Question 1

Simple Interest — Time proportionality: At the same rate, what is the ratio of simple interest earned on the same amount for 12 years and for 18 years?

Accepted Answer

Introduction / Context: Under simple interest, interest is directly proportional to time when principal and rate are fixed. Ratios of interests for different durations equal the ratios of those durations. Given Data / Assumptions: Same principal and same annual rate Two time periods: 12 years and 18 years Concept / Approach: Let SI1 = P * r * 12 and SI2 = P * r * 18. Then SI1 : SI2 = 12 : 18, which simplifies by dividing both terms by 6. Step-by-Step Solution: SI1 : SI2 = (P * r * 12) : (P * r * 18) = 12 : 18.Reduce 12 : 18 by 6 → 2 : 3. Verification / Alternative check: Pick a trial P and r to confirm numerically (e.g., P = 100, r = 10%): SI1 = 120; SI2 = 180 → 2 : 3. Why Other Options Are Wrong: 1 : 3, 2 : 5, 3 : 1, 5 : 2 do not match the proportionality of 12 and 18. Common Pitfalls: Applying compound interest thinking; in SI, proportionality with time is exact. Final Answer: 2 : 3.

Question 2

Simple Interest — Two-part split with equal interests: Rs 2600 is split into two parts so that interest on one part at 10% for 5 years equals the interest on the other part at 9% for 6 years. What is the amount invested at 10%?

Accepted Answer

Introduction / Context: When two sums earn equal simple interest over different time–rate combinations, the equality condition directly relates the two principals. With the total known, we can solve for each part. Given Data / Assumptions: Total = Rs 2600 Part X at 10% for 5 years Part (2600 − X) at 9% for 6 years Interests are equal Concept / Approach: Set X * 0.10 * 5 = (2600 − X) * 0.09 * 6 and solve for X. This is a linear equation in X. Step-by-Step Solution: 0.5 X = 0.54 (2600 − X)0.5 X = 1404 − 0.54 X1.04 X = 1404 ⇒ X = 1404 / 1.04 = Rs 1350. Verification / Alternative check: Interest on X: 1350 * 0.10 * 5 = 675. Interest on other: 1250 * 0.09 * 6 = 675 (matches). Why Other Options Are Wrong: 1150, 1200, 1250, 1450 do not satisfy the equal-interest constraint with the given total and rates. Common Pitfalls: Mistaking 9% for 0.9 in decimal; correct is 0.09. Forgetting total is fixed at 2600. Final Answer: Rs. 1350.

Question 3

Simple Interest — Symbolic condition where rate% equals time: On a certain principal, the simple interest equals 9/16 of the principal. If the numerical values of rate percent per annum and time in years are equal, what is the time?

Accepted Answer

Introduction / Context: Symbolic SI problems test fluency with variables and percent conversion. When the rate in percent and the time (in years) are equal numerically, the product r * t must be formed carefully using r in decimal form (percent divided by 100). Given Data / Assumptions: SI = (9/16) * P Let rate percent = time in years = n (so r = n/100, t = n) Concept / Approach: Under SI: SI = P * r * t = P * (n/100) * n = P * n^2 / 100. Equate to (9/16) * P and solve for n. Cancel P > 0 from both sides. Step-by-Step Solution: n^2 / 100 = 9 / 16n^2 = 900 / 16 = 56.25n = 7.5 ⇒ time = 7.5 years = 7 1/2 years. Verification / Alternative check: Compute SI: P * (7.5/100) * 7.5 = P * 56.25 / 100 = (9/16) P (consistent). Why Other Options Are Wrong: Other durations do not satisfy n^2 = 56.25. Common Pitfalls: Using n (percent) directly without dividing by 100. Final Answer: 7 1/2 years.

Question 4

Simple Interest — Piecewise annual rates over 9 years: A sum earns SI at 4% p.a. for the first 2 years, 6% p.a. for the next 4 years, and 8% p.a. for the period beyond 6 years. If total SI in 9 years is Rs 1120, find the principal.

