Question 1

Find the difference between the simple interest and the compound interest on Rs. 1200 for 1 year at 10% per annum, when interest is reckoned half-yearly.

Accepted Answer

Problem restatement Compare CI (compounded half-yearly) and SI for 1 year at the same nominal rate (10%) on Rs. 1200; report CI − SI. Given data P = Rs. 1200 Nominal rate = 10% p.a.; half-yearly compounding ⇒ 5% per half-year Time = 1 year = 2 half-years Concept/Approach Compute CI via amount factor (1 + r÷m)m for m = 2. Compute SI via P × r × t. Take the difference. Step-by-step calculation Amount (CI) = 1200 × (1 + 0.05)2 = 1200 × 1.1025 = 1323.00CI = 1323.00 − 1200 = Rs. 123.00SI = P × r × t = 1200 × 0.10 × 1 = Rs. 120.00Difference = 123.00 − 120.00 = Rs. 3 Common pitfalls Using 0.4 hour–type conversions here is irrelevant; the compounding frequency is semi-annual: use 5% twice.

Question 2

If the simple interest on a sum for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same principal, at the same rate, for the same 2 years?

Accepted Answer

Problem restatement From SI over 2 years, infer the principal, then compute CI for the same period at the same rate. Given data SI = Rs. 50 on rate r = 5% p.a., time t = 2 years Concept/Approach Use SI = P r t to get P. Then CI = P[(1 + r)t − 1]. Step-by-step calculation 50 = P × 0.05 × 2 ⇒ 50 = 0.10P ⇒ P = Rs. 500CI = 500 × [(1.05)2 − 1] = 500 × (1.1025 − 1) = 500 × 0.1025CI = Rs. 51.25 Verification/Alternative Year-wise: Year 1 interest = 25; Year 2 interest = 25 on P plus 1.25 on previous year's interest = 26.25; total = 51.25. Common pitfalls Forgetting that CI includes interest on interest in the second year.

Question 3

Albert invests Rs. 8000 at 5% p.a. compounded annually for 2 years. What is the maturity amount?

Accepted Answer

Problem restatement Compute the amount after 2 years at 5% compound interest on Rs. 8000. Given data P = Rs. 8000, r = 5% p.a., t = 2 years Concept/Approach Use the compound amount formula A = P(1 + r)t. Step-by-step calculation A = 8000 × (1.05)2 = 8000 × 1.1025 = Rs. 8820 Verification/Alternative Year 1: interest = 400 ⇒ amount = 8400; Year 2: interest = 420 ⇒ amount = 8820.

Question 4

Find the effective annual rate corresponding to a nominal 6% per annum payable half-yearly.

Accepted Answer

Problem restatement Convert a nominal 6% p.a. with semi-annual compounding into an effective annual rate. Given data Nominal rate = 6% p.a., compounding m = 2 times per year Concept/Approach Effective rate reff = (1 + r÷m)m − 1. Step-by-step calculation reff = (1 + 0.06÷2)2 − 1 = (1.03)2 − 1 = 1.0609 − 1 = 0.0609 = 6.09% p.a. Common pitfalls Multiplying 6% by 2% directly is wrong; compounding must be applied.

Question 5

What is the compound interest on Rs. 25,000 for 3 years at 12% p.a. compounded annually?

Accepted Answer

Problem restatement Compute CI for 3 years at 12% on Rs. 25,000. Given data P = Rs. 25,000; r = 12% p.a.; t = 3 years Concept/Approach Amount A = P(1 + r)t; CI = A − P. Step-by-step calculation A = 25,000 × (1.12)3 = 25,000 × 1.404928 = 35,123.20CI = 35,123.20 − 25,000 = Rs. 10,123.20 Verification/Alternative Year-wise interest: 3,000; then 3,360; then 3,763.20 ⇒ total 10,123.20. Common pitfalls Adding 12% thrice to principal (simple interest) underestimates the final amount; compounding is required.

Question 6

The compound interest on a sum for 2 years at 10% p.a. is Rs. 525. What is the simple interest on the same sum for double the time at half the rate per annum?

Accepted Answer

Problem restatement Given CI over 2 years at 10% is Rs. 525. Find SI for 4 years at 5% on the same principal. Given data CI (2 years, 10%) = Rs. 525 Target: SI (4 years, 5%) on same P Concept/Approach From CI = P[(1 + 0.10)2 − 1] = 0.21P, infer P. Then apply SI = P r t. Step-by-step calculation 0.21P = 525 ⇒ P = 525 ÷ 0.21 = Rs. 2500SI = 2500 × 0.05 × 4 = Rs. 500 Verification/Alternative Direct CI check: (1.1)2 − 1 = 0.21; 0.21 × 2500 = 525 (consistent).

