Question 1

Compound vs Simple Interest — Use 2-year difference to recover principal: For a certain principal, the difference between the 2-year compound interest and the 2-year simple interest at 8% per annum is ₹ 32. Find the principal.

Accepted Answer

Introduction / Context: The difference between compound interest (CI) and simple interest (SI) over exactly two years at the same annual rate is a classic identity. It isolates the “extra” interest-on-interest that CI accumulates. This lets us back-solve the principal without separately knowing the amount. Given Data / Assumptions: Annual rate r = 8% = 0.08 Time t = 2 years CI − SI over 2 years = ₹ 32 Principal P unknown Concept / Approach: For two years at rate r (decimal), CI − SI = P * r^2. This follows from CI for 2 years = P[(1 + r)^2 − 1] = P(2r + r^2), SI for 2 years = P(2r), hence the difference is P r^2. Step-by-Step Solution: Use identity: CI − SI = P r^2.Substitute: 32 = P * (0.08)^2 = P * 0.0064.Solve P = 32 / 0.0064 = 5000. Verification / Alternative check: Check numerically: On ₹ 5000, SI(2y, 8%) = 5000 * 0.16 = 800; CI(2y) adds r^2 * P = 0.0064 * 5000 = 32 more ⇒ 832. Difference 32 (matches). Why Other Options Are Wrong: ₹ 5250, ₹ 5550, ₹ 6000, ₹ 4800 do not satisfy 32 = P * 0.0064. Common Pitfalls: Using r as “8” instead of 0.08 in the formula. Applying the identity to durations other than 2 years; it is specific to 2 years in this neat form. Final Answer: ₹ 5000.

Question 2

Compound Interest — Recover principal from total CI over 2 years: At 5% per annum compounded yearly, a sum earns ₹ 410 as compound interest in 2 years. Find the principal.

Accepted Answer

Introduction / Context: When the total compound interest over a fixed number of years is known, and the annual rate is given, we can express that interest directly in terms of the principal and solve for the principal. For 2 years, the growth factor is (1 + r)^2 and the CI equals P[(1 + r)^2 − 1]. Given Data / Assumptions: Annual rate r = 5% = 0.05 Time t = 2 years with annual compounding Total CI over 2 years = ₹ 410 Principal P unknown Concept / Approach: CI(2y) = P[(1 + r)^2 − 1] = P(2r + r^2) = P(0.10 + 0.0025) = 0.1025 P. Set 0.1025 P = 410 and solve for P. Step-by-Step Solution: Compute factor: (1.05)^2 − 1 = 0.1025.Equation: 0.1025 * P = 410.P = 410 / 0.1025 = 4000. Verification / Alternative check: Forward check: On ₹ 4000, CI(2y) = 4000 * 0.1025 = ₹ 410 (matches). Why Other Options Are Wrong: ₹ 8000, ₹ 21000, ₹ 42000, ₹ 4100 do not satisfy 410 = 0.1025 * P. Common Pitfalls: Treating 5% as 5 (forgetting to convert to 0.05). Confusing total amount with interest; here ₹ 410 is interest only. Final Answer: ₹ 4000.

Question 3

Compound Interest — Identify rate from “27 times in 3 years”: At what annual compound interest rate will a sum become 27 times itself in 3 years?

Accepted Answer

Introduction / Context: When a principal multiplies by a known factor over an integer number of years with annual compounding, we can take the matching root of that factor to recover the per-year multiplier. This is a direct application of P(1 + r)^t = final amount. Given Data / Assumptions: Multiplier over 3 years = 27 Annual compounding Find r (per annum) Concept / Approach: If P becomes 27P in 3 years, then (1 + r)^3 = 27. Take the real cube root: 1 + r = 3. Then subtract 1 to isolate r, and express as a percent. Step-by-Step Solution: (1 + r)^3 = 27.1 + r = 27^(1/3) = 3.r = 3 − 1 = 2 ⇒ 200% per annum. Verification / Alternative check: Forward check: (1 + 2)^3 = 3^3 = 27 (matches). Why Other Options Are Wrong: 100%, 150%, 75%, 50% yield three-year multipliers far less than 27 under compounding. Common Pitfalls: Using simple interest logic; compounding is essential for exponential growth like 27× in 3 years. Final Answer: 200% per annum (compounded annually).