Accepted Answer

Introduction / Context: With simple interest, different flat rates over disjoint time blocks add linearly. Summing each block’s contribution gives the overall SI factor times the principal. Given Data / Assumptions: Years 1–2: 4% p.a. ⇒ 0.04 * 2 Years 3–6: 6% p.a. ⇒ 0.06 * 4 Years 7–9: 8% p.a. ⇒ 0.08 * 3 Total SI = Rs 1120 over 9 years Concept / Approach: Total SI factor = 0.04*2 + 0.06*4 + 0.08*3. Multiply this factor by P to equal Rs 1120, then solve for P. Step-by-Step Solution: Factor = 0.08 + 0.24 + 0.24 = 0.56.0.56 * P = 1120 ⇒ P = 1120 / 0.56 = Rs 2000. Verification / Alternative check: Block-wise SI on 2000: 160 + 480 + 480 = 1120 (matches). Why Other Options Are Wrong: 1500, 1800, 2500, 4000 produce total SI unequal to Rs 1120 using the same block factors. Common Pitfalls: Misallocating years across blocks or mixing up rates. Final Answer: Rs 2000.

Question 5

Simple Interest — Relation among three principals: If A is the simple interest on B and B is the simple interest on C at the same rate and time, what is the relation among A, B, and C?

Accepted Answer

Introduction / Context: Symbolic SI relationships test algebraic manipulation. If two statements about simple interest share the same rate and time, we can express each principal in terms of the others and eliminate the common factors. Given Data / Assumptions: A is SI on principal B at rate r and time t ⇒ A = B * r * t B is SI on principal C at the same rate r and time t ⇒ B = C * r * t Concept / Approach: From B = C r t, substitute into A = B r t to get A = (C r t) r t = C (r t)^2. Use these to connect A, B, C by eliminating r t. Step-by-Step Solution: Given B = C r t.Given A = B r t = (C r t) r t = C (r t)^2.Compute A C = [C (r t)^2] * C = C^2 (r t)^2; and B^2 = (C r t)^2 = C^2 (r t)^2.Hence B^2 = A C. Verification / Alternative check: Take C = 100, r t = 0.2 ⇒ B = 20, A = 4; indeed B^2 = 400 and A C = 400. Why Other Options Are Wrong: A2 = B C or C2 = A B do not match the algebra. A B C = 1 and A = B = C are unrelated to SI scaling. Common Pitfalls: Treating r as percent value instead of decimal in algebra; only the product r t matters. Final Answer: B^2 = A C.

Question 6

Simple Interest — Mixed rates over 8 years: Rate is 5% p.a. for the first 2 years, 8% p.a. for the next 3 years, and 10% p.a. thereafter. If the total simple interest in 8 years is Rs 1280, find the principal.

Accepted Answer

Introduction / Context: Piecewise constant rates under simple interest sum to an overall factor multiplying the principal. Carefully allocate the correct years to each rate block and add the contributions. Given Data / Assumptions: Years 1–2: 5% ⇒ 0.05 * 2 Years 3–5: 8% ⇒ 0.08 * 3 Years 6–8: 10% ⇒ 0.10 * 3 Total SI over 8 years = Rs 1280 Concept / Approach: Total factor = 0.05*2 + 0.08*3 + 0.10*3. Multiply by P to equal 1280; solve for P. Step-by-Step Solution: Factor = 0.10 + 0.24 + 0.30 = 0.64.0.64 * P = 1280 ⇒ P = 1280 / 0.64 = Rs 2000. Verification / Alternative check: Block SI on 2000: 100 + 480 + 600 = 1180? (Correct recomputation:) 0.05*2*2000 = 200; 0.08*3*2000 = 480; 0.10*3*2000 = 600; total = 1280 (matches). Why Other Options Are Wrong: Other values of P do not satisfy 0.64 * P = 1280. Common Pitfalls: An arithmetic slip on the first block (ensure 0.05 * 2 * 2000 = 200). Final Answer: Rs 2000.

