Q: Back out the annual rate from consecutive CI amounts: A sum becomes ₹ 9680 in 2 years and ₹ 10648 in 3 years under compound interest (annual compounding). What is the annual rate of interest?

Introduction / Context:When two successive-year amounts are known under annual compounding, their ratio directly gives (1 + r). This avoids computing the principal and provides a quick path to the annual rate. Given Data / Assumptions: A(2 years) = ₹ 9680 A(3 years) = ₹ 10648 Annual compounding at fixed r Concept / Approach:A(3) / A(2) = (P * (1 + r)^3) / (P * (1 + r)^2) = 1 + r. Hence r = (A3 / A2) − 1. Step-by-Step Solution:1 + r = 10648 / 9680 = 1.10r = 0.10 = 10% per annum Verification / Alternative check:If r = 10%, then A(3) = A(2) * 1.10 = 9680 * 1.10 = ₹ 10648 (exact). Why Other Options Are Wrong:5%, 15%, and 20% do not satisfy the observed ratio; 12% would give 1.12 rather than 1.10. Common Pitfalls:Attempting to solve for P first, or confusing CI with SI where amounts grow linearly rather than multiplicatively. Final Answer:10%

Q: Successive-year CI known — find the principal (annual compounding): A sum is invested at compound interest payable annually. The interest in two successive years is ₹ 500 and ₹ 540. What is the principal?

Introduction / Context:Under annual compounding, the second-year interest equals first-year interest multiplied by (1 + r). From the ratio of successive interests, we can find r, and then recover P from I1 = P * r. Given Data / Assumptions: I1 = ₹ 500 (Year-1 interest) I2 = ₹ 540 (Year-2 interest) Annual compounding at fixed r Concept / Approach:I2 / I1 = 1 + r → 540 / 500 = 1.08 → r = 8%. Then use I1 = P * r to get the principal P. Step-by-Step Solution:1 + r = 540 / 500 = 1.08 → r = 8%P = I1 / r = 500 / 0.08 = ₹ 6250 Verification / Alternative check:Year-1 interest on ₹ 6250 at 8% is ₹ 500. Amount after Year-1 = 6250 + 500 = ₹ 6750. Year-2 interest = 6750 * 8% = ₹ 540 (checks). Why Other Options Are Wrong:₹ 3750, ₹ 5000, or ₹ 5600 do not produce the given pair (500, 540) at 8%. Common Pitfalls:Using I2 = P * r directly without accounting for growth in the amount from year 1 to year 2 under compounding. Final Answer:₹ 6250

Q: Find the annual rate from CI − SI over 2 years: The difference between compound interest and simple interest on ₹ 5000 for 2 years is ₹ 72. What is the annual rate of interest (compounded yearly)?

Introduction / Context:For 2 years at a fixed rate with annual compounding, the excess of CI over SI equals P * r^2, where r is expressed as a decimal. This comes from the second-year “interest on interest.” Given Data / Assumptions: P = ₹ 5000 CI − SI = ₹ 72 t = 2 years, annual compounding Concept / Approach:Use CI − SI = P * r^2. Solve for r: r = sqrt((CI − SI)/P). Convert decimal r to percent by multiplying by 100. Step-by-Step Solution:P * r^2 = 72 → r^2 = 72 / 5000 = 0.0144r = sqrt(0.0144) = 0.12Annual rate = 0.12 × 100% = 12% Verification / Alternative check:At 12%, CI − SI should be 5000 * 0.12^2 = 5000 * 0.0144 = ₹ 72, exactly as given. Why Other Options Are Wrong:10%, 8%, 6%, or 9% do not satisfy P * r^2 = 72 for P = 5000. Common Pitfalls:Using 2 * r or r * t in the difference; the 2-year CI − SI shortcut works only with P * r^2 for annual compounding at a fixed r. Final Answer:12%

Q: Recover the principal from two CI snapshots 3 years apart: A sum becomes ₹ 6690 after 3 years and ₹ 10035 after 6 years at compound interest (same annual rate). What is the original principal (sum)?

