Interest is payment for using money over time. Principal P is the starting amount, rate r is usually expressed per year, and time t must be expressed in the same period as the rate. The distinction between simple and compound interest is the balance on which each period's interest is calculated.
1. Simple interest
With annual rate r% and time t years, simple interest SI=Prt/100 and amount A=P+SI. Every year's interest is calculated on the original principal, so the yearly addition is constant. If the question gives months, use t=months/12 for an annual rate unless a different convention is specified.
Worked example 1. At 8% per year, ₹7,500 earns simple interest for 18 months. Time is 1.5 years, so SI=7,500×8×1.5/100=₹900. The amount is ₹8,400. Using 18 as years would inflate the answer twelvefold.
When any three of SI, P, r and t are known, rearrange the formula for the fourth. Check units and reasonableness: 10% for one year adds one-tenth of P; 10% for three simple-interest years adds three-tenths of P.
2. Compound interest
Under annual compounding, each year's interest joins the balance and earns interest thereafter. The amount after n whole years is A=P(1+r/100)^n, and compound interest CI=A−P. Unlike simple interest, the interest added each year increases at a positive rate. Show an amount table when the exponent is small; it makes the changed base visible.
Worked example 2. ₹10,000 at 10% compounded annually becomes ₹11,000 after one year and ₹12,100 after two. CI is ₹2,100. Simple interest for the same two years would be ₹2,000; the extra ₹100 is interest on the first year's interest.
3. Compounding frequency
If an annual nominal rate is compounded half-yearly, divide the annual rate by two and double the number of years expressed as periods. For 12% per annum over one year with half-yearly compounding, use 6% for each of two periods: A=P(1.06)^2. Quarterly compounding uses four periods per year at one-quarter of the annual nominal rate. Do not divide the time and the rate in the same direction.
Worked example 3. ₹5,000 at 12% per annum compounded half-yearly for one year becomes 5,000×(1.06)^2=₹5,618. The CI is ₹618, compared with ₹600 for simple interest at 12% over one year.
4. Changing rates and partial years
When rates differ by year, use a separate factor for each period. A principal P at 10% in year one and 12% in year two becomes P×1.10×1.12. No single arithmetic average of the rates reproduces the exact result unless the question explicitly accepts an approximation.
For a fractional year, compounding convention matters. A common school-exam convention compounds for complete years and uses simple interest on the resulting amount for the remaining months. Apply that only if the question or its exam convention indicates it; an institution may use more frequent compounding. If the wording is silent and options make different conventions possible, the item is ambiguous and should be rejected rather than guessed.
5. Difference between CI and SI
For two years at annual rate r%, CI−SI=P(r/100)^2. It is the second year's interest on the first year's interest. For three years, additional terms appear; do not reuse the two-year formula. A binomial expansion or year-by-year table is safer.
Worked example 4. If the two-year difference is ₹80 at 10% annually, 80=P×0.10²=P/100, so P=₹8,000. Check: SI=₹1,600; CI=8,000×1.1²−8,000=₹1,680.
6. Growth, depreciation and reverse amounts
Compound-interest factors also describe repeated growth and depreciation. A machine losing 15% of its value yearly retains 0.85^n of its value after n years. To recover a starting value, divide the final value by the compounded factor. A decrease followed by an equal increase does not restore the original, because the second change uses a smaller base.
Recall before practice
1. Explain the amount on which year-two interest is calculated in SI and CI. 2. Write the factor for 12% nominal annual rate compounded half-yearly for one year. 3. Derive the two-year CI−SI difference from a year-by-year table. 4. State why a partial-year problem needs a compounding convention.
Chapter 9 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. Find the simple interest on ₹3,250 at 6% per annum for 6 years.
A. ₹14,040 B. ₹4,420 C. ₹1,170 D. ₹585
2. Find the compound interest on ₹12,800 at 10% per annum for 1 year, the interest being compounded half-yearly.
A. ₹1,312 B. ₹640 C. ₹1,280 D. ₹2,688
3. Bank A offers a loan of ₹1,00,000 at 4.5% per annum simple interest and bank B offers the same loan at 8% per annum. What is the difference in the interest to be paid to the two banks after 2 years?
