COMPOUND INTEREST (Stage 1) - Questions & Answers
EXERCISE 2 (A)1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) For a particular year the simple interest at 10% is ₹ 800. The compound interest for the next year at the same rate is :
(i) ₹ 880 (ii) ₹ 800 (iii) ₹ 720 (iv) ₹ 968
Step 1: Simple interest for the first year = ₹ 800.
Step 2: Interest earned on this simple interest for the next year = 10% of ₹ 800 = (10 / 100) × 800 = ₹ 80.
Step 3: Compound interest for the next year = Simple interest + Interest on simple interest.
Step 4: C.I. for the next year = 800 + 80 = ₹ 880.
Step 5: Correct option is (i) ₹ 880.
Step 2: Interest earned on this simple interest for the next year = 10% of ₹ 800 = (10 / 100) × 800 = ₹ 80.
Step 3: Compound interest for the next year = Simple interest + Interest on simple interest.
Step 4: C.I. for the next year = 800 + 80 = ₹ 880.
Step 5: Correct option is (i) ₹ 880.
(b) The compound interest on ₹ 5,000 at 10% per annum and in 6 months amounts to :
(i) ₹ 5,500 (ii) ₹ 250 (iii) ₹ 600 (iv) ₹ 2,500
Step 1: Principal (P) = ₹ 5,000, Rate (R) = 10% p.a., Time (T) = 6 months = 1/2 year.
Step 2: Since it is only for 6 months, compound interest is the same as simple interest.
Step 3: C.I. = (P × R × T) / 100 = (5000 × 10 × 0.5) / 100.
Step 4: C.I. = ₹ 250.
Step 5: Correct option is (ii) ₹ 250.
Step 2: Since it is only for 6 months, compound interest is the same as simple interest.
Step 3: C.I. = (P × R × T) / 100 = (5000 × 10 × 0.5) / 100.
Step 4: C.I. = ₹ 250.
Step 5: Correct option is (ii) ₹ 250.
(c) The compound interest on ₹ 5,000 at 10% per annum and in one year compounded half-yearly is :
(i) ₹ 1,050 (ii) ₹ 525 (iii) ₹ 512.50 (iv) ₹ 5,512.50
Step 1: Principal (P) = ₹ 5,000, Annual Rate = 10%, Time = 1 year.
Step 2: Half-yearly rate = 10% / 2 = 5% per half-year. Number of periods (n) = 2.
Step 3: Amount = P(1 + R/100)^n = 5000(1 + 5/100)².
Step 4: Amount = 5000(105/100)² = 5000(1.05)² = 5000 × 1.1025 = ₹ 5,512.50.
Step 5: C.I. = Amount - Principal = 5512.50 - 5000 = ₹ 512.50.
Step 6: Correct option is (iii) ₹ 512.50.
Step 2: Half-yearly rate = 10% / 2 = 5% per half-year. Number of periods (n) = 2.
Step 3: Amount = P(1 + R/100)^n = 5000(1 + 5/100)².
Step 4: Amount = 5000(105/100)² = 5000(1.05)² = 5000 × 1.1025 = ₹ 5,512.50.
Step 5: C.I. = Amount - Principal = 5512.50 - 5000 = ₹ 512.50.
Step 6: Correct option is (iii) ₹ 512.50.
(d) A sum of ₹ 20,000 is lent at 12% compound interest compounded yearly. The compound interest accrued in the second year will be :
(i) ₹ 4,800 (ii) ₹ 288 (iii) ₹ 2,688 (iv) ₹ 5,088
Step 1: Principal (P) = ₹ 20,000, Rate = 12%.
Step 2: Interest for the 1st year = 12% of 20000 = ₹ 2,400.
Step 3: Amount at the end of 1st year = 20000 + 2400 = ₹ 22,400.
Step 4: This amount becomes the principal for the 2nd year.
Step 5: Interest accrued in the 2nd year = 12% of 22400 = (12 / 100) × 22400 = ₹ 2,688.
Step 6: Correct option is (iii) ₹ 2,688.
Step 2: Interest for the 1st year = 12% of 20000 = ₹ 2,400.
Step 3: Amount at the end of 1st year = 20000 + 2400 = ₹ 22,400.
Step 4: This amount becomes the principal for the 2nd year.
Step 5: Interest accrued in the 2nd year = 12% of 22400 = (12 / 100) × 22400 = ₹ 2,688.
Step 6: Correct option is (iii) ₹ 2,688.
(e) During the year 2022, the interest accrued at the rate of 5% is ₹ 1,250. The compound interest accrued at the same rate during the year 2023 is :
(i) ₹ 1,312.50 (ii) ₹ 62.50 (iii) ₹ 6,250 (iv) ₹ 2,000
Step 1: Interest for the year 2022 = ₹ 1,250.
Step 2: Rate of interest = 5%.
Step 3: C.I. for the next year (2023) = Interest of 2022 + (5% of Interest of 2022).
Step 4: C.I. for 2023 = 1250 + (5/100) × 1250 = 1250 + 62.50.
Step 5: C.I. for 2023 = ₹ 1,312.50.
Step 6: Correct option is (i) ₹ 1,312.50.
Step 2: Rate of interest = 5%.
Step 3: C.I. for the next year (2023) = Interest of 2022 + (5% of Interest of 2022).
Step 4: C.I. for 2023 = 1250 + (5/100) × 1250 = 1250 + 62.50.
Step 5: C.I. for 2023 = ₹ 1,312.50.
Step 6: Correct option is (i) ₹ 1,312.50.
(f) Rates of interest for two consecutive years are 10% and 12% respectively. The percentage increase during these two years is :
(i) 22% (ii) 23.2% (iii) 123.2% (iv) 122%
Step 1: Let the principal be ₹ 100.
Step 2: Amount after 1st year at 10% = 100 × (1 + 10/100) = ₹ 110.
Step 3: Amount after 2nd year at 12% = 110 × (1 + 12/100) = 110 × 1.12 = ₹ 123.20.
Step 4: Total percentage increase = Total increase on ₹ 100 = 123.20 - 100 = 23.20.
Step 5: Therefore, the percentage increase is 23.2%.
Step 6: Correct option is (ii) 23.2%.
Step 2: Amount after 1st year at 10% = 100 × (1 + 10/100) = ₹ 110.
Step 3: Amount after 2nd year at 12% = 110 × (1 + 12/100) = 110 × 1.12 = ₹ 123.20.
Step 4: Total percentage increase = Total increase on ₹ 100 = 123.20 - 100 = 23.20.
Step 5: Therefore, the percentage increase is 23.2%.
Step 6: Correct option is (ii) 23.2%.
2. ₹ 16,000 is invested at 5% compound interest compounded per annum. Use the table, given below, to find the amount in 4 years.
Step 1: The rate of interest is 5% which is calculated on the Initial amount of each year.
Step 2: Year 1: Initial = 16,000, Interest = 5% of 16,000 = 800. Final = 16,000 + 800 = 16,800.
Step 3: Year 2: Initial = 16,800, Interest = 5% of 16,800 = 840. Final = 16,800 + 840 = 17,640.
Step 4: Year 3: Initial = 17,640, Interest = 5% of 17,640 = 882. Final = 17,640 + 882 = 18,522.
Step 5: Year 4: Initial = 18,522, Interest = 5% of 18,522 = 926.10. Final = 18,522 + 926.10 = 19,448.10.
Step 6: Year 5: Initial = 19,448.10, Interest = 5% of 19,448.10 = 972.41. Final = 19,448.10 + 972.41 = 20,420.51.
Step 2: Year 1: Initial = 16,000, Interest = 5% of 16,000 = 800. Final = 16,000 + 800 = 16,800.
Step 3: Year 2: Initial = 16,800, Interest = 5% of 16,800 = 840. Final = 16,800 + 840 = 17,640.
Step 4: Year 3: Initial = 17,640, Interest = 5% of 17,640 = 882. Final = 17,640 + 882 = 18,522.
Step 5: Year 4: Initial = 18,522, Interest = 5% of 18,522 = 926.10. Final = 18,522 + 926.10 = 19,448.10.
Step 6: Year 5: Initial = 19,448.10, Interest = 5% of 19,448.10 = 972.41. Final = 19,448.10 + 972.41 = 20,420.51.
3. Calculate the amount and the compound interest on ₹ 8,000 in 2½ years at 15% per annum.
Step 1: Principal (P) = ₹ 8,000, Rate (R) = 15% p.a., Time (T) = 2½ years.
Step 2: Amount for 2 full years = P(1 + R/100)² = 8000(1 + 15/100)².
Step 3: Amount for 2 years = 8000(115/100)² = 8000(1.15)² = 8000 × 1.3225 = ₹ 10,580.
Step 4: This amount becomes the principal for the remaining half year.
Step 5: Interest for the remaining half year = (10580 × 15 × 0.5) / 100 = ₹ 793.50.
Step 6: Total Amount after 2½ years = 10580 + 793.50 = ₹ 11,373.50.
Step 7: Compound Interest = Amount - Principal = 11373.50 - 8000 = ₹ 3,373.50.
Step 2: Amount for 2 full years = P(1 + R/100)² = 8000(1 + 15/100)².
Step 3: Amount for 2 years = 8000(115/100)² = 8000(1.15)² = 8000 × 1.3225 = ₹ 10,580.
Step 4: This amount becomes the principal for the remaining half year.
Step 5: Interest for the remaining half year = (10580 × 15 × 0.5) / 100 = ₹ 793.50.
Step 6: Total Amount after 2½ years = 10580 + 793.50 = ₹ 11,373.50.
Step 7: Compound Interest = Amount - Principal = 11373.50 - 8000 = ₹ 3,373.50.
4. Calculate the amount and the compound interest on : ₹ 4,600 in 2 years when the rates of interest of successive years are 10% and 12% respectively.
Step 1: Principal (P) = ₹ 4,600. Rates r₁ = 10%, r₂ = 12%.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100).
Step 3: A = 4600(1 + 10/100)(1 + 12/100) = 4600 × (1.10) × (1.12).
Step 4: A = 4600 × 1.232 = ₹ 5,667.20.
Step 5: Compound Interest = Amount - Principal = 5667.20 - 4600 = ₹ 1,067.20.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100).
Step 3: A = 4600(1 + 10/100)(1 + 12/100) = 4600 × (1.10) × (1.12).
Step 4: A = 4600 × 1.232 = ₹ 5,667.20.
Step 5: Compound Interest = Amount - Principal = 5667.20 - 4600 = ₹ 1,067.20.