Question 4

Compound Interest — Determine time from total CI on a known principal: The compound interest on ₹ 30000 at 7% per annum for a certain time is ₹ 4347. Find the time (compounded yearly).

Accepted Answer

Introduction / Context: Total compound interest equals P[(1 + r)^t − 1]. If P, r, and CI are known, we can solve for t by matching the factor (1 + r)^t to 1 + CI/P. With standard rates and times, this often collapses to recognizing a tabulated power. Given Data / Assumptions: P = ₹ 30000 r = 7% = 0.07 per year CI = ₹ 4347 Annual compounding; find t Concept / Approach: Compute 1 + CI/P = 1 + 4347/30000 = 1 + 0.1449 = 1.1449. Recognize that (1.07)^2 = 1.1449. Therefore, t = 2 years exactly. Step-by-Step Solution: CI = P[(1 + r)^t − 1] ⇒ (1 + r)^t = 1 + CI/P.1 + CI/P = 1 + 4347/30000 = 1.1449.(1.07)^2 = 1.1449 ⇒ t = 2. Verification / Alternative check: Forward check: CI on ₹ 30000 for 2 years = 30000[(1.07)^2 − 1] = 30000 * 0.1449 = ₹ 4347. Why Other Options Are Wrong: Other times do not produce the factor 1.1449 at 7% with annual compounding. Common Pitfalls: Mistaking amount (principal + interest) for “interest only”. Final Answer: 2 years.

Question 5

Compound Frequency Effect — Half-yearly vs quarterly (same nominal rate, 1 year): Find the difference between compound interest on ₹ 800 for 1 year at a nominal 20% p.a. when compounded half-yearly versus quarterly.

Accepted Answer

Introduction / Context: For a fixed nominal annual rate, increasing compounding frequency increases the effective annual yield. Comparing half-yearly (2 times) to quarterly (4 times) over one year highlights the frequency effect while keeping the nominal rate unchanged. Given Data / Assumptions: P = ₹ 800 Nominal rate = 20% p.a. Case H: half-yearly compounding (2 periods at 10% each) Case Q: quarterly compounding (4 periods at 5% each) Time = 1 year in both cases Concept / Approach: Compute amounts separately, then subtract to get the difference in interest. With n compounding periods: A = P (1 + r/n)^(n t). For 1 year, t = 1. Step-by-Step Solution: Half-yearly: A_H = 800 * (1 + 0.20/2)^2 = 800 * (1.10)^2 = 800 * 1.21 = 968; CI_H = 168.Quarterly: A_Q = 800 * (1 + 0.20/4)^4 = 800 * (1.05)^4 ≈ 800 * 1.21550625 ≈ 972.405; CI_Q ≈ 172.405.Difference = CI_Q − CI_H ≈ 172.405 − 168 = ₹ 4.405 ≈ ₹ 4.40. Verification / Alternative check: Recompute with more decimals to confirm ≈ ₹ 4.40. The sign is positive since quarterly compounds more frequently. Why Other Options Are Wrong: Nil, ₹ 2.50, ₹ 6.60, ₹ 3.20 do not match the computed frequency effect at 20% nominal on ₹ 800 over one year. Common Pitfalls: Using 20% for each subperiod instead of dividing by the number of periods. Comparing amounts instead of interest and then forgetting to subtract principal. Final Answer: Rs. 4.40.

Question 6

Compound Interest with Mixed Time — 2 years + 73 days at 6 1/4% p.a.: Find the compound interest on ₹ 20480 at 6 1/4% per annum for 2 years and 73 days (assume 365-day year and annual compounding with simple interest on the fractional year).

