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COMPOUND INTEREST (Stage 2) [Applications] - Questions & Answers


EXERCISE 3 (A)

1. Multiple Choice Type :

Choose the correct answer from the options given below.
(a) If P = sum lent, r = rate of interest per year, n = number of years and amounts = A, then :
(i) A = P(1 + r/100)n (ii) A - P = (1 + r/100)n (iii) A = P(1 - r/100)n (iv) A - P = (1 - r/100)n
Step 1: The formula for Amount (A) under compound interest is principal times the growth factor raised to the number of years.
Answer: (i) A = P(1 + r/100)n

(b) If the letters used have usual meanings; r1% and r2% are rates of interests for two consecutive years then :
(i) A = P(1 + r1/100)(1 - r2/100) (ii) A = P(1 + r1/100)(1 + r2/100) (iii) A - P = (1 + r1/100)(1 + r2/100) (iv) P = A(1 - r1/100)(1 - r2/100)
Step 1: For successive different interest rates, the growth factors are multiplied for each year sequentially.
Answer: (ii) A = P(1 + r1/100)(1 + r2/100)

(c) Compound interest on ₹ 6,000 in 2 years at 5% per annum is :
(i) ₹ 615 (ii) ₹ 630 (iii) ₹ 600 (iv) ₹ 690
Step 1: P = 6000, r = 5%, n = 2.
Step 2: A = 6000(1 + 5/100)2 = 6000(1.05)2.
Step 3: A = 6000 × 1.1025 = 6615.
Step 4: C.I. = A - P = 6615 - 6000 = 615.
Answer: (i) ₹ 615

(d) A sum of money, lent at 10% C.I. compounded yearly becomes ₹ 6,050 in 2 years. The sum lent is :
(i) ₹ 7,260 (ii) ₹ 4,000 (iii) ₹ 5,000 (iv) ₹ 7,320.50
Step 1: A = 6050, r = 10%, n = 2.
Step 2: 6050 = P(1 + 10/100)2 = P(1.1)2.
Step 3: 6050 = P(1.21).
Step 4: P = 6050 / 1.21 = 5000.
Answer: (iii) ₹ 5,000

(e) On a certain sum, the compound interest accrued in one year is ₹ 550. If the rate of interest is 10%, the sum is :
(i) ₹ 5,000 (ii) ₹ 6,000 (iii) ₹ 4,500 (iv) ₹ 5,500
Step 1: For the first year, compound interest equals simple interest (S.I.).
Step 2: C.I. = P × r × t / 100.
Step 3: 550 = P × 10 × 1 / 100.
Step 4: 550 = P / 10 ⇒ P = 5500.
Answer: (iv) ₹ 5,500

(f) ₹ 4,000 amounts to ₹ 4,600 in one year at compound interest compounded yearly. The rate of interest is :
(i) 15% (ii) 12% (iii) 10% (iv) 20%
Step 1: P = 4000, A = 4600, n = 1.
Step 2: A = P(1 + r/100)n ⇒ 4600 = 4000(1 + r/100)1.
Step 3: 4600 / 4000 = 1 + r/100 ⇒ 1.15 = 1 + r/100.
Step 4: r/100 = 0.15 ⇒ r = 15%.
Answer: (i) 15%

(g) ₹ 4,000 amounts to ₹ 5017.60 in two months at compound interest compounded per month. The rate of interest per month is :
(i) 12% (ii) 15% (iii) 10% (iv) 20%
Step 1: P = 4000, A = 5017.60, n = 2 months.
Step 2: 5017.60 = 4000(1 + r/100)2.
Step 3: (1 + r/100)2 = 5017.60 / 4000 = 1.2544.
Step 4: 1 + r/100 = √1.2544 = 1.12.
Step 5: r/100 = 0.12 ⇒ r = 12%.
Answer: (i) 12%