Question 7

Simple Interest — Tiered rates over 12 years: Anil borrows at simple interest: 5% p.a. for the first 3 years, 8% p.a. for the next 2 years, 10% p.a. for the next 4 years, and 12% p.a. for the final 3 years. If total interest over 12 years is Rs 8560…

Accepted Answer

Introduction / Context: Aggregating simple interest across multiple fixed-rate intervals is additive. Summing the block contributions yields a single multiplier times principal equal to total interest paid. Given Data / Assumptions: Years 1–3: 5% ⇒ 0.05 * 3 = 0.15 Years 4–5: 8% ⇒ 0.08 * 2 = 0.16 Years 6–9: 10% ⇒ 0.10 * 4 = 0.40 Years 10–12: 12% ⇒ 0.12 * 3 = 0.36 Total SI = Rs 8560 (12 years) Concept / Approach: Total SI factor = 0.15 + 0.16 + 0.40 + 0.36 = 1.07; hence 1.07 * P = 8560 ⇒ P = 8560 / 1.07. Step-by-Step Solution: Sum factor = 1.07.P = 8560 / 1.07 = Rs 8000. Verification / Alternative check: Compute SI on 8000 blockwise: 1200 + 1280 + 3200 + 2880 = 8560 (matches). Why Other Options Are Wrong: 6000, 5000, 7000, 9000 do not satisfy 1.07 * P = 8560. Common Pitfalls: Forgetting the last 3-year block at 12%. Final Answer: Rs 8000.

Question 8

Simple Interest — Borrow low, lend high arbitrage: Prabhat borrows at 8% p.a. SI and lends the same sum to Ashish at 12% p.a. SI. If his profit in 2 years is Rs 320, how much did he borrow?

Accepted Answer

Introduction / Context: Arbitrage under SI arises from lending at a higher rate than the borrowing rate. The net gain equals principal times the rate spread times time. Given Data / Assumptions: Borrowing rate = 8% p.a. Lending rate = 12% p.a. Time = 2 years Profit = Rs 320 Concept / Approach: Profit = P * (rate spread) * t = P * (0.12 − 0.08) * 2 = 0.08 * P. Solve 0.08 * P = 320 for P. Step-by-Step Solution: 0.08 * P = 320 ⇒ P = 320 / 0.08 = Rs 4000. Verification / Alternative check: Interest received = 4000 * 12% * 2 = 960; paid = 4000 * 8% * 2 = 640; profit = 320 (matches). Why Other Options Are Wrong: 2000, 3000, 5000, 6000 do not yield a spread-based profit of 320 over 2 years. Common Pitfalls: Using 12% − 8% = 2% instead of 4%; the spread is 4% per annum. Final Answer: Rs 4000.

Question 9

Simple Interest — Tiered rates; interest exceeds principal: Pratap borrows at simple interest: 12% p.a. for 3 years, 16% p.a. for the next 5 years, and 20% p.a. thereafter. Over 11 years, the total SI exceeds the principal by Rs 6080. Find the princ…

Accepted Answer

Introduction / Context: When SI over a period is stated to exceed the principal by a known amount, we form an equation where (total SI factor) * P = P + excess, and solve for P after computing the piecewise factor. Given Data / Assumptions: Years 1–3: 12% ⇒ 0.12 * 3 = 0.36 Years 4–8: 16% ⇒ 0.16 * 5 = 0.80 Years 9–11: 20% ⇒ 0.20 * 3 = 0.60 Total SI = P + Rs 6080 (i.e., exceeds principal by 6080) Concept / Approach: Compute SI factor f = 0.36 + 0.80 + 0.60 = 1.76. Set f * P = P + 6080 ⇒ (1.76 − 1) P = 6080 ⇒ 0.76 P = 6080 ⇒ P = 8000. Step-by-Step Solution: f = 1.76; f * P = P + 60800.76 P = 6080 ⇒ P = 8000. Verification / Alternative check: Total SI on 8000: 2880 + 6400 + 4800 = 14080. Indeed 14080 = 8000 + 6080. Why Other Options Are Wrong: Other principals do not satisfy (1.76 − 1) P = 6080. Common Pitfalls: Interpreting “exceeds by” as “equals”. Final Answer: Rs 8000.