A. ₹16,000 B. ₹9,000 C. ₹7,000 D. ₹3,500
4. What is the difference between the compound interest and the simple interest on ₹26,800 for 2 years at 10% per annum?
A. ₹536 B. ₹5,360 C. ₹268 D. ₹2,680
5. Find the compound interest on ₹62,500 at 16% per annum for 1 year, the interest being compounded half-yearly.
A. ₹10,000 B. ₹21,600 C. ₹5,000 D. ₹10,400
6. The simple interest on a certain sum at 11% per annum for 2 years is ₹880. What is the sum?
A. ₹8,000 B. ₹4,000 C. ₹44,000 D. ₹1,760
7. Bank A offers a loan of ₹3,00,000 at 6% per annum simple interest and bank B offers the same loan at 8.5% per annum. What is the difference in the interest to be paid to the two banks after 2 years?
A. ₹7,500 B. ₹51,000 C. ₹36,000 D. ₹15,000
8. Find the compound interest on ₹5,056 for 2 years at 25% per annum, compounded annually.
A. ₹1,264 B. ₹7,900 C. ₹2,528 D. ₹2,844
9. Find the compound interest on ₹35,000 at 12% per annum for 1 year, the interest being compounded half-yearly.
A. ₹4,200 B. ₹2,100 C. ₹8,904 D. ₹4,326
10. In how many years will a sum of ₹7,000 yield a simple interest of ₹2,800 at 8% per annum?
A. 6 years B. 5 years C. 10 years D. 4 years
11. Find the compound interest on ₹70,800 for 2 years at 5% per annum, compounded annually.
A. ₹78,057 B. ₹7,080 C. ₹7,257 D. ₹3,540
12. In how many years will a sum of ₹8,000 yield a simple interest of ₹2,880 at 4.5% per annum?
A. 8 years B. 16 years C. 7 years D. 9 years
13. What amount will ₹73,712 become in 2 years at 25% per annum compound interest, compounded annually?
A. ₹41,463 B. ₹92,140 C. ₹1,10,568 D. ₹1,15,175
14. Find the simple interest on ₹2,000 at 7% per annum for 6 years.
A. ₹10,080 B. ₹420 C. ₹840 D. ₹2,840
15. A sum of ₹11,500 amounts to ₹14,375 in 2 years at simple interest. At the same rate, what simple interest will ₹19,000 earn in 8 years?
A. ₹4,750 B. ₹38,000 C. ₹19,000 D. ₹20,520
16. Suresh lent ₹22,500 in two parts, one at 5% per annum and the other at 12% per annum simple interest. The total interest after 3 years was ₹5,475. What was the part lent at 5%?
A. ₹11,250 B. ₹12,500 C. ₹10,000 D. ₹36,500
17. Ravi lent ₹14,500 in two parts, one at 4% per annum and the other at 12% per annum simple interest. The total interest after 2 years was ₹2,040. What was the part lent at 4%?
A. ₹5,500 B. ₹25,500 C. ₹7,250 D. ₹9,000
18. A loan at 25% per annum compound interest is repaid in 2 equal annual instalments of ₹800 each. What was the amount of the loan?
A. ₹1,200 B. ₹1,152 C. ₹1,280 D. ₹1,600
19. The simple interest on a sum for 2 years at 10% per annum is ₹8,020. What is the compound interest on the same sum for the same period at the same rate?
A. ₹8,822 B. ₹8,421 C. ₹8,020 D. ₹8,431
20. Bank A offers a loan of ₹3,50,000 at 5.5% per annum simple interest and bank B offers the same loan at 9% per annum. What is the difference in the interest to be paid to the two banks after 2 years?
