Question 1

Compound Interest – Half-yearly compounding in one year: Raja invests ₹ 15,000 at 10% per annum, compounded half-yearly. What is the amount at the end of 1 year?

Accepted Answer

Introduction / Context:With half-yearly compounding, the annual rate is split into two periods. Each half-year applies r/2, and there are 2 periods in one full year. Given Data / Assumptions: P = ₹ 15,000 Nominal annual rate = 10% Compounding = semiannual → rate per period = 10%/2 = 5% Number of periods in 1 year = 2 Concept / Approach:Amount A = P * (1 + r/m)^(m*t), where m = 2 for half-yearly. Step-by-Step Solution:A = 15000 * (1 + 0.10/2)^(2*1)A = 15000 * (1.05)^2 = 15000 * 1.1025A = ₹ 16,537.50 Verification / Alternative check:Sequential: After 6 months = 15000 * 1.05 = 15750; after 12 months = 15750 * 1.05 = 16537.50. Why Other Options Are Wrong:₹ 18,000, ₹ 19,000.50, and ₹ 20,000 assume larger rates or extra periods. Common Pitfalls:Applying 10% once (simple interest) or missing the second compounding within the year. Final Answer:₹ 16537.50

Question 2

Compound Interest – Quarterly compounding with fractional years: Find the amount on ₹ 15,000 at 12% per annum compounded quarterly for 1¼ years (i.e., 5 quarters).

Accepted Answer

Introduction / Context:When compounding is quarterly, we convert the time into quarters and the rate into a quarterly rate. Fractional years are handled by counting exact compounding periods. Given Data / Assumptions: P = ₹ 15,000 Annual r = 12% → per quarter = 12%/4 = 3% Time = 1.25 years = 5 quarters Concept / Approach:Amount A = P * (1 + r/4)^(4t) = P * (1.03)^5. Step-by-Step Solution:A = 15000 * (1.03)^5(1.03)^5 ≈ 1.159274…A ≈ 15000 * 1.159274… = ₹ 17,389.12 (rounded to paise) Verification / Alternative check:Step compounding per quarter produces the same result within rounding tolerance. Why Other Options Are Wrong:Rs 16,596.88 and Rs 16,789.08 are too low; Rs 17,630.77 is too high for 5 quarters at 3%. Common Pitfalls:Using 1.25 as “quarters” directly, or applying 12% for 1.25 years without dividing by 4. Final Answer:Rs 17389.12

Question 3

Compound Interest – Find time from interest earned: On ₹ 30,000 at 7% per annum compound interest, the total compound interest is ₹ 4,347. Find the time in years (annual compounding).

Accepted Answer

Introduction / Context:Given principal, rate, and compound interest earned, we can infer the time by comparing amount factors (1 + r)^t. Given Data / Assumptions: P = ₹ 30,000 r = 7% per annum Compound interest (CI) = ₹ 4,347 Compounding yearly Concept / Approach:CI = A − P, so A = P + CI = 304347. Compare A/P to (1 + r)^t to solve for t. Step-by-Step Solution:A = 30000 + 4347 = 34347A/P = 34347 / 30000 = 1.1449(1 + 0.07)^2 = 1.1449 → t = 2 years Verification / Alternative check:Direct CI for 2 years: P[(1.07)^2 − 1] = 30000(1.1449 − 1) = 30000 * 0.1449 = 4347 (matches). Why Other Options Are Wrong:2.5, 3, 4 years give factors higher than 1.1449 at 7%. Common Pitfalls:Using simple interest 2 * 7% = 14% (which would give 4200, not 4347). Final Answer:2 yr

Question 4

Compound vs Simple Interest – Difference over 2 years: For ₹ 1,250 at 4% per annum, what is the difference between compound interest (annual compounding) and simple interest for 2 years?