5. Meenal lends ₹ 75,000 at C.I. for 3 years. If the rate of interest for the first two years is 15% per year and for the third year it is 16%, calculate the sum Meenal will get at the end of the third year.
Step 1: Principal (P) = ₹ 75,000. Rates: r₁ = 15%, r₂ = 15%, r₃ = 16%.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: A = 75000(1 + 15/100)(1 + 15/100)(1 + 16/100).
Step 4: A = 75000 × 1.15 × 1.15 × 1.16.
Step 5: A = 75000 × 1.3225 × 1.16 = 75000 × 1.5341 = ₹ 1,15,057.50.
Step 6: Meenal will get ₹ 1,15,057.50 at the end of the third year.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: A = 75000(1 + 15/100)(1 + 15/100)(1 + 16/100).
Step 4: A = 75000 × 1.15 × 1.15 × 1.16.
Step 5: A = 75000 × 1.3225 × 1.16 = 75000 × 1.5341 = ₹ 1,15,057.50.
Step 6: Meenal will get ₹ 1,15,057.50 at the end of the third year.
6. Calculate the amount and the compound interest on ₹ 16,000 in 3 years, when the rates of the interest for successive years are 10%, 14% and 15% respectively.
Step 1: Principal (P) = ₹ 16,000. Rates: r₁ = 10%, r₂ = 14%, r₃ = 15%.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: A = 16000(1 + 10/100)(1 + 14/100)(1 + 15/100).
Step 4: A = 16000 × 1.10 × 1.14 × 1.15.
Step 5: A = 16000 × 1.4421 = ₹ 23,073.60.
Step 6: Compound Interest = Amount - Principal = 23073.60 - 16000 = ₹ 7,073.60.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: A = 16000(1 + 10/100)(1 + 14/100)(1 + 15/100).
Step 4: A = 16000 × 1.10 × 1.14 × 1.15.
Step 5: A = 16000 × 1.4421 = ₹ 23,073.60.
Step 6: Compound Interest = Amount - Principal = 23073.60 - 16000 = ₹ 7,073.60.
7. Calculate the compound interest for the second year on ₹ 8,000/- invested for 3 years at 10% per annum.
Step 1: Principal for 1st year = ₹ 8,000. Rate = 10%.
Step 2: Interest for the 1st year = (8000 × 10 × 1) / 100 = ₹ 800.
Step 3: Amount at the end of 1st year = 8000 + 800 = ₹ 8,800.
Step 4: This amount is the principal for the 2nd year.
Step 5: Compound interest for the 2nd year = (8800 × 10 × 1) / 100 = ₹ 880.
Step 2: Interest for the 1st year = (8000 × 10 × 1) / 100 = ₹ 800.
Step 3: Amount at the end of 1st year = 8000 + 800 = ₹ 8,800.
Step 4: This amount is the principal for the 2nd year.
Step 5: Compound interest for the 2nd year = (8800 × 10 × 1) / 100 = ₹ 880.
8. Find the compound interest, correct to the nearest rupee, on ₹ 2,400 for 2½ years at 5 percent per annum.
Step 1: Principal = ₹ 2,400, Rate = 5% p.a., Time = 2½ years.
Step 2: Amount for 2 years = P(1 + R/100)² = 2400(1 + 5/100)².
Step 3: Amount for 2 years = 2400(1.05)² = 2400 × 1.1025 = ₹ 2,646.
Step 4: Interest for the remaining half year = (2646 × 5 × 0.5) / 100 = ₹ 66.15.
Step 5: Total Amount = 2646 + 66.15 = ₹ 2,712.15.
Step 6: Total Compound Interest = 2712.15 - 2400 = ₹ 312.15.
Step 7: C.I. rounded to the nearest rupee is ₹ 312.
Step 2: Amount for 2 years = P(1 + R/100)² = 2400(1 + 5/100)².
Step 3: Amount for 2 years = 2400(1.05)² = 2400 × 1.1025 = ₹ 2,646.
Step 4: Interest for the remaining half year = (2646 × 5 × 0.5) / 100 = ₹ 66.15.
Step 5: Total Amount = 2646 + 66.15 = ₹ 2,712.15.
Step 6: Total Compound Interest = 2712.15 - 2400 = ₹ 312.15.
Step 7: C.I. rounded to the nearest rupee is ₹ 312.
9. A borrowed ₹ 2,500 from B at 12% per annum compound interest. After 2 years, A gave ₹ 2,936 and a watch to B to clear the account. Find the cost of the watch.
Step 1: Principal (P) = ₹ 2,500, Rate = 12% p.a., Time = 2 years.
Step 2: Amount to be paid after 2 years = P(1 + R/100)² = 2500(1 + 12/100)².
Step 3: Amount = 2500(1.12)² = 2500 × 1.2544 = ₹ 3,136.
Step 4: Value given in cash = ₹ 2,936.
Step 5: The remaining balance was cleared by giving the watch.
Step 6: Cost of the watch = Total amount due - Cash given = 3136 - 2936 = ₹ 200.
Step 2: Amount to be paid after 2 years = P(1 + R/100)² = 2500(1 + 12/100)².
Step 3: Amount = 2500(1.12)² = 2500 × 1.2544 = ₹ 3,136.
Step 4: Value given in cash = ₹ 2,936.
Step 5: The remaining balance was cleared by giving the watch.
Step 6: Cost of the watch = Total amount due - Cash given = 3136 - 2936 = ₹ 200.
10. How much will ₹ 50,000 amount to in 3 years, compounded yearly, if the rates for the successive years are 6%, 8% and 10% respectively.
Step 1: Principal (P) = ₹ 50,000. Rates: r₁ = 6%, r₂ = 8%, r₃ = 10%.
Step 2: Amount = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: Amount = 50000(1 + 6/100)(1 + 8/100)(1 + 10/100).
Step 4: Amount = 50000 × 1.06 × 1.08 × 1.10.
Step 5: Amount = 50000 × 1.25928 = ₹ 62,964.
Step 2: Amount = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: Amount = 50000(1 + 6/100)(1 + 8/100)(1 + 10/100).
Step 4: Amount = 50000 × 1.06 × 1.08 × 1.10.
Step 5: Amount = 50000 × 1.25928 = ₹ 62,964.
11. Govind borrows ₹ 18,000 at 10% simple interest. He immediately invests the money borrowed at 10% compound interest compounded half-yearly. How much money does Govind gain in one year ?
Step 1: Simple Interest (S.I.) to be paid by Govind = (18000 × 10 × 1) / 100 = ₹ 1,800.
Step 2: He invests at 10% compounded half-yearly, so rate = 5% per half-year. Number of periods = 2.
Step 3: Amount he receives after 1 year = P(1 + R/100)² = 18000(1 + 5/100)².
Step 4: Amount = 18000(1.05)² = 18000 × 1.1025 = ₹ 19,845.
Step 5: Compound Interest (C.I.) earned by Govind = 19845 - 18000 = ₹ 1,845.
Step 6: Govind's gain = C.I. earned - S.I. paid = 1845 - 1800 = ₹ 45.
Step 2: He invests at 10% compounded half-yearly, so rate = 5% per half-year. Number of periods = 2.
Step 3: Amount he receives after 1 year = P(1 + R/100)² = 18000(1 + 5/100)².
Step 4: Amount = 18000(1.05)² = 18000 × 1.1025 = ₹ 19,845.
Step 5: Compound Interest (C.I.) earned by Govind = 19845 - 18000 = ₹ 1,845.
Step 6: Govind's gain = C.I. earned - S.I. paid = 1845 - 1800 = ₹ 45.
12. Find the compound interest on ₹ 4,000 accrued in three years, when the rate of interest is 8% for the first year and 10% per year for the second and the third years.
Step 1: Principal (P) = ₹ 4,000. Rates: r₁ = 8%, r₂ = 10%, r₃ = 10%.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: A = 4000(1 + 8/100)(1 + 10/100)(1 + 10/100).
Step 4: A = 4000 × 1.08 × 1.10 × 1.10.
Step 5: A = 4000 × 1.3068 = ₹ 5,227.20.
Step 6: Compound Interest = Amount - Principal = 5227.20 - 4000 = ₹ 1,227.20.
Step 2: Amount (A) = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).
Step 3: A = 4000(1 + 8/100)(1 + 10/100)(1 + 10/100).
Step 4: A = 4000 × 1.08 × 1.10 × 1.10.
Step 5: A = 4000 × 1.3068 = ₹ 5,227.20.
Step 6: Compound Interest = Amount - Principal = 5227.20 - 4000 = ₹ 1,227.20.
EXERCISE 2 (B)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The difference between C.I. and S.I. in one year on ₹ 5,000 at the rate of 10% per annum is :
(i) ₹ 00 (ii) ₹ 500 (iii) ₹ 5,500 (iv) ₹ 5,250
Step 1: For the first year (or conversion period), compound interest and simple interest are exactly the same.
Step 2: Therefore, the difference between them is ₹ 00.
Step 3: Correct option is (i) ₹ 00.
Step 2: Therefore, the difference between them is ₹ 00.
Step 3: Correct option is (i) ₹ 00.
(b) ₹ 2,000 is saved during the year 2022 and deposited in a bank at the beginning of year 2023 at 8% compound interest. During 2023, ₹ 3,000 more is saved and deposited in the same bank at the beginning of 2024 and at the same rate of interest. The C.I. earned upto the end of 2024 is :
(i) ₹ 560 (ii) ₹ 572.80 (iii) ₹ 5,400 (iv) ₹ 400
Step 1: In 2023, Principal = ₹ 2,000. Interest for 2023 = 8% of 2000 = ₹ 160.
Step 2: Amount at the end of 2023 = 2000 + 160 = ₹ 2,160.
Step 3: In 2024, new Principal = Amount from 2023 + New Deposit = 2160 + 3000 = ₹ 5,160.
Step 4: Interest for 2024 = 8% of 5160 = ₹ 412.80.
Step 5: Total C.I. earned = Interest of 2023 + Interest of 2024 = 160 + 412.80 = ₹ 572.80.
Step 6: Correct option is (ii) ₹ 572.80.
Step 2: Amount at the end of 2023 = 2000 + 160 = ₹ 2,160.
Step 3: In 2024, new Principal = Amount from 2023 + New Deposit = 2160 + 3000 = ₹ 5,160.
Step 4: Interest for 2024 = 8% of 5160 = ₹ 412.80.