A. ₹24,500 B. ₹38,500 C. ₹12,250 D. ₹63,000
21. Find the compound interest on ₹23,750 at 8% per annum for 1 year, the interest being compounded half-yearly.
A. ₹1,900 B. ₹1,938 C. ₹950 D. ₹3,952
22. A sum of ₹1,440 is divided between two sons so that, invested at 25% per annum compound interest, the elder receives his share after 9 years and the younger after 10 years, both getting equal amounts. What is the younger son's share?
A. ₹800 B. ₹672 C. ₹720 D. ₹640
23. Find the simple interest on ₹36,500 at 6% per annum from 21 August to 3 October of the same (non-leap) year.
A. ₹365 B. ₹252 C. ₹264 D. ₹258
24. A sum of ₹2,000 amounts to ₹2,320 in 4 years at simple interest. At the same rate, what simple interest will ₹16,000 earn in 10 years?
A. ₹6,400 B. ₹22,400 C. ₹2,560 D. ₹8,000
25. What amount will ₹78,125 become in 3 years at 8% per annum compound interest, compounded annually?
A. ₹91,125 B. ₹98,415 C. ₹20,290 D. ₹96,875
26. Sanjay lent ₹16,000 in two parts, one at 4% per annum and the other at 10% per annum simple interest. The total interest after 2 years was ₹2,360. What was the part lent at 4%?
A. ₹7,000 B. ₹8,000 C. ₹29,500 D. ₹9,000
27. A sum of money becomes ₹37,268 in 3 years at 10% per annum compound interest. What is the sum?
A. ₹28,500 B. ₹29,400 C. ₹30,800 D. ₹28,000
28. Find the simple interest on ₹6,500 at 9% per annum for 20 months.
A. ₹11,700 B. ₹1,950 C. ₹975 D. ₹7,475
29. A sum amounts to ₹24,500 in 5 years at 15% per annum simple interest. What is the simple interest on the same sum for 1 year at the same rate?
A. ₹10,500 B. ₹4,200 C. ₹2,100 D. ₹3,675
30. A sum of money amounts to ₹7,800 in 5 years at 4% per annum simple interest. What is the sum?
A. ₹7,150 B. ₹6,500 C. ₹39,000 D. ₹6,240
31. Find the compound interest on ₹15,000 for 2 years at 4% per annum, compounded annually.
A. ₹1,200 B. ₹16,224 C. ₹600 D. ₹1,224
32. A sum of money becomes ₹1,331 in 3 years at 10% per annum compound interest. What is the sum?
A. ₹1,500 B. ₹1,050 C. ₹1,100 D. ₹1,000
33. A sum amounts to ₹8,700 in 2 years at 8% per annum simple interest. What is the simple interest on the same sum for 1 year at the same rate?
A. ₹900 B. ₹1,200 C. ₹696 D. ₹600
34. In how many years will ₹4,608 become ₹6,561 at 12.5% per annum compound interest, compounded annually?
A. 5 years B. 3 years C. 4 years D. 2 years
35. The difference between the compound interest and the simple interest on ₹50,512 for 2 years is ₹3,157. What is the rate of interest per annum?
A. 50% B. 3⅛% C. 25% D. 6.25%
36. A sum of money becomes ₹73,002 in 3 years at 15% per annum compound interest. What is the sum?
A. ₹45,600 B. ₹55,200 C. ₹48,000 D. ₹48,500
37. In how many years will ₹19,875 become ₹34,344 at 20% per annum compound interest, compounded annually?
A. 2 years B. 5 years C. 3 years D. 4 years
38. The simple interest on a certain sum for 3½ years at 15% per annum exceeds the simple interest on the same sum for 2½ years at 8% per annum by ₹4,225. Find the sum.
A. ₹13,000 B. ₹10,400 C. ₹26,000 D. ₹6,500
39. In how many years will a sum of ₹1,60,000 yield a simple interest of ₹64,000 at 10% per annum?
A. 3 years B. 5 years C. 4 years D. 8 years
40. A loan of ₹3,825 at 4% per annum compound interest is to be repaid in 2 equal annual instalments. What is the value of each instalment?