Introduction / Context:Knowing amounts at two times that are exactly 3 years apart lets us find (1 + r)^3 from their ratio. Dividing the earlier amount by this factor gives the principal directly, avoiding separate rate computation. Given Data / Assumptions: A(3 years) = ₹ 6690 A(6 years) = ₹ 10035 Annual compounding at fixed r Concept / Approach:A(6)/A(3) = (1 + r)^3. Therefore (1 + r)^3 = 10035/6690 = 1.5. Hence the principal P = A(3) / (1 + r)^3 = A(3) / 1.5. Step-by-Step Solution:(1 + r)^3 = 10035 / 6690 = 1.5P = 6690 / 1.5 = ₹ 4460 Verification / Alternative check:If P = 4460, then A(3) = 4460 * 1.5 = 6690 (given), and A(6) = 6690 * 1.5 = 10035 (consistent). Why Other Options Are Wrong:₹ 4400, ₹ 4445, or ₹ 4520 do not maintain the exact 1.5 growth from year 3 to year 6 when multiplied by (1 + r)^3. Common Pitfalls:Trying to compute r via cube root first; while valid, it is unnecessary when the question only asks for the principal. Final Answer:₹ 4460

Q: Half-yearly compounding for 1.5 years at 10% nominal: What is the compound interest on ₹ 8000 at 10% per annum for 1.5 years when interest is compounded half-yearly?

Introduction / Context:When interest is credited more frequently than yearly, use the periodic rate and number of periods. For half-yearly compounding, divide the nominal annual rate by 2 and multiply years by 2 to get the number of half-years. Given Data / Assumptions: P = ₹ 8000 Nominal r = 10% p.a.; half-yearly rate = 10%/2 = 5% Time = 1.5 years → n = 3 half-years Concept / Approach:Amount A = P * (1 + i)^n, where i = 0.05 and n = 3. Then CI = A − P. Step-by-Step Solution:A = 8000 * (1.05)^3 = 8000 * 1.157625 = ₹ 9261CI = 9261 − 8000 = ₹ 1261 Verification / Alternative check:Effective 1.5-year multiplier at 10% nominal, half-yearly: (1.05)^3 = 1.157625, so a 15.7625% gain on ₹ 8000 is ₹ 1261 (rounded to whole rupees here exact). Why Other Options Are Wrong:₹ 1385 and ₹ 1480 assume higher periodic rates or counts; ₹ 1255 is slightly low; ₹ 1200 ignores compounding beyond simple 15%. Common Pitfalls:Using 1.10^(1.5) (annual compounding) instead of half-yearly periods; forgetting that 1.5 years equals 3 half-years. Final Answer:₹ 1261

Q: Second-year interest at 8% CI given first-year interest: At 8% per annum compound interest, the interest in the first year on a principal is ₹ 48. What will be the interest for the second year?

Introduction / Context:Under compound interest, second-year interest accrues on the first-year amount, not just the principal. Hence I2 = (P * (1 + r)) * r. If we know I1 = P * r, we can compute I2 quickly from I1 without finding P explicitly. Given Data / Assumptions: I1 = ₹ 48 at r = 8% = 0.08 Annual compounding Concept / Approach:I2 = I1 * (1 + r) because I2 = (P * (1 + r)) * r = (P * r) * (1 + r) = I1 * (1 + r). Step-by-Step Solution:I2 = 48 * (1 + 0.08) = 48 * 1.08 = ₹ 51.84 Verification / Alternative check:We can recover P: P = I1 / r = 48 / 0.08 = ₹ 600. Amount after Year-1 = 600 * 1.08 = 648. I2 = 648 * 0.08 = 51.84 (same). Why Other Options Are Wrong:₹ 52.55 and ₹ 53.04 assume higher r; ₹ 58.60 is too large; ₹ 50 flatly ignores the compounding uplift. Common Pitfalls:Reusing I1 as I2 (SI logic) or forgetting to multiply by (1 + r) for the second year under compounding. Final Answer:₹ 51.84

Question 1

From simple to compound — same rate (10%) and time (2 years): The simple interest on a certain principal for 2 years at 10% per annum is ₹ 90. What is the corresponding compound interest for the same time and rate?

Accepted Answer

Introduction / Context:This problem converts a known simple-interest outcome into compound interest (same rate and duration). First deduce the principal from SI, then compute CI using the compounding formula. Given Data / Assumptions: SI over 2 years at 10% p.a. = ₹ 90 Rate r = 10% = 0.10 Time t = 2 years Principal P = ? Concept / Approach:Simple interest SI = P * r * t. Hence P = SI / (r * t). After getting P, compute CI via A = P * (1 + r)^t and CI = A − P. Step-by-Step Solution:P = 90 / (0.10 * 2) = 90 / 0.20 = ₹ 450A = 450 * (1.10)^2 = 450 * 1.21 = ₹ 544.50CI = A − P = 544.50 − 450 = ₹ 94.50 Verification / Alternative check:Second-year uplift equals “interest on interest” = P * r^2 = 450 * 0.01 = ₹ 4.50; SI for 2 years would be ₹ 90; CI must therefore be ₹ 94.50 (consistent). Why Other Options Are Wrong:₹ 99 and ₹ 108 overstate compounding; ₹ 95.60 does not match exact (1.10)^2; ₹ 92.00 undercounts by ignoring second-year interest on interest. Common Pitfalls:Mistaking SI = P * r (forgetting time), or using 2 * 10% as a compounded rate rather than multiplying by (1.10)^2. Final Answer:₹ 94.50

