Simple Interest (SI) and Compound Interest (CI) Formulas, Problems, and Solutions

Simple Interest (SI) Answers:

1. What is the simple interest on ₹5000 at 8% per annum for 3 years?
   • Answer: $$\text{SI} = \frac{5000 \times 8 \times 3}{100} = ₹1200$$SI=1005000×8×3​=₹1200
2. A sum of ₹2000 is invested at 5% simple interest per annum for 4 years. What will be the total interest?
   • Answer: $$\text{SI} = \frac{2000 \times 5 \times 4}{100} = ₹400$$SI=1002000×5×4​=₹400
3. Find the principal if the simple interest on it at 10% for 5 years is ₹2000.
   • Answer: $$\text{SI} = \frac{P \times R \times T}{100} \implies 2000 = \frac{P \times 10 \times 5}{100}$$SI=100P×R×T​⟹2000=100P×10×5​ $$P = \frac{2000 \times 100}{10 \times 5} = ₹4000$$P=10×52000×100​=₹4000
4. A person invests ₹1000 at 6% simple interest per annum. How much will he have in total after 5 years?
   • Answer: $$\text{SI} = \frac{1000 \times 6 \times 5}{100} = ₹300$$SI=1001000×6×5​=₹300 Total Amount = Principal + SI = ₹1000 + ₹300 = ₹1300
5. The simple interest on a certain amount is ₹900 for 3 years at 12% per annum. What is the principal?
   • Answer: $$\text{SI} = \frac{P \times 12 \times 3}{100} \implies 900 = \frac{P \times 36}{100}$$SI=100P×12×3​⟹900=100P×36​ $$P = \frac{900 \times 100}{36} = ₹2500$$P=36900×100​=₹2500

Compound Interest (CI) Answers:

6. Find the compound interest on ₹2500 for 2 years at 10% per annum, compounded annually.
   • Answer: $$A = 2500 \times \left(1 + \frac{10}{100}\right)^2 = 2500 \times (1.1)^2 = 2500 \times 1.21 = ₹3025$$A=2500×(1+10010​)2=2500×(1.1)2=2500×1.21=₹3025 Compound Interest = $A – P = 3025 – 2500 = ₹525$A−P=3025−2500=₹525
7. If ₹1000 is invested at 8% compound interest for 3 years, find the total amount at the end of the period.
   • Answer: $$A = 1000 \times \left(1 + \frac{8}{100}\right)^3 = 1000 \times (1.08)^3 = 1000 \times 1.2597 = ₹1259.70$$A=1000×(1+1008​)3=1000×(1.08)3=1000×1.2597=₹1259.70
8. A sum of ₹3000 is invested at 5% compound interest for 2 years. What is the compound interest at the end of 2 years?
   • Answer: $$A = 3000 \times \left(1 + \frac{5}{100}\right)^2 = 3000 \times (1.05)^2 = 3000 \times 1.1025 = ₹3307.50$$A=3000×(1+1005​)2=3000×(1.05)2=3000×1.1025=₹3307.50 Compound Interest = $A – P = 3307.50 – 3000 = ₹307.50$A−P=3307.50−3000=₹307.50
9. The principal is ₹5000, and the rate of interest is 10% per annum, compounded annually. Find the amount after 2 years.
   • Answer: $$A = 5000 \times \left(1 + \frac{10}{100}\right)^2 = 5000 \times (1.1)^2 = 5000 \times 1.21 = ₹6050$$A=5000×(1+10010​)2=5000×(1.1)2=5000×1.21=₹6050
10. A sum of money doubles itself in 5 years at 20% compound interest compounded annually. What is the principal amount?
    • Answer:
      Let the Principal be $P$P.
      Since the amount doubles, $$A = 2P$$A=2P $$2P = P \times \left(1 + \frac{20}{100}\right)^5 \implies 2 = (1.2)^5 \implies P = \frac{P}{2}$$2P=P×(1+10020​)5⟹2=(1.2)5⟹P=2P​ Solving gives Principal = ₹1000 (this is the principal to double to 2000)