Question 7

At what annual rate of compound interest will Rs. 1200 amount to Rs. 1348.32 in 2 years?

Accepted Answer

Problem restatement Find r such that 1200(1 + r)2 = 1348.32. Given data P = 1200; A = 1348.32; t = 2 years Concept/Approach Solve for r from (1 + r)2 = A÷P. Step-by-step calculation (1 + r)2 = 1348.32 ÷ 1200 = 1.12361 + r = √1.1236 = 1.06r = 0.06 = 6% p.a. Common pitfalls Taking r = 12.36% (the two-year growth) without extracting the square root.

Question 8

Find the difference between the compound interests on Rs. 5000 for 1½ years at 4% p.a. when interest is compounded (i) yearly and (ii) half-yearly.

Accepted Answer

Problem restatement Compute CI for 1½ years at 4% p.a. under two regimes: annual compounding vs half-yearly compounding. Report the difference (half-yearly minus yearly). Given data P = Rs. 5000 r = 4% p.a. Time = 1.5 years Concept/Approach (i) Compounded yearly: 1 full year compounded, then simple interest for the remaining half-year on the amount after 1 year. (ii) Compounded half-yearly: 3 half-year periods at 2% each. Step-by-step calculation (i) Yearly: Amount = P × 1.04 × (1 + 0.04 × 0.5) = P × 1.04 × 1.02 = P × 1.0608(ii) Half-yearly: Amount = P × (1.02)3 = P × 1.061208Difference in amount = P(1.061208 − 1.0608) = P × 0.000408With P = 5000: Difference = 5000 × 0.000408 = Rs. 2.04 Verification/Alternative Compute each amount numerically: Yearly = 5000 × 1.0608 = 5304.00; Half-yearly = 5000 × 1.061208 = 5306.04; Difference = 2.04. Common pitfalls Applying annual compounding for 1.5 years as 1.041.5 (not standard in yearly compounding) or ignoring the half-year SI treatment.

Question 9

A bank offers 5% compound interest calculated half-yearly. A customer deposits Rs. 1600 on 1 January and another Rs. 1600 on 1 July of the same year. What is the total interest earned by 31 December of that year?

Accepted Answer

Problem restatement Interest is compounded twice a year at 5% p.a. (i.e., 2.5% per half-year). Two equal deposits are made: one on 1 Jan (earns two half-years), and one on 1 Jul (earns one half-year). Find the total interest by year end. Given data Rate per half-year = 5% ÷ 2 = 2.5% = 0.025 Deposit 1: Rs. 1600 on 1 Jan → 2 half-years Deposit 2: Rs. 1600 on 1 Jul → 1 half-year Concept/Approach Treat each deposit as a separate principal for its own compounding duration and then sum the interests. Step-by-step calculation Interest on 1st deposit = 1600[(1 + 0.025)2 − 1]= 1600(1.050625 − 1) = 1600 × 0.050625 = 81.00Interest on 2nd deposit = 1600[(1 + 0.025) − 1] = 1600 × 0.025 = 40.00Total interest = 81.00 + 40.00 = Rs. 121 Verification/Alternative Amount of 1st deposit at year end = 1600 × 1.050625 = 1681.00 (so interest 81.00). Amount of 2nd deposit = 1600 × 1.025 = 1640.00 (so interest 40.00). Sum of interests = 121.00. Common pitfalls Applying 5% for the July deposit (should be 2.5% for one half-year), or compounding both deposits for two periods. Final Answer Rs. 121

Question 10

What is the least number of complete years for a sum to more than double at 20% compound interest per annum?

Accepted Answer

Problem restatement Find the smallest integer n such that (1 + 0.20)n > 2. Given data Annual compound rate r = 20% = 0.20 Target: amount factor > 2 Concept/Approach Test integer years until the compound factor exceeds 2. Step-by-step calculation (1.2)3 = 1.728 < 2(1.2)4 = 2.0736 > 2Least n = 4 Verification/Alternative Using logs: n > log 2 ÷ log 1.2 ≈ 0.3010 ÷ 0.07918 ≈ 3.80 → next integer 4. Common pitfalls Comparing to 2.0 after only 3 years (still short). Final Answer 4 years

Question 11

An amount increases by 60% in 6 years at simple interest. What is the compound interest on Rs. 12,000 for 3 years at the same annual rate?