Accepted Answer

Introduction / Context: When interest is compounded annually and the time contains whole years plus a fractional part (days), the standard convention is to compound for the whole years, then apply simple interest on the amount for the fractional remainder at the same annual rate (using fraction of a year). This avoids “partial compounding periods.” Given Data / Assumptions: P = ₹ 20480 r = 6 1/4% = 6.25% = 0.0625 Time = 2 years + 73 days; take 73/365 = 0.2 year Annual compounding; SI on fractional part Concept / Approach: Step 1: Compound for 2 whole years: A2 = P(1 + r)^2. Step 2: For the remaining 0.2 year, add simple interest on A2: extra = A2 * r * 0.2. Total amount = A2 + extra. CI equals total amount minus principal. Step-by-Step Solution: A2 = 20480 * (1.0625)^2 = 20480 * 1.12890625 = ₹ 23120.Fractional extra = 23120 * 0.0625 * 0.2 = 23120 * 0.0125 = ₹ 289.Total amount = 23120 + 289 = ₹ 23409 ⇒ CI = 23409 − 20480 = ₹ 2929. Verification / Alternative check: Reasonableness: 6.25% of ~₹ 23k for 0.2 year is ≈ ₹ 287.5, close to exact ₹ 289; totals align. Why Other Options Are Wrong: ₹ 3000, ₹ 3131, ₹ 3636, ₹ 2880 do not match the mixed-period calculation. Common Pitfalls: Compounding the 73-day portion instead of using simple interest after annual periods. Using a 360-day year; the question implies 365 days by specifying 73 days explicitly. Final Answer: Rs. 2929.

Question 7

CI–SI Comparison — Find rate from 2-year difference and SI total: For a sum, the difference between compound and simple interest for 2 years is ₹ 160. The simple interest for 2 years is ₹ 2880. Find the annual rate (percent per annum).

Accepted Answer

Introduction / Context: Two years at the same annual rate produces neat identities: SI(2y) = 2 P r and CI − SI = P r^2, where r is the annual rate in decimal and P is the principal. Having both the 2-year SI and the 2-year difference allows elimination of P to solve directly for r, then convert it to a percent value. Given Data / Assumptions: SI over 2 years = ₹ 2880 ⇒ 2 P r = 2880 CI − SI over 2 years = ₹ 160 ⇒ P r^2 = 160 r is an annual rate (decimal), P > 0 Concept / Approach: Divide the two equations to eliminate P: (P r^2)/(2 P r) = r/2 = 160/2880. This quickly yields r. Finally, express r as a mixed fraction percentage for clarity. Step-by-Step Solution: From 2 P r = 2880, we have P r = 1440.From P r^2 = 160, compute r = (P r^2)/(P r) = 160 / 1440 = 1/9.Therefore r = 1/9 ≈ 0.111… ⇒ 11 1/9% per annum. Verification / Alternative check: Check: With r = 1/9, CI − SI (2y) = P r^2 = P/81. Also, SI(2y) = 2 P r = 2P/9. Ratio (difference)/(SI) = (P/81)/(2P/9) = 1/18 = 160/2880 (consistent). Why Other Options Are Wrong: 9%, 10%, 12 1/2%, 55/9% do not satisfy both equations simultaneously; only 11 1/9% works. Common Pitfalls: Treating r as “11.1” instead of 0.111…; always convert properly between decimal and percent. Final Answer: 11 1/9% per annum.

Question 8

CI vs SI — Use 2-year difference to infer rate, then extend to 3 years: For a sum, CI for 2 years is ₹ 832 while SI for 2 years is ₹ 800. What is the difference (CI − SI) for 3 years at the same annual rate?

Accepted Answer

Introduction / Context: Differences between CI and SI have closed-form expressions: for 2 years, CI − SI = P r^2; for 3 years, CI − SI = P r^2 (3 + r), where r is the annual rate in decimal and P the principal. Given the 2-year CI and SI, we can find r (or P and r) and compute the 3-year difference directly. Given Data / Assumptions: CI(2y) = ₹ 832; SI(2y) = ₹ 800 Thus, CI − SI (2y) = ₹ 32 Annual rate r same for all years Concept / Approach: From SI(2y) = 2 P r = 800 ⇒ P r = 400. From the difference identity, P r^2 = 32. Divide to get r = (P r^2)/(P r) = 32/400 = 0.08 = 8%. Then apply the 3-year formula for the difference: P r^2 (3 + r). Compute P via P = (P r)/r = 400/0.08 = 5000. Step-by-Step Solution: r = 0.08; P = 5000.CI − SI (3y) = P * r^2 * (3 + r) = 5000 * 0.0064 * 3.08 = 5000 * 0.019712 = ₹ 98.56. Verification / Alternative check: Direct calculation with amounts confirms the same extra interest-on-interest over 3 years. Why Other Options Are Wrong: ₹ 48, ₹ 66.56, and other figures do not match P r^2 (3 + r) at r = 8% and P = 5000. Common Pitfalls: Using 2-year identity (P r^2) again for 3 years; the 3-year expression includes the (3 + r) factor. Final Answer: Rs. 98.56.