2. Find the amount and the compound interest on ₹ 12,000 in 3 years at 5%; interest being compounded annually.
Step 1: P = 12000, r = 5%, n = 3 years.
Step 2: A = P(1 + r/100)n = 12000(1 + 5/100)3.
Step 3: A = 12000(21/20)3 = 12000 × (9261/8000) = 1.5 × 9261 = 13891.50.
Step 4: C.I. = A - P = 13891.50 - 12000 = 1891.50.
Answer: Amount = ₹ 13,891.50, Compound Interest = ₹ 1,891.50

3. Calculate the amount, if ₹ 15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% p.a. and 10% p.a. respectively.
Step 1: P = 15000, r1 = 8%, r2 = 10%.
Step 2: A = P(1 + r1/100)(1 + r2/100).
Step 3: A = 15000(1 + 8/100)(1 + 10/100) = 15000(1.08)(1.10).
Step 4: A = 15000 × 1.188 = 17820.
Answer: Amount = ₹ 17,820

4. Calculate the compound interest accrued on ₹ 6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.
Step 1: P = 6000, r1 = 5%, r2 = 8%, r3 = 10%.
Step 2: A = 6000(1 + 5/100)(1 + 8/100)(1 + 10/100).
Step 3: A = 6000 × 1.05 × 1.08 × 1.10 = 6000 × 1.2474 = 7484.40.
Step 4: C.I. = A - P = 7484.40 - 6000 = 1484.40.
Answer: Compound Interest = ₹ 1,484.40

5. What sum of money will amount to ₹ 5,445 in 2 years at 10% per annum compound interest?
Step 1: A = 5445, r = 10%, n = 2 years.
Step 2: 5445 = P(1 + 10/100)2.
Step 3: 5445 = P(1.1)2 = P(1.21).
Step 4: P = 5445 / 1.21 = 4500.
Answer: Sum of money = ₹ 4,500

6. On what sum of money will the compound interest for 2 years at 5 per cent per annum amount to ₹ 768.75 ?
Step 1: C.I. = 768.75, r = 5%, n = 2 years.
Step 2: C.I. = P[(1 + r/100)n - 1].
Step 3: 768.75 = P[(1.05)2 - 1] = P[1.1025 - 1] = P(0.1025).
Step 4: P = 768.75 / 0.1025 = 7500.
Answer: Sum of money = ₹ 7,500

7. Find the sum on which the compound interest for 3 years at 10% per annum amounts to ₹ 1,655.
Step 1: C.I. = 1655, r = 10%, n = 3 years.
Step 2: C.I. = P[(1 + 10/100)3 - 1].
Step 3: 1655 = P[(1.1)3 - 1] = P[1.331 - 1] = P(0.331).
Step 4: P = 1655 / 0.331 = 5000.
Answer: Sum = ₹ 5,000

8. At what rate per cent per annum will ₹ 6,000 amount to ₹ 6,615 in 2 years when interest is compounded annually ?
Step 1: P = 6000, A = 6615, n = 2 years.
Step 2: 6615 = 6000(1 + r/100)2.
Step 3: (1 + r/100)2 = 6615 / 6000 = 1.1025.
Step 4: 1 + r/100 = √1.1025 = 1.05.
Step 5: r/100 = 0.05 ⇒ r = 5%.
Answer: Rate = 5% p.a.

9. What principal will amount to ₹ 9,856 in two years, if the rates of interest for successive years are 10% and 12% respectively ?
Step 1: A = 9856, r1 = 10%, r2 = 12%.
Step 2: 9856 = P(1 + 10/100)(1 + 12/100) = P(1.10)(1.12).
Step 3: 9856 = P(1.232).
Step 4: P = 9856 / 1.232 = 8000.
Answer: Principal = ₹ 8,000

10. On a certain sum, the compound interest in 2 years amounts to ₹ 4,240. If the rates of interest for successive years are 10% and 15% respectively, find the sum.
Step 1: C.I. = 4240, r1 = 10%, r2 = 15%.
Step 2: C.I. = P[(1 + 10/100)(1 + 15/100) - 1].
Step 3: 4240 = P[(1.10)(1.15) - 1] = P[1.265 - 1] = P(0.265).
Step 4: P = 4240 / 0.265 = 16000.
Answer: Sum = ₹ 16,000