A. ₹1,825 B. ₹2,434 C. ₹2,028 D. ₹1,989
41. Mohan invests ₹4,500 and Vikas invests ₹14,500 at the same rate of simple interest. After 5 years, Vikas receives ₹5,500 more interest than Mohan. What is the rate of interest per annum?
A. 12% B. 11% C. 55% D. 10%
42. A sum amounts to ₹16,400 in 5 years at 12% per annum simple interest. What is the simple interest on the same sum for 8 years at the same rate?
A. ₹6,150 B. ₹9,840 C. ₹15,744 D. ₹9,890
43. A sum of ₹3,000 amounts to ₹3,900 in 5 years at simple interest. At the same rate, what simple interest will ₹31,000 earn in 2 years?
A. ₹3,720 B. ₹34,720 C. ₹9,300 D. ₹4,340
44. The compound interest on a certain sum for 2 years at 5% per annum, compounded annually, is ₹1,353. What is the sum?
A. ₹13,530 B. ₹13,200 C. ₹14,553 D. ₹6,765
45. The simple interest on a certain sum for 3½ years at 15% per annum exceeds the simple interest on the same sum for 2½ years at 6% per annum by ₹1,875. Find the sum.
A. ₹10,000 B. ₹5,000 C. ₹5,250 D. ₹2,500
46. Find the amount on ₹24,400 for 6 months at 20% per annum, the interest being compounded quarterly.
A. ₹26,901 B. ₹26,840 C. ₹26,951 D. ₹29,280
47. The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is ₹1,891. Find the sum.
A. ₹30,500 B. ₹1,89,100 C. ₹61,000 D. ₹1,22,000
48. A television is available for ₹5,500 cash, or for ₹2,500 cash down followed by a single payment of ₹3,135 after 3 months. What rate of simple interest per annum is charged under the instalment plan?
A. 18% B. 20% C. 19% D. 4.5%
49. A sum invested for 2 years at 20% per annum compound interest earns ₹13,376 as interest. If twice that sum is invested for 2 years at 12.5% per annum compound interest, what compound interest will it earn?
A. ₹15,200 B. ₹16,720 C. ₹26,752 D. ₹16,150
50. Find the amount on ₹50,000 for 3 years if the rates of interest for the first, second and third years are 5%, 6% and 6% per annum respectively.
A. ₹58,989 B. ₹58,500 C. ₹59,089 D. ₹8,989
Chapter 9 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. C. Simple interest = P × R × T/100 with P = ₹3,250, R = 6% and T = 6 years: ₹3,250 × 6 × 6/100 = ₹1,170.
APAR26-09-01 | Simple interest (direct) | Easy
2. A. Half-yearly: rate per period = 10/2 = 5%, periods = 1 × 2 = 2. A = P(1 + 5/100)^2 = ₹12,800 × (21/20)^2 = ₹14,112. CI = ₹14,112 − ₹12,800 = ₹1,312. (Compounding annually would give CI ₹1,280; simple interest ₹1,280.)
APAR26-09-28 | Half-yearly compounding | Easy
3. C. SI = P × R × T/100. Bank A: ₹1,00,000 × 4.5 × 2/100 = ₹9,000; bank B: ₹1,00,000 × 8 × 2/100 = ₹16,000. Difference = ₹16,000 − ₹9,000 = ₹7,000. Shortcut: P × (R₂ − R₁) × T/100 = ₹1,00,000 × 3.5 × 2/100 = ₹7,000.
APAR26-09-06 | Difference in SI at two rates | Easy
4. C. For 2 years, CI − SI = P(r/100)² = ₹26,800 × (10/100)² = ₹26,800 × 1/100 = ₹268. (It is the one year's interest on the first year's interest ₹2,680: ₹2,680 × 10/100 = ₹268.)
