📘 Compound Interest
Basic Terms
Principal (P): Initial amount
Rate (R): Rate of interest per year
Time (T): Time period in years
Amount (A): Final amount
Compound Interest Formula
\[
A = P\left(1 + \frac{R}{100}\right)^T
\]
\[
\text{Compound Interest (CI)} = A - P
\]
Compounding More Than Once a Year
\[
A = P\left(1 + \frac{R}{100n}\right)^{nT}
\]
Where \(n\) = number of times interest is compounded per year.
🔥 Exam Shortcuts & Tricks
Shortcut 1:
For 2 years: \[ \text{CI} = P\left(\frac{R}{100}\right)\left(2 + \frac{R}{100}\right) \]
For 2 years: \[ \text{CI} = P\left(\frac{R}{100}\right)\left(2 + \frac{R}{100}\right) \]
Shortcut 2:
Difference between CI and SI for 2 years: \[ = P\left(\frac{R}{100}\right)^2 \]
Difference between CI and SI for 2 years: \[ = P\left(\frac{R}{100}\right)^2 \]
Shortcut 3:
If rate is same every year, use direct formula instead of year-wise calculation.
If rate is same every year, use direct formula instead of year-wise calculation.
📘 Solved Examples
Example 1:
Find CI on ₹1000 at 10% per annum for 2 years. \[ A = 1000\left(1 + \frac{10}{100}\right)^2 = 1000(1.1)^2 = 1210 \] \[ \text{CI} = 1210 - 1000 = 210 \]
Find CI on ₹1000 at 10% per annum for 2 years. \[ A = 1000\left(1 + \frac{10}{100}\right)^2 = 1000(1.1)^2 = 1210 \] \[ \text{CI} = 1210 - 1000 = 210 \]
Example 2:
Find amount on ₹2000 at 5% per annum for 3 years. \[ A = 2000\left(1 + \frac{5}{100}\right)^3 = 2000(1.05)^3 \approx 2315.25 \]
Find amount on ₹2000 at 5% per annum for 3 years. \[ A = 2000\left(1 + \frac{5}{100}\right)^3 = 2000(1.05)^3 \approx 2315.25 \]
Example 3:
Difference between CI and SI on ₹1000 at 10% for 2 years. \[ \text{Difference} = 1000\left(\frac{10}{100}\right)^2 = 10 \]
Difference between CI and SI on ₹1000 at 10% for 2 years. \[ \text{Difference} = 1000\left(\frac{10}{100}\right)^2 = 10 \]
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