Step 5: Total C.I. earned = Interest of 2023 + Interest of 2024 = 160 + 412.80 = ₹ 572.80.
Step 6: Correct option is (ii) ₹ 572.80.
(c) ₹ 1,000 is borrowed at 10% per annum C.I. If ₹ 300 is repaid at the end of each year, the amount of loan outstanding at the end of 2nd year is :
(i) ₹ 880 (ii) ₹ 610 (iii) ₹ 580 (iv) ₹ 484
Step 1: Principal for 1st year = ₹ 1,000.
Step 2: Amount at end of 1st year = 1000 + 10% of 1000 = 1000 + 100 = ₹ 1,100.
Step 3: Balance after repayment = 1100 - 300 = ₹ 800. This is the Principal for 2nd year.
Step 4: Amount at end of 2nd year = 800 + 10% of 800 = 800 + 80 = ₹ 880.
Step 5: Balance after 2nd repayment = 880 - 300 = ₹ 580.
Step 6: Correct option is (iii) ₹ 580.
Step 2: Amount at end of 1st year = 1000 + 10% of 1000 = 1000 + 100 = ₹ 1,100.
Step 3: Balance after repayment = 1100 - 300 = ₹ 800. This is the Principal for 2nd year.
Step 4: Amount at end of 2nd year = 800 + 10% of 800 = 800 + 80 = ₹ 880.
Step 5: Balance after 2nd repayment = 880 - 300 = ₹ 580.
Step 6: Correct option is (iii) ₹ 580.
(d) The difference between C.I. and S.I. in 2 years on ₹ 4,000 at 10% per annum is :
(i) ₹ 840 (ii) ₹ 800 (iii) ₹ 400 (iv) ₹ 40
Step 1: Formula for difference between C.I. and S.I. for 2 years is P(R/100)².
Step 2: Difference = 4000 × (10/100)² = 4000 × (1/10)² = 4000 × 1/100.
Step 3: Difference = ₹ 40.
Step 4: Correct option is (iv) ₹ 40.
Step 2: Difference = 4000 × (10/100)² = 4000 × (1/10)² = 4000 × 1/100.
Step 3: Difference = ₹ 40.
Step 4: Correct option is (iv) ₹ 40.
(e) ₹ 10 is the difference between compound interest and simple interest in 2 years and at 5% per annum. The principal amount is:
(i) ₹ 4,400 (ii) ₹ 4,100 (iii) ₹ 4,000 (iv) none of these
Step 1: Formula for difference for 2 years is Difference = P(R/100)².
Step 2: 10 = P × (5/100)² = P × (1/20)³ = P × (1/400).
Step 3: P = 10 × 400 = ₹ 4,000.
Step 4: Correct option is (iii) ₹ 4,000.
Step 2: 10 = P × (5/100)² = P × (1/20)³ = P × (1/400).
Step 3: P = 10 × 400 = ₹ 4,000.
Step 4: Correct option is (iii) ₹ 4,000.
2. Calculate the difference between the simple interest and the compound interest on ₹ 4,000 in 2 years at 8% per annum compounded yearly.
Step 1: Simple Interest (S.I.) = (P × R × T) / 100 = (4000 × 8 × 2) / 100 = ₹ 640.
Step 2: Amount at C.I. = P(1 + R/100)² = 4000(1 + 8/100)² = 4000(1.08)² = 4000 × 1.1664 = ₹ 4,665.60.
Step 3: Compound Interest (C.I.) = Amount - Principal = 4665.60 - 4000 = ₹ 665.60.
Step 4: Difference = C.I. - S.I. = 665.60 - 640 = ₹ 25.60.
Step 2: Amount at C.I. = P(1 + R/100)² = 4000(1 + 8/100)² = 4000(1.08)² = 4000 × 1.1664 = ₹ 4,665.60.
Step 3: Compound Interest (C.I.) = Amount - Principal = 4665.60 - 4000 = ₹ 665.60.
Step 4: Difference = C.I. - S.I. = 665.60 - 640 = ₹ 25.60.
3. A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by ₹ 96, find the sum of money.
Step 1: Let the sum of money be P.
Step 2: Interest for the 1st year = 8% of P = 0.08 P.
Step 3: The difference in interest between the 2nd year and 1st year is the interest calculated on the 1st year's interest.
Step 4: Therefore, 8% of (Interest for 1st year) = ₹ 96.
Step 5: 8% of (0.08 P) = 96.
Step 6: (8/100) × (8/100) × P = 96.
Step 7: P = (96 × 100 × 100) / (8 × 8) = 960000 / 64 = ₹ 15,000.
Step 2: Interest for the 1st year = 8% of P = 0.08 P.
Step 3: The difference in interest between the 2nd year and 1st year is the interest calculated on the 1st year's interest.
Step 4: Therefore, 8% of (Interest for 1st year) = ₹ 96.
Step 5: 8% of (0.08 P) = 96.
Step 6: (8/100) × (8/100) × P = 96.
Step 7: P = (96 × 100 × 100) / (8 × 8) = 960000 / 64 = ₹ 15,000.
4. A man invests ₹ 5,600 at 14% per annum compound interest for 2 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year, correct to the nearest rupee.
Step 1: (i) Interest for 1st year = (5600 × 14 × 1) / 100 = ₹ 784.
Step 2: (ii) Amount at the end of 1st year = 5600 + 784 = ₹ 6,384.
Step 3: (iii) Interest for 2nd year = (6384 × 14 × 1) / 100 = ₹ 893.76.
Step 4: Interest rounded to nearest rupee = ₹ 894.
Step 2: (ii) Amount at the end of 1st year = 5600 + 784 = ₹ 6,384.
Step 3: (iii) Interest for 2nd year = (6384 × 14 × 1) / 100 = ₹ 893.76.
Step 4: Interest rounded to nearest rupee = ₹ 894.
5. A man saves ₹ 3,000 every year and invests it at the end of the year at 10% compound interest. Calculate the total amount of his savings at the end of the third year.
Step 1: Savings at the end of 1st year = ₹ 3,000. He invests this, so it becomes the principal for the 2nd year.
Step 2: Amount at the end of 2nd year (before adding new savings) = 3000 + 10% of 3000 = 3000 + 300 = ₹ 3,300.
Step 3: He saves another ₹ 3,000 at the end of 2nd year. Total principal for 3rd year = 3300 + 3000 = ₹ 6,300.
Step 4: Amount at the end of 3rd year (before adding new savings) = 6300 + 10% of 6300 = 6300 + 630 = ₹ 6,930.
Step 5: He saves another ₹ 3,000 at the end of 3rd year. Total amount = 6930 + 3000 = ₹ 9,930.
Step 2: Amount at the end of 2nd year (before adding new savings) = 3000 + 10% of 3000 = 3000 + 300 = ₹ 3,300.
Step 3: He saves another ₹ 3,000 at the end of 2nd year. Total principal for 3rd year = 3300 + 3000 = ₹ 6,300.
Step 4: Amount at the end of 3rd year (before adding new savings) = 6300 + 10% of 6300 = 6300 + 630 = ₹ 6,930.
Step 5: He saves another ₹ 3,000 at the end of 3rd year. Total amount = 6930 + 3000 = ₹ 9,930.
6. A man lends ₹ 12,500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. of the first year and the compound interest for the third year.
Step 1: Principal for 1st year = ₹ 12,500.
Step 2: C.I. for 1st year = 12% of 12500 = ₹ 1,500.
Step 3: Amount at the end of 1st year = 12500 + 1500 = ₹ 14,000.
Step 4: Principal for 2nd year = ₹ 14,000.
Step 5: C.I. for 2nd year = 15% of 14000 = ₹ 2,100.
Step 6: Amount at the end of 2nd year = 14000 + 2100 = ₹ 16,100.
Step 7: Principal for 3rd year = ₹ 16,100.
Step 8: C.I. for 3rd year = 18% of 16100 = ₹ 2,898.
Step 9: Difference between C.I. of 3rd year and 1st year = 2898 - 1500 = ₹ 1,398.
Step 2: C.I. for 1st year = 12% of 12500 = ₹ 1,500.
Step 3: Amount at the end of 1st year = 12500 + 1500 = ₹ 14,000.
Step 4: Principal for 2nd year = ₹ 14,000.
Step 5: C.I. for 2nd year = 15% of 14000 = ₹ 2,100.
Step 6: Amount at the end of 2nd year = 14000 + 2100 = ₹ 16,100.
Step 7: Principal for 3rd year = ₹ 16,100.
Step 8: C.I. for 3rd year = 18% of 16100 = ₹ 2,898.
Step 9: Difference between C.I. of 3rd year and 1st year = 2898 - 1500 = ₹ 1,398.
7. A man borrows ₹ 6,000 at 5 percent C.I. per annum. If he repays ₹ 1,200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.
Step 1: Principal for 1st year = ₹ 6,000.
Step 2: Interest for 1st year = 5% of 6000 = ₹ 300.
Step 3: Amount at the end of 1st year = 6000 + 300 = ₹ 6,300.
Step 4: Loan outstanding after 1st repayment = 6300 - 1200 = ₹ 5,100 (This is Principal for 2nd year).
Step 5: Interest for 2nd year = 5% of 5100 = ₹ 255.
Step 6: Amount at the end of 2nd year = 5100 + 255 = ₹ 5,355.
Step 7: Loan outstanding after 2nd repayment = 5355 - 1200 = ₹ 4,155.
Step 8: Outstanding loan at the beginning of the 3rd year = ₹ 4,155.
Step 2: Interest for 1st year = 5% of 6000 = ₹ 300.
Step 3: Amount at the end of 1st year = 6000 + 300 = ₹ 6,300.
Step 4: Loan outstanding after 1st repayment = 6300 - 1200 = ₹ 5,100 (This is Principal for 2nd year).
Step 5: Interest for 2nd year = 5% of 5100 = ₹ 255.
Step 6: Amount at the end of 2nd year = 5100 + 255 = ₹ 5,355.
Step 7: Loan outstanding after 2nd repayment = 5355 - 1200 = ₹ 4,155.
Step 8: Outstanding loan at the beginning of the 3rd year = ₹ 4,155.
8. A man borrows ₹ 5,000 at 12 percent compound interest payable every six months. He repays ₹ 1,800 at the end of every six months. Calculate the third payment he has to make at the end of 18 months in order to clear the entire loan.
Step 1: Principal = ₹ 5,000. Rate = 12% p.a. = 6% per six months.
