Simple and Compound Interest

For 3 years at 5%, the difference between compound and simple interest is P[(1.05)³ - 1 - 0.15] = P[1.157625 - 1.15] = 0.007625P. Given this equals Rs. 11.40, we get P = 11.40/0.007625 ≈ Rs. 1495.

Let the amount deposited in each bank be P. Simple interest at 8% for 5 years = P × 8 × 5 / 100 = 0.4P. Simple interest at 12% for 5 years = P × 12 × 5 / 100 = 0.6P. The difference is 0.6P - 0.4P = 0.2P = 2850. Solving: P = 2850 / 0.2 = Rs 14250.

For the first 3 years, interest is on Rs. 7000. For the next 2 years, principal is Rs. 4000. Total payment is 3000 + 5450 = Rs. 8450. Interest = 8450 - 7000 = Rs. 1450. Equation: (7000 × r × 3)/100 + (4000 × r × 2)/100 = 1450. This gives 210r + 80r = 1450, so r = 5%.

If money becomes 3 times in 8 years at compound interest, then (1+r)^8 = 3, where r is the annual rate. We want the time t for money to become 243 times. Since 243 = 3^5, we have (1+r)^t = 3^5 = ((1+r)^8)^5. Therefore: (1+r)^t = (1+r)^40, which gives t = 40 years.

For a loan paid in equal annual installments with compound interest, each installment = Principal × r/(1 - (1+r)^-n) where r = 0.2, n = 2. Installment = 555500 × 0.2/(1 - 1/1.44) = 111100/(0.3056) ≈ 363600. Alternatively: 555500 = x/1.2 + x/1.44 = x(0.833 + 0.694) = 1.527x. Solving: x = 555500/1.527 ≈ 363600. Answer A is correct.

When interest is compounded every 6 months at 10% p.a., the rate per period is 5%. Principal = Rs. 15000. After first 6 months: 15000 × 1.05 = 15750. After second 6 months: 15750 × 1.05 = 16537.50. The formula A = P(1 + r/n)^(nt) where r=0.10, n=2, t=1 gives A = 15000(1 + 0.05)^2 = 15000 × 1.1025 = 16537.50.

For compound interest, Amount = Principal × (1 + rate/100)^time. At 5% for n years, this becomes P × (21/20)^n. The ratio of Amount to Principal is (21)^n : (20)^n.

Using SI = P×R×T/100: 942 = P + (P×9.5×6)/100 = P + 0.57P = 1.57P. So P = 942/1.57 = 600. Alternatively: Interest = 942 - P, and Interest = P×0.57, giving 942 - P = 0.57P.

Simple Interest = Principal × Rate × Time. Here: SI = 1600 × 0.05 × 6 = 1600 × 0.30 = Rs 480. Option D (2080) would be the amount (principal + interest), not just the interest earned.

Let P be principal and R be annual interest. Amount after 2 years = P + 2PR = 2100. Amount after 5 years = P + 5PR = 2250. Subtracting: 3PR = 150, so PR = 50. From first eq: P + 2(50) = 2100, so P = 2000. Rate = (50/2000) × 100 = 2.5%.

For CI, the amount after n years is P(1 + r/100)^n. After 1 year: 18000 × 1.15 = 20700. After 2 years: 18000 × 1.15² = 23805. After 3 years: 18000 × 1.15³ = 27375.75. CI for 2nd year = 23805 - 20700 = 3105. CI for 3rd year = 27375.75 - 23805 = 3570.75. Difference = 3570.75 - 3105 = 465.75.

CI for 2 years: 15000 × 1.16² - 15000 = 15000(1.3456 - 1) = 5184. CI for 3 years: 15000 × 1.16³ - 15000 = 15000(1.560896 - 1) = 8384. The difference is 8384 - 5184 = 3200.

Let P be principal, R be rate. Amount after 5 years: P + 5PR = 2500. Amount after 10 years: P + 10PR = 3500. Subtracting: 5PR = 1000, so PR = 200. Substituting back: P + 1000 = 2500, P = 1500. If rate doubles (2R), simple interest for 2 years = 2P(2R) = 2 × 1500 × 2R = 6000R. Since PR = 200, R = 200/1500. So SI = 6000 × 200/1500 = 800. Principal = 1500, SI = 800.

SI difference = P × (R2 - R1) × T/100. 54000 = P × (9.5 - 8) × 4/100 = P × 1.5 × 4/100 = 6P/100. Therefore P = 54000 × 100/6 = Rs.9,00,000. The same amount was deposited in both banks at different rates.

For compound interest compounded biannually at 10% per annum, the rate per half-year period is 5% and there are 8 half-year periods in 4 years. The difference between compound interest and simple interest for 4 years is given as Rs.32.60. Using the formula for this difference and solving for the principal P, we get P = 32.60 × 10000 / 400 = Rs.815. Option C is correct.