APAR26-09-27 | CI − SI for 2 years | Easy
5. D. Half-yearly: rate per period = 16/2 = 8%, periods = 1 × 2 = 2. A = P(1 + 8/100)^2 = ₹62,500 × (27/25)^2 = ₹72,900. CI = ₹72,900 − ₹62,500 = ₹10,400. (Compounding annually would give CI ₹10,000; simple interest ₹10,000.)
APAR26-09-31 | Half-yearly compounding | Easy
6. B. P = SI × 100/(R × T) = ₹880 × 100/(11 × 2) = ₹88,000/22 = ₹4,000.
APAR26-09-05 | Principal from SI | Easy
7. D. SI = P × R × T/100. Bank A: ₹3,00,000 × 6 × 2/100 = ₹36,000; bank B: ₹3,00,000 × 8.5 × 2/100 = ₹51,000. Difference = ₹51,000 − ₹36,000 = ₹15,000. Shortcut: P × (R₂ − R₁) × T/100 = ₹3,00,000 × 2.5 × 2/100 = ₹15,000.
APAR26-09-03 | Difference in SI at two rates | Easy
8. D. A = P(1 + 25/100)^2 = ₹5,056 × (5/4)^2 = ₹7,900 (year by year: ₹6,320 → ₹7,900). CI = A − P = ₹7,900 − ₹5,056 = ₹2,844. (SI would be ₹2,528.)
APAR26-09-26 | Compound interest (annual) | Easy
9. D. Half-yearly: rate per period = 12/2 = 6%, periods = 1 × 2 = 2. A = P(1 + 6/100)^2 = ₹35,000 × (53/50)^2 = ₹39,326. CI = ₹39,326 − ₹35,000 = ₹4,326. (Compounding annually would give CI ₹4,200; simple interest ₹4,200.)
APAR26-09-29 | Half-yearly compounding | Easy
10. B. T = SI × 100/(P × R) = ₹2,800 × 100/(₹7,000 × 8) = ₹2,80,000/₹56,000 = 5 years.
APAR26-09-07 | Find the time | Easy
11. C. A = P(1 + 5/100)^2 = ₹70,800 × (21/20)^2 = ₹78,057 (year by year: ₹74,340 → ₹78,057). CI = A − P = ₹78,057 − ₹70,800 = ₹7,257. (SI would be ₹7,080.)
APAR26-09-32 | Compound interest (annual) | Easy
12. A. T = SI × 100/(P × R) = ₹2,880 × 100/(₹8,000 × 4.5) = ₹2,88,000/₹36,000 = 8 years.
APAR26-09-02 | Find the time | Easy
13. D. A = P(1 + 25/100)^2 = ₹73,712 × (5/4)^2 = ₹1,15,175. (Simple interest would give only ₹1,10,568; the extra ₹4,607 is interest on interest.)
APAR26-09-30 | Amount at compound interest | Easy
14. C. Simple interest = P × R × T/100 with P = ₹2,000, R = 7% and T = 6 years: ₹2,000 × 7 × 6/100 = ₹840.
APAR26-09-04 | Simple interest (direct) | Easy
15. C. SI on ₹11,500 = ₹14,375 − ₹11,500 = ₹2,875 in 2 years, so R = ₹2,875 × 100/(₹11,500 × 2) = 12.5% p.a. Then SI on ₹19,000 for 8 years = ₹19,000 × 12.5 × 8/100 = ₹19,000.
APAR26-09-19 | Rate from one investment applied to another | Medium
16. B. Let the part at 5% be x; the rest is ₹22,500 − x. Interest: x × 5 × 3/100 + (₹22,500 − x) × 12 × 3/100 = ₹5,475. If the whole sum were at 12%, interest = ₹8,100; the shortfall ₹2,625 comes from x earning 7% × 3 less: x = ₹2,625 × 100/21 = ₹12,500.
APAR26-09-20 | Sum lent in two parts, total interest known | Medium
17. D. Let the part at 4% be x; the rest is ₹14,500 − x. Interest: x × 4 × 2/100 + (₹14,500 − x) × 12 × 2/100 = ₹2,040. If the whole sum were at 12%, interest = ₹3,480; the shortfall ₹1,440 comes from x earning 8% × 2 less: x = ₹1,440 × 100/16 = ₹9,000.