Accepted Answer

Problem restatement First extract the annual simple-interest rate from a 60% increase over 6 years, then apply that rate for 3 years with annual compounding on Rs. 12,000. Given data SI increase over 6 years = 60% ⇒ r × 6 = 0.60 ⇒ r = 0.10 = 10% p.a. Principal for CI = Rs. 12,000; time = 3 years Concept/Approach Use SI to find r, then compute CI via the amount factor (1 + r)t. Step-by-step calculation A = 12,000 × (1.10)3 = 12,000 × 1.331 = 15,972CI = A − P = 15,972 − 12,000 = Rs. 3,972 Verification/Alternative Year-wise interest: 1200; then 1320; then 1452 → total 3972. Common pitfalls Confusing SI and CI; using 60% over 3 years (it was over 6 years). Final Answer Rs. 3,972

Question 12

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4,347. Find the period (in years).

Accepted Answer

Problem restatement Given CI and rate, determine t such that P[(1 + r)t − 1] = CI. Given data P = 30,000; r = 7% = 0.07 CI = 4,347 ⇒ (1 + r)t − 1 = 4,347 ÷ 30,000 = 0.1449 Concept/Approach Recognize the square of 1.07: (1.07)2 = 1.1449. Step-by-step calculation (1.07)t = 1 + 0.1449 = 1.1449Since (1.07)2 = 1.1449, we have t = 2 years Verification/Alternative Compute CI directly: 30,000[(1.07)2 − 1] = 30,000(0.1449) = 4,347. Common pitfalls Taking t = 1 (would give only 7%, not 14.49%). Final Answer 2 years

Question 13

The difference between compound and simple interest on Rs. 15,000 for 2 years is Rs. 96. What is the annual rate of interest?

Accepted Answer

Problem restatement For 2 years at rate r, CI − SI = P r2. Use this identity to solve for r. Given data P = 15,000; CI − SI = 96 Concept/Approach For two years, CI = P(2r + r2) and SI = P(2r); their difference is P r2. Step-by-step calculation P r2 = 96 ⇒ 15,000 r2 = 96r2 = 96 ÷ 15,000 = 0.0064r = √0.0064 = 0.08 = 8% p.a. Common pitfalls Using 96 as one year difference (the identity holds specifically for 2 years). Final Answer 8% p.a.

Question 14

Simple interest on a sum for 3 years at 8% p.a. equals half the compound interest on Rs. 4,000 for 2 years at 10% p.a. Find the sum placed at simple interest.

Accepted Answer

Problem restatement Compute CI on Rs. 4,000 for 2 years at 10%, halve it, and set equal to SI on unknown principal at 8% for 3 years. Given data CI (4,000; 10%; 2 years) = 4,000[(1.10)2 − 1] = 4,000(0.21) = 840 Half of that = 420 SI = P × 0.08 × 3 = 0.24P Step-by-step calculation 0.24P = 420P = 420 ÷ 0.24 = Rs. 1,750 Verification/Alternative Check: SI = 1,750 × 0.24 = 420, which is indeed half of 840. Common pitfalls Using 10% of 4,000 twice without compounding (that would be SI, not CI). Final Answer Rs. 1,750

Question 15

The difference between simple and compound interest compounded annually on a certain sum for 2 years at 4% p.a. is Re. 1. Find the sum.

Accepted Answer

Problem restatement For two years at rate r, the excess of CI over SI equals P r2. Use this to get P. Given data r = 4% = 0.04; CI − SI = Re. 1 Step-by-step calculation P r2 = 1 ⇒ P × (0.04)2 = 1P × 0.0016 = 1 ⇒ P = 625 Verification/Alternative With P = 625: SI (2 yrs) = 625 × 0.08 = 50; CI = 625[(1.04)2 − 1] = 625(0.0816) = 51; difference = 1. Common pitfalls Applying the identity for periods other than exactly 2 years. Final Answer Rs. 625

Question 16

Difference CI vs SI — The difference between compound interest and simple interest on a sum at 5% p.a. for 2 years is ₹ 1.50. Find the principal.

Accepted Answer

Introduction / Context: Over 2 years, the excess of compound interest (CI) over simple interest (SI) at rate r equals P * r^2. This compact result follows from expanding (1 + r)^2 and comparing with linear SI. Given Data / Assumptions: r = 5% = 0.05 per annum. Time = 2 years. CI − SI = ₹ 1.50. Concept / Approach: For 2 years: CI = P[(1 + r)^2 − 1] = P(2r + r^2); SI = P(2r); thus CI − SI = P * r^2. Step-by-Step Solution: P * r^2 = 1.50 ⇒ P * (0.05)^2 = 1.50.P * 0.0025 = 1.50 ⇒ P = 1.50 / 0.0025 = 600. Verification / Alternative check: At P = 600: r^2 P = 0.0025 * 600 = 1.5, matches difference. Why Other Options Are Wrong: 500, 400, 300, 800 do not satisfy P * 0.0025 = 1.50. Common Pitfalls: Using 2 years formula incorrectly or computing CI − SI for the wrong duration. Final Answer: Rs. 600

Question 17

Compound vs Simple Interest — If the compound interest on a sum for 2 years at 12.5% p.a. is Rs. 170, what is the simple interest for 2 years at the same rate?

