Compound Interest: Concepts & Shortcuts
What is Compound Interest?
As we discussed in Simple Interest, interest is the extra amount paid for borrowing money. While Simple Interest is calculated only on the principal, Compound Interest is calculated on the Principal PLUS the Interest earned in previous periods. It is essentially "Interest on Interest."
The Standard Formula
To calculate the Total Amount ($A$), we use:
$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$
To calculate only the Interest ($I$) earned:
$$ I = P \left[ \left(1 + \frac{r}{n}\right)^{nt} - 1 \right] $$
Where:
- A: Final Amount
- P: Principal (Initial Sum)
- r: Annual Interest Rate
- n: Compounds per Year
- t: Time (in Years)
Compound Interest Interactive Practice Set
Click an option to check your answer. By Wisdom Helps Mahesh Sir
Q1. In how many years will ₹2,000 amount to ₹2,420 at 10% per annum compound interest? [cite: 3]
Solution: $2420 = 2000(1.1)^n \Rightarrow 1.21 = (1.1)^n$. Since $1.1^2 = 1.21$, n = 2 years.
Q2. In what time will ₹1,000 become ₹1,331 at 10% per annum compounded annually? [cite: 4]
Solution: $1331 = 1000(1.1)^n \Rightarrow 1.331 = (1.1)^n$. Since $1.1^3 = 1.331$, n = 3 years.
Q3. The principal, which will amount to ₹270.40 in 2 years at 4% p.a. compound interest? [cite: 5]
Solution: $270.40 = P(1.04)^2 \Rightarrow P = 270.40 / 1.0816 = \mathbf{₹250}$.
Q4. A sum amounts to ₹10,648 in 3 years and ₹9,680 in 2 years. Rate of interest per annum? [cite: 6]
Solution: Interest for 3rd year = $10648 - 9680 = 968$. Rate = $(968/9680) \times 100 = \mathbf{10\%}$.
Q5. At what rate per cent per annum will ₹2,304 amount to ₹2,500 in 2 years? [cite: 7]
Solution: $\sqrt{2500/2304} = 1 + R/100 \Rightarrow 50/48 - 1 = R/100 \Rightarrow R = 100/24 = \mathbf{4 \frac{1}{6}\%}$.
Q6. A sum becomes ₹1,352 in 2 years at 4% p.a. compound interest. The sum is? [cite: 8]
Solution: $P = 1352 / (1.04)^2 = 1352 / 1.0816 = \mathbf{₹1,250}$.
Q7. If C.I. for 2 years at 4% is ₹102, find S.I. at the same rate for 2 years? [cite: 9]
Solution: Effective CI rate = $4+4+(4 \times 4)/100 = 8.16\%$. $SI = (102 / 8.16) \times 8 = \mathbf{₹100}$.
Q8. Difference between S.I. and C.I. for 2 years at 4% is ₹4. The sum is? [cite: 1]
Solution: $Diff = P(R/100)^2 \Rightarrow 4 = P(4/100)^2 \Rightarrow P = 4 \times 625 = \mathbf{₹2,500}$.
Q9. Difference between C.I. and S.I. for 2 years at 8% is ₹768. The sum is? [cite: 1]
Solution: $768 = P(8/100)^2 \Rightarrow P = (768 \times 10000) / 64 = \mathbf{₹1,20,000}$.
Q10. Difference between C.I. and S.I. for 3 years at 5% is ₹15.25. The sum is? [cite: 10]
Solution: $Diff = P(R/100)^2 \times (300+R)/100 \Rightarrow 15.25 = P(1/400) \times (305/100) \Rightarrow P = \mathbf{₹2,000}$.
Q11. A sum doubles itself in 5 years. In how many years will it be 8 times? [cite: 11, 12]
Solution: $2^1$ in 5 years. $8 = 2^3$. So, $5 \times 3 = \mathbf{15 years}$.
Q12. A sum doubles itself in 4 years. In how many years will it be 4 times? [cite: 13, 14]
Solution: $2^1$ in 4 years. $4 = 2^2$. So, $4 \times 2 = \mathbf{8 years}$.
Q13. Borrowed ₹2,550, 4% p.a., 2 equal yearly instalments. Amount of each instalment? [cite: 15]
Solution: $2550 = x/1.04 + x/(1.04)^2 \Rightarrow 2550 = 25x/26 + 625x/676 \Rightarrow x = \mathbf{₹1,352}$.
Q14. Difference between C.I. and S.I. at 10% for 2 years is ₹464. Find Sum. [cite: 16]
Solution: $464 = P(10/100)^2 \Rightarrow P = 464 \times 100 = \mathbf{₹46,400}$.
Q15. C.I. on ₹16,000 for 9 months at 20% p.a., payable quarterly? [cite: 17]
Solution: $R=5\%$ (quarterly), $n=3$ periods. $A = 16000(1.05)^3 = 18522$. $CI = 18522 - 16000 = \mathbf{2522}$.
Q16. What sum will become ₹5,618 at 6% p.a. compound interest in 2 years? [cite: 18]
Solution: $P = 5618 / (1.06)^2 = 5618 / 1.1236 = \mathbf{₹5,000}$.
By Wisdom Helps Mahesh Sir