Question 9

Loan Amortization under Compound Interest — Three equal end-of-year payments: A man borrows ₹ 4000 at 7 1/2% p.a. compound interest. At the end of every year he pays ₹ 1500 (covering interest and part principal). How much does he still owe after thr…

Accepted Answer

Introduction / Context: This is a straightforward amortization with annual compounding: each year the outstanding balance accrues interest, then the borrower pays a fixed amount, reducing the balance. Repeating the process year by year yields the remaining debt after the specified number of payments. Given Data / Assumptions: Initial principal P0 = ₹ 4000 Annual interest rate i = 7.5% = 0.075 End-of-year payment each year = ₹ 1500 Number of payments = 3 Concept / Approach: Iterate: after interest, subtract the payment. Algebraically: B1 = P0(1 + i) − 1500; B2 = B1(1 + i) − 1500; B3 = B2(1 + i) − 1500. The remaining balance after the 3rd payment is owed to the bank. Step-by-Step Solution: After year 1: 4000 * 1.075 = 4300; pay 1500 ⇒ B1 = 2800.After year 2: 2800 * 1.075 = 3010; pay 1500 ⇒ B2 = 1510.After year 3: 1510 * 1.075 = 1623.25; pay 1500 ⇒ B3 = ₹ 123.25. Verification / Alternative check: Annuity-present-value equivalence gives the same residual when reversed; iterative method is clearer here. Why Other Options Are Wrong: ₹ 125 and ₹ 469.18 are near-miss computations typically arising from rounding or misordering “interest then payment”. ₹ 400 and ₹ 0 do not reflect the calculated amortization path. Common Pitfalls: Subtracting the payment before applying interest; the problem states end-of-year payments (interest applies first). Final Answer: Rs. 123.25.

Question 10

Equal Annual Instalments under Compound Interest — Two-year repayment at 20%: A loan of ₹ 550 is to be repaid in two equal annual instalments with 20% compound interest per annum. Find the value of each instalment.

Accepted Answer

Introduction / Context: Equal-instalment repayment under compound interest uses time-value equivalence: the present value of all instalments discounted at the loan rate equals the principal borrowed. For annual instalments over two years at rate i, PV = A/(1 + i) + A/(1 + i)^2. Set this equal to the principal and solve for A (the instalment size). Given Data / Assumptions: Principal P = ₹ 550 Annual rate i = 20% = 0.20 Two equal end-of-year instalments = A each Concept / Approach: Present value equation: 550 = A/(1.2) + A/(1.2)^2. Compute the discount sum, then isolate A. This avoids stepwise amortization while yielding the exact same result. Step-by-Step Solution: Compute discount factors: 1/1.2 ≈ 0.833333..., 1/1.2^2 ≈ 0.694444...Sum ≈ 0.833333... + 0.694444... = 1.527777...Solve A = 550 / 1.527777... = ₹ 360 (exact). Verification / Alternative check: Amortization route gives same A: After first year, 550*1.2 − 360 = 300; after second, 300*1.2 − 360 = 0. Why Other Options Are Wrong: ₹ 421, ₹ 396, ₹ 350, ₹ 380 do not satisfy the PV equality at 20% for two periods. Common Pitfalls: Discounting at 20 (instead of 0.20) or forgetting the second discount. Final Answer: Rs. 360.

Question 11

CI vs SI on Same Sum and Rate — Recover principal from 2-year totals: The simple interest on a certain sum for 2 years is ₹ 40, while the compound interest on the same sum for the same rate and 2 years is ₹ 45. Find the principal.

