Simple and Compound Interest

Let principal be P and rate be R% per annum. Simple interest for 3 years = P × R × 3/100. At (R+4)% rate, interest = P × (R+4) × 3/100. Difference = 408 = P × 4 × 3/100 = 12P/100. Therefore P = 408 × 100/12 = Rs. 3400.

Calculate interest for first 3 years on original principal (Pr × 3), then for next 3 years on doubled principal (2Pr × 3). Total = 3Pr + 6Pr = 9Pr. Given Pr = 270, total SI = 9 × 270 = Rs. 2430.

Let P be principal and R be rate. Given: SI after 12 years = P × R × 12 = 2940. After 6 years, amount becomes 5P, so SI = 4P = P × R × 6. From this, 6R = 4, giving R = 2/3%. Substituting: P × (2/3) × 12 = 2940, so 8P = 2940 and P = 367.5. Total interest after 12 years = 2940 (from SI calculation) + additional from principal growth = 8820. Option A (8720) is close but incorrect. Options C, D, E are arithmetic errors.

Let P be the amount. SI1 = P × 3.5 × 12/100 = 0.42P. SI2 = P × 3.5 × 8.5/100 = 0.2975P. Difference = 0.42P - 0.2975P = 0.1225P = 189.875. Solving: P = 189.875/0.1225 = 1550.

Compound Interest formula: A = P(1 + r/100)ⁿ. Here P = 8500, r = 15%, n = 3. A = 8500(1.15)³ = 8500 × 1.520875 = 12927.44. CI = A - P = 12927.44 - 8500 = 4427.43. Round to two decimal places as specified. The key is applying compound interest formula correctly for 3 years.

Principal = Rs. 8000, annual rate = 20%, time = 1 year, compounded quarterly. Quarterly rate = 20/4 = 5%. Number of quarters = 4. Amount = 8000 × (1 + 5/100)⁴ = 8000 × (1.05)⁴ = 8000 × 1.21550625 = Rs. 9724.05. Compound Interest = 9724.05 - 8000 = Rs. 1674.05.

For a principal P at 12% for 2 years: the difference between CI and SI is P(1+0.12)² - P - (P×0.12×2) = P(1.2544 - 1 - 0.24) = P(0.0144) = 0.0144P. Given this difference is Rs. 90, we get P = 90/0.0144 = Rs. 6250. Amount after 3 years at CI: 6250(1.12)³ = 6250 × 1.404928 = Rs. 8780.80. Option D is correct. Options A, B, and C are incorrect calculations from wrong principals or formulas.

Simple interest formula: SI = P×R×T/100. Amount trebles means 3P = P + SI, so SI = 2P. If this happens in 15.5 years (15 years 6 months), then 2P = P×R×15.5/100, giving R = 200/15.5 = 12.9%. For amount to double: 2P = P + SI, SI = P. So P = P×12.9×T/100, giving T = 100/12.9 = 7.75 years = 7 years 9 months.

Simple interest = P × R × T = 12000 × 0.12 × 3 = Rs. 4320. Compound interest = P[(1+R)^T - 1] = 12000[(1.12)^3 - 1] = 12000[1.404928 - 1] = Rs. 4859.136. The difference is 4859.136 - 4320 = Rs. 539.136.

If an amount doubles in 10 years under compound interest, it will become fourfold (2×2) in 20 years. This is because compound interest grows exponentially - doubling once takes 10 years, and doubling again (to reach 4x) takes another 10 years. The formula A = P(1 + r)^t shows that the time to multiply by a factor n follows logarithmic growth.

Compound Interest = P(1 + r/100)^n - P = 275000(1.16)^4 - 275000. (1.16)^4 ≈ 1.8106. Amount = 275000 × 1.8106 ≈ 497,915. CI = 497,915 - 275,000 = 222,915 (approx). Option B (222,925.80) matches.

Using simple interest formula SI = P×R×T/100. Total interest = 1788. So (1540×R×5/100) + (1800×R×4/100) = 1788. This simplifies to 77R + 72R = 1788, giving 149R = 1788, so R = 1788/149 = 12%. Verify: 1540×12×5/100 = 924, 1800×12×4/100 = 864, total = 1788.

Simple Interest formula: SI = P × R × T ÷ 100. Total Amount = P + SI = P(1 + RT/100). Here, R = 15%, T = 3 years, so RT/100 = 45/100 = 0.45. Total = 1.45P = 7250. Therefore, P = 7250 ÷ 1.45 = Rs. 5000. Option B (4500) would give total 6525. Option C (6500) would give total 9425.

Let the sum at 10% be x. Then interest = x×10×5/100 = x/2. Remaining sum = 2600-x at 9% for 6 years gives interest = (2600-x)×9×6/100 = 54(2600-x)/100. Equating: x/2 = 54(2600-x)/100. Solving gives x = 1350. Option C is correct.

Let original deposit be P. At 5%, income = 0.05P. With Rs. 2000 deposit at 4%, income = 0.04 × 2000 = Rs. 80. Setting equal: 0.05P = 80, so P = 80/0.05 = Rs. 8000. The original deposit was Rs. 8000.