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Number - Integers, Fractions, Decimals, and Percentages

Grade 8IGCSE

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

🔑Concepts

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Directed Numbers: Understanding operations with positive and negative integers using the number line.

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HCF and LCM: Finding the Highest Common Factor and Lowest Common Multiple using prime factorization.

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BODMAS/BIDMAS: The order of operations—Brackets, Indices, Division/Multiplication, Addition/Subtraction.

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Equivalent FDP: 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: Calculating the new value after a percentage change or finding the percentage change itself.

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Reverse Percentages: Finding the original value after a percentage increase or decrease has occurred.

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Rounding and Significant Figures: Rounding numbers to a specific number of decimal places (d.p.) or significant figures (s.f.).

📐Formulae

Percentage Change=Actual ChangeOriginal Value×100%\text{Percentage Change} = \frac{\text{Actual Change}}{\text{Original Value}} \times 100\%

New Value=Original Value×(1±Percentage100)\text{New Value} = \text{Original Value} \times (1 \pm \frac{\text{Percentage}}{100})

Original Value=New ValueMultiplier\text{Original Value} = \frac{\text{New Value}}{\text{Multiplier}}

Fraction to Percentage=NumeratorDenominator×100\text{Fraction to Percentage} = \frac{\text{Numerator}}{\text{Denominator}} \times 100

💡Examples

Problem 1:

Calculate 35+14÷12\frac{3}{5} + \frac{1}{4} \div \frac{1}{2}.

Solution:

35+(14×21)=35+24=35+12=610+510=1110=1110\frac{3}{5} + (\frac{1}{4} \times \frac{2}{1}) = \frac{3}{5} + \frac{2}{4} = \frac{3}{5} + \frac{1}{2} = \frac{6}{10} + \frac{5}{10} = \frac{11}{10} = 1\frac{1}{10}

Explanation:

Follow BIDMAS: perform the division first by multiplying by the reciprocal, then find a common denominator (10) to add the fractions.

Problem 2:

A laptop is sold for 540aftera540 after a 10%$ discount. Find the original price.

Solution:

Multiplier=1−0.10=0.90\text{Multiplier} = 1 - 0.10 = 0.90. Original Price=540÷0.90=600\text{Original Price} = 540 \div 0.90 = 600.

Explanation:

This is a reverse percentage problem. Since there was a 10%10\% discount, 540540 represents 90%90\% of the original price. Divide the sale price by the multiplier (0.90.9) to find the 100%100\% value.

Problem 3:

The population of a town increased from 15,000 to 18,000. Calculate the percentage increase.

Solution:

Change=18,000−15,000=3,000\text{Change} = 18,000 - 15,000 = 3,000. Percentage Increase=3,00015,000×100=20%\text{Percentage Increase} = \frac{3,000}{15,000} \times 100 = 20\%.

Explanation:

Find the actual increase first, then divide by the original population and multiply by 100 to get the percentage.