11. At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?
Step 1: Let Principal = P. Amount = 1.44P, n = 2 years.
Step 2: 1.44P = P(1 + r/100)2.
Step 3: 1.44 = (1 + r/100)2.
Step 4: 1 + r/100 = √1.44 = 1.2.
Step 5: r/100 = 0.2 ⇒ r = 20%.
Answer: Rate = 20%

12. At what rate per cent will a sum of ₹ 4,000 yield ₹ 1,324 as compound interest in 3 years ?
Step 1: P = 4000, C.I. = 1324, n = 3 years.
Step 2: A = P + C.I. = 4000 + 1324 = 5324.
Step 3: 5324 = 4000(1 + r/100)3.
Step 4: (1 + r/100)3 = 5324 / 4000 = 1331 / 1000 = 1.331.
Step 5: 1 + r/100 = ³√1.331 = 1.1.
Step 6: r/100 = 0.1 ⇒ r = 10%.
Answer: Rate = 10%

13. A person invests ₹ 5,000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to ₹ 6,272. Calculate :
(i) the rate of interest per annum.
(ii) the amount at the end of the third year.
Step 1: P = 5000, A (after 2 yrs) = 6272.
Step 2: 6272 = 5000(1 + r/100)2.
Step 3: (1 + r/100)2 = 6272 / 5000 = 1.2544.
Step 4: 1 + r/100 = √1.2544 = 1.12 ⇒ r = 12%.
Step 5: (ii) Amount at end of 3rd year = Amount after 2 yrs × (1 + r/100).
Step 6: A3 = 6272 × 1.12 = 7024.64.
Answer: (i) Rate = 12% p.a., (ii) Amount = ₹ 7,024.64

14. In how many years will ₹ 7,000 amount to ₹ 9,317 at 10 per cent per annum compound interest ?
Step 1: P = 7000, A = 9317, r = 10%.
Step 2: 9317 = 7000(1 + 10/100)n.
Step 3: (1.1)n = 9317 / 7000 = 1.331.
Step 4: 1.331 = (1.1)3.
Step 5: Equating powers, n = 3.
Answer: 3 years

15. Find the time, in years, in which ₹ 4,000 will produce ₹ 630.50 as compound interest at 5 per cent p.a. interest being compounded annually.
Step 1: P = 4000, C.I. = 630.50, r = 5%.
Step 2: A = P + C.I. = 4000 + 630.50 = 4630.50.
Step 3: 4630.50 = 4000(1 + 5/100)n.
Step 4: (1.05)n = 4630.50 / 4000 = 1.157625.
Step 5: 1.157625 = (1.05)3 ⇒ n = 3.
Answer: 3 years

16. Divide ₹ 28,730 between A and B so that when their shares are lent out at 10 per cent compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.
Step 1: Let A's share be x and B's share be (28730 - x).
Step 2: A's Amount after 3 yrs = x(1 + 10/100)3 = x(1.1)3.
Step 3: B's Amount after 5 yrs = (28730 - x)(1 + 10/100)5 = (28730 - x)(1.1)5.
Step 4: Equating both: x(1.1)3 = (28730 - x)(1.1)5.
Step 5: x = (28730 - x)(1.1)2 = (28730 - x)(1.21).
Step 6: x = 34763.30 - 1.21x ⇒ 2.21x = 34763.30.
Step 7: x = 34763.30 / 2.21 = 15730.
Step 8: B's share = 28730 - 15730 = 13000.
Answer: A's share = ₹ 15,730, B's share = ₹ 13,000