Step 2: Interest for 1st six months = 6% of 5000 = ₹ 300. Amount = 5000 + 300 = ₹ 5,300.
Step 3: Balance after 1st repayment = 5300 - 1800 = ₹ 3,500.
Step 4: Interest for 2nd six months = 6% of 3500 = ₹ 210. Amount = 3500 + 210 = ₹ 3,710.
Step 5: Balance after 2nd repayment = 3710 - 1800 = ₹ 1,910.
Step 6: Interest for 3rd six months = 6% of 1910 = ₹ 114.60. Amount = 1910 + 114.60 = ₹ 2,024.60.
Step 7: The third payment to clear the loan = ₹ 2,024.60.
Step 2: Interest for 1st six months = 6% of 5000 = ₹ 300. Amount = 5000 + 300 = ₹ 5,300.
Step 3: Balance after 1st repayment = 5300 - 1800 = ₹ 3,500.
Step 4: Interest for 2nd six months = 6% of 3500 = ₹ 210. Amount = 3500 + 210 = ₹ 3,710.
Step 5: Balance after 2nd repayment = 3710 - 1800 = ₹ 1,910.
Step 6: Interest for 3rd six months = 6% of 1910 = ₹ 114.60. Amount = 1910 + 114.60 = ₹ 2,024.60.
Step 7: The third payment to clear the loan = ₹ 2,024.60.
9. On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is ₹ 180/-. Find the sum lent out, if the rate of interest in both the cases is 10% per annum.
Step 1: Let the sum be P. Rate = 10% p.a.
Step 2: S.I. for 1 year = (P × 10 × 1) / 100 = 0.10 P.
Step 3: For C.I. half-yearly, rate = 5% per half-year, n = 2 periods.
Step 4: Amount = P(1 + 5/100)² = P(1.05)² = 1.1025 P.
Step 5: C.I. = Amount - P = 1.1025 P - P = 0.1025 P.
Step 6: Difference = C.I. - S.I. = 0.1025 P - 0.10 P = 0.0025 P.
Step 7: We are given difference = 180. So, 0.0025 P = 180.
Step 8: P = 180 / 0.0025 = 1800000 / 25 = ₹ 72,000.
Step 2: S.I. for 1 year = (P × 10 × 1) / 100 = 0.10 P.
Step 3: For C.I. half-yearly, rate = 5% per half-year, n = 2 periods.
Step 4: Amount = P(1 + 5/100)² = P(1.05)² = 1.1025 P.
Step 5: C.I. = Amount - P = 1.1025 P - P = 0.1025 P.
Step 6: Difference = C.I. - S.I. = 0.1025 P - 0.10 P = 0.0025 P.
Step 7: We are given difference = 180. So, 0.0025 P = 180.
Step 8: P = 180 / 0.0025 = 1800000 / 25 = ₹ 72,000.
10. A manufacturer estimates that his machine depreciates by 15% of its value at the beginning of the year. Find the original value (cost) of the machine, if it depreciates by ₹ 5,355 during the second year.
Step 1: Let the original value of the machine be P.
Step 2: Depreciation in 1st year = 15% of P = 0.15 P.
Step 3: Value at the beginning of 2nd year = P - 0.15 P = 0.85 P.
Step 4: Depreciation in 2nd year = 15% of value at beginning of 2nd year.
Step 5: Depreciation in 2nd year = 15% of 0.85 P = 0.15 × 0.85 P = 0.1275 P.
Step 6: We are given depreciation in 2nd year = 5355. So, 0.1275 P = 5355.
Step 7: P = 5355 / 0.1275 = ₹ 42,000.
Step 2: Depreciation in 1st year = 15% of P = 0.15 P.
Step 3: Value at the beginning of 2nd year = P - 0.15 P = 0.85 P.
Step 4: Depreciation in 2nd year = 15% of value at beginning of 2nd year.
Step 5: Depreciation in 2nd year = 15% of 0.85 P = 0.15 × 0.85 P = 0.1275 P.
Step 6: We are given depreciation in 2nd year = 5355. So, 0.1275 P = 5355.
Step 7: P = 5355 / 0.1275 = ₹ 42,000.
11. A man borrows ₹ 10,000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year and 42% of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt?
Step 1: Sum borrowed = ₹ 10,000. Rate = 5% p.a.
Step 2: Amount at end of 1st yr = 10000 + 5% of 10000 = 10000 + 500 = ₹ 10,500.
Step 3: Repayment at end of 1st yr = 35% of sum borrowed = 35% of 10000 = ₹ 3,500.
Step 4: Balance for 2nd yr = 10500 - 3500 = ₹ 7,000.
Step 5: Amount at end of 2nd yr = 7000 + 5% of 7000 = 7000 + 350 = ₹ 7,350.
Step 6: Repayment at end of 2nd yr = 42% of sum borrowed = 42% of 10000 = ₹ 4,200.
Step 7: Balance for 3rd yr = 7350 - 4200 = ₹ 3,150.
Step 8: Amount to pay at end of 3rd yr = 3150 + 5% of 3150 = 3150 + 157.50 = ₹ 3,307.50.
Step 2: Amount at end of 1st yr = 10000 + 5% of 10000 = 10000 + 500 = ₹ 10,500.
Step 3: Repayment at end of 1st yr = 35% of sum borrowed = 35% of 10000 = ₹ 3,500.
Step 4: Balance for 2nd yr = 10500 - 3500 = ₹ 7,000.
Step 5: Amount at end of 2nd yr = 7000 + 5% of 7000 = 7000 + 350 = ₹ 7,350.
Step 6: Repayment at end of 2nd yr = 42% of sum borrowed = 42% of 10000 = ₹ 4,200.
Step 7: Balance for 3rd yr = 7350 - 4200 = ₹ 3,150.
Step 8: Amount to pay at end of 3rd yr = 3150 + 5% of 3150 = 3150 + 157.50 = ₹ 3,307.50.
EXERCISE 2 (C)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) ₹ 300 and ₹ 360 are the compound interest for two consecutive years. The rate of interest is :
(i) 1.2% (ii) 12% (iii) 120% (iv) 20%
Step 1: Difference in compound interest = 360 - 300 = ₹ 60.
Step 2: This difference is the interest on the C.I. of the preceding year for 1 year.
Step 3: Rate = (Difference × 100) / (C.I. of preceding year) = (60 × 100) / 300.
Step 4: Rate = 6000 / 300 = 20%.
Step 5: Correct option is (iv) 20%.
Step 2: This difference is the interest on the C.I. of the preceding year for 1 year.
Step 3: Rate = (Difference × 100) / (C.I. of preceding year) = (60 × 100) / 300.
Step 4: Rate = 6000 / 300 = 20%.
Step 5: Correct option is (iv) 20%.
(b) A certain sum of money amounts to ₹ 5,000 at the end of 5th year and to ₹ 6,000 at the end of 6th year. The rate of interest is :
(i) 120% (ii) 20% (iii) 1.2% (iv) 12%
Step 1: The amount at the end of the 5th year acts as the principal for the 6th year.
Step 2: Principal for 6th year = ₹ 5,000. Amount at end of 6th year = ₹ 6,000.
Step 3: Interest for 6th year = 6000 - 5000 = ₹ 1,000.
Step 4: Rate = (Interest × 100) / Principal = (1000 × 100) / 5000 = 20%.
Step 5: Correct option is (ii) 20%.
Step 2: Principal for 6th year = ₹ 5,000. Amount at end of 6th year = ₹ 6,000.
Step 3: Interest for 6th year = 6000 - 5000 = ₹ 1,000.
Step 4: Rate = (Interest × 100) / Principal = (1000 × 100) / 5000 = 20%.
Step 5: Correct option is (ii) 20%.
(c) At the end of 2020, the compound interest amounted to ₹ 3,850 at 10% C.I. The C.I. on the same sum and at the same rate amounted at the end of 2019 was :
(i) ₹ 3,500 (ii) ₹ 4,235 (iii) ₹ 3,181 (iv) ₹ 3,182
Step 1: Let the C.I. at the end of 2019 be X.
Step 2: C.I. for the next year (2020) = C.I. of 2019 + Interest on C.I. of 2019.
Step 3: C.I. of 2020 = X + (10% of X) = 1.10 X.
Step 4: We know C.I. of 2020 is 3850, so 1.10 X = 3850.
Step 5: X = 3850 / 1.10 = ₹ 3,500.
Step 6: Correct option is (i) ₹ 3,500.
Step 2: C.I. for the next year (2020) = C.I. of 2019 + Interest on C.I. of 2019.
Step 3: C.I. of 2020 = X + (10% of X) = 1.10 X.
Step 4: We know C.I. of 2020 is 3850, so 1.10 X = 3850.
Step 5: X = 3850 / 1.10 = ₹ 3,500.
Step 6: Correct option is (i) ₹ 3,500.
(d) A sum of money, lent out at C.I., amounts to ₹ 4,500 in 6 years. If rate of C.I. is 12%, the same money will amount in 7 years to rupees :
(i) 540 (ii) 5,040 (iii) 5,400 (iv) 4,725
Step 1: Amount at the end of 6 years = ₹ 4,500.
Step 2: This acts as the principal for the 7th year.
Step 3: Interest for 7th year = 12% of 4500 = ₹ 540.
Step 4: Amount in 7 years = 4500 + 540 = ₹ 5,040.
Step 5: Correct option is (ii) 5,040.
Step 2: This acts as the principal for the 7th year.
Step 3: Interest for 7th year = 12% of 4500 = ₹ 540.
Step 4: Amount in 7 years = 4500 + 540 = ₹ 5,040.
Step 5: Correct option is (ii) 5,040.
(e) For two consecutive years, a sum lent out at C.I. earns ₹ 600 and ₹ 690 respectively. The rate of C.I. is :
(i) 12% (ii) 18% (iii) 15% (iv) 5%
Step 1: Difference in C.I. = 690 - 600 = ₹ 90.
Step 2: Rate = (Difference × 100) / (C.I. of preceding year) = (90 × 100) / 600.
Step 3: Rate = 9000 / 600 = 15%.
Step 4: Correct option is (iii) 15%.
Step 2: Rate = (Difference × 100) / (C.I. of preceding year) = (90 × 100) / 600.
Step 3: Rate = 9000 / 600 = 15%.
Step 4: Correct option is (iii) 15%.
(f) For two consecutive years a sum of money lent out at C.I. amounts to ₹ 2,400 and ₹ 2,760 respectively. The rate of interest is:
(i) 5% (ii) 15% (iii) 18% (iv) 10%
Step 1: The amount at the end of the first year (₹ 2,400) becomes the principal for the next year.