Accepted Answer

Introduction / Context:For 2 years, the excess of CI over SI equals P * (r^2)/10000 (with r in percent), because CI includes “interest on interest” for the second year. Given Data / Assumptions: P = ₹ 1,250 r = 4% per annum t = 2 years Concept / Approach:Difference (CI − SI) for 2 years: P * r^2 / 10000. Step-by-Step Solution:Difference = 1250 * (4^2) / 10000= 1250 * 16 / 10000 = 1250 * 0.0016= ₹ 2.00 Verification / Alternative check:Compute SI = 1250 * 4% * 2 = ₹ 100; CI for 2 years at 4% = 1250[(1.04)^2 − 1] = 1250(0.0816) = ₹ 102; difference = ₹ 2. Why Other Options Are Wrong:₹ 3, ₹ 4, ₹ 8 do not match the exact formula outcome. Common Pitfalls:Using P * r / 100 again instead of r^2/10000 for the difference or mixing percent and decimal. Final Answer:₹ 2

Question 5

Compound vs Simple Interest – Difference over 3 years: Find the difference between compound interest and simple interest on ₹ 10,000 for 3 years at 5% per annum (annual compounding).

Accepted Answer

Introduction / Context:Over 3 years, CI grows as P(1 + r)^3 − P, while SI is P * r * t. Their difference quantifies the extra “interest on interest”. Given Data / Assumptions: P = ₹ 10,000 r = 5% per annum t = 3 years Concept / Approach:Compute SI and CI separately, then subtract: (CI − SI) = P[(1 + r)^3 − 1] − P * r * t. Step-by-Step Solution:SI = 10000 * 0.05 * 3 = ₹ 1,500CI = 10000[(1.05)^3 − 1] = 10000(1.157625 − 1) = ₹ 1,576.25Difference = 1576.25 − 1500 = ₹ 76.25 Verification / Alternative check:Year-wise addition confirms the same excess due to compounding in years 2 and 3. Why Other Options Are Wrong:₹ 76, ₹ 76.50, ₹ 76.75 are near but not exact. Common Pitfalls:Using the 2-year shortcut P * r^2 / 10000 (only valid for exactly 2 years). Final Answer:₹ 76.25

Question 6

Compound Interest – Present value of equal annual instalments: A borrowed sum is repaid in two equal annual instalments of ₹ 121 each at 10% per annum compound interest. What was the sum borrowed (present value)?

Accepted Answer

Introduction / Context:When a loan is repaid by equal instalments under compound interest, the principal equals the present value (PV) of all future instalments discounted at the loan rate. Given Data / Assumptions: Two instalments: ₹ 121 at the end of Year 1 and Year 2 Annual rate = 10% (discount rate for PV) Concept / Approach:PV = 121/(1.10) + 121/(1.10)^2. This sums the present values of each instalment to today (time 0). Step-by-Step Solution:PV = 121/1.10 + 121/1.21= 121(0.909090… + 0.826446…)= 121 * 1.735537… = ₹ 210 (exact) Verification / Alternative check:Forward check: Borrow ₹ 210. After 1 year it becomes 210 * 1.10 = 231; pay 121 → 110 remains. After 2nd year: 110 * 1.10 = 121; pay 121 → 0 outstanding. Why Other Options Are Wrong:₹ 200 and ₹ 217.80/₹ 220 do not satisfy the PV equality at 10%. Common Pitfalls:Discounting at 10% once for both instalments or forgetting to discount the second instalment twice. Final Answer:₹ 210

Question 7

Compound Interest – Amount after half the time: If ₹ 3,000 becomes ₹ 4,329 at compound interest in a certain time, what will ₹ 3,000 amount to in half of that time (same rate, same compounding)?