Question 2

Identify principal from second-year compound interest: Under 10% per annum compound interest (annual compounding), the interest earned in the second year alone is ₹ 132. What is the principal?

Accepted Answer

Introduction / Context:In compound interest, the second-year interest equals the first year’s amount multiplied by the rate. That is effectively P * r * (1 + r). Knowing the second-year interest lets us back-out P directly. Given Data / Assumptions: Annual rate r = 10% = 0.10 Second-year interest I2 = ₹ 132 Compounding annually Concept / Approach:First-year amount = P * (1 + r). Second-year interest = (P * (1 + r)) * r = P * r * (1 + r). Hence P = I2 / (r * (1 + r)). Step-by-Step Solution:I2 = P * 0.10 * 1.10 = 0.11P132 = 0.11P → P = 132 / 0.11 = ₹ 1200 Verification / Alternative check:Year-1 interest = 1200 * 0.10 = 120; amount after Year-1 = 1320. Year-2 interest = 1320 * 0.10 = 132 (matches). Why Other Options Are Wrong:₹ 1000, ₹ 1320, or ₹ 1100 do not yield a second-year interest of ₹ 132 when multiplied by 0.11. Common Pitfalls:Using I2 = P * r instead of P * r * (1 + r); mixing SI and CI logic will misestimate the second-year interest. Final Answer:₹ 1200

Question 3

Convert SI to CI at 4% for 2 years — same principal: Simple interest at 4% per annum for 2 years on a certain sum is ₹ 80. What is the compound interest on the same principal for the same period and rate?

Accepted Answer

Introduction / Context:Given SI over a duration, we can determine the principal; then compute CI with annual compounding for the same rate and time. The CI will be slightly larger due to interest on interest in the second year. Given Data / Assumptions: SI = ₹ 80 for 2 years at 4% p.a. r = 4% = 0.04; t = 2 years Concept / Approach:SI = P * r * t → P = SI / (r * t). CI uses A = P * (1 + r)^t; CI = A − P. Alternatively, CI − SI over 2 years equals P * r^2. Step-by-Step Solution:P = 80 / (0.04 * 2) = 80 / 0.08 = ₹ 1000A = 1000 * (1.04)^2 = 1000 * 1.0816 = ₹ 1081.60CI = 1081.60 − 1000 = ₹ 81.60 Verification / Alternative check:CI − SI = P * r^2 = 1000 * 0.0016 = ₹ 1.60. Since SI was ₹ 80, CI must be ₹ 81.60 (consistent). Why Other Options Are Wrong:₹ 160 and ₹ 1081.60 confuse total amount or double the period incorrectly; ₹ 79.20 understates the compounding effect; “None” is unnecessary because we have an exact value. Common Pitfalls:Confusing amount A with CI, or forgetting that CI − SI for 2 years equals P * r^2 only when compounding annually at a fixed r. Final Answer:₹ 81.60

Question 4

Minimum years to more-than-double at 20% compound interest: Find the least number of complete years for which an amount invested at 20% per annum, compounded annually, will exceed double the principal.

Accepted Answer

Introduction / Context:Doubling under compound interest depends on the growth factor (1 + r)^n. We seek the smallest integer n with (1 + r)^n > 2. This is a classic time-to-double check without logarithms by testing small n values. Given Data / Assumptions: r = 20% per annum, compounded yearly Need the least integer n such that amount > 2P Concept / Approach:Compute powers of 1.20: 1.20^3 and 1.20^4; identify the first to exceed 2. Exact doubling time is ln 2 / ln 1.2 ≈ 3.80, so the least complete year is the ceiling of this value. Step-by-Step Solution:1.20^3 = 1.728 (less than 2)1.20^4 = 2.0736 (exceeds 2)Therefore, the least complete n is 4 years. Verification / Alternative check:Using logs: n = ln(2)/ln(1.2) ≈ 0.6931/0.1823 ≈ 3.80 → round up to 4 full years for “more than double.” Why Other Options Are Wrong:3 years yields 1.728P < 2P; 5 or 6 years are not the least; 2 years is far below doubling at this rate. Common Pitfalls:Answering 3 years by comparing to simple interest or by stopping at “approximately doubled” instead of strictly exceeding 2P. Final Answer:4 years

Question 5

Back out the annual rate from consecutive CI amounts: A sum becomes ₹ 9680 in 2 years and ₹ 10648 in 3 years under compound interest (annual compounding). What is the annual rate of interest?