17. A sum of ₹ 44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10 percent per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of ₹ 44,200?
(ii) What will each receive, when 16 years old ?
Step 1: John's investment time = 16 - 12 = 4 years. Smith's time = 16 - 14 = 2 years.
Step 2: Let John's share be x, Smith's share be (44200 - x).
Step 3: x(1.1)4 = (44200 - x)(1.1)2.
Step 4: x(1.1)2 = 44200 - x ⇒ 1.21x = 44200 - x.
Step 5: 2.21x = 44200 ⇒ x = 20000. (John's share)
Step 6: Smith's share = 44200 - 20000 = 24200.
Step 7: (ii) Amount they receive = 24200(1.1)2 = 24200 × 1.21 = 29282.
Answer: (i) John's share = ₹ 20,000, Smith's share = ₹ 24,200; (ii) Each receives = ₹ 29,282

18. The simple interest on a certain sum of money at 10% per annum is ₹ 6,000 in 2 years. Find:
(i) the sum.
(ii) the amount due at the end of 3 years and at the same rate of interest compounded annually.
(iii) the compound interest earned in 3 years.
Step 1: (i) Sum (P) = (S.I. × 100) / (r × t) = (6000 × 100) / (10 × 2) = 30000.
Step 2: (ii) Amount = P(1 + r/100)n = 30000(1 + 10/100)3.
Step 3: Amount = 30000(1.1)3 = 30000 × 1.331 = 39930.
Step 4: (iii) C.I. = Amount - P = 39930 - 30000 = 9930.
Answer: (i) Sum = ₹ 30,000 (ii) Amount = ₹ 39,930 (iii) C.I. = ₹ 9,930

19. Find the difference between compound interest and simple interest on ₹ 8,000 in 2 years and at 5% per annum.
Step 1: P = 8000, r = 5%, n = 2.
Step 2: S.I. = P × r × t / 100 = 8000 × 5 × 2 / 100 = 800.
Step 3: C.I. = P[(1 + 5/100)2 - 1] = 8000[1.1025 - 1] = 8000 × 0.1025 = 820.
Step 4: Difference = 820 - 800 = 20.
Answer: ₹ 20

EXERCISE 3 (B)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) On ₹ 6,000, the difference between C.I. and S.I. in 2 years and at 10% compound interest, compounded per year, is :
(i) ₹ 120 (ii) ₹ 600 (iii) ₹ 60 (iv) ₹ 180
Step 1: Difference for 2 years = P(r/100)2.
Step 2: Difference = 6000(10/100)2 = 6000(0.01) = 60.
Answer: (iii) ₹ 60

(b) ₹ 10,000 amounts to ₹ 12,500 in one year. The rate of interest per year is :
(i) 15% (ii) 12.5% (iii) 20% (iv) 25%
Step 1: A = P(1 + r/100)n ⇒ 12500 = 10000(1 + r/100)1.
Step 2: 12500 / 10000 = 1.25 = 1 + r/100 ⇒ r/100 = 0.25 ⇒ r = 25%.
Answer: (iv) 25%

(c) The C.I. on ₹ 16,000 in 2 years at the rate of 20% per annum is :
(i) ₹ 19,360 (ii) ₹ 7,040 (iii) ₹ 23,040 (iv) ₹ 22,400
Step 1: A = 16000(1 + 20/100)2 = 16000(1.2)2 = 16000(1.44) = 23040.
Step 2: C.I. = 23040 - 16000 = 7040.
Answer: (ii) ₹ 7,040

(d) Simple interest, at the same rate for the same period as given above in part (c) is:
(i) ₹ 7,040 (ii) ₹ 6,400 (iii) ₹ 3,200 (iv) ₹ 1,280
Step 1: S.I. = P × r × t / 100 = 16000 × 20 × 2 / 100 = 6400.
Answer: (ii) ₹ 6,400

(e) The difference between C.I. and S.I. in 2 years as given above for parts (c) and (d) is:
(i) ₹ 640 (ii) ₹ 3,840 (iii) ₹ 1,920 (iv) ₹ 1,280
Step 1: Difference = C.I. - S.I. = 7040 - 6400 = 640.
Answer: (i) ₹ 640

2. The difference between simple interest and compound interest on a certain sum is ₹ 54.40 for 2 years at 8 percent per annum. Find the sum.
Step 1: For 2 years, Difference = P(r/100)2.
Step 2: 54.40 = P(8/100)2 = P(0.0064).
Step 3: P = 54.40 / 0.0064 = 8500.
Answer: Sum = ₹ 8,500