Accepted Answer

Introduction / Context: For two years at the same annual rate r (decimal), SI(2y) = 2 P r, and CI(2y) = 2 P r + P r^2. The difference isolates the interest-on-interest term P r^2, letting us solve for r and/or P using the two totals provided. Given Data / Assumptions: SI over 2 years = ₹ 40 ⇒ 2 P r = 40 CI over 2 years = ₹ 45 ⇒ 2 P r + P r^2 = 45 Same sum and same rate for both measures Concept / Approach: Subtract to get P r^2 = 5. From 2 P r = 40 infer P r = 20. Then r = (P r^2)/(P r) = 5 / 20 = 0.25 (25%). Finally, compute P = (P r)/r = 20 / 0.25 = 80. Step-by-Step Solution: P r = 20; P r^2 = 5; r = 0.25.Principal P = 20 / 0.25 = ₹ 80.Check: SI = 2 * 80 * 0.25 = ₹ 40; CI = SI + P r^2 = 40 + 5 = ₹ 45. Verification / Alternative check: Forward compute amounts to confirm interest breakdowns match the totals. Why Other Options Are Wrong: ₹ 70, ₹ 85, ₹ 90, ₹ 75 do not simultaneously satisfy SI = 40 and CI = 45 at one common rate. Common Pitfalls: Mistaking the 5 difference as the rate itself; it represents P r^2, not r. Final Answer: Rs. 80.

Question 12

Compound Interest on Periodic Savings — Annuity due (deposits at the beginning of each year): Neeraj saves ₹ 400 at the beginning of each year and lends each deposit at 5% per annum, compounded yearly. What will his savings be worth at the end of 3 …

Accepted Answer

Introduction / Context: When equal deposits are made every period and earn compound interest, the situation is an annuity. If deposits occur at the beginning of each period, it is an annuity due; each payment earns one extra period of interest compared with end-of-period deposits. The future value (FV) of an annuity due at interest rate i for n years with payment A is FV = A * [(1 + i) + (1 + i)^2 + ... + (1 + i)^n]. Given Data / Assumptions: Payment each year A = ₹ 400 Interest rate i = 5% = 0.05 per annum Number of years n = 3 Deposits at the beginning of each year (annuity due); evaluate value at end of year 3 Concept / Approach: The k-th deposit (counting from 1) compounds for (n − k + 1) years to the end. Thus FV = 400 * [(1.05)^3 + (1.05)^2 + (1.05)]. This is equivalent to the annuity-due formula FV = A * [((1 + i)^n − 1)/i] * (1 + i). Step-by-Step Solution: Compute powers: (1.05)^3 = 1.157625; (1.05)^2 = 1.1025; (1.05) = 1.05.Sum the factors: S = 1.157625 + 1.1025 + 1.05 = 3.310125.Future value: FV = 400 * 3.310125 = ₹ 1324.05. Verification / Alternative check: Using annuity-due formula: FV = 400 * [((1.05)^3 − 1)/0.05] * 1.05 = 400 * (0.157625/0.05) * 1.05 = 400 * 3.1525 * 1.05 = ₹ 1324.05 (matches). Why Other Options Are Wrong: ₹ 1261.00 corresponds to end-of-year deposits (ordinary annuity), which is not the stated timing here. ₹ 1312.50, ₹ 1284, ₹ 1315 are near-miss values from rounding or mixed timing assumptions. Common Pitfalls: Confusing beginning vs end of year deposits; timing changes the compounding periods. Rounding intermediate steps too early. Final Answer: Rs. 1324.05.

Question 13

Compound Growth Factor Reasoning — From 3× in 3 years to 9× total: If a sum at compound interest becomes 3 times itself in 3 years, in how many years will it become 9 times at the same annual rate?