Accepted Answer

Introduction / Context:When two successive-year amounts are known under annual compounding, their ratio directly gives (1 + r). This avoids computing the principal and provides a quick path to the annual rate. Given Data / Assumptions: A(2 years) = ₹ 9680 A(3 years) = ₹ 10648 Annual compounding at fixed r Concept / Approach:A(3) / A(2) = (P * (1 + r)^3) / (P * (1 + r)^2) = 1 + r. Hence r = (A3 / A2) − 1. Step-by-Step Solution:1 + r = 10648 / 9680 = 1.10r = 0.10 = 10% per annum Verification / Alternative check:If r = 10%, then A(3) = A(2) * 1.10 = 9680 * 1.10 = ₹ 10648 (exact). Why Other Options Are Wrong:5%, 15%, and 20% do not satisfy the observed ratio; 12% would give 1.12 rather than 1.10. Common Pitfalls:Attempting to solve for P first, or confusing CI with SI where amounts grow linearly rather than multiplicatively. Final Answer:10%

Question 6

Successive-year CI known — find the principal (annual compounding): A sum is invested at compound interest payable annually. The interest in two successive years is ₹ 500 and ₹ 540. What is the principal?

Accepted Answer

Introduction / Context:Under annual compounding, the second-year interest equals first-year interest multiplied by (1 + r). From the ratio of successive interests, we can find r, and then recover P from I1 = P * r. Given Data / Assumptions: I1 = ₹ 500 (Year-1 interest) I2 = ₹ 540 (Year-2 interest) Annual compounding at fixed r Concept / Approach:I2 / I1 = 1 + r → 540 / 500 = 1.08 → r = 8%. Then use I1 = P * r to get the principal P. Step-by-Step Solution:1 + r = 540 / 500 = 1.08 → r = 8%P = I1 / r = 500 / 0.08 = ₹ 6250 Verification / Alternative check:Year-1 interest on ₹ 6250 at 8% is ₹ 500. Amount after Year-1 = 6250 + 500 = ₹ 6750. Year-2 interest = 6750 * 8% = ₹ 540 (checks). Why Other Options Are Wrong:₹ 3750, ₹ 5000, or ₹ 5600 do not produce the given pair (500, 540) at 8%. Common Pitfalls:Using I2 = P * r directly without accounting for growth in the amount from year 1 to year 2 under compounding. Final Answer:₹ 6250

Question 7

Find the annual rate from CI − SI over 2 years: The difference between compound interest and simple interest on ₹ 5000 for 2 years is ₹ 72. What is the annual rate of interest (compounded yearly)?

Accepted Answer

Introduction / Context:For 2 years at a fixed rate with annual compounding, the excess of CI over SI equals P * r^2, where r is expressed as a decimal. This comes from the second-year “interest on interest.” Given Data / Assumptions: P = ₹ 5000 CI − SI = ₹ 72 t = 2 years, annual compounding Concept / Approach:Use CI − SI = P * r^2. Solve for r: r = sqrt((CI − SI)/P). Convert decimal r to percent by multiplying by 100. Step-by-Step Solution:P * r^2 = 72 → r^2 = 72 / 5000 = 0.0144r = sqrt(0.0144) = 0.12Annual rate = 0.12 × 100% = 12% Verification / Alternative check:At 12%, CI − SI should be 5000 * 0.12^2 = 5000 * 0.0144 = ₹ 72, exactly as given. Why Other Options Are Wrong:10%, 8%, 6%, or 9% do not satisfy P * r^2 = 72 for P = 5000. Common Pitfalls:Using 2 * r or r * t in the difference; the 2-year CI − SI shortcut works only with P * r^2 for annual compounding at a fixed r. Final Answer:12%

Question 8

Recover the principal from two CI snapshots 3 years apart: A sum becomes ₹ 6690 after 3 years and ₹ 10035 after 6 years at compound interest (same annual rate). What is the original principal (sum)?

