Simple and Compound Interest

Let each installment be x. Using the concept that each installment earns interest for the remaining period at 10% simple interest: Total amount = x + x(1 + 0.10×1) + x(1 + 0.10×2) + x(1 + 0.10×3) + x(1 + 0.10×4) = x(1 + 1.1 + 1.2 + 1.3 + 1.4) = x(6) = 12000. Therefore x = 2000.

Difference between CI and SI for 2 years at rate r is P(r/100)². Here, 360 = P(6/100)² = P(0.0036). Solving: P = 360/0.0036 = 100000. Wait, this gives P = 100000, but 360 = P × 36/10000 means P = 360 × 10000/36 = 10000. Option B matches.

Compound interest formula: A = P(1 + r)^t. For 2 years: A = 5000(1.1)² = 5000 × 1.21 = 6050. For remaining 6 months: A = 6050 × 1.05 = 6352.50. CI = 6352.50 - 5000 = 1352.50. Option D is correct.

Let the amount be A and rate be r%. In compound interest, A(1+r/100)^3 = 80000 and A(1+r/100)^5 = 115200. Dividing: (1+r/100)^2 = 115200/80000 = 1.44, so 1+r/100 = √1.44 = 1.2, meaning r = 20%. Therefore, the rate is 20%, which matches option D.

2 paise per rupee per month = 2% per month. Principal = Rs. 471. Monthly interest = 471 × 0.02 = Rs. 9.42. For 7 months: 9.42 × 7 = Rs. 65.94.

Simple Interest for 3 years = P × 5% × 3 = 0.15P. Simple Interest for 4 years = P × 5% × 4 = 0.20P. The difference is 0.20P - 0.15P = 0.05P, which equals Rs. 42. Therefore, P = 42 / 0.05 = Rs. 840. The key insight is that the difference in interest for consecutive years at simple interest is just the interest for one year.

SI = P×R×T/100 = 2600×15×3/100 = 1170. Amount after 3 years with CI: A = P(1 + R/100)^T = 2600(1.15)³ = 2600×1.520875 = 3954.275. CI = 3954.275 - 2600 = 1354.275. Difference = CI - SI = 1354.275 - 1170 = 184.275 ≈ 184.

Let parts be x and (1200 - x). SI1 = x×6×2/100 = 12x/100. SI2 = (1200-x)×8×3/100 = 24(1200-x)/100. Setting equal: 12x = 24(1200-x), 12x = 28800 - 24x, 36x = 28800, x = 800. So parts are 800 and 400.

Using SI = P x R x T, where P is principal, R is rate (9%), and T is time (5 years): 4275 = P x 0.09 x 5, so P = 4275/0.45 = Rs. 9500. For compound interest: Amount = P(1 + r)^n = 9500(1 + 0.09)^2 = 9500 x 1.09 x 1.09 = Rs. 11286.95. CI = Amount - Principal = 11286.95 - 9500 = Rs. 1786.95. The key is finding principal from simple interest first, then applying compound interest formula.

Using the compound interest formula: CI = P[(1 + r/100)^n - 1]. Here 2496 = 15000[(1 + r/100)^2 - 1]. Solving gives (1 + r/100)^2 = 1.1664, so 1 + r/100 = 1.08, hence r = 8%. Option A is correct.

Simple interest = P*r*t = 4000*0.05*2 = Rs. 400. Compound interest = 4000*(1.05)^2 - 4000 = Rs. 410. Difference = Rs. 10. Since Rs. 10 is not in options A-D, 'None of these' is correct.

Let A's share be x, then B's share is (25230-x). A invests for 13 years (40-27), B for 15 years (40-25). With 5% CI, future values must equal: x(1.05)^13 = (25230-x)(1.05)^15. Simplifying: x = (25230-x)(1.05)^2 = (25230-x)(1.1025). Solving: x = 13230. A receives Rs. 13230.

Principal = amount in 3 years - interest for 3 years. Interest for 2 years (years 4 and 5) = 700 - 620 = 80. So annual interest = 40. Principal = 620 - 120 = 500. Rate = (40/500) × 100 = 8% per annum.

Compound interest formula: A = P(1 + r/100)^t. Amount = 18000 × (1.10)^3 = 18000 × 1.331 = 23958. Compound Interest = Amount - Principal = 23958 - 18000 = 5958. The interest earned over 3 years at 10% compounded annually is Rs. 5958.

For principal P at rate r% for 2 years, the difference between compound and simple interest is P × (r/100)². Here: P × 0.06² = 232.56, so P = 232.56/0.0036 = 64600. Option B is correct.