Accepted Answer

Introduction / Context: Compound growth over whole years follows Amount = Principal * (1 + r)^t. If we know the multiplier for a certain time, we can scale to other targets using exponents rather than recomputing rates explicitly. Given Data / Assumptions: (1 + r)^3 = 3 (triples in 3 years) Target: (1 + r)^t = 9 Same annual rate r; find t Concept / Approach: Observe that 9 = 3^2. Since (1 + r)^3 = 3, taking powers yields (1 + r)^6 = 3^2 = 9. Therefore, t = 6 years gives the ninefold increase. Step-by-Step Solution: Given: (1 + r)^3 = 3.Square both sides: (1 + r)^6 = 3^2 = 9.Hence, t = 6 years. Verification / Alternative check: After 3 years: 3×; after another 3 years: multiply by 3 again ⇒ 9× total. Why Other Options Are Wrong: 5, 7, 9, 12 years do not align with exponent scaling from a fixed 3-year triple. Common Pitfalls: Using simple interest logic; compounding multiplies factors, not adds linearly. Final Answer: 6 years.

Question 14

Compare SI vs CI — Extra interest earned at 6% over 3 years: A person invests at 6% per annum. The simple interest over 3 years is ₹ 900. If interest were compounded annually at the same rate for 3 years on the same principal, how much more interest…

Accepted Answer

Introduction / Context: Simple interest (SI) grows linearly with time, whereas compound interest (CI) includes interest on previously earned interest. The “extra” comes from this compounding effect. With the same rate and time, comparing SI and CI on the same principal cleanly quantifies that extra amount. Given Data / Assumptions: SI over 3 years = ₹ 900 at 6% p.a. Thus P * 0.06 * 3 = 900 ⇒ P = 900 / 0.18 = ₹ 5000 CI rate r = 6% per annum, t = 3 years (annual compounding) Concept / Approach: Compute CI over 3 years: CI = P[(1 + r)^3 − 1]. The extra interest = CI − SI. Plug P = 5000, r = 0.06, t = 3 to get the numeric difference. Step-by-Step Solution: Compute growth factor: (1.06)^3 = 1.191016.CI = 5000 * (1.191016 − 1) = 5000 * 0.191016 = ₹ 955.08.SI = ₹ 900 ⇒ Extra = 955.08 − 900 = ₹ 55.08. Verification / Alternative check: Direct term-by-term compounding confirms the same result; differences arise from interest on prior interest. Why Other Options Are Wrong: ₹ 38.13, ₹ 25.33, ₹ 35.30, ₹ 45.00 do not match the computed extra at 6% over 3 years on ₹ 5000. Common Pitfalls: Using 6 as 6% (0.06) confusion; always convert to decimal. Forgetting to subtract SI from CI to isolate the “extra”. Final Answer: Rs. 55.08.

Question 15

Find Principal from CI vs SI with Semiannual Compounding — 2-year horizon at 10% p.a. (half-yearly compounding): The difference between the compound interest (compounded every 6 months) and simple interest at 10% per annum over 2 years is ₹ 124.05. …

Accepted Answer

Introduction / Context: When compounding frequency increases (semiannual here), the effective amount exceeds simple interest by a calculable margin. If this margin is given for a known horizon, we can back out the principal by comparing the two amount factors over the same time span. Given Data / Assumptions: Nominal annual rate = 10%. Compounded half-yearly ⇒ per half-year rate = 10%/2 = 5%, number of periods over 2 years = 4. Simple interest over 2 years amount factor = 1 + rt = 1 + 0.10*2 = 1.20. Difference in interest (CI − SI) over 2 years = ₹ 124.05. Concept / Approach: Amount with semiannual compounding: A_CI = P(1 + 0.05)^4. Amount with SI: A_SI = P(1.20). The difference in interest equals P[(1.05)^4 − 1] − P(0.20) = P[(1.05)^4 − 1.20]. Solve P from the given difference. Step-by-Step Solution: Compute (1.05)^4 = 1.21550625.Difference factor = 1.21550625 − 1.20 = 0.01550625.Given difference = 124.05 = P * 0.01550625 ⇒ P = 124.05 / 0.01550625 = ₹ 8000. Verification / Alternative check: Check: 8000 * 0.01550625 = 124.05 (exact). Why Other Options Are Wrong: ₹ 10000, ₹ 6000, ₹ 12000, ₹ 9000 do not satisfy the exact difference factor at 10% with semiannual compounding for 2 years. Common Pitfalls: Using 10% per half-year instead of 5%. Subtracting principals instead of interest portions. Final Answer: Rs. 8000.