Accepted Answer

Introduction / Context:Knowing amounts at two times that are exactly 3 years apart lets us find (1 + r)^3 from their ratio. Dividing the earlier amount by this factor gives the principal directly, avoiding separate rate computation. Given Data / Assumptions: A(3 years) = ₹ 6690 A(6 years) = ₹ 10035 Annual compounding at fixed r Concept / Approach:A(6)/A(3) = (1 + r)^3. Therefore (1 + r)^3 = 10035/6690 = 1.5. Hence the principal P = A(3) / (1 + r)^3 = A(3) / 1.5. Step-by-Step Solution:(1 + r)^3 = 10035 / 6690 = 1.5P = 6690 / 1.5 = ₹ 4460 Verification / Alternative check:If P = 4460, then A(3) = 4460 * 1.5 = 6690 (given), and A(6) = 6690 * 1.5 = 10035 (consistent). Why Other Options Are Wrong:₹ 4400, ₹ 4445, or ₹ 4520 do not maintain the exact 1.5 growth from year 3 to year 6 when multiplied by (1 + r)^3. Common Pitfalls:Trying to compute r via cube root first; while valid, it is unnecessary when the question only asks for the principal. Final Answer:₹ 4460

Question 9

Half-yearly compounding for 1.5 years at 10% nominal: What is the compound interest on ₹ 8000 at 10% per annum for 1.5 years when interest is compounded half-yearly?

Accepted Answer

Introduction / Context:When interest is credited more frequently than yearly, use the periodic rate and number of periods. For half-yearly compounding, divide the nominal annual rate by 2 and multiply years by 2 to get the number of half-years. Given Data / Assumptions: P = ₹ 8000 Nominal r = 10% p.a.; half-yearly rate = 10%/2 = 5% Time = 1.5 years → n = 3 half-years Concept / Approach:Amount A = P * (1 + i)^n, where i = 0.05 and n = 3. Then CI = A − P. Step-by-Step Solution:A = 8000 * (1.05)^3 = 8000 * 1.157625 = ₹ 9261CI = 9261 − 8000 = ₹ 1261 Verification / Alternative check:Effective 1.5-year multiplier at 10% nominal, half-yearly: (1.05)^3 = 1.157625, so a 15.7625% gain on ₹ 8000 is ₹ 1261 (rounded to whole rupees here exact). Why Other Options Are Wrong:₹ 1385 and ₹ 1480 assume higher periodic rates or counts; ₹ 1255 is slightly low; ₹ 1200 ignores compounding beyond simple 15%. Common Pitfalls:Using 1.10^(1.5) (annual compounding) instead of half-yearly periods; forgetting that 1.5 years equals 3 half-years. Final Answer:₹ 1261

Question 10

Second-year interest at 8% CI given first-year interest: At 8% per annum compound interest, the interest in the first year on a principal is ₹ 48. What will be the interest for the second year?

Accepted Answer

Introduction / Context:Under compound interest, second-year interest accrues on the first-year amount, not just the principal. Hence I2 = (P * (1 + r)) * r. If we know I1 = P * r, we can compute I2 quickly from I1 without finding P explicitly. Given Data / Assumptions: I1 = ₹ 48 at r = 8% = 0.08 Annual compounding Concept / Approach:I2 = I1 * (1 + r) because I2 = (P * (1 + r)) * r = (P * r) * (1 + r) = I1 * (1 + r). Step-by-Step Solution:I2 = 48 * (1 + 0.08) = 48 * 1.08 = ₹ 51.84 Verification / Alternative check:We can recover P: P = I1 / r = 48 / 0.08 = ₹ 600. Amount after Year-1 = 600 * 1.08 = 648. I2 = 648 * 0.08 = 51.84 (same). Why Other Options Are Wrong:₹ 52.55 and ₹ 53.04 assume higher r; ₹ 58.60 is too large; ₹ 50 flatly ignores the compounding uplift. Common Pitfalls:Reusing I1 as I2 (SI logic) or forgetting to multiply by (1 + r) for the second year under compounding. Final Answer:₹ 51.84

Question 11

CI known for 2 years at 5% — find corresponding SI for 2 years: If the compound interest on a certain principal for 2 years at 5% per annum is ₹ 328, what is the simple interest on the same principal for the same time at the same rate?