Question 10

Simple Interest — Three parts with equal interest earnings: Rs 1440 is lent in three parts so that the interest on the first at 2% for 3 years, the second at 3% for 4 years, and the third at 4% for 5 years are equal. What is the difference between t…

Accepted Answer

Introduction / Context: Equal SI on different parts over different time–rate products implies those parts are inversely proportional to the respective (rate * time) values. With the total known, we apportion by the reciprocal weights and compare extremes. Given Data / Assumptions: Part1: r * t = 0.02 * 3 = 0.06 Part2: r * t = 0.03 * 4 = 0.12 Part3: r * t = 0.04 * 5 = 0.20 Sum of parts = Rs 1440 All three interests equal Concept / Approach: If I1 = I2 = I3, then P1 * 0.06 = P2 * 0.12 = P3 * 0.20 = k. Hence P1 : P2 : P3 = 1/0.06 : 1/0.12 : 1/0.20 = 16.666... : 8.333... : 5 = 50 : 25 : 15 upon scaling. Step-by-Step Solution: Total ratio units = 50 + 25 + 15 = 90.Each unit = 1440 / 90 = 16.P1 = 50 * 16 = 800; P2 = 25 * 16 = 400; P3 = 15 * 16 = 240.Difference (largest − smallest) = 800 − 240 = 560. Verification / Alternative check: Check equal interests: 0.06*800 = 48; 0.12*400 = 48; 0.20*240 = 48. Why Other Options Are Wrong: Other numeric differences do not match the proportional allocation. Common Pitfalls: Proportioning directly with r or t alone instead of r * t product. Final Answer: ₹ 560.

Question 11

Simple Interest — Mixed fractions of a fund at different rates: One-third of a sum is at 3% p.a., one-sixth at 6% p.a., and the rest at 8% p.a. If total SI for 2 years is ₹ 600, find the original sum.

Accepted Answer

Introduction / Context: This is a weighted-average SI problem across sub-parts of one principal. Each fraction contributes SI proportional to its share, rate, and time; the total is the sum of these contributions over 2 years. Given Data / Assumptions: Total principal S (unknown) 1/3 at 3% p.a.; 1/6 at 6% p.a.; remainder at 8% p.a. Time = 2 years; total SI = ₹ 600 Concept / Approach: Total SI = S * [ (1/3)*0.03*2 + (1/6)*0.06*2 + (1 − 1/3 − 1/6)*0.08*2 ]. Evaluate the bracket and equate to 600 to solve for S. Step-by-Step Solution: Bracket = (1/3)*0.06 + (1/6)*0.12 + (1/2)*0.16 = 0.02 + 0.02 + 0.08 = 0.12.0.12 * S = 600 ⇒ S = 600 / 0.12 = ₹ 5000. Verification / Alternative check: Compute SI parts on ₹ 5000: 5000*(1/3)*0.03*2 = 100; 5000*(1/6)*0.06*2 = 100; 5000*(1/2)*0.08*2 = 400; total = 600. Why Other Options Are Wrong: Other sums do not satisfy the 0.12 * S = 600 condition. Common Pitfalls: Forgetting that the leftover fraction is 1 − 1/3 − 1/6 = 1/2. Final Answer: ₹ 5000.

Question 12

Simple Interest — Deferred balance with SI on the remainder: A music system costs ₹ 8000. A buyer pays ₹ 3500 now and the balance after 18 months with simple interest at 8% p.a. Find the total amount paid.

Accepted Answer

Introduction / Context: When a balance is carried forward with simple interest, the interest accrues only on the unpaid remainder for the specified time. The total outlay is the upfront payment plus the future payment including its interest. Given Data / Assumptions: Cash price = ₹ 8000 Paid now = ₹ 3500 → balance principal = ₹ 4500 Interest on balance: r = 8% p.a.; t = 18 months = 1.5 years Concept / Approach: Interest on balance = 4500 * 0.08 * 1.5. Total paid = 3500 + 4500 + interest. Step-by-Step Solution: Interest = 4500 * 0.08 * 1.5 = ₹ 540.Total = 3500 + 4500 + 540 = ₹ 8540. Verification / Alternative check: Balance with SI equals ₹ 5040 after 18 months; add the cash payment to confirm total. Why Other Options Are Wrong: Other totals imply incorrect interest on the balance or wrong time conversion. Common Pitfalls: Treating 18 months as 18 years or forgetting to convert to 1.5 years. Final Answer: ₹ 8540.