Accepted Answer

Introduction / Context:Compound growth over time is multiplicative. Knowing the full-period amount factor allows us to take the square root to get the half-period factor (for equal subperiods). Given Data / Assumptions: P = ₹ 3,000 Amount in full time = ₹ 4,329 → growth factor G = 4329/3000 = 1.443 Half the time → factor = √G Concept / Approach:For equal halves of time at the same rate, the mid-time amount is P * √G. Step-by-Step Solution:√1.443 ≈ 1.20125Amount at half time ≈ 3000 * 1.20125 = ₹ 3,603.75Closest listed option = ₹ 3,600 Verification / Alternative check:Squaring 1.20125 gives ≈ 1.443 (returns the full-period factor as required). Why Other Options Are Wrong:₹ 3,520 and ₹ 3,400 are too low; ₹ 38,000 is clearly a typo/misfit. Common Pitfalls:Halving the rupee interest from the full period instead of taking the square root of the factor. Final Answer:₹ 3600

Question 8

Population Growth – Reverse compounding over 3 years: A city’s population grows at 5% per annum. If the present population is 185,220, what was the population 3 years ago?

Accepted Answer

Introduction / Context:Population growth under a fixed yearly percentage is a compound process. To find a past population, we divide by the forward growth factor (1 + r)^t. Given Data / Assumptions: Current population = 185,220 Growth rate r = 5% per annum Years back t = 3 Concept / Approach:Past population = Present / (1 + r)^t. Step-by-Step Solution:Growth factor over 3 years = (1.05)^3 = 1.157625Past population = 185220 / 1.157625= 160,000 (exact match) Verification / Alternative check:Forward check: 160000 * (1.05)^3 = 160000 * 1.157625 = 185,220 (given). Why Other Options Are Wrong:181,500 and 183,433 are too close to the present; 127,783 is far too low. Common Pitfalls:Subtracting 5% of the present thrice (a simple, not compound, approach). Final Answer:160000

Question 9

Compound Interest – Effect of increasing the rate: ₹ 400 amounts to ₹ 441 in 2 years at some rate r (annual compounding). If the rate is increased by 5% per annum, what will be the new amount after 2 years?

Accepted Answer

Introduction / Context:We first infer the original rate from the known amount, then add 5% to that rate and recompute the 2-year amount. Given Data / Assumptions: 400 grows to 441 in 2 years (1 + r)^2 = 441/400 = 1.1025 → 1 + r = 1.05 → r = 5% New rate = 5% + 5% = 10% Concept / Approach:Amount with new rate: A_new = 400 * (1.10)^2. Step-by-Step Solution:A_new = 400 * 1.21 = ₹ 484 Verification / Alternative check:Original A = 400 * 1.1025 = 441 (checks), so r = 5% is correct before adding 5%. Why Other Options Are Wrong:₹ 560, ₹ 512, ₹ 600 correspond to higher rates or extra years. Common Pitfalls:Adding 5% to the amount instead of the rate, or applying simple interest. Final Answer:₹ 484

Question 10

Depreciation – Successive annual reduction: An LCD TV costing ₹ 1,00,000 depreciates at 10% per annum. What will be its value after 3 years (annual depreciation compounded)?

Accepted Answer

Introduction / Context:Depreciation at a fixed percentage each year mirrors compound decrease. Each year’s value is multiplied by (1 − d), where d is the depreciation rate. Given Data / Assumptions: Initial value = ₹ 1,00,000 d = 10% per annum t = 3 years Concept / Approach:Value after t years: V = 100000 * (0.90)^t. Step-by-Step Solution:V = 100000 * (0.9)^3= 100000 * 0.729 = ₹ 72,900 Verification / Alternative check:Year-wise: 100000 → 90000 → 81000 → 72900. Why Other Options Are Wrong:₹ 80,000/₹ 90,000 imply single-year reductions; ₹ 85,000 is not a valid 3-year compounding outcome at 10%. Common Pitfalls:Subtracting 10% of the original each year instead of 10% of the current value. Final Answer:₹ 72900

Question 11

Compound Interest – Repeated periods from a known multiplier: A sum becomes 2½ times in 6 years at compound interest. What multiple of the principal will it become in 18 years at the same rate?