Step 2: Interest earned in the next year = 2760 - 2400 = ₹ 360.
Step 3: Rate = (Interest × 100) / Principal = (360 × 100) / 2400.
Step 4: Rate = 36000 / 2400 = 15%.
Step 5: Correct option is (ii) 15%.
Step 2: Interest earned in the next year = 2760 - 2400 = ₹ 360.
Step 3: Rate = (Interest × 100) / Principal = (360 × 100) / 2400.
Step 4: Rate = 36000 / 2400 = 15%.
Step 5: Correct option is (ii) 15%.
2. A certain sum amounts to ₹ 5,292 in two years and ₹ 5,556.60 in three years, interest being compounded annually. Find:
(i) the rate of interest
(ii) the original sum.
Step 1: Amount in 2 years = ₹ 5,292. Amount in 3 years = ₹ 5,556.60.
Step 2: Interest for 3rd year = Amount in 3 years - Amount in 2 years = 5556.60 - 5292 = ₹ 264.60.
Step 3: (i) Rate of interest = (Interest for 3rd yr × 100) / Amount in 2 yrs = (264.60 × 100) / 5292 = 26460 / 5292 = 5%.
Step 4: (ii) Let the original sum be P. Amount = P(1 + R/100)².
Step 5: 5292 = P(1 + 5/100)² = P(1.05)².
Step 6: P = 5292 / (1.05)² = 5292 / 1.1025 = ₹ 4,800.
Step 2: Interest for 3rd year = Amount in 3 years - Amount in 2 years = 5556.60 - 5292 = ₹ 264.60.
Step 3: (i) Rate of interest = (Interest for 3rd yr × 100) / Amount in 2 yrs = (264.60 × 100) / 5292 = 26460 / 5292 = 5%.
Step 4: (ii) Let the original sum be P. Amount = P(1 + R/100)².
Step 5: 5292 = P(1 + 5/100)² = P(1.05)².
Step 6: P = 5292 / (1.05)² = 5292 / 1.1025 = ₹ 4,800.
3. Mohit invests ₹ 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 9,440. Calculate :
(i) the rate of interest per annum.
(ii) the amount at the end of the second year.
(iii) the interest accrued in the third year.
Step 1: Principal = ₹ 8,000. Amount end of 1st yr = ₹ 9,440.
Step 2: Interest for 1st yr = 9440 - 8000 = ₹ 1,440.
Step 3: (i) Rate of interest = (Interest × 100) / Principal = (1440 × 100) / 8000 = 18%.
Step 4: (ii) Amount at end of 2nd yr = Amount end 1st yr + Interest on it.
Step 5: Amount at end of 2nd yr = 9440 + 18% of 9440 = 9440 + 1699.20 = ₹ 11,139.20.
Step 6: (iii) Interest accrued in 3rd yr = 18% of Amount end 2nd yr.
Step 7: Interest 3rd yr = 18% of 11139.20 = ₹ 2,005.06.
Step 2: Interest for 1st yr = 9440 - 8000 = ₹ 1,440.
Step 3: (i) Rate of interest = (Interest × 100) / Principal = (1440 × 100) / 8000 = 18%.
Step 4: (ii) Amount at end of 2nd yr = Amount end 1st yr + Interest on it.
Step 5: Amount at end of 2nd yr = 9440 + 18% of 9440 = 9440 + 1699.20 = ₹ 11,139.20.
Step 6: (iii) Interest accrued in 3rd yr = 18% of Amount end 2nd yr.
Step 7: Interest 3rd yr = 18% of 11139.20 = ₹ 2,005.06.
4. The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1,089 and for the third year it is ₹ 1,197.90. Calculate the rate of interest and the sum of money.
Step 1: Difference in interest = 1197.90 - 1089 = ₹ 108.90.
Step 2: Rate of interest = (Difference × 100) / C.I. of 2nd year = (108.90 × 100) / 1089 = 10%.
Step 3: Let C.I. for 1st year be x. Then C.I. for 2nd year = x + 10% of x = 1.10x.
Step 4: 1.10x = 1089 => x = 1089 / 1.10 = ₹ 990 (This is C.I. for 1st yr).
Step 5: Let the sum be P. C.I. for 1st yr = 10% of P = 0.10 P.
Step 6: 0.10 P = 990 => P = 990 / 0.10 = ₹ 9,900.
Step 2: Rate of interest = (Difference × 100) / C.I. of 2nd year = (108.90 × 100) / 1089 = 10%.
Step 3: Let C.I. for 1st year be x. Then C.I. for 2nd year = x + 10% of x = 1.10x.
Step 4: 1.10x = 1089 => x = 1089 / 1.10 = ₹ 990 (This is C.I. for 1st yr).
Step 5: Let the sum be P. C.I. for 1st yr = 10% of P = 0.10 P.
Step 6: 0.10 P = 990 => P = 990 / 0.10 = ₹ 9,900.
5. A sum is invested at compound interest compounded yearly. If the interest for two successive years be ₹ 5,700 and ₹ 7,410, calculate the rate of interest.
Step 1: Difference in interest for successive years = 7410 - 5700 = ₹ 1,710.
Step 2: The difference is the interest calculated on the previous year's interest.
Step 3: Rate = (Difference × 100) / Interest of previous year.
Step 4: Rate = (1710 × 100) / 5700 = 171000 / 5700 = 30%.
Step 2: The difference is the interest calculated on the previous year's interest.
Step 3: Rate = (Difference × 100) / Interest of previous year.
Step 4: Rate = (1710 × 100) / 5700 = 171000 / 5700 = 30%.
6. The cost of a machine depreciated by ₹ 4,000 during the first year and by ₹ 3,600 during the second year. Calculate :
(i) the rate of depreciation.
(ii) the original cost of the machine.
(iii) its cost at the end of the third year.
Step 1: Difference in depreciation = 4000 - 3600 = ₹ 400.
Step 2: (i) Rate of depreciation = (Difference × 100) / Depreciation of 1st year = (400 × 100) / 4000 = 10%.
Step 3: (ii) Let original cost be P. Depreciation 1st yr = 10% of P = 4000. So, P = 4000 / 0.10 = ₹ 40,000.
Step 4: (iii) Value at end of 1st yr = 40000 - 4000 = 36000.
Step 5: Value at end of 2nd yr = 36000 - 3600 = 32400.
Step 6: Depreciation in 3rd yr = 10% of 32400 = 3240.
Step 7: Cost at end of 3rd yr = 32400 - 3240 = ₹ 29,160.
Step 2: (i) Rate of depreciation = (Difference × 100) / Depreciation of 1st year = (400 × 100) / 4000 = 10%.
Step 3: (ii) Let original cost be P. Depreciation 1st yr = 10% of P = 4000. So, P = 4000 / 0.10 = ₹ 40,000.
Step 4: (iii) Value at end of 1st yr = 40000 - 4000 = 36000.
Step 5: Value at end of 2nd yr = 36000 - 3600 = 32400.
Step 6: Depreciation in 3rd yr = 10% of 32400 = 3240.
Step 7: Cost at end of 3rd yr = 32400 - 3240 = ₹ 29,160.
7. Ramesh invests ₹ 12,800 for three years at the rate of 10% per annum compound interest. Find:
(i) the sum due to Ramesh at the end of the first year.
(ii) the interest he earns for the second year.
(iii) the total amount due to him at the end of the third year.
Step 1: (i) Sum due at end 1st yr = Principal + Interest = 12800 + 10% of 12800 = 12800 + 1280 = ₹ 14,080.
Step 2: (ii) Interest for 2nd yr = 10% of 14080 = ₹ 1,408.
Step 3: Amount at end 2nd yr = 14080 + 1408 = ₹ 15,488.
Step 4: (iii) Interest for 3rd yr = 10% of 15488 = ₹ 1,548.80.
Step 5: Total amount due at end 3rd yr = 15488 + 1548.80 = ₹ 17,036.80.
Step 2: (ii) Interest for 2nd yr = 10% of 14080 = ₹ 1,408.
Step 3: Amount at end 2nd yr = 14080 + 1408 = ₹ 15,488.
Step 4: (iii) Interest for 3rd yr = 10% of 15488 = ₹ 1,548.80.
Step 5: Total amount due at end 3rd yr = 15488 + 1548.80 = ₹ 17,036.80.
8. A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are ₹ 650 and ₹ 760.50; find the rate of interest.
Step 1: Difference in interest for successive half-years = 760.50 - 650 = ₹ 110.50.
Step 2: Rate per half-year = (Difference × 100) / Interest of previous half-year = (110.50 × 100) / 650.
Step 3: Rate per half-year = 11050 / 650 = 17%.
Step 4: Annual rate of interest = 17% × 2 = 34%.
Step 2: Rate per half-year = (Difference × 100) / Interest of previous half-year = (110.50 × 100) / 650.
Step 3: Rate per half-year = 11050 / 650 = 17%.
Step 4: Annual rate of interest = 17% × 2 = 34%.
9. Geeta borrowed ₹ 15,000 for 18 months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to ₹ 15,600; calculate :
(i) the rate of interest per annum.
(ii) the total amount of money that Geeta must pay at the end of 18 months in order to clear the account.
Step 1: Principal = ₹ 15,000. Amount after 6 months = ₹ 15,600.
Step 2: Interest for 1st six months = 15600 - 15000 = ₹ 600.
Step 3: Rate per half-year = (Interest × 100) / Principal = (600 × 100) / 15000 = 4%.
Step 4: (i) Rate of interest per annum = 4% × 2 = 8%.
Step 5: (ii) Total periods (18 months) = 3 half-years.
Step 6: Amount at end of 18 months = P(1 + R_half/100)³ = 15000(1 + 4/100)³.
Step 7: Amount = 15000(1.04)³ = 15000 × 1.124864 = ₹ 16,872.96.
Step 2: Interest for 1st six months = 15600 - 15000 = ₹ 600.
Step 3: Rate per half-year = (Interest × 100) / Principal = (600 × 100) / 15000 = 4%.
Step 4: (i) Rate of interest per annum = 4% × 2 = 8%.
Step 5: (ii) Total periods (18 months) = 3 half-years.
Step 6: Amount at end of 18 months = P(1 + R_half/100)³ = 15000(1 + 4/100)³.
Step 7: Amount = 15000(1.04)³ = 15000 × 1.124864 = ₹ 16,872.96.
