Edexcel Mathematics (4XMAH) -Unit 2 - 1.6 Percentages- Study Notes- New Syllabus
Edexcel Mathematics (4XMAH) -Unit 2 – 1.6 Percentages- Study Notes- New syllabus
Edexcel Mathematics (4XMAH) -Unit 2 – 1.6 Percentages- Study Notes -Edexcel iGCSE Mathematics – per latest Syllabus.
Key Concepts:
A use repeated percentage change
Calculate the total percentage increase when a 30% increase is followed by a 20% decrease
B solve compound interest problems
Edexcel iGCSE Mathematics -Concise Summary Notes- All Topics
Repeated Percentage Change
When a percentage increase or decrease happens more than once, we do not simply add or subtract the percentages.
Instead, each change acts on the new value.
Multipliers
A percentage change can be written as a multiplier:
- Increase of \( p\% \) → multiply by \( 1+\dfrac{p}{100} \)
- Decrease of \( p\% \) → multiply by \( 1-\dfrac{p}{100} \)
Total Change
Multiply the multipliers together.
\( \mathrm{Final\ value}=\mathrm{Original}\times(\text{multiplier}_1)\times(\text{multiplier}_2)\dots \)
Example 1:
Calculate the overall percentage change when an increase of \( 30\% \) is followed by a decrease of \( 20\% \).
▶️ Answer/Explanation
Increase \( 30\% \Rightarrow \times1.30 \)
Decrease \( 20\% \Rightarrow \times0.80 \)
\( 1.30\times0.80=1.04 \)
Final multiplier \( =1.04 \)
So the value increased by \( 4\% \).
Conclusion: Total increase \( =4\% \).
Example 2:
A price of \( £200 \) increases by \( 10\% \) and then increases again by \( 5\% \). Find the final price.
▶️ Answer/Explanation
\( 200\times1.10\times1.05 \)
\( =200\times1.155=231 \)
Conclusion: \( £231 \).
Example 3:
A population decreases by \( 12\% \) each year for 2 years. What fraction of the original population remains?
▶️ Answer/Explanation
Decrease \( 12\% \Rightarrow \times0.88 \)
\( 0.88\times0.88=0.7744 \)
Fraction remaining \( =0.7744=\dfrac{7744}{10000}=\dfrac{1936}{2500} \)
Conclusion: \( \dfrac{1936}{2500} \) of the original population remains.
Compound Interest
Compound interest means interest is added to the amount and then future interest is calculated on the new total.
So the money grows each year.
Compound Interest Formula
\( A=P\left(1+\dfrac{r}{100}\right)^n \)
where:
- \( P \) = initial amount (principal)
- \( r \) = interest rate (%)
- \( n \) = number of years
- \( A \) = final amount
Interest Earned
\( \mathrm{Interest}=A-P \)
Example 1:
£500 is invested at \( 4\% \) per year for 3 years. Find the final amount.
▶️ Answer/Explanation
\( A=500(1.04)^3 \)
\( A=500(1.124864)=562.43 \)
Conclusion: \( £562.43 \) (to nearest penny).
Example 2:
£800 is invested for 2 years at \( 5\% \) compound interest. Find the interest earned.
▶️ Answer/Explanation
\( A=800(1.05)^2 \)
\( A=800(1.1025)=882 \)
Interest \( =882-800=£82 \)
Conclusion: \( £82 \).
Example 3:
How many years will £1000 take to exceed £1210 at \( 10\% \) compound interest?
▶️ Answer/Explanation
\( A=1000(1.10)^n \)
Try values:
Year 1: \( 1000(1.10)=1100 \)
Year 2: \( 1000(1.10)^2=1210 \)
It reaches £1210 after 2 years and exceeds it after 3 years.
Conclusion: \( 3 \) years.