Question 13

Simple Interest — Two sums at 6% and 7% with a linking condition: A man invests one sum at 6% p.a. SI and another at 7% p.a. SI. In 2 years, the combined interest is ₹ 792. Also, half of the first sum equals one-third of the second sum. Find the tot…

Accepted Answer

Introduction / Context: Multiple investments with a relationship between principals can be solved by converting the relation into a single variable and using the total interest condition under SI. This avoids solving a full two-variable system from scratch. Given Data / Assumptions: P at 6% p.a., Q at 7% p.a., both for 2 years Total interest in 2 years = ₹ 792 (1/2) P = (1/3) Q ⇒ Q = (3/2) P Concept / Approach: Total SI = P * 0.06 * 2 + Q * 0.07 * 2 = 0.12 P + 0.14 Q. Substitute Q = 1.5 P to get a single equation in P; then compute Q and P + Q. Step-by-Step Solution: 0.12 P + 0.14 (1.5 P) = 0.12 P + 0.21 P = 0.33 P = 792.P = 792 / 0.33 = 2400; Q = 1.5 * 2400 = 3600.Total = 2400 + 3600 = ₹ 6000. Verification / Alternative check: Interest check: 2400*0.12 + 3600*0.14 = 288 + 504 = 792 (over 2 years combined factors already applied). Why Other Options Are Wrong: 5800, 5900, 6100, 6200 do not satisfy both the interest total and the linkage. Common Pitfalls: Mistaking (1/2)P = (1/3)Q into Q = (2/3)P; correct is Q = (3/2)P. Final Answer: ₹ 6000.

Question 14

Simple Interest — Borrow at 9%, add own money, lend at 12%: A borrows ₹ 4800 at 9% p.a. SI for 3 years. He adds some of his own money and lends the total at 12% p.a. SI for the same 3 years. If his net gain is ₹ 720, how much did he add?

Accepted Answer

Introduction / Context: Net profit in a borrow-and-lend setup equals interest earned on the lent total minus interest paid on the borrowed amount. With equal terms and simple interest, it becomes a straightforward linear equation in the added amount. Given Data / Assumptions: Borrowed principal = ₹ 4800 at 9% for 3 years Let added amount = m; lent principal = 4800 + m at 12% for 3 years Net gain over 3 years = ₹ 720 Concept / Approach: Profit = (4800 + m) * 0.12 * 3 − 4800 * 0.09 * 3. Set equal to 720 and solve for m. Step-by-Step Solution: Profit = (4800 + m)*0.36 − 4800*0.27 = 1728 + 0.36 m − 1296 = 432 + 0.36 m.432 + 0.36 m = 720 ⇒ 0.36 m = 288 ⇒ m = 800. Verification / Alternative check: Interest received: (5600)*0.36 = 2016; paid: 1296; difference = 720 (matches). Why Other Options Are Wrong: 500, 640, 740, 900 do not balance the profit equation to 720. Common Pitfalls: Mixing rates or durations; both are 3 years so factors are comparable. Final Answer: ₹ 800.

Question 15

Simple Interest — Reverse present worth via SI discounting: A person closes an investment by withdrawing ₹ 10000 today, ₹ 6000 one year ago, and ₹ 5000 two years ago. Three years ago there was no withdrawal. Assuming 10% p.a. simple interest since o…

Accepted Answer

Introduction / Context: With simple interest, the present worth of withdrawals at different times (counted from opening) uses linear discounting: PV at opening = Withdrawal / (1 + r * t). Summing the present worths of all withdrawals reconstructs the initial deposit, assuming the scheme accrued SI uniformly from the opening on the original principal. Given Data / Assumptions: Annual simple interest r = 10% = 0.10 Opening was 4 years ago Withdrawals: ₹ 5000 at t = 2 years, ₹ 6000 at t = 3 years, ₹ 10000 at t = 4 years Goal: approximate initial deposit (no other cash flows) Concept / Approach: Under SI, accumulation to a future date uses A = P * (1 + r * t). Conversely, discounting a known future amount W back to opening uses PV = W / (1 + r * t). The initial principal is the sum of the PVs of all withdrawals. Step-by-Step Solution: PV(₹ 5000 at t = 2) = 5000 / (1 + 0.10 * 2) = 5000 / 1.20 ≈ 4166.67PV(₹ 6000 at t = 3) = 6000 / (1 + 0.10 * 3) = 6000 / 1.30 ≈ 4615.38PV(₹ 10000 at t = 4) = 10000 / (1 + 0.10 * 4) = 10000 / 1.40 ≈ 7142.86Initial deposit ≈ 4166.67 + 4615.38 + 7142.86 ≈ ₹ 15924.91. Verification / Alternative check: Rounding to the nearest option, ₹ 15600 is closest (diff ≈ ₹ 325), consistent with “approximately”. Why Other Options Are Wrong: ₹ 16500 and ₹ 17280 deviate farther from the computed present worth. ₹ 15000 underestimates the required initial deposit. Common Pitfalls: Using compound discounting; SI requires linear discount factors 1 + r * t. Mistiming the withdrawals (ensure t = 2, 3, 4 from opening). Final Answer: Approximately ₹ 15600.