10. ₹ 8,000 is lent out at 7% compound interest for 2 years. At the end of the first year ₹ 3,560 are returned. Calculate :
(i) the interest paid for the second year.
(ii) the total interest paid in two years
(iii) the total amount of money paid in two years to clear the debt.
Step 1: Principal = ₹ 8,000. Rate = 7%.
Step 2: Interest 1st yr = 7% of 8000 = ₹ 560. Amount end 1st yr = 8000 + 560 = ₹ 8,560.
Step 3: Balance after return = 8560 - 3560 = ₹ 5,000. (Principal for 2nd yr).
Step 4: (i) Interest for 2nd yr = 7% of 5000 = ₹ 350.
Step 5: (ii) Total interest paid in two years = Int 1st yr + Int 2nd yr = 560 + 350 = ₹ 910.
Step 6: (iii) Total amount paid = (Returned end 1st yr) + (Amount to clear debt end 2nd yr).
Step 7: Amount to clear debt end 2nd yr = 5000 + 350 = ₹ 5,350.
Step 8: Total money paid = 3560 + 5350 = ₹ 8,910.
Step 2: Interest 1st yr = 7% of 8000 = ₹ 560. Amount end 1st yr = 8000 + 560 = ₹ 8,560.
Step 3: Balance after return = 8560 - 3560 = ₹ 5,000. (Principal for 2nd yr).
Step 4: (i) Interest for 2nd yr = 7% of 5000 = ₹ 350.
Step 5: (ii) Total interest paid in two years = Int 1st yr + Int 2nd yr = 560 + 350 = ₹ 910.
Step 6: (iii) Total amount paid = (Returned end 1st yr) + (Amount to clear debt end 2nd yr).
Step 7: Amount to clear debt end 2nd yr = 5000 + 350 = ₹ 5,350.
Step 8: Total money paid = 3560 + 5350 = ₹ 8,910.
11. Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by ₹ 252.
Step 1: Let the sum be P = 100x.
Step 2: Interest for 1st yr = 10% of 100x = 10x.
Step 3: Amount end 1st yr = 110x.
Step 4: Interest for 2nd yr = 10% of 110x = 11x.
Step 5: Amount end 2nd yr = 121x.
Step 6: Interest for 3rd yr = 10% of 121x = 12.1x.
Step 7: Difference = Interest 3rd yr - Interest 1st yr = 12.1x - 10x = 2.1x.
Step 8: We are given difference = 252. So, 2.1x = 252.
Step 9: x = 252 / 2.1 = 120.
Step 10: Original sum = 100x = 100 × 120 = ₹ 12,000.
Step 2: Interest for 1st yr = 10% of 100x = 10x.
Step 3: Amount end 1st yr = 110x.
Step 4: Interest for 2nd yr = 10% of 110x = 11x.
Step 5: Amount end 2nd yr = 121x.
Step 6: Interest for 3rd yr = 10% of 121x = 12.1x.
Step 7: Difference = Interest 3rd yr - Interest 1st yr = 12.1x - 10x = 2.1x.
Step 8: We are given difference = 252. So, 2.1x = 252.
Step 9: x = 252 / 2.1 = 120.
Step 10: Original sum = 100x = 100 × 120 = ₹ 12,000.
12. A man borrows ₹ 10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 30% of the sum borrowed. How much money is left unpaid just after the second year ?
Step 1: Sum borrowed = ₹ 10,000. Yearly payment = 30% of 10000 = ₹ 3,000.
Step 2: Amount at end of 1st yr = 10000 + 10% of 10000 = 11,000.
Step 3: Balance after 1st payment = 11000 - 3000 = ₹ 8,000.
Step 4: Amount at end of 2nd yr = 8000 + 10% of 8000 = 8,800.
Step 5: Balance after 2nd payment = 8800 - 3000 = ₹ 5,800.
Step 6: Money left unpaid just after the 2nd year is ₹ 5,800.
Step 2: Amount at end of 1st yr = 10000 + 10% of 10000 = 11,000.
Step 3: Balance after 1st payment = 11000 - 3000 = ₹ 8,000.
Step 4: Amount at end of 2nd yr = 8000 + 10% of 8000 = 8,800.
Step 5: Balance after 2nd payment = 8800 - 3000 = ₹ 5,800.
Step 6: Money left unpaid just after the 2nd year is ₹ 5,800.
13. A man borrows ₹ 10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year ?
Step 1: Sum borrowed = ₹ 10,000.
Step 2: Amount at end of 1st yr = 10000 + 10% of 10000 = ₹ 11,000.
Step 3: Payment end 1st yr = 20% of 11000 = ₹ 2,200.
Step 4: Balance for 2nd yr = 11000 - 2200 = ₹ 8,800.
Step 5: Amount at end of 2nd yr = 8800 + 10% of 8800 = 8800 + 880 = ₹ 9,680.
Step 6: Payment end 2nd yr = 20% of 9680 = ₹ 1,936.
Step 7: Money left unpaid = 9680 - 1936 = ₹ 7,744.
Step 2: Amount at end of 1st yr = 10000 + 10% of 10000 = ₹ 11,000.
Step 3: Payment end 1st yr = 20% of 11000 = ₹ 2,200.
Step 4: Balance for 2nd yr = 11000 - 2200 = ₹ 8,800.
Step 5: Amount at end of 2nd yr = 8800 + 10% of 8800 = 8800 + 880 = ₹ 9,680.
Step 6: Payment end 2nd yr = 20% of 9680 = ₹ 1,936.
Step 7: Money left unpaid = 9680 - 1936 = ₹ 7,744.
TEST YOURSELF
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) Certain sum is lent at 10% compound interest per annum. If the interest accrued during the year 2024 was ₹ 1,331 then the interest accrued during the year 2022, was:
(i) ₹ 1,210 (ii) ₹ 1,100 (iii) ₹ 1,464.10 (iv) ₹ 1,610.51
Step 1: Interest grows by the rate percent each year.
Step 2: Let Interest in 2022 be I.
Step 3: Interest in 2023 = I × (1 + 10/100) = 1.10 I.
Step 4: Interest in 2024 = 1.10 I × 1.10 = 1.21 I.
Step 5: 1.21 I = 1331 => I = 1331 / 1.21 = ₹ 1,100.
Step 6: Correct option is (ii) ₹ 1,100.
Step 2: Let Interest in 2022 be I.
Step 3: Interest in 2023 = I × (1 + 10/100) = 1.10 I.
Step 4: Interest in 2024 = 1.10 I × 1.10 = 1.21 I.
Step 5: 1.21 I = 1331 => I = 1331 / 1.21 = ₹ 1,100.
Step 6: Correct option is (ii) ₹ 1,100.
(b) During 2023, the population of a small village was 45,000 which increased every year by 5%. The population during the year 2024 was :
(i) 45,000 + 5% of 45,000 (ii) 45,000 - 5% of 45,000 (iii) 45,000 × 5% of 45,000 (iv) 45,000 ÷ 5% of 45,000
Step 1: Population in 2023 = 45,000.
Step 2: Increase in 2024 = 5% of Population in 2023 = 5% of 45,000.
Step 3: Population in 2024 = 45,000 + 5% of 45,000.
Step 4: Correct option is (i).
Step 2: Increase in 2024 = 5% of Population in 2023 = 5% of 45,000.
Step 3: Population in 2024 = 45,000 + 5% of 45,000.
Step 4: Correct option is (i).
(c) In how many years will a sum of money double itself at 10% C.I. :
(i) 5 years (ii) 10 years (iii) 8 years (iv) none of these
Step 1: We want P(1 + 10/100)^n = 2P => (1.1)^n = 2.
Step 2: Calculate consecutive powers: 1.1^7 ≈ 1.948, 1.1^8 ≈ 2.143.
Step 3: The sum will double itself during the 8th year.
Step 4: Correct option is (iii) 8 years.
Step 2: Calculate consecutive powers: 1.1^7 ≈ 1.948, 1.1^8 ≈ 2.143.
Step 3: The sum will double itself during the 8th year.
Step 4: Correct option is (iii) 8 years.
(d) The cost of a machine depreciates every year by 10%; the percentage decrease during two years will be :
(i) 20% (ii) 18% (iii) 19% (iv) 21%
Step 1: Let initial cost be 100.
Step 2: Value after 1st year = 100 - 10% of 100 = 90.
Step 3: Value after 2nd year = 90 - 10% of 90 = 81.
Step 4: Decrease in 2 years = 100 - 81 = 19.
Step 5: Percentage decrease = 19%.
Step 6: Correct option is (iii) 19%.
Step 2: Value after 1st year = 100 - 10% of 100 = 90.
Step 3: Value after 2nd year = 90 - 10% of 90 = 81.
Step 4: Decrease in 2 years = 100 - 81 = 19.
Step 5: Percentage decrease = 19%.
Step 6: Correct option is (iii) 19%.
(e) Assertion (A) : On a certain sum and at a certain rate,
C.I. for 3rd year = Amount in 3rd year - Amount in 2nd year
Reason (R) : Amount in 3 years = Principal + C.I. of 3 years and Amount in 2 years = Principal + C.I. of 2 years => Amount in 3 years - Amount in 2 years = C.I. in 3rd year
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 1: Assertion is correct since the interest generated in a specific year is the difference between amounts.
Step 2: Reason correctly proves it mathematically: (P + C.I. of 3 yrs) - (P + C.I. of 2 yrs) = C.I. of 3rd year.
Step 3: Both are true and Reason is correct.
Step 4: Correct option is (iii).
Step 2: Reason correctly proves it mathematically: (P + C.I. of 3 yrs) - (P + C.I. of 2 yrs) = C.I. of 3rd year.
Step 3: Both are true and Reason is correct.
Step 4: Correct option is (iii).
(f) Assertion (A) : At compound interest, interest of 5th year = Interest in 5 years - Interest in 4 years
Reason (R) : Interest of 5th year = Amount in 5 years - Amount in 4 years
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 1: Assertion is true, the interest for the 5th year alone is the total interest in 5 years minus total interest in 4 years.
Step 2: Reason is also a true statement defining the interest of the 5th year.
Step 3: Both are true and Reason correctly explains the relationship.
Step 4: Correct option is (iii).
Step 2: Reason is also a true statement defining the interest of the 5th year.
Step 3: Both are true and Reason correctly explains the relationship.
Step 4: Correct option is (iii).
