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Number - Fractions, Decimals, and Percentages

Grade 9IGCSE

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

🔑Concepts

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Equivalent Fractions and Simplifying: Finding fractions with the same value by multiplying or dividing the numerator and denominator by the same number.

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Operations with Fractions: Adding and subtracting (requires common denominators), multiplying (top times top, bottom times bottom), and dividing (invert and multiply).

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FDP Conversion: Converting between fractions, decimals, and percentages (e.g., 1/4=0.25=25%1/4 = 0.25 = 25\%).

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Percentage Increase/Decrease: Using multipliers to calculate new values (e.g., a 15%15\% increase uses a multiplier of 1.151.15).

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Reverse Percentages: Finding the original value after a percentage change has occurred.

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Recurring Decimals: Converting repeating decimals into fractions using algebraic methods.

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Compound Interest: Calculating the total value of an investment over time where interest is earned on previous interest.

📐Formulae

Percentage Change=Actual ChangeOriginal Value×100\text{Percentage Change} = \frac{\text{Actual Change}}{\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 ValueMultiplier\text{Original Value} = \frac{\text{New Value}}{\text{Multiplier}}

Compound Interest: A=P(1+r100)n\text{Compound Interest: } A = P(1 + \frac{r}{100})^n

Simple Interest: I=P×R×T100\text{Simple Interest: } I = \frac{P \times R \times T}{100}

💡Examples

Problem 1:

Calculate 135÷231 \frac{3}{5} \div \frac{2}{3}. Give your answer as a mixed number.

Solution:

135÷23=85×32=2410=125=2251 \frac{3}{5} \div \frac{2}{3} = \frac{8}{5} \times \frac{3}{2} = \frac{24}{10} = \frac{12}{5} = 2 \frac{2}{5}

Explanation:

First, convert the mixed number to an improper fraction (135=851 \frac{3}{5} = \frac{8}{5}). To divide, multiply by the reciprocal of the second fraction (flip 23\frac{2}{3} to 32\frac{3}{2}). Multiply the numerators and denominators, then simplify and convert back to a mixed number.

Problem 2:

A car's value depreciates by 12%12\% each year. If the car is worth $13,200 now, what was its value one year ago?

Solution:

Multiplier=1−0.12=0.88\text{Multiplier} = 1 - 0.12 = 0.88. Original Value=13,200÷0.88=15,000\text{Original Value} = 13,200 \div 0.88 = 15,000.

Explanation:

This is a reverse percentage problem. A 12%12\% decrease means the current value is 88%88\% (0.880.88) of the original. Divide the current value by the multiplier to find the original price.

Problem 3:

Convert the recurring decimal 0.4˙7˙0.\dot{4}\dot{7} to a fraction.

Solution:

Let x=0.4747...x = 0.4747.... Then 100x=47.4747...100x = 47.4747.... 100x−x=47⇒99x=47⇒x=4799100x - x = 47 \Rightarrow 99x = 47 \Rightarrow x = \frac{47}{99}.

Explanation:

Multiply the decimal by a power of 10 to shift the decimal point past one full repeating cycle. Subtract the original equation from the new one to cancel out the infinite repeating part, then solve for xx.