Question 16

Hire Purchase / Instalment Plan under CI — Recover equal instalment: A refrigerator costs ₹ 7044 (cash price). A customer pays ₹ 2000 in cash and finances the balance via 3 equal annual instalments at 5% per annum compound interest. What is each ins…

Accepted Answer

Introduction / Context: For equal instalments under compound interest, the present value (PV) of all instalments discounted at the contract rate equals the financed amount. This avoids iterative amortization and yields the exact instalment size directly. Given Data / Assumptions: Cash price = ₹ 7044; down payment = ₹ 2000 ⇒ financed amount = ₹ 5044. Annual rate i = 5% = 0.05. Number of annual instalments n = 3 (end-of-year). Concept / Approach: Let each instalment be A. Present value at time 0: PV = A/(1.05) + A/(1.05)^2 + A/(1.05)^3. Set PV = 5044 and solve A = 5044 / [v + v^2 + v^3], where v = 1/1.05. Step-by-Step Solution: v = 1/1.05 ≈ 0.95238095; v^2 ≈ 0.90702948; v^3 ≈ 0.86383760.Sum S ≈ 0.95238095 + 0.90702948 + 0.86383760 ≈ 2.72324803.A = 5044 / 2.72324803 ≈ ₹ 1852.20. Verification / Alternative check: Forward check by compounding the financed balance and subtracting amortized portions confirms the same equal instalment. Why Other Options Are Wrong: ₹ 1868.28 and ₹ 1865 are rounding/alt-frequency artifacts. ₹ 1752, ₹ 1900 do not satisfy PV equality at 5% over 3 years. Common Pitfalls: Using simple interest discounting instead of compound discounting. Omitting one of the instalment discount terms. Final Answer: Rs. 1852.20.

Question 17

Compound Interest on Loan at “2.5 paise per rupee per annum” — 3-year horizon: Vijay obtains a loan of ₹ 64000. The rate is 2.5 paise per rupee per annum (i.e., 2.5% p.a.) compounded yearly. Find the compound interest payable after 3 years.

Accepted Answer

Introduction / Context: Rates sometimes appear as “paise per rupee per annum.” Since 100 paise = 1 rupee, 2.5 paise per rupee equals 2.5/100 = 0.025 = 2.5% per annum. Use standard compound interest with annual compounding to compute the total interest for the given horizon. Given Data / Assumptions: Principal P = ₹ 64000 Annual rate r = 2.5% = 0.025 Time t = 3 years; annual compounding Concept / Approach: Amount after 3 years: A = P(1 + r)^3. The compound interest (CI) equals A − P = P[(1 + r)^3 − 1]. Plug in r = 0.025 and compute the factor precisely. Step-by-Step Solution: (1.025)^3 = 1.076890625.Interest factor = 1.076890625 − 1 = 0.076890625.CI = 64000 * 0.076890625 = ₹ 4921 (exact to the rupee). Verification / Alternative check: Compute A = 64000 * 1.076890625 = ₹ 68921.00; A − P = 4921.00. Why Other Options Are Wrong: ₹ 4920 is rounding-down; ₹ 5020, ₹ 4821, ₹ 5000 do not match exact compounding. Common Pitfalls: Interpreting 2.5 paise per rupee as 2.5% points incorrectly. Final Answer: Rs. 4921.

Question 18

Target-Equality under CI — Split ₹ 3903 so that A after 7 years equals B after 9 years (4% p.a.): Divide ₹ 3903 between A and B such that A's amount after 7 years at 4% compound equals B's amount after 9 years at the same rate. Find the two shares.