Accepted Answer

Introduction / Context:Under compound interest, equal time blocks multiply the amount by the same factor each time. If 6 years multiplies by 2.5, then 18 years (three such blocks) multiplies by (2.5)^3. Given Data / Assumptions: Multiplier over 6 years M = 2.5 18 years = 3 × 6 years Concept / Approach:Overall multiplier over 18 years = M^3 = (2.5)^3 = 15.625. Step-by-Step Solution:(2.5)^2 = 6.25(2.5)^3 = 6.25 * 2.5 = 15.62515.625 = 125/8 Verification / Alternative check:Express 2.5 as 5/2; (5/2)^3 = 125/8 (same result). Why Other Options Are Wrong:5/2 and 25/4 reflect 1 and 2 blocks; 625/16 = (2.5)^4 (too many blocks). Common Pitfalls:Adding multipliers linearly instead of multiplying them across equal periods. Final Answer:125/8

Question 12

Compound vs Simple Interest – Given difference to find SI: On a certain principal, the difference between compound interest and simple interest at 5% per annum for 2 years is ₹ 16. Find the simple interest for 2 years.

Accepted Answer

Introduction / Context:For exactly 2 years, CI − SI = P * r^2 / 10000 (r in percent). Using this, we recover P, then compute SI = P * r * t / 100. Given Data / Assumptions: CI − SI = ₹ 16 r = 5% t = 2 years Concept / Approach:16 = P * 5^2 / 10000 = P * 25 / 10000 = 0.0025 P → P = 6400. SI = P * r * t / 100. Step-by-Step Solution:P = 16 / 0.0025 = 6400SI = 6400 * 5 * 2 / 100 = 6400 * 0.10 = ₹ 640 Verification / Alternative check:CI for 2 years at 5% on 6400 is 640 + 16 = 656; difference is 16, as given. Why Other Options Are Wrong:₹ 520, ₹ 840, ₹ 932 arise from using wrong r or t. Common Pitfalls:Applying the 2-year shortcut to other durations, or mixing percent with decimal. Final Answer:₹ 640

Question 13

Compound vs Simple Interest – Difference given to find principal: At 8% per annum, compounded annually, the compound interest exceeds the simple interest by ₹ 6.40 over 2 years. Find the principal.

Accepted Answer

Introduction / Context:For 2 years, CI − SI = P * r^2 / 10000. Knowing r and the difference, we can solve directly for P and then sanity-check. Given Data / Assumptions: CI − SI = ₹ 6.40 r = 8% per annum t = 2 years Concept / Approach:6.40 = P * (8^2) / 10000 = P * 64 / 10000 = 0.0064 P → P = 6.40 / 0.0064. Step-by-Step Solution:P = 1000(Optional) SI = 1000 * 8% * 2 = ₹ 160; CI = 160 + 6.40 = ₹ 166.40 Verification / Alternative check:Compute CI via A = 1000(1.08)^2 − 1000 = 1000(1.1664 − 1) = 166.40; difference with SI = 6.40. Why Other Options Are Wrong:₹ 800 and ₹ 1800 give differences 5.12 and 11.52, not ₹ 6.40; ₹ 10,000 is off by a factor of 10. Common Pitfalls:Using 2 * 8% = 16% of P as the difference (that is SI, not the CI-SI gap). Final Answer:₹ 1000

Question 14

Compound Interest – Rate from consecutive-year amounts: A sum amounts to ₹ 12,960 in 2 years and to ₹ 13,176 in 3 years at the same compound rate (annual). Find the rate per cent per annum.