3. Pramod and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand received ₹ 15 more than Pramod. Find the amount of money lent by each and the interest received.
Step 1: Difference (C.I. - S.I.) = ₹ 15.
Step 2: 15 = P(5/100)2 = P(0.0025) ⇒ P = 15 / 0.0025 = 6000.
Step 3: Pramod's Interest (S.I.) = 6000 × 5 × 2 / 100 = 600.
Step 4: Anand's Interest (C.I.) = 600 + 15 = 615.
Answer: Money lent by each = ₹ 6,000, Pramod's Interest = ₹ 600, Anand's Interest = ₹ 615

4. Simple interest on a sum of money for 2 years at 4% is ₹ 450. Find compound interest on the same sum and at the same rate for 2 years.
Step 1: Sum (P) = (450 × 100) / (4 × 2) = 5625.
Step 2: C.I. = 5625[(1 + 4/100)2 - 1] = 5625[(1.04)2 - 1].
Step 3: C.I. = 5625(1.0816 - 1) = 5625 × 0.0816 = 459.
Answer: Compound Interest = ₹ 459

5. Compound interest on a certain sum of money at 5% per annum for two years is ₹ 246. Calculate simple interest on the same sum for 3 years at 6% per annum.
Step 1: 246 = P[(1.05)2 - 1] = P(1.1025 - 1) = P(0.1025).
Step 2: P = 246 / 0.1025 = 2400.
Step 3: New S.I. = 2400 × 6 × 3 / 100 = 432.
Answer: Simple Interest = ₹ 432

6. A sum of money, invested at compound interest, amounts to ₹ 19,360 in 2 years and to ₹ 23,425.60 in 4 years. Find the rate percent and the original sum of money.
Step 1: A4 = P(1 + r/100)4 = 23425.60 and A2 = P(1 + r/100)2 = 19360.
Step 2: Divide A4 by A2: (1 + r/100)2 = 23425.60 / 19360 = 1.21.
Step 3: 1 + r/100 = √1.21 = 1.1 ⇒ r = 10%.
Step 4: P(1.1)2 = 19360 ⇒ P(1.21) = 19360 ⇒ P = 19360 / 1.21 = 16000.
Answer: Rate = 10%, Original sum = ₹ 16,000

7. A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.
Step 1: A = P(1 + r/100)8 = 3P ⇒ (1 + r/100)8 = 3.
Step 2: We need Time (T) for Amount to be 27P. So, (1 + r/100)T = 27 = 33.
Step 3: Substitute 3 from Step 1: ((1 + r/100)8)3 = (1 + r/100)24.
Step 4: Equating powers, T = 24 years.
Answer: 24 years

8. On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on ₹ 9,430 for 10 years, both at the rate of 5 percent per annum ?
Step 1: S.I. on 9430 = 9430 × 5 × 10 / 100 = 4715.
Step 2: C.I. = P[(1.05)2 - 1] = P(0.1025).
Step 3: P(0.1025) = 4715 ⇒ P = 4715 / 0.1025 = 46000.
Answer: Sum = ₹ 46,000

9. Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228. Find the sum.
Step 1: S.I. = P × 4 × 4 / 100 = 0.16P.
Step 2: C.I. = P[(1.05)3 - 1] = P(1.157625 - 1) = 0.157625P.
Step 3: 0.16P - 0.157625P = 228.
Step 4: 0.002375P = 228 ⇒ P = 228 / 0.002375 = 96000.
Answer: Sum = ₹ 96,000

10. A certain sum of money amounts to ₹ 23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest. (HOTS)
Step 1: Amount at S.I. = P + P × r × t / 100 ⇒ 23400 = P(1 + 10×3/100) = P(1.3).
Step 2: P = 23400 / 1.3 = 18000.
Step 3: Amount at C.I. = 18000(1 + 10/100)2 = 18000(1.1)2.
Step 4: Amount = 18000 × 1.21 = 21780.
Answer: Amount = ₹ 21,780

11. Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12,600 at the end of the first year and ₹ 17,640 at the end of the second year. Find the sum borrowed.
Step 1: Let P be the total sum borrowed. It equals the present value of both payments.
Step 2: P = P1 (for 1st payment) + P2 (for 2nd payment).
Step 3: 12600 = P1(1.05)1 ⇒ P1 = 12600 / 1.05 = 12000.
Step 4: 17640 = P2(1.05)2 ⇒ P2 = 17640 / 1.1025 = 16000.
Step 5: Total sum = 12000 + 16000 = 28000.
Answer: Sum borrowed = ₹ 28,000

EXERCISE 3 (C)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) If the letters have usual meanings, the formula for finding compound interest, compounded yearly for the given time is :
(i) 1 1/2 years :
(1) P[(1+r/100)1/2 - 1] (2) P(1+r/100)(1+r/200) - P
(3) P(1+r/200)3 - P (4) P(1+r/100)3(1+r/200) - P
Step 1: For a fractional year, compound full years yearly, then simple interest for fraction.
Step 2: Amount = P(1+r/100)(1+(1/2)r/100). Interest = Amount - P.
Answer: (2) P(1+r/100)(1+r/200) - P

(ii) 1 year :
(1) P(1+r/100)1 - P (2) P(1+r/200) - P (3) P(1+r/200)2 - P (4) P(1-r/100)2 - P
Step 1: Yearly compounding for 1 year is standard formula.
Answer: (1) P(1+r/100)1 - P

(iii) 2 1/2 years :
(1) P(1+r/100)5/2 - P (2) P(1+r/100)5/2
(3) P(1+r/100)2(1+r/200) - P (4) P(1+r/100)2(1+r/200) - P
Step 1: 2 full years yearly + half year interest.
Answer: (3) P(1+r/100)2(1+r/200) - P

(b) When interest is compounded half-yearly then the formula for C.I. for the given time is:
(i) 1 year :
(1) P(1+r/200)2 - P (2) P(1+r/100)2 - P (3) P(1+r/200)2 (4) P(1+r/100)2
Step 1: Half yearly for 1 yr implies n=2 periods, rate=r/2.
Answer: (1) P(1+r/200)2 - P

(ii) 1 1/2 years :
(1) P(1+r/200)3 - P (2) P(1+r/200)3 - P (3) P(1+r/100)3 - P (4) P(1+r/100)3/2 - P
Step 1: 1.5 yrs is 3 half-years.
Answer: (1) P(1+r/200)3 - P

(iii) 2 years :
(1) P(1+r/100)2 - P (2) P(1+r/200)2 - P (3) P(1+r/200)4 - P (4) P(1+r/100)4 - P
Step 1: 2 yrs is 4 half-years.
Answer: (3) P(1+r/200)4 - P

(c) If ₹ 6,000 earns C.I. = ₹ 1,200 in 6 months; then the rate of interest per year is :
(i) 40% (ii) 15% (iii) 20% (iv) 24%
Step 1: For a single period (6 months), C.I. = S.I.
Step 2: 1200 = 6000 × rhalf / 100 ⇒ rhalf = 20%.
Step 3: Rate per year = 20% × 2 = 40%.
Answer: (i) 40%

(d) On a certain sum, the S.I. for 2 years is ₹ 2,400. If the rate of interest is 10% p.a., then :
(i) C.I. for 1st year is :
(1) ₹ 1,200 (2) ₹ 1,320 (3) ₹ 1,800 (4) ₹ 2,640
Step 1: S.I. for 1 year = 2400 / 2 = 1200. First year C.I. = S.I.
Answer: (1) ₹ 1,200

(ii) C.I. for 2nd year is :
(1) ₹ 1,320 (2) ₹ 2,640 (3) ₹ 1,980 (4) ₹ 2,400
Step 1: 2nd year C.I. = 1st yr C.I. + Interest on 1st yr C.I.
Step 2: 1200 + 10% of 1200 = 1200 + 120 = 1320.
Answer: (1) ₹ 1,320