APAR26-09-12 | Sum lent in two parts, total interest known | Medium
18. B. Present value of the instalments must equal the loan: ₹800 × (4/5) + ₹800 × (4/5)^2 = ₹640 + ₹512 = ₹1,152. (Dividing the loan by 2 or using simple interest ignores discounting each instalment by (1 + r)ᵏ.)
APAR26-09-34 | Equal annual instalments at CI | Medium
19. B. For 2 years, CI = SI × (1 + r/200) = SI × (2 + 10/100)/2, i.e. CI/SI = 21/20. CI = ₹8,020 × 21/20 = ₹8,421. (Check: P = ₹40,100, yearly SI ₹4,010, extra CI ₹401 = interest on ₹4,010.)
APAR26-09-36 | CI from SI (2 years) | Medium
20. A. SI = P × R × T/100. Bank A: ₹3,50,000 × 5.5 × 2/100 = ₹38,500; bank B: ₹3,50,000 × 9 × 2/100 = ₹63,000. Difference = ₹63,000 − ₹38,500 = ₹24,500. Shortcut: P × (R₂ − R₁) × T/100 = ₹3,50,000 × 3.5 × 2/100 = ₹24,500.
APAR26-09-14 | Difference in SI at two rates | Medium
21. B. Half-yearly: rate per period = 8/2 = 4%, periods = 1 × 2 = 2. A = P(1 + 4/100)^2 = ₹23,750 × (26/25)^2 = ₹25,688. CI = ₹25,688 − ₹23,750 = ₹1,938. (Compounding annually would give CI ₹1,900; simple interest ₹1,900.)
APAR26-09-42 | Half-yearly compounding | Medium
22. D. Let the shares be x (elder) and y (younger). Equal amounts: x(1 + r)^9 = y(1 + r)^10 ⇒ x/y = (1 + r)^1 = (5/4) = 5/4. So the sum is split in the ratio 5 : 4: elder = ₹1,440 × 5/9 = ₹800, younger = ₹640. (The share invested for fewer years must be the larger one.)
APAR26-09-40 | Sum divided for equal amounts at different times | Medium
23. D. Count the days excluding the starting day: August 10 + September 30 + October 3 = 43 days. T = 43/365 year. SI = ₹36,500 × 6 × 43/(100 × 365) = ₹258.
APAR26-09-10 | SI between two dates | Medium
24. A. SI on ₹2,000 = ₹2,320 − ₹2,000 = ₹320 in 4 years, so R = ₹320 × 100/(₹2,000 × 4) = 4% p.a. Then SI on ₹16,000 for 10 years = ₹16,000 × 4 × 10/100 = ₹6,400.
APAR26-09-11 | Rate from one investment applied to another | Medium
25. B. A = P(1 + 8/100)^3 = ₹78,125 × (27/25)^3 = ₹98,415. (Simple interest would give only ₹96,875; the extra ₹1,540 is interest on interest.)
APAR26-09-41 | Amount at compound interest | Medium
26. A. Let the part at 4% be x; the rest is ₹16,000 − x. Interest: x × 4 × 2/100 + (₹16,000 − x) × 10 × 2/100 = ₹2,360. If the whole sum were at 10%, interest = ₹3,200; the shortfall ₹840 comes from x earning 6% × 2 less: x = ₹840 × 100/12 = ₹7,000.
APAR26-09-08 | Sum lent in two parts, total interest known | Medium
27. D. A = P(1 + 10/100)^3 ⇒ P = ₹37,268 ÷ (11/10)^3 = ₹37,268 × 1000/1331 = ₹28,000. (Dividing by (1 + 10 × 3/100), the SI factor, gives ₹28,667.69.)
APAR26-09-37 | Principal from amount | Medium
28. C. Simple interest = P × R × T/100 with P = ₹6,500, R = 9% and T = 20/12 years: ₹6,500 × 9 × 20/12/100 = ₹975. (Time must be in years: 20 months = 20/12 year.)