(g) Statement (1) : Rate of C.I. accrued in 3rd year = ((Amount of 3 years - Amount of 2 years) / Amount in 2 years) × 100%
Statement (2) : Amount at the end of 3 years = Amount at the end of 2nd year + Interest on it
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: Statement 1 gives the correct formula to find the rate from consecutive amounts.
Step 2: Statement 2 correctly defines how compound amounts are formed.
Step 3: Both statements are true.
Step 4: Correct option is (i).
Step 2: Statement 2 correctly defines how compound amounts are formed.
Step 3: Both statements are true.
Step 4: Correct option is (i).
(h) Statement (1) : If P is the sum invested for 2 years at 20% rate of interest, then Amount in 2 years = P × (20/100) × (20/100)
Statement (2) : Interest accrued in 2 years = ₹ (P × (120/100) × (120/100) - P)
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: Statement 1 is false, the formula for amount is P(1 + 20/100)² = P(120/100)², not P(20/100)².
Step 2: Statement 2 is true, it correctly calculates C.I. = Amount - Principal.
Step 3: Statement 1 is false, statement 2 is true.
Step 4: Correct option is (iv).
Step 2: Statement 2 is true, it correctly calculates C.I. = Amount - Principal.
Step 3: Statement 1 is false, statement 2 is true.
Step 4: Correct option is (iv).
2. What sum will amount to ₹ 6,593.40 in 2 years at C.I., if the rates are 10 percent and 11 percent for the two successive years ?
Step 1: Let the sum be P.
Step 2: Amount = P(1 + r₁/100)(1 + r₂/100).
Step 3: 6593.40 = P(1 + 10/100)(1 + 11/100).
Step 4: 6593.40 = P(1.10)(1.11) = P(1.221).
Step 5: P = 6593.40 / 1.221 = ₹ 5,400.
Step 2: Amount = P(1 + r₁/100)(1 + r₂/100).
Step 3: 6593.40 = P(1 + 10/100)(1 + 11/100).
Step 4: 6593.40 = P(1.10)(1.11) = P(1.221).
Step 5: P = 6593.40 / 1.221 = ₹ 5,400.
3. The value of a machine depreciated by 10% per year during the first two years and 15% per year during the third year. Express the total depreciation of the machine, as percent, during the three years
Step 1: Let the original value of the machine = 100.
Step 2: Value after 1st year = 100 - (10% of 100) = 90.
Step 3: Value after 2nd year = 90 - (10% of 90) = 81.
Step 4: Value after 3rd year = 81 - (15% of 81) = 81 - 12.15 = 68.85.
Step 5: Total depreciation = 100 - 68.85 = 31.15.
Step 6: Total depreciation percentage = 31.15%.
Step 2: Value after 1st year = 100 - (10% of 100) = 90.
Step 3: Value after 2nd year = 90 - (10% of 90) = 81.
Step 4: Value after 3rd year = 81 - (15% of 81) = 81 - 12.15 = 68.85.
Step 5: Total depreciation = 100 - 68.85 = 31.15.
Step 6: Total depreciation percentage = 31.15%.
4. Rachna borrows ₹ 12,000 at 10 per cent per annum interest compounded half-yearly. She repays ₹ 4,000 at the end of every six months. Calculate the third payment she has to make at the end of 18 months in order to clear the entire loan.
Step 1: Principal = ₹ 12,000, Rate = 5% per six months.
Step 2: Interest 1st half-yr = 5% of 12000 = ₹ 600. Amount = ₹ 12,600.
Step 3: Balance after 1st payment = 12600 - 4000 = ₹ 8,600.
Step 4: Interest 2nd half-yr = 5% of 8600 = ₹ 430. Amount = 8600 + 430 = ₹ 9,030.
Step 5: Balance after 2nd payment = 9030 - 4000 = ₹ 5,030.
Step 6: Interest 3rd half-yr = 5% of 5030 = ₹ 251.50. Amount = 5030 + 251.50 = ₹ 5,281.50.
Step 7: The third payment to clear the loan is ₹ 5,281.50.
Step 2: Interest 1st half-yr = 5% of 12000 = ₹ 600. Amount = ₹ 12,600.
Step 3: Balance after 1st payment = 12600 - 4000 = ₹ 8,600.
Step 4: Interest 2nd half-yr = 5% of 8600 = ₹ 430. Amount = 8600 + 430 = ₹ 9,030.
Step 5: Balance after 2nd payment = 9030 - 4000 = ₹ 5,030.
Step 6: Interest 3rd half-yr = 5% of 5030 = ₹ 251.50. Amount = 5030 + 251.50 = ₹ 5,281.50.
Step 7: The third payment to clear the loan is ₹ 5,281.50.
5. On a certain sum of money, invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is ₹ 2,652. Find the sum.
Step 1: Let the sum be 100x.
Step 2: Interest 1st yr = 10% of 100x = 10x.
Step 3: Amount 1st yr = 110x.
Step 4: Interest 2nd yr = 10% of 110x = 11x. Amount 2nd yr = 121x.
Step 5: Interest 3rd yr = 10% of 121x = 12.1x.
Step 6: Sum of 1st and 3rd year interest = 10x + 12.1x = 22.1x.
Step 7: We are given 22.1x = 2652 => x = 2652 / 22.1 = 120.
Step 8: The original sum = 100x = 100 × 120 = ₹ 12,000.
Step 2: Interest 1st yr = 10% of 100x = 10x.
Step 3: Amount 1st yr = 110x.
Step 4: Interest 2nd yr = 10% of 110x = 11x. Amount 2nd yr = 121x.
Step 5: Interest 3rd yr = 10% of 121x = 12.1x.
Step 6: Sum of 1st and 3rd year interest = 10x + 12.1x = 22.1x.
Step 7: We are given 22.1x = 2652 => x = 2652 / 22.1 = 120.
Step 8: The original sum = 100x = 100 × 120 = ₹ 12,000.
6. During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates by ₹ 2,640 during the second financial year of its purchase.
Step 1: Let original cost be P.
Step 2: Value after 1st yr = P - 0.12 P = 0.88 P.
Step 3: Depreciation during 2nd yr = 12% of 0.88 P = 0.12 × 0.88 P = 0.1056 P.
Step 4: Given 0.1056 P = 2640.
Step 5: P = 2640 / 0.1056 = ₹ 25,000.
Step 2: Value after 1st yr = P - 0.12 P = 0.88 P.
Step 3: Depreciation during 2nd yr = 12% of 0.88 P = 0.12 × 0.88 P = 0.1056 P.
Step 4: Given 0.1056 P = 2640.
Step 5: P = 2640 / 0.1056 = ₹ 25,000.
7. Find the sum on which the difference between the simple interest and the compound interest at the rate of 8% per annum compounded annually be ₹ 64 in 2 years.
Step 1: Difference between C.I. and S.I. for 2 years = P(R/100)².
Step 2: 64 = P(8/100)² = P(64/10000).
Step 3: P = (64 × 10000) / 64 = ₹ 10,000.
Step 2: 64 = P(8/100)² = P(64/10000).
Step 3: P = (64 × 10000) / 64 = ₹ 10,000.
8. A sum of ₹ 13,500 is invested at 16% per annum compound interest for 5 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year, correct to the nearest rupee.
Step 1: (i) Interest for 1st yr = 16% of 13500 = ₹ 2,160.
Step 2: (ii) Amount at end of 1st yr = 13500 + 2160 = ₹ 15,660.
Step 3: (iii) Interest for 2nd yr = 16% of 15660 = ₹ 2,505.60.
Step 4: Interest rounded to nearest rupee = ₹ 2,506.
Step 2: (ii) Amount at end of 1st yr = 13500 + 2160 = ₹ 15,660.
Step 3: (iii) Interest for 2nd yr = 16% of 15660 = ₹ 2,505.60.
Step 4: Interest rounded to nearest rupee = ₹ 2,506.
9. Saurabh invests ₹ 48,000 for 7 years at 10% per annum compound interest. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the second year.
(iii) the interest for the third year.
Step 1: (i) Interest for 1st yr = 10% of 48000 = ₹ 4,800.
Step 2: (ii) Amount at end of 2nd yr = P(1 + R/100)² = 48000(1.10)² = 48000 × 1.21 = ₹ 58,080.
Step 3: (iii) Interest for 3rd yr = 10% of Principal for 3rd yr (Amount at end of 2nd yr).
Step 4: Interest for 3rd yr = 10% of 58080 = ₹ 5,808.
Step 2: (ii) Amount at end of 2nd yr = P(1 + R/100)² = 48000(1.10)² = 48000 × 1.21 = ₹ 58,080.
Step 3: (iii) Interest for 3rd yr = 10% of Principal for 3rd yr (Amount at end of 2nd yr).
Step 4: Interest for 3rd yr = 10% of 58080 = ₹ 5,808.
10. Ashok borrowed ₹ 12,000 at some rate per cent compound interest. After a year, he paid back ₹ 4,000. If compound interest for the second year be ₹ 920, find :
(i) the rate of interest charged
(ii) the amount of debt at the end of the second year.
Step 1: Let the rate be R%.
Step 2: Amount at end of 1st yr = 12000 + (12000 × R / 100) = 12000 + 120R.
Step 3: Balance after repayment = (12000 + 120R) - 4000 = 8000 + 120R. This is Principal for 2nd yr.
Step 4: C.I. for 2nd yr = (8000 + 120R) × R / 100 = 80R + 1.2R².
Step 5: Given C.I. for 2nd yr = 920. So, 1.2R² + 80R - 920 = 0.
Step 6: Multiply by 10 to clear decimal: 12R² + 800R - 9200 = 0, divide by 4: 3R² + 200R - 2300 = 0.
Step 7: Factorizing: 3R² + 230R - 30R - 2300 = 0 => R(3R + 230) - 10(3R + 230) = 0 => (R - 10)(3R + 230) = 0. R = 10%.
Step 8: (i) The rate of interest charged is 10%.
Step 9: (ii) Principal for 2nd yr = 8000 + 120(10) = ₹ 9,200.
Step 10: Amount of debt at end of 2nd yr = 9200 + 920 = ₹ 10,120.
Step 2: Amount at end of 1st yr = 12000 + (12000 × R / 100) = 12000 + 120R.
Step 3: Balance after repayment = (12000 + 120R) - 4000 = 8000 + 120R. This is Principal for 2nd yr.
Step 4: C.I. for 2nd yr = (8000 + 120R) × R / 100 = 80R + 1.2R².