Question 16

Simple Interest — Same duration at two rates; recover P and time: A certain sum becomes ₹ 600 at 6% p.a. SI in a certain time and ₹ 200 at 1% p.a. SI in the same time. Find the principal and the time.

Accepted Answer

Introduction / Context: Two simple-interest amounts with the same time but different rates provide two linear equations in principal P and time t. Solving them simultaneously yields both the initial sum and the duration, even when the time is unusually long. Given Data / Assumptions: A1 = P * (1 + 0.06 * t) = ₹ 600 A2 = P * (1 + 0.01 * t) = ₹ 200 Same t in both cases Concept / Approach: Divide equations or subtract to eliminate P or t. One direct approach is to solve as a 2×2 system for P and t in real numbers under SI linearity. Step-by-Step Solution: From A2: P = 200 / (1 + 0.01 t).Substitute in A1: 200 * (1 + 0.06 t) / (1 + 0.01 t) = 600 ⇒ (1 + 0.06 t) / (1 + 0.01 t) = 3.Cross-multiply: 1 + 0.06 t = 3 + 0.03 t ⇒ 0.03 t = 2 ⇒ t = 66.666... years.Then P = 200 / (1 + 0.6666...) = 200 / 1.6666... = ₹ 120. Verification / Alternative check: Check A1: 120 * (1 + 0.06 * 66.666...) = 120 * 5 = 600; A2: 120 * (1 + 0.01 * 66.666...) = 120 * 1.6666... = 200. Why Other Options Are Wrong: Other principals do not satisfy both amounts simultaneously for one common t. Common Pitfalls: Assuming realistic times; math permits long durations under SI. Final Answer: ₹ 120 and 66 2/3 years.

Question 17

Simple Interest — Compare two scenarios to find principal: SI on a sum for 2 1/2 years at 12% p.a. is ₹ 20 less than the SI on the same sum for 3 1/2 years at 10% p.a. Find the sum.

Accepted Answer

Introduction / Context: Comparative SI statements produce a single linear equation in the principal when time and rate differ. Careful reading of “less than” ensures the equation’s direction is correct. Given Data / Assumptions: SI1 (2.5 years at 12%) is ₹ 20 less than SI2 (3.5 years at 10%). Let principal be P. Concept / Approach: Write P * 0.12 * 2.5 + 20 = P * 0.10 * 3.5 and solve for P. All quantities are linear in P under SI. Step-by-Step Solution: Left side = 0.30 P + 20; right side = 0.35 P.0.30 P + 20 = 0.35 P ⇒ 20 = 0.05 P ⇒ P = ₹ 400. Verification / Alternative check: SI1 = 0.30 * 400 = ₹ 120; SI2 = 0.35 * 400 = ₹ 140; indeed SI1 is ₹ 20 less than SI2. Why Other Options Are Wrong: Other values of P fail to keep a consistent ₹ 20 gap between the two computed SIs. Common Pitfalls: Reversing “less than” and writing 0.30 P = 0.35 P + 20, which yields a negative P. Final Answer: ₹ 400.

Question 18

Simple Interest — A sum is lent for a fixed time. At 10% per annum simple interest it amounts to ₹ 400, and at 4% per annum for the same time it amounts to ₹ 200. Find the original principal (sum).