Accepted Answer

Introduction / Context:We can convert CI over 2 years into the principal using the compounding factor, then compute the equivalent SI over the same period from the recovered principal. CI exceeds SI because of the interest-on-interest term. Given Data / Assumptions: CI (2 years, 5% p.a.) = ₹ 328 Annual compounding, fixed rate Concept / Approach:For 2 years, A = P * (1.05)^2 = 1.1025P; CI = A − P = 0.1025P. Hence P = 328 / 0.1025. Once P is known, SI over 2 years is P * 0.05 * 2 = 0.10P. Step-by-Step Solution:P = 328 / 0.1025 = ₹ 3200SI (2 years) = 3200 * 0.10 = ₹ 320 Verification / Alternative check:CI − SI for 2 years equals P * r^2 = 3200 * 0.0025 = ₹ 8. Thus, CI = SI + 8 → SI = 328 − 8 = ₹ 320 (consistent). Why Other Options Are Wrong:₹ 322, ₹ 325, or ₹ 326 do not fit the exact factors (1.1025 and 0.10). Common Pitfalls:Using 2 * 5% as a compounded rate; failing to invert 0.1025 correctly when solving for P. Final Answer:₹ 320

Question 12

Least complete years to exceed double at 20% CI: Find the minimum number of complete years for a sum to become more than twice itself at 20% per annum compounded annually.

Accepted Answer

Introduction / Context:This repeats the classic doubling-time logic under compounding. “More than doubled” means the amount strictly exceeds 2P, not merely reaches or approximates it. We therefore need the smallest integer n with (1.20)^n > 2. Given Data / Assumptions: r = 20% per annum, annual compounding We require amount > 2P Concept / Approach:Compute powers: 1.2^2 = 1.44, 1.2^3 = 1.728, 1.2^4 = 2.0736. The first power strictly above 2 is at n = 4. Step-by-Step Solution:Check n = 3 → 1.728P < 2P (not enough)Check n = 4 → 2.0736P > 2P (succeeds)Hence, least complete years = 4 Verification / Alternative check:Log method: n = ln(2)/ln(1.2) ≈ 3.80 → need full years, so n = 4. Why Other Options Are Wrong:2 or 3 years are insufficient; “Data inadequate” is incorrect because the rate is fully specified; 5 years is not the minimum. Common Pitfalls:Answering 3 years based on SI thinking or misreading “more than doubled” as “at least doubled.” Final Answer:4 years

Question 13

Find the annual rate from two consecutive CI amounts: A sum grows to ₹ 578.40 in 2 years and to ₹ 614.55 in 3 years under compound interest with annual compounding. Determine the annual rate of interest.

Accepted Answer

Introduction / Context:Two successive annual amounts let us take their ratio to get 1 + r immediately. This avoids principal calculations and provides a clean way to read off the annual rate under CI (annual compounding). Given Data / Assumptions: A(2 years) = ₹ 578.40 A(3 years) = ₹ 614.55 Annual compounding Concept / Approach:(1 + r) = A(3)/A(2). Convert the decimal to a recognisable fraction/percentage if possible. Then r = (A3/A2 − 1) × 100% per annum. Step-by-Step Solution:1 + r = 614.55 / 578.40 = 1.0625r = 0.0625 = 6.25% = 6¼% p.a. Verification / Alternative check:1.0625 = 17/16. Over one year, that’s exactly 6.25% (since 16 grows to 17). Multiplying ₹ 578.40 by 1.0625 gives ₹ 614.55 exactly. Why Other Options Are Wrong:6%, 6½%, or 6¾% do not match the precise ratio; 5% is far too small. Common Pitfalls:Rounding the ratio too early and picking 6% or 6.5% by approximation rather than exact calculation. Final Answer:6¼%

Question 14

CI vs SI for 1 year at 10% when reckoned half-yearly: Find the difference between compound interest and simple interest on ₹ 600 for 1 year at 10% per annum when compounding (for CI) is half-yearly.

Accepted Answer

Introduction / Context:For one year, simple interest at 10% is straightforward. For CI with half-yearly crediting, apply 5% twice. The CI will be slightly higher, and the small difference demonstrates frequency effects within a single year. Given Data / Assumptions: P = ₹ 600 Nominal r = 10% p.a. Compounding frequency = half-yearly (i = 5% per half-year, n = 2) Concept / Approach:SI(1 year) = P * 0.10. CI amount with half-yearly = P * (1.05)^2. The difference is CI − SI. Step-by-Step Solution:SI = 600 * 0.10 = ₹ 60.00A_CI = 600 * (1.05)^2 = 600 * 1.1025 = ₹ 661.50 → CI = ₹ 61.50Difference = 61.50 − 60.00 = ₹ 1.50 Verification / Alternative check:Effective annual rate under half-yearly compounding at 10% nominal is 1.05^2 − 1 = 10.25%; 10.25% of 600 is ₹ 61.50 (CI). Why Other Options Are Wrong:₹ 6.60 and ₹ 4.40 are far too large; zero would only occur with no compounding or zero rate; ₹ 3.00 is not exact. Common Pitfalls:Using 10% twice (giving 20%) or treating half-yearly as still 10% flat with no compounding. Final Answer:₹ 1.50

Question 15

Reverse depreciation — find value 3 years ago at 10% yearly drop: A machine depreciates at 10% each year on its beginning-of-year value. If its present value is ₹ 729, what was its worth 3 years ago?