APAR26-09-13 | Simple interest (direct) | Medium
29. C. A = P(1 + RT/100) ⇒ P = ₹24,500 × 100/(100 + 15 × 5) = ₹24,500 × 100/175 = ₹14,000. SI for 1 year = ₹14,000 × 15 × 1/100 = ₹2,100. (Applying the rate to the amount ₹24,500 gives ₹3,675, which is wrong.)
APAR26-09-16 | SI for another period from a known amount | Medium
30. B. Amount = P(1 + RT/100) = P × (100 + 4 × 5)/100 = P × 120/100. So P = ₹7,800 × 100/120 = ₹6,500. (Treating ₹7,800 as the principal and subtracting interest gives ₹6,240, which is wrong.)
APAR26-09-15 | Principal from amount | Medium
31. D. A = P(1 + 4/100)^2 = ₹15,000 × (26/25)^2 = ₹16,224 (year by year: ₹15,600 → ₹16,224). CI = A − P = ₹16,224 − ₹15,000 = ₹1,224. (SI would be ₹1,200.)
APAR26-09-35 | Compound interest (annual) | Medium
32. D. A = P(1 + 10/100)^3 ⇒ P = ₹1,331 ÷ (11/10)^3 = ₹1,331 × 1000/1331 = ₹1,000. (Dividing by (1 + 10 × 3/100), the SI factor, gives ₹1,023.85.)
APAR26-09-45 | Principal from amount | Medium
33. D. A = P(1 + RT/100) ⇒ P = ₹8,700 × 100/(100 + 8 × 2) = ₹8,700 × 100/116 = ₹7,500. SI for 1 year = ₹7,500 × 8 × 1/100 = ₹600. (Applying the rate to the amount ₹8,700 gives ₹696, which is wrong.)
APAR26-09-17 | SI for another period from a known amount | Medium
34. B. A/P = ₹6,561/₹4,608 = 729/512 = (9/8)^3, and 1 + 12.5/100 = 9/8. Matching powers, n = 3 years. (The SI estimate (A − P) × 100/(P × r) = 3.39 is not the CI answer.)
APAR26-09-33 | Find the time | Medium
35. C. CI − SI (2 years) = P(r/100)² ⇒ (r/100)² = ₹3,157/₹50,512 = 1/16 ⇒ r/100 = 1/4 ⇒ r = 25%. (Formula: r = 100 × √(difference/P). Forgetting the square root gives 6.25%.)
APAR26-09-39 | Rate from CI − SI of 2 years | Medium
36. C. A = P(1 + 15/100)^3 ⇒ P = ₹73,002 ÷ (23/20)^3 = ₹73,002 × 8000/12167 = ₹48,000. (Dividing by (1 + 15 × 3/100), the SI factor, gives ₹50,346.21.)
APAR26-09-43 | Principal from amount | Medium
37. C. A/P = ₹34,344/₹19,875 = 216/125 = (6/5)^3, and 1 + 20/100 = 6/5. Matching powers, n = 3 years. (The SI estimate (A − P) × 100/(P × r) = 3.64 is not the CI answer.)
APAR26-09-44 | Find the time | Medium
38. A. SI₁ − SI₂ = P × (15 × 42/12 − 8 × 30/12)/100 = P × (52.5 − 20)/100 = P × 32.5/100 = ₹4,225. So P = ₹4,225 × 100/32.5 = ₹13,000. (Convert months to years before multiplying.)
APAR26-09-09 | Sum from SI difference of two (rate, time) pairs | Medium
39. C. T = SI × 100/(P × R) = ₹64,000 × 100/(₹1,60,000 × 10) = ₹64,00,000/₹16,00,000 = 4 years.
APAR26-09-18 | Find the time | Medium
40. C. Present value of the instalments must equal the loan: ₹2,028 × (25/26) + ₹2,028 × (25/26)^2 = ₹1,950 + ₹1,875 = ₹3,825, which gives each instalment = ₹2,028. (Dividing the loan by 2 or using simple interest ignores discounting each instalment by (1 + r)ᵏ.)