Accepted Answer

Introduction / Context:Successive compound amounts differ by a factor of (1 + r). Thus, the ratio of 3-year to 2-year amounts immediately yields 1 + r. Given Data / Assumptions: A(2 years) = ₹ 12,960 A(3 years) = ₹ 13,176 Compounded annually Concept / Approach:(1 + r) = A3 / A2. Step-by-Step Solution:1 + r = 13176 / 12960 = 1.016666…r = 0.016666… = 1 2/3 % Verification / Alternative check:Check: A3 = A2 * (1 + r) = 12960 * 1.016666… = 13176 (given). Why Other Options Are Wrong:3 2/3 %, 4 %, and 7 % are larger than the ratio implies. Common Pitfalls:Computing r from (A3 − A2)/P instead of from A3/A2. Final Answer:1 2/3 %

Question 15

Simple vs Compound Interest – Given SI and CI to find the rate: On a certain sum, the simple interest for 2 years is ₹ 120, while the compound interest for the same period is ₹ 129. Find the annual rate of interest.

Accepted Answer

Introduction / Context:Relating SI and CI over 2 years allows solving for the rate using two equations: SI = P * r * t / 100, and CI = SI + P * r^2 / 10000. Given Data / Assumptions: SI (2 years) = ₹ 120 CI (2 years) = ₹ 129 t = 2 years Concept / Approach:Let P be principal and r the rate (% per annum). From SI: 2Pr/100 = 120 → Pr = 6000. From CI − SI: P * r^2 / 10000 = 9. Step-by-Step Solution:Pr = 6000 → r = 6000 / PP * r^2 / 10000 = 9 → (P * (6000/P)^2) / 10000 = 936,000,000 / (10000 * P) = 9 → 3600 / P = 9 → P = 400r = 6000 / 400 = 15% per annum Verification / Alternative check:At P = 400, SI = 400 * 15% * 2 = 120; CI − SI = 400 * 15^2 / 10000 = 9 → CI = 129 (matches). Why Other Options Are Wrong:12% and 14% do not satisfy both SI and CI simultaneously. Common Pitfalls:Forgetting the CI − SI relation for 2 years or mixing percent and decimal. Final Answer:15%

Question 16

Growth Factor – Annual increase by one-eighth: A tree’s height increases each year by 1/8 of its current height (i.e., multiplied by 9/8 annually). If its present height is 64 cm, what will be the height after 2 years?

Accepted Answer

Introduction / Context:This is a compound growth problem with multiplier 1 + 1/8 = 9/8 per year. Two years apply the multiplier twice. Given Data / Assumptions: Present height H0 = 64 cm Annual multiplier = 9/8 Years = 2 Concept / Approach:Future height H = H0 * (9/8)^2. Step-by-Step Solution:(9/8)^2 = 81/64H = 64 * 81/64 = 81 cm Verification / Alternative check:Year 1: 64 * 9/8 = 72. Year 2: 72 * 9/8 = 81. Why Other Options Are Wrong:76, 80, 84 cm arise from simple (non-compound) additions or wrong multipliers. Common Pitfalls:Adding 1/8 of the original each year instead of 1/8 of the current height. Final Answer:81 cm

Question 17

Compound vs Simple Interest – Extra ₹ 10 after 2 years at 5%: At 5% per annum, the amount at compound interest exceeds that at simple interest by ₹ 10 after 2 years. Find the principal.

Accepted Answer

Introduction / Context:The 2-year CI–SI shortcut gives a direct route to the principal when the excess is known. Given Data / Assumptions: CI − SI = ₹ 10 r = 5% per annum t = 2 years Concept / Approach:CI − SI over 2 years = P * r^2 / 10000. Solve for P. Step-by-Step Solution:10 = P * 25 / 10000 → P = 10 / 0.0025P = ₹ 4,000 Verification / Alternative check:SI = 4000 * 5% * 2 = 400; CI = 4000[(1.05)^2 − 1] = 400 + 10 = 410; difference = 10. Why Other Options Are Wrong:₹ 4050/₹ 5000/₹ 5500 yield excesses different from ₹ 10 at 5% for 2 years. Common Pitfalls:Forgetting the square on r in the 2-year CI–SI formula. Final Answer:₹ 4000

Question 18

Compound Interest – Interest earned in the second year: What principal will generate ₹ 208 as compound interest in the second year at 4% per annum (annual compounding)?

