×100%=12.5%\text{Percentage Change} = \frac{150}{1200} \times 100\% = 12.5\%

Explanation:

We use the percentage change formula: DifferenceOriginal×100%\frac{\text{Difference}}{\text{Original}} \times 100\%. The difference is 150 and the original value is 1200, resulting in a 12.5%12.5\% increase.

Problem 3:

A pair of shoes is sold for Rs 72 after a 20%20\% discount. Find the original price of the shoes.

Solution:

Multiplier=1−20100=0.80\text{Multiplier} = 1 - \frac{20}{100} = 0.80 Original Price=720.80=90\text{Original Price} = \frac{72}{0.80} = 90

Explanation:

This is a reverse percentage problem. A 20%20\% discount means the sale price is 80%80\% of the original price. Dividing the sale price (Rs 72) by the multiplier (0.800.80) gives the original price of Rs 90.