Accepted Answer

Introduction / Context:Depreciation at a fixed percentage each year is a compound decrease. To find the earlier value from a current value, divide by the compounded decay factor over the elapsed years. Given Data / Assumptions: Current value V_now = ₹ 729 Yearly depreciation = 10% → yearly multiplier = 0.90 We need value 3 years ago (reverse 3 years of depreciation) Concept / Approach:Forward: V_now = V_past * (0.90)^3. Hence V_past = V_now / (0.90)^3. Since 0.9^3 = 0.729, this inversion is clean. Step-by-Step Solution:(0.90)^3 = 0.729V_past = 729 / 0.729 = ₹ 1000 Verification / Alternative check:Apply depreciation forward: 1000 → 900 → 810 → 729 (exact). Why Other Options Are Wrong:₹ 947.10 or ₹ 750.87 are not exact reversals for 10% three times; ₹ 800 and ₹ 900 are intermediate-year values, not 3 years ago. Common Pitfalls:Subtracting 10% three times from the original instead of compounding multiplicatively, which would misestimate past value. Final Answer:₹ 1000

Question 16

Growth by one-eighth annually — compute height after 2 years (Recovery-First applied): A tree increases annually by 1/8 of its height. If today it is 64 cm high, what will be its height after 2 years?

Accepted Answer

Introduction / Context:(Recovery-First) The original stem asked “by how much will it increase,” but the options list final heights (e.g., 81 cm). To make the item solvable without changing its core meaning, we restate it to ask for height after 2 years, consistent with the options. Annual growth is a compound increase of 12.5% (1/8) each year. Given Data / Assumptions: Initial height H0 = 64 cm Yearly growth rate = 1/8 = 12.5% Compounded annually for 2 years Concept / Approach:Height after n years: Hn = H0 * (1 + r)^n. Here, r = 1/8 so (1 + r) = 9/8. After 2 years, multiply by (9/8)^2. Step-by-Step Solution:H1 = 64 * (9/8) = 72 cmH2 = 72 * (9/8) = 81 cmHence the height after 2 years is 81 cm Verification / Alternative check:ΔH over 2 years = 81 − 64 = 17 cm. This is the “increase” the original stem likely intended, but since 17 cm is not an option, final height framing is appropriate. Why Other Options Are Wrong:72 cm is after 1 year only; 74 cm and 75 cm reflect incorrect additive growth; 80 cm undercounts the compound step. Common Pitfalls:Treating growth as linear (adding 1/8 of the original only once) rather than compounding year-on-year. Final Answer:81 cm

Question 17

Present value of two equal annual instalments at 10% CI: A loan is repaid by two annual instalments of ₹ 121 each at 10% per annum compounded annually. What was the sum borrowed (present value at the time of borrowing)?

Accepted Answer

Introduction / Context:When a loan is cleared by equal annual instalments under compound interest, the borrowed amount equals the sum of the present values of each instalment discounted at the annual rate back to the start date. Given Data / Assumptions: Two instalments of ₹ 121 each Annual rate r = 10% = 0.10 Payments at the end of Year-1 and Year-2 Concept / Approach:Present value PV = 121 / (1.10)^1 + 121 / (1.10)^2. Compute each term and add to get the principal originally borrowed. Step-by-Step Solution:PV1 = 121 / 1.10 = ₹ 110.00 (actually ₹ 110 is incorrect; see below)Correct PV1 = 121 / 1.10 = ₹ 110.00? No → 1.10 × 110 = 121 → so PV1 = ₹ 110.00 is exactPV2 = 121 / (1.10)^2 = 121 / 1.21 = ₹ 100.00Total PV = 110 + 100 = ₹ 210 Verification / Alternative check:Forward compute: ₹ 210 borrowed grows to 231 at end of Year-1; pay 121, balance 110. End of Year-2 balance becomes 121; final payment 121 clears the loan (consistent). Why Other Options Are Wrong:₹ 200 and ₹ 205 understate PV; ₹ 217.80 belongs to a different rate/term; ₹ 280 is far too high. Common Pitfalls:Adding undiscounted instalments (121 + 121) or using SI logic instead of discounting each payment by (1 + r)^t. Final Answer:₹ 210

Question 18

Repeated doubling under CI — from 5 years to 20 years: ₹ 12000 deposited at compound interest doubles in 5 years. What will it become after 20 years (same rate maintained)?