Accepted Answer

Introduction / Context: When two parts of a sum will be left to grow for different durations at the same compound rate, equalizing their future amounts defines a proportional split today. This is a typical present-value balancing problem under compound interest. Given Data / Assumptions: Total sum S = ₹ 3903 Annual rate = 4% = 0.04 A grows for 7 years; B grows for 9 years Condition: A's future amount = B's future amount Concept / Approach: Let A's share be x; B's share is 3903 − x. Equality at maturity: x*(1.04)^7 = (3903 − x)*(1.04)^9. Divide both sides by (1.04)^7 to solve for x. Step-by-Step Solution: x = (3903 − x)*(1.04)^2 = (3903 − x)*1.0816.x + 1.0816x = 3903 * 1.0816 ⇒ 2.0816x = 4221.4848.x = 4221.4848 / 2.0816 = ₹ 2028 ⇒ B = 3903 − 2028 = ₹ 1875. Verification / Alternative check: Forward check: 2028*(1.04)^7 equals 1875*(1.04)^9 (same value). Why Other Options Are Wrong: All other splits do not satisfy x*(1.04)^7 = (3903 − x)*(1.04)^9. Common Pitfalls: Forgetting to divide out common growth before solving; mixing SI and CI. Final Answer: Rs. 2028 for A and Rs. 1875 for B.

Question 19

Arbitrage of SI vs CI at same nominal rate — Net gain over 3 years: Akash borrows ₹ 65000 at 10% per annum simple interest for 3 years and lends the same principal at 10% per annum compound interest for 3 years. What is his gain after 3 years?

Accepted Answer

Introduction / Context: Borrowing at simple interest and lending at compound interest at the same nominal rate creates a spread due to interest-on-interest on the lending side. The net gain equals CI received minus SI paid on the identical principal over the same time horizon. Given Data / Assumptions: Principal P = ₹ 65000 SI outflow: r = 10% = 0.10, t = 3 years ⇒ SI_paid = P * r * t. CI inflow: same r and t with annual compounding ⇒ CI_received = P[(1 + r)^t − 1]. Concept / Approach: Compute both interest totals and subtract: Gain = CI_received − SI_paid. This isolates the compounding advantage over 3 years at 10%. Step-by-Step Solution: SI_paid = 65000 * 0.10 * 3 = ₹ 19500.(1.10)^3 = 1.331 ⇒ CI_received = 65000 * (1.331 − 1) = 65000 * 0.331 = ₹ 21515.Gain = 21515 − 19500 = ₹ 2015. Verification / Alternative check: Direct computation of amounts confirms the same spread due purely to compounding. Why Other Options Are Wrong: ₹ 1330, ₹ 1300 underestimate the compounding effect; “None of these” is incorrect. Common Pitfalls: Forgetting that SI does not compound; only CI side gains interest on interest. Final Answer: ₹ 2015.

Question 20

Compound vs Simple Interest — A father divides ₹ 25,000 between his two sons A and B. A invests his share at 8% per annum compounded annually, and B invests his share at 10% per annum simple interest. After 2 years, B's interest exceeds A's interest…

Accepted Answer

Introduction / Context: This problem mixes compound interest (for A) and simple interest (for B) for the same 2-year period. We are told how much more interest B earns than A, which lets us set up a single linear equation in A's share. Given Data / Assumptions: Total property = ₹ 25,000. A invests at 8% p.a. compounded annually for 2 years. B invests at 10% p.a. simple interest for 2 years. After 2 years, Interest(B) − Interest(A) = ₹ 1,336. Concept / Approach: Let A's share be x ⇒ B's share = 25,000 − x. A's 2-year compound interest on x is x * [(1.08)^2 − 1] = 0.1664x. B's 2-year simple interest on (25,000 − x) is (25,000 − x) * 0.10 * 2 = 5,000 − 0.20x. Their difference is given. Step-by-Step Solution: (5,000 − 0.20x) − 0.1664x = 1,3365,000 − 0.3664x = 1,3360.3664x = 3,664 ⇒ x = 10,000 Verification / Alternative check: Interest(A) = 0.1664 * 10,000 = ₹ 1,664. Interest(B) = 0.20 * 15,000 = ₹ 3,000. Difference = ₹ 1,336 ✔ Why Other Options Are Wrong: Any value other than ₹ 10,000 fails the linear relation and cannot produce a difference of ₹ 1,336 over 2 years with the stated rates. Common Pitfalls: Applying simple interest to A's share (instead of compound) or compounding B's share by mistake; both lead to incorrect equations. Final Answer: ₹ 10,000

Compound Interest Questions