Step 5: Given C.I. for 2nd yr = 920. So, 1.2R² + 80R - 920 = 0.
Step 6: Multiply by 10 to clear decimal: 12R² + 800R - 9200 = 0, divide by 4: 3R² + 200R - 2300 = 0.
Step 7: Factorizing: 3R² + 230R - 30R - 2300 = 0 => R(3R + 230) - 10(3R + 230) = 0 => (R - 10)(3R + 230) = 0. R = 10%.
Step 8: (i) The rate of interest charged is 10%.
Step 9: (ii) Principal for 2nd yr = 8000 + 120(10) = ₹ 9,200.
Step 10: Amount of debt at end of 2nd yr = 9200 + 920 = ₹ 10,120.
11. On a certain sum of money, lent out at C.I., interests for first, second and third years are ₹ 1,500; ₹ 1,725 and ₹ 2,070 respectively. Find the rate of interest for the (i) second year (ii) third year.
Step 1: The increase in interest from one year to the next is the interest generated on the previous year's interest.
Step 2: (i) Difference in interest (1st to 2nd yr) = 1725 - 1500 = ₹ 225.
Step 3: Rate for 2nd year = (Difference × 100) / Interest of 1st year.
Step 4: Rate for 2nd year = (225 × 100) / 1500 = 15%.
Step 5: (ii) Difference in interest (2nd to 3rd yr) = 2070 - 1725 = ₹ 345.
Step 6: Rate for 3rd year = (Difference × 100) / Interest of 2nd year.
Step 7: Rate for 3rd year = (345 × 100) / 1725 = 20%.
Step 2: (i) Difference in interest (1st to 2nd yr) = 1725 - 1500 = ₹ 225.
Step 3: Rate for 2nd year = (Difference × 100) / Interest of 1st year.
Step 4: Rate for 2nd year = (225 × 100) / 1500 = 15%.
Step 5: (ii) Difference in interest (2nd to 3rd yr) = 2070 - 1725 = ₹ 345.
Step 6: Rate for 3rd year = (Difference × 100) / Interest of 2nd year.
Step 7: Rate for 3rd year = (345 × 100) / 1725 = 20%.
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What term describes the money borrowed from a bank or agency for a specified period?
Answer
Principal
Question
What is the extra money paid to a lender for using their money called?
Answer
Interest
Question
What is the total money paid back to a lender at the end of a specified period called?
Answer
Amount
Question
State the fundamental formula for calculating the total Amount (A) using Principal (P) and Interest (I).
Answer
A = P + I
Question
Interest that is calculated on the original principal throughout the entire loan period is known as _____ interest.
Answer
Simple
Question
What is the mathematical formula for Simple Interest (I)?
Answer
I = \frac{P \times R \times T}{100}
Question
If a financial problem mentions 'interest' without any further description, which type of interest is always implied?
Answer
Simple Interest
Question
In Compound Interest, what happens to the interest due at the end of a fixed period?
Answer
It is added to the principal to form the new principal for the next period.
Question
How is the total Compound Interest (C.I.) determined using the final amount and the original principal?
Answer
C.I. = \text{Final Amount} - \text{Original Principal}
Question
For the first year of a loan, how does Compound Interest compare to Simple Interest given the same principal and rate?
Answer
They are equal (C.I. = S.I.)
Question
From the second year onwards, why is Compound Interest higher than Simple Interest for the same sum and rate?
Answer
Because the interest is calculated on a principal that grows each year.
Question
What is the specific name for the time interval after which interest is added to the principal and the principal changes?
Answer
Conversion period
Question
If interest is compounded half-yearly, what is the length of the conversion period?
Answer
Six months
Question
In Compound Interest calculations, the amount at the end of the first year becomes the _____ for the second year.
Answer
Principal
Question
Calculate the Simple Interest on ₹ 1,000 at 10\% per annum for 3 years.
Answer
₹ 300
Question
If the Compound Interest for the 1st, 2nd, and 3rd years are ₹ 100, ₹ 110, and ₹ 121 respectively, what is the total C.I. for 3 years?
Answer
₹ 331
Question
When interest is compounded half-yearly, what value for Time (T) is used to calculate interest for each conversion period?
Answer
T = \frac{1}{2} year
Question
If ₹ 8,000 is lent at 5\% compound interest for 2 years, what is the interest for the first year?
Answer
₹ 400
Question
What is the amount at the end of the first year for ₹ 8,000 at 5\% compound interest?
Answer
₹ 8,400
Question
Calculate the interest for the second year if the principal at the start of that year is ₹ 8,400 and the rate is 5\%.
Answer
₹ 420
Question
What is the total Compound Interest on ₹ 16,000 in 3 years if the successive annual rates are 10\%, 12\%, and 15\%?
Answer
₹ 6,668.80
Question
To calculate C.I. for 2 \frac{1}{2} years compounded annually, what time value is used for the interest calculation of the final period?
Answer
T = \frac{1}{2} year
Question
Find the difference between C.I. and S.I. on ₹ 4,000 at 8\% per annum for 2 years.
Answer
₹ 25.60
Question
What is the formula for finding the Rate of interest (R) when the C.I. of two consecutive periods is known?
Answer
R = \frac{\text{Difference in C.I.} \times 100}{\text{C.I. of preceding period} \times \text{Time}}
Question
What is the formula for finding the Rate of interest (R) when the Amounts (A) of two consecutive periods are known?
Answer
R = \frac{\text{Difference in Amounts} \times 100}{\text{Preceding Amount} \times \text{Time}}
Question
Term: Depreciation
Answer
Definition: The reduction in the value of an asset (like a machine) over time due to use or age.
Question
How is the value of a machine at the beginning of the second year calculated if it depreciates by 10\% in the first year?
Answer
\text{Original Cost} - (10\% \text{ of Original Cost})
Question
In a repayment problem, if a borrower pays back ₹ 2,500 at the end of the first year, how is the principal for the second year determined?
Answer
\text{Amount at the end of 1st year} - ₹ 2,500
Question
If the Simple Interest on a sum for 3 years is ₹ 600, what is the Simple Interest for just the first year?
Answer
₹ 200
Question
For any conversion period after the first, why is the Compound Interest always more than the C.I. of the previous period?
Answer
Because the interest is calculated on a principal that has increased by the previous period's interest.
Question
If the C.I. of the 1st year is x, what is the C.I. for the 2nd year at rate r\%?
Answer
x + (r\% \text{ of } x)
Question
The difference between the compound interests for any two consecutive conversion periods is the interest on the _____ of the preceding period.
Answer
Interest
Question
The difference between the amounts of any two consecutive conversion periods is the interest on the _____ of the preceding period.
Answer
Amount
Question
If an asset worth ₹ 100 depreciates by 10\% in the first year and 10\% of the remaining value in the second year, what is its value after 2 years?
Answer
₹ 81
Question
If interest is compounded half-yearly, how many conversion periods are there in 1.5 years?
Answer
Three
Question
Find the interest on ₹ 10,000 for \frac{1}{2} year at 8\% per annum.
Answer
₹ 400
Question
If the amount after 2 years at C.I. is ₹ 6,272 and after 3 years is ₹ 7,024.64, what is the interest for the 3rd year?
Answer
₹ 752.64
Question
In Example 9, what assumed principal (P) is used to solve the problem by the unitary method?
Answer
₹ 100
Question
If a machine depreciates by 10\% per annum and the depreciation during the second year is ₹ 9, what was the value at the start of the second year?
Answer
₹ 90
Question
If the amount at the end of the 2nd year is ₹ 13,860 and an additional ₹ 6,000 is invested at the start of the 3rd year, what is the principal for the 3rd year?
Answer
₹ 19,860
Question
Find the interest on ₹ 6,000 for 1 year at 10\% per annum.
Answer
₹ 600
Question
If a sum grows to ₹ 22,400 after 1 year and ₹ 8,400 is repaid, what is the new principal for the 2nd year?
Answer
₹ 14,000
Question
Find the interest on ₹ 5,900 for \frac{1}{2} year at 10\% per annum.
Answer
₹ 295
Question
If the difference between C.I. of the 1st and 3rd year is ₹ 0.5125 for a principal of ₹ 100, what is the principal for a difference of ₹ 61.50?
Answer
₹ 12,000
Question
What is the Simple Interest on ₹ 46,875 at 4\% for 1 year?
Answer
₹ 1,875
Question
If the C.I. for two successive years is ₹ 2,700 and ₹ 2,880, calculate the rate of interest.
Answer
6 \frac{2}{3} \%
Question
How does the Simple Interest on a fixed sum at a constant rate change over multiple years?
Answer
It remains the same every year.
Question
If the S.I. for 3 years is ₹ 600 and the C.I. for 2 years is ₹ 410 on the same sum and rate, what is the interest for the 1st year?
Answer
₹ 200 (for both S.I. and C.I.)
Question
In half-yearly compounding at 10\% p.a., what is the interest on ₹ 8,000 for the first six months?
Answer
₹ 400
Question
If a machine depreciates by 15\% of its value each year, and its initial value is ₹ V, what is the depreciation amount in the first year?
Answer
0.15 V
Question
If the principal for the 2nd year is ₹ 14,000 and the rate is 12\%, what is the interest for the 2nd year?
Answer
₹ 1,680
Question
Calculate the amount for the 3rd year if the principal is ₹ 19,712 and the rate is 15\%.
Answer
₹ 22,668.80
Question
What is the C.I. for the 3rd year in Problem 19 if the interest for the 2nd year is ₹ 1,056 and the rate is 10\%?
Answer
₹ 1,161.60
Question
If the C.I. for the 5th year is ₹ 665.50 and the rate is 10\%, what is the C.I. for the 6th year?
Answer
₹ 732.05
Question
In Example 18, if the amount in 5 years is ₹ 8,100 and in 6 years is ₹ 8,748, what is the interest for the 6th year?
Answer
₹ 648
Question
What is the rate percent in Example 18 if the interest is ₹ 648 on a sum of ₹ 8,100?
Answer
8\%
Question
If a machine's original cost is ₹ 100 and it depreciates by 10\% annually, what is its value at the beginning of the 3rd year?
Answer
₹ 81
Question
For ₹ 6,000 at 10\% C.I., find the interest for the second year.
Answer
₹ 660
Question
If the principal at the start of the final half-year is ₹ 7,260 and the rate is 10\%, what is the interest for that half-year?
Answer
₹ 363
Question
In Example 11, find the interest for the 3rd year on ₹ 50,700 at 4\%.
Answer
₹ 2,028