APAR26-09-38 | Equal annual instalments at CI | Medium
41. B. The extra interest comes only from the extra principal ₹14,500 − ₹4,500 = ₹10,000. So ₹10,000 × R × 5/100 = ₹5,500 ⇒ R = ₹5,500 × 100/(₹10,000 × 5) = 11%. (Using ₹14,500 instead of the difference is the usual slip.)
APAR26-09-22 | Two investors, find the rate | Difficult
42. B. A = P(1 + RT/100) ⇒ P = ₹16,400 × 100/(100 + 12 × 5) = ₹16,400 × 100/160 = ₹10,250. SI for 8 years = ₹10,250 × 12 × 8/100 = ₹9,840. (Applying the rate to the amount ₹16,400 gives ₹15,744, which is wrong.)
APAR26-09-24 | SI for another period from a known amount | Difficult
43. A. SI on ₹3,000 = ₹3,900 − ₹3,000 = ₹900 in 5 years, so R = ₹900 × 100/(₹3,000 × 5) = 6% p.a. Then SI on ₹31,000 for 2 years = ₹31,000 × 6 × 2/100 = ₹3,720.
APAR26-09-23 | Rate from one investment applied to another | Difficult
44. B. CI = P[(1 + 5/100)^2 − 1] = P × [(21/20)^2 − 1] = P × 41/400. So P = ₹1,353 × 400/41 = ₹13,200. (Using the SI formula P = CI × 100/(r × n) gives ₹13,530, which is wrong.)
APAR26-09-47 | Principal from CI | Difficult
45. B. SI₁ − SI₂ = P × (15 × 42/12 − 6 × 30/12)/100 = P × (52.5 − 15)/100 = P × 37.5/100 = ₹1,875. So P = ₹1,875 × 100/37.5 = ₹5,000. (Convert months to years before multiplying.)
APAR26-09-21 | Sum from SI difference of two (rate, time) pairs | Difficult
46. A. Quarterly: rate per quarter = 20/4 = 5%, number of quarters = 6/3 = 2. A = ₹24,400 × (1 + 5/100)^2 = ₹24,400 × (21/20)^2 = ₹26,901; CI = ₹26,901 − ₹24,400 = ₹2,501. (Simple interest for 6 months would be ₹2,440.)
APAR26-09-46 | Quarterly compounding | Difficult
47. C. For 3 years, CI − SI = P(r/100)²(3 + r/100) = P × 31/1000. So P = ₹1,891 × 1000/31 = ₹61,000. (The 2-year formula would give ₹1,89,100, which is wrong.)
APAR26-09-48 | Sum from CI − SI of 3 years | Difficult
48. A. Balance owed = ₹5,500 − ₹2,500 = ₹3,000. Interest charged = ₹3,135 − ₹3,000 = ₹135 for 3 months. Rate = ₹135 × 100 × 12/(₹3,000 × 3) = 18% p.a. (Interest is on the balance, not on the full cash price.)
APAR26-09-25 | Rate charged under a cash-down scheme | Difficult
49. D. CI for 2 years = P[(1 + r/100)² − 1]. At 20%: factor = (6/5)^2 − 1 = 11/25, so P = ₹13,376 × 25/11 = ₹30,400. New sum = ₹60,800; CI at 12.5% = ₹60,800 × [(9/8)^2 − 1] = ₹16,150. (Scaling ₹13,376 by twice and the rate ratio is not valid for CI.)
APAR26-09-49 | Principal from one CI, reinvested elsewhere | Difficult
50. A. A = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100) = ₹50,000 × 105/100 × 106/100 × 106/100 = ₹58,989. CI = ₹58,989 − ₹50,000 = ₹8,989. (Adding the rates to 17% and applying once gives ₹8,500 of interest, which ignores compounding.)
APAR26-09-50 | Different rates in successive years | Difficult