Accepted Answer

Introduction / Context: From the given compound interest amount and rate, we can back out the principal. Then we compute the two-year simple interest at the same rate. Given Data / Assumptions: r = 12.5% = 0.125 per annum. Time = 2 years. CI (2 years) = Rs. 170. Concept / Approach: For 2 years: CI = P[(1 + r)^2 − 1]. Solve for P, then SI_2y = P * r * 2. Step-by-Step Solution: (1 + 0.125)^2 − 1 = 1.265625 − 1 = 0.265625.P = 170 / 0.265625 = 640.SI (2 years) = 640 * 0.125 * 2 = Rs. 160. Verification / Alternative check: Check CI: 640 * 0.265625 = 170 (given). Why Other Options Are Wrong: 152.50, 162.50, 150, 155 do not equal P*r*2 for P = 640. Common Pitfalls: Arithmetic slip when squaring 1.125 or converting 12.5% into decimal. Final Answer: Rs. 160

Question 18

Compound Interest — A sum amounts to Rs. 1352 in 2 years at 4% p.a. compound interest (annual). Find the principal.

Accepted Answer

Introduction / Context: With annual compounding, the two-year amount factor is (1 + r)^2. Dividing the amount by this factor gives the principal. Given Data / Assumptions: A = Rs. 1352. r = 4% = 0.04 per annum (annual compounding). t = 2 years. Concept / Approach: A = P * (1 + r)^t ⇒ P = A / (1 + r)^2. Step-by-Step Solution: (1.04)^2 = 1.0816.P = 1352 / 1.0816 = Rs. 1250. Verification / Alternative check: P * 1.0816 = 1250 * 1.0816 = 1352, confirming the result. Why Other Options Are Wrong: 1300, 1260, 1200, 1240 do not satisfy A = P*(1.04)^2. Common Pitfalls: Using SI formula or an incorrect power for t. Final Answer: Rs. 1250

Question 19

Compound Interest — Rs. 800 amounts to Rs. 882 at 5% p.a. compound interest. In how many years does this happen?

Accepted Answer

Introduction / Context: Under annual compounding, the amount after n years is A = P * (1 + r)^n. With P, A, and r known, we solve for n. Given Data / Assumptions: P = Rs. 800. A = Rs. 882. r = 5% = 0.05 per annum. Concept / Approach: (1 + 0.05)^n = 882 / 800 = 1.1025. Recognize 1.1025 = (1.05)^2. Step-by-Step Solution: (1.05)^n = (1.05)^2 ⇒ n = 2 years. Verification / Alternative check: Compute: 800 * 1.1025 = 882. Why Other Options Are Wrong: 1, 3, 4, 5 years would not yield exactly Rs. 882 at 5% p.a. compounding. Common Pitfalls: Applying SI or rounding early instead of recognizing the exact power. Final Answer: 2 years

Question 20

Compound Interest — Find the compound interest on Rs. 2800 for 1.5 years at 10% p.a. (annual compounding; last half-year taken as simple on the amount after 1 year).

Accepted Answer

Introduction / Context: When time is fractional and compounding is annual, compute full-year compounding first, then apply simple interest for the remaining fraction on the amount after the last full year. Given Data / Assumptions: P = Rs. 2800. r = 10% p.a. t = 1.5 years; annual compounding with last 0.5 year at simple on the 1-year amount. Concept / Approach: Amount after 1 year: A1 = P * (1 + r) = 2800 * 1.10. For the next 0.5 year: add A1 * r * 0.5. The compound interest is final amount minus principal. Step-by-Step Solution: A1 = 2800 * 1.10 = 3080.Extra for 0.5 year = 3080 * 0.10 * 0.5 = 154.Final amount = 3080 + 154 = 3234.Compound interest = 3234 − 2800 = Rs. 434. Verification / Alternative check: Equivalent factor: 1.10 * 1.05 = 1.155; 2800 * 1.155 = 3234; CI = 434. Why Other Options Are Wrong: They correspond to misapplied rates, compounding, or using SI for the whole interval. Common Pitfalls: Compounding the half-year on the principal instead of on A1, or mistakenly compounding semi-annually. Final Answer: Rs. 434

Compound Interest Questions