Accepted Answer

Introduction / Context: When the same principal is invested for the same time under simple interest at two different rates, the two final amounts give two linear equations in the principal and time. Solving those equations yields the principal directly. Given Data / Assumptions: Let principal be P and time be t years (same in both cases). At 10% p.a., amount A1 = ₹ 400. At 4% p.a., amount A2 = ₹ 200. Simple interest applies (linear growth). Concept / Approach: Under simple interest: A = P * (1 + r * t), with r as a decimal per year. Form two equations and solve for P and t. Step-by-Step Solution: P * (1 + 0.10 t) = 400 … (1)P * (1 + 0.04 t) = 200 … (2)Divide (1) by (2): (1 + 0.10 t) / (1 + 0.04 t) = 2.1 + 0.10 t = 2 + 0.08 t ⇒ 0.02 t = 1 ⇒ t = 50 years.From (2): P = 200 / (1 + 0.04 * 50) = 200 / 3 = ₹ 200/3. Verification / Alternative check: At 10% for 50 years: A1 = (200/3) * (1 + 5) = 200/3 * 6 = 400. At 4%: A2 = (200/3) * (1 + 2) = 200/3 * 3 = 200. Both match. Why Other Options Are Wrong: ₹ 100, ₹ 400/3, ₹ 500/3 do not satisfy both amounts simultaneously with a single t under simple interest. Common Pitfalls: Confusing simple interest with compound interest or assuming different times; both scenarios share the same t here. Final Answer: ₹ 200/3

Question 19

Simple Interest — Equal annual payments of ₹ 160 are made at the end of each year for 5 years. At 5% per annum simple interest, these payments together will discharge a debt of how much (amount due at the end of year 5)?

Accepted Answer

Introduction / Context: With simple interest, the future value at a common focal date (here, end of year 5) of each payment is found by adding linear interest for the remaining years to the focal date. The sum of these accumulated values equals the debt amount due at that focal date. Given Data / Assumptions: Payment each year (end): ₹ 160. Number of payments: 5 (at ends of years 1 to 5). Rate: 5% per annum, simple interest. Focal date: end of year 5. Concept / Approach: Accumulate each installment to end of year 5 using simple interest: FV_k = 160 * (1 + r * (5 - k)), for k = 1…5. Sum all five future values to get the total amount these payments can discharge at the focal date. Step-by-Step Solution: Accumulation factors to end year 5: for k=1: (1+0.05*4)=1.20; k=2: 1.15; k=3: 1.10; k=4: 1.05; k=5: 1.00.Sum of factors = 1.20 + 1.15 + 1.10 + 1.05 + 1.00 = 5.50.Total future value = 160 * 5.50 = ₹ 880. Verification / Alternative check: Each payment grows linearly by 5% per year for the remaining years; arithmetic series of factors ensures a neat sum. Why Other Options Are Wrong: ₹ 980 and ₹ 1,100 overstate the accumulation; ₹ 440 and ₹ 220 undercount or ignore time value. Common Pitfalls: Discounting (present value) instead of accumulating to a future date, or incorrectly compounding under SI. Final Answer: ₹ 880

Question 20

Simple Interest — What equal annual payment will discharge a debt of ₹ 1092 due at the end of 2 years, if money earns 12% per annum simple interest?

Accepted Answer

Introduction / Context: To discharge a debt due at a given future date using equal yearly payments under simple interest, accumulate each installment to the due date and set their sum equal to the amount owed. Given Data / Assumptions: Debt due at end of year 2: ₹ 1092. Equal payments A at ends of years 1 and 2. Rate r = 12% per annum, simple interest. Concept / Approach: Accumulate payments to end of year 2. First payment (end year 1) earns SI for 1 year: A*(1 + 0.12). Second payment occurs at focal date, so it contributes A. Equation: A*(1 + 0.12) + A = 1092. Step-by-Step Solution: A*(2 + 0.12) = 1092 ⇒ 2.12A = 1092.A = 1092 / 2.12 = ₹ 515. Verification / Alternative check: A = 515 gives 515*1.12 + 515 = 576.8 + 515 = 1091.8 ≈ 1092 (exact with paise rounding ignored in options). Why Other Options Are Wrong: ₹ 725, ₹ 900 overshoot; ₹ 325 and ₹ 545 fail to match the accumulation equation. Common Pitfalls: Using compound interest instead of SI or forgetting that the second payment does not accrue further interest. Final Answer: ₹ 515

Simple Interest Questions