Simple & Compound Interest MCQs for UPSC Prelims

Same principal P grows to the same amount Q in one year under both schemes, so P(1 + R/200)^2 = P(1 + S/100). Expanding: 1 + R/100 + R^2/40000 = 1 + S/100, hence S = R + R^2/400. Since R > 0, S exceeds R, i.e. R < S - option (c). Intuitively, half-yearly compounding earns interest on interest within the year, so a smaller nominal rate R suffices to match the annually compounded rate S. Option (a) ignores intra-year compounding, option (b) reverses the inequality, and option (d) wrongly allows equality, which would need R = 0.

Interest rate 12.5% = 1/8, so each year the balance is multiplied by 1.125. Year 1: 22,800 x 1.125 = 25,650; after paying 8,650, balance = 17,000. Year 2: 17,000 x 1.125 = 19,125; after paying 9,125, balance = 10,000. Year 3: 10,000 x 1.125 = 11,250, which must be paid in full to clear the debt. Answer is Rs. 11,250.

Simple interest for 3 years = 8000 x 10 x 3 / 100 = Rs. 2,400. Compound amount = 8000 x (1.1)^3 = 8000 x 1.331 = Rs. 10,648, so CI = 2,648. Difference = 2,648 - 2,400 = Rs. 248. This matches the formula for the 3-year difference: P(r/100)^2 x (3 + r/100) = 8000 x 0.01 x 3.1 = 248.

Under compound interest, the amount grows by the same factor each year. The interest earned in the third year = 2662 - 2420 = 242, and this is the interest on the amount at the end of year 2 (Rs. 2420). So the rate = 242 / 2420 = 0.10 = 10%. Verify: 2420 x 1.10 = 2662, correct. Option 12% gives 2420 x 1.12 = 2710.4, wrong. Option 8% gives 2613.6, wrong. Option 11% gives 2686.2, wrong. The rate is 10%.

For 2 years, the difference between compound and simple interest equals P x (r/100)^2. Here (r/100)^2 = (8/100)^2 = 0.0064. So P x 0.0064 = 32, giving P = 32 / 0.0064 = 5000. Checking: SI for 2 years = 5000 x 8 x 2/100 = 800; CI = 5000(1.08^2 - 1) = 5000 x 0.1664 = 832; difference = 32. Correct.

For 2 years, the difference between compound and simple interest equals P x (r/100)^2. So 40 = P x (8/100)^2 = P x 0.0064. Therefore P = 40 / 0.0064 = Rs. 6,250.

Simple interest = 8000 x 5 x 3 / 100 = 1200. For compound interest, amount = 8000 x (1.05)^3 = 8000 x 1.157625 = 9261, so CI = 1261. Difference = 1261 - 1200 = 61. (The standard 3-year formula P x r^2 x (300 + r)/10^6 = 8000 x 25 x 305 / 1000000 = 61 confirms this.)

Let each installment be X. The present value of the two installments must equal the loan. So X/1.1 + X/(1.1)^2 = 6300. Multiplying through: X(1.1 + 1)/(1.21) = 6300, i.e., X x (2.1/1.21) = 6300. Thus X = 6300 x 1.21 / 2.1 = 7623 / 2.1 = 3630. Each installment is Rs. 3,630.