Mixed Questions (SI & CI) Answers:

11. A sum of ₹5000 is invested at 8% per annum for 4 years. Find the difference between the simple interest and compound interest at the end of 4 years.
    • Answer:
      SI = $\frac{5000 \times 8 \times 4}{100} = ₹1600$1005000×8×4​=₹1600
      $A = 5000 \times (1.08)^4 = 5000 \times 1.3605 = ₹6802.50$A=5000×(1.08)4=5000×1.3605=₹6802.50
      CI = $A – P = 6802.50 – 5000 = ₹1802.50$A−P=6802.50−5000=₹1802.50
      Difference = $1802.50 – 1600 = ₹202.50$1802.50−1600=₹202.50
12. A person invests ₹1200 at 10% simple interest per annum for 3 years. Find the total amount at the end of 3 years.
    • Answer: $$\text{SI} = \frac{1200 \times 10 \times 3}{100} = ₹360$$SI=1001200×10×3​=₹360 Total Amount = Principal + SI = ₹1200 + ₹360 = ₹1560
13. What will be the compound interest on ₹3000 at 12% per annum for 1 year, compounded half-yearly?
    • Answer: $$A = 3000 \times \left(1 + \frac{12}{200}\right)^2 = 3000 \times (1.06)^2 = 3000 \times 1.1236 = ₹3370.80$$A=3000×(1+20012​)2=3000×(1.06)2=3000×1.1236=₹3370.80 CI = $A – P = 3370.80 – 3000 = ₹370.80$A−P=3370.80−3000=₹370.80
14. If ₹5000 is invested at 12% compound interest per annum for 3 years, what will be the compound interest if the interest is compounded quarterly?
    • Answer: $$A = 5000 \times \left(1 + \frac{12}{400}\right)^{12} = 5000 \times (1.03)^{12} = 5000 \times 1.4256 = ₹7128$$A=5000×(1+40012​)12=5000×(1.03)12=5000×1.4256=₹7128 CI = $A – P = 7128 – 5000 = ₹2128$A−P=7128−5000=₹2128
15. A person borrows ₹4000 at 15% compound interest for 2 years, compounded annually. What will be the total amount he has to repay at the end of 2 years?
    • Answer: $$A = 4000 \times \left(1 + \frac{15}{100}\right)^2 = 4000 \times (1.15)^2 = 4000 \times 1.3225 = ₹5290$$A=4000×(1+10015​)2=4000×(1.15)2=4000×1.3225=₹5290 Total Amount = ₹5290

Advanced Compound Interest Problems Answers:

Answer: $$A = 1500 \times \left(1 + \frac{8}{400}\right)^{8} = 1500 \times (1.02)^8 = 1500 \times 1.3728 = ₹2059.20$$A=1500×(1+4008​)8=1500×(1.02)8=1500×1.3728=₹2059.20 CI = $A – P = 2059.20 – 1500 = ₹559.20$A−P=2059.20−1500=₹559.20

A sum of ₹4000 is invested at 5% compound interest for 2 years, compounded annually. Find the compound interest for the second year.

Answer: $$A = 4000 \times \left(1 + \frac{5}{100}\right)^2 = 4000 \times 1.1025 = ₹4400$$A=4000×(1+1005​)2=4000×1.1025=₹4400 CI for second year = ₹4400 – ₹4000 = ₹400

If ₹1200 is invested for 3 years at 10% compound interest, compounded annually, find the compound interest after 3 years.

Answer: $$A = 1200 \times \left(1 + \frac{10}{100}\right)^3 = 1200 \times (1.1)^3 = 1200 \times 1.331 = ₹1597.20$$A=1200×(1+10010​)3=1200×(1.1)3=1200×1.331=₹1597.20 CI = $A – P = 1597.20 – 1200 = ₹397.20$A−P=1597.20−1200=₹397.20

Find the compound interest on ₹1500 at 8% per annum for 2 years, compounded quarterly.