Accepted Answer

Introduction / Context:If an amount doubles in a fixed interval under compound interest, then over k such intervals it multiplies by 2^k. Recognizing the number of equal intervals between the two horizons gives the final multiplier immediately. Given Data / Assumptions: Principal P0 = ₹ 12000 It doubles in 5 years We need the amount after 20 years at the same rate Concept / Approach:20 years contains 4 blocks of 5 years. Therefore, amount after 20 years is P0 * 2^4 = P0 * 16. Step-by-Step Solution:A(20) = 12000 * 16 = ₹ 192000 Verification / Alternative check:Timeline view: 0→5 yrs: 12000→24000; 10 yrs: 48000; 15 yrs: 96000; 20 yrs: 192000 (four doublings). Why Other Options Are Wrong:₹ 120000 and ₹ 96000 reflect incorrect counts of doublings; ₹ 124000 assumes linear growth; ₹ 240000 is not a power-of-two multiple of the principal for 20 years. Common Pitfalls:Using SI reasoning (adding 4 * 12000) rather than multiplying by 2^4; miscounting the number of 5-year blocks in 20 years. Final Answer:₹ 192000

Question 19

Present value of two instalments of ₹ 1089 each at 10% CI: A loan is repaid in two equal annual instalments of ₹ 1089 each. If the rate is 10% per annum compounded annually, what was the sum borrowed (present value)?

Accepted Answer

Introduction / Context:The amount borrowed equals the sum of present values of future instalments discounted at the loan’s compound rate. With annual compounding and two end-of-year payments, discount each by (1 + r)^t with t = 1 and 2 respectively. Given Data / Assumptions: Two instalments: ₹ 1089 each Annual rate r = 10% = 0.10 Payments at end of Year-1 and Year-2 Concept / Approach:PV = 1089/1.10 + 1089/(1.10)^2. Noting 1089 = 1.10 * 990 = 1.21 * 900 gives exact integers after division, simplifying arithmetic greatly. Step-by-Step Solution:PV1 = 1089 / 1.10 = ₹ 990PV2 = 1089 / 1.21 = ₹ 900Total PV = 990 + 900 = ₹ 1890 Verification / Alternative check:Forward: Borrow 1890 → after 1 year: 2079; pay 1089, balance 990. After Year-2: 990 * 1.10 = 1089; pay 1089 → loan cleared. Why Other Options Are Wrong:₹ 1840, ₹ 1850, or ₹ 1860 understate PV; ₹ 1900 is close but not exact given the neat divisibility above. Common Pitfalls:Forgetting to discount the second payment by (1.10)^2, or mixing SI with CI discounting. Final Answer:₹ 1890

Question 20

Compound Interest – Maturity amount over 2 years: Vandana invests ₹ 8,000 in a fixed deposit for 2 years at 5% per annum, compounded annually. What total maturity amount will she receive at the end of 2 years?

Accepted Answer

Introduction / Context:Fixed deposits commonly use compound interest. The maturity amount for annual compounding is found by multiplying the principal by (1 + r)^t, where r is the annual rate and t is time in years. Given Data / Assumptions: Principal P = ₹ 8,000 Annual rate r = 5% = 0.05 Time t = 2 years Compounding frequency = yearly (once per year) Concept / Approach:For yearly compounding: Amount A = P * (1 + r)^t. Interest is automatically added to the base each year, so the second year earns interest on the first year’s interest as well (the hallmark of compounding). Step-by-Step Solution:A = 8000 * (1 + 0.05)^2A = 8000 * (1.05)^2 = 8000 * 1.1025A = ₹ 8,820 Verification / Alternative check:Year-wise: End of Year 1 = 8000 * 1.05 = 8400. End of Year 2 = 8400 * 1.05 = 8820. Matches the direct formula. Why Other Options Are Wrong:₹ 8890 and ₹ 8888 imply incorrect rounding or rates; ₹ 8000 ignores interest entirely. Common Pitfalls:Using simple interest formula P * r * t or forgetting to apply the exponent on (1 + r). Final Answer:₹ 8820

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