Accepted Answer

Introduction / Context:Under annual compounding, the interest in the second year equals the first-year amount times r, i.e., P * (1 + r) * r. This lets us back-solve for P from the known second-year interest. Given Data / Assumptions: Second-year interest = ₹ 208 r = 4% = 0.04 Compounded annually Concept / Approach:Second-year CI = P * (1 + r) * r = P * 1.04 * 0.04 = 0.0416 P. Step-by-Step Solution:0.0416 P = 208 → P = 208 / 0.0416 = ₹ 5,000 Verification / Alternative check:Year 1 interest = 5000 * 4% = 200 → amount = 5200. Year 2 interest = 5200 * 4% = 208 (matches). Why Other Options Are Wrong:₹ 6,500 or ₹ 10,000 produce second-year interest larger than ₹ 208; ₹ 1,30,000 is far off. Common Pitfalls:Using P * r (first-year interest) instead of P * (1 + r) * r for the second year. Final Answer:₹ 5000

Question 19

Compound Interest – Varying rates across years: Find the compound interest on ₹ 9,375 in 2 years when the rate is 2% for the first year and 4% for the second year (annual compounding, rate changes per year).

Accepted Answer

Introduction / Context:When rates vary by year, we apply each year’s multiplier sequentially. The amount after 2 years is P * (1 + r1) * (1 + r2); CI is amount minus principal. Given Data / Assumptions: P = ₹ 9,375 Year-1 rate r1 = 2% = 0.02 Year-2 rate r2 = 4% = 0.04 Concept / Approach:Amount A = 9375 * 1.02 * 1.04. Compound interest CI = A − 9375. Step-by-Step Solution:A = 9375 * 1.02 = 9562.50A = 9562.50 * 1.04 = ₹ 9,945.00CI = 9945 − 9375 = ₹ 570 Verification / Alternative check:The order of yearly multipliers does not change the 2-year result (commutative multiplication). Why Other Options Are Wrong:₹ 670/₹ 760/₹ 770 imply higher rates or additional years. Common Pitfalls:Adding rates (2% + 4% = 6%) and applying simple interest, which ignores compounding. Final Answer:₹ 570

Question 20

Compound Interest — Recover annual growth rate from 3-year population multiplier: A country's population is 100 crore. It is expected to become 133.1 crore in 3 years. What is the annual percentage growth rate (compounded yearly) that leads to this …

Accepted Answer

Introduction / Context:Population growth that “becomes” a larger value after several years is naturally modeled using compound interest (or compound growth). If P0 is today's population and it becomes P after t years at an annual compounding rate r (per year), then P = P0 * (1 + r)^t. This question asks us to recover the annual percentage growth from a known 3-year multiplier. Given Data / Assumptions: Initial population P0 = 100 crore Population after 3 years P = 133.1 crore Annual compounding; t = 3 years Goal: find r as a percent per annum Concept / Approach:The ratio P / P0 equals (1 + r)^t. Thus r = (P / P0)^(1/t) − 1. Because 133.1 / 100 = 1.331, and 1.331 is a well-known cube (1.1^3), the rate appears to be exactly 10% per year when compounded annually. Step-by-Step Solution: Compute multiplier: M = P / P0 = 133.1 / 100 = 1.331.Solve for r from M = (1 + r)^3.Recognize 1.331 = 1.1^3 ⇒ 1 + r = 1.1 ⇒ r = 0.10 = 10%. Verification / Alternative check: Forward check: 100 * (1.10)^3 = 100 * 1.331 = 133.1 (matches). Why Other Options Are Wrong: 8%, 9%, 12.7%, 15% do not cube to 1.331 when added to 1. Their three-year multipliers are not 1.331. Common Pitfalls: Accidentally using simple interest (linear growth) instead of compound growth. Rounding too early; here the cube equality is exact with 10%. Final Answer:10% per annum (compounded annually).

Compound Interest Questions

Questions

Practice smarter