(iii) the sum is :
(1) ₹ 12,000 (2) ₹ 26,400 (3) ₹ 13,200 (4) ₹ 24,000
Step 1: Sum (P) = (S.I. × 100) / (r × t) = (2400 × 100) / (10 × 2) = 12000.
Answer: (1) ₹ 12,000

(iv) the amount in 2 years, at compound interest, is :
(1) ₹ 14,520 (2) ₹ 12,000 (3) ₹ 12,120 (4) ₹ 24,000
Step 1: Amount = Sum + C.I.1st yr + C.I.2nd yr.
Step 2: 12000 + 1200 + 1320 = 14520.
Answer: (1) ₹ 14,520

2. If the interest is compounded half-yearly, calculate the amount when principal is ₹ 7,400; the rate of interest is 5% per annum and the duration is one year.
Step 1: P = 7400, r = 5%/2 = 2.5% per half-yr, n = 2 periods.
Step 2: A = 7400(1 + 2.5/100)2 = 7400(1.025)2.
Step 3: A = 7400 × 1.050625 = 7774.625.
Answer: Amount = ₹ 7,774.63

3. Find the difference between the compound interest compounded yearly and half-yearly on ₹ 10,000 for 18 months at 10% per annum.
Step 1: Yearly: A1 = 10000(1 + 10/100)(1 + 5/100) = 10000(1.1)(1.05) = 11550.
Step 2: C.I.yearly = 11550 - 10000 = 1550.
Step 3: Half-yearly: A2 = 10000(1 + 5/100)3 = 10000(1.157625) = 11576.25.
Step 4: C.I.half-yearly = 11576.25 - 10000 = 1576.25.
Step 5: Difference = 1576.25 - 1550 = 26.25.
Answer: ₹ 26.25

4. A man borrowed ₹ 16,000 for 3 years under the following terms :
20% simple interest for the first 2 years.
20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.
Find the total amount to be paid at the end of three years.
Step 1: S.I. for first 2 yrs = 16000 × 20 × 2 / 100 = 6400.
Step 2: Amount due after 2 yrs = 16000 + 6400 = 22400.
Step 3: C.I. half-yearly for 1 yr on 22400 at 20%: P = 22400, r = 10% per half-yr, n = 2 periods.
Step 4: Final Amount = 22400(1 + 10/100)2 = 22400(1.21) = 27104.
Answer: Total amount = ₹ 27,104

5. What sum of money will amount to ₹ 27,783 in one and a half years at 10% per annum compounded half-yearly ?
Step 1: A = 27783, r = 5% per half-yr, n = 3 half-yrs.
Step 2: 27783 = P(1.05)3 = P(1.157625).
Step 3: P = 27783 / 1.157625 = 24000.
Answer: Sum of money = ₹ 24,000

6. Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets ₹ 33 more than Ashok in 18 months, calculate the money invested by each. (HOTS)
Step 1: Let P be the sum. 18 months = 1.5 years.
Step 2: Ashok's Amount = P(1 + 20/100)(1 + 10/100) = P(1.2)(1.1) = 1.32P.
Step 3: Geeta's Amount = P(1 + 10/100)3 = P(1.1)3 = 1.331P.
Step 4: Difference = 1.331P - 1.32P = 0.011P.
Step 5: 0.011P = 33 ⇒ P = 33 / 0.011 = 3000.
Answer: Money invested by each = ₹ 3,000

7. At what rate of interest per annum will a sum of ₹ 50,000 earn a compound interest of ₹ 5,100 in one year ? The interest is to be compounded half-yearly.
Step 1: P = 50000, C.I. = 5100, A = 55100, n = 2 half-yrs.
Step 2: 55100 = 50000(1 + r/200)2.
Step 3: (1 + r/200)2 = 55100 / 50000 = 1.1025.
Step 4: 1 + r/200 = √1.1025 = 1.05 ⇒ r/200 = 0.05 ⇒ r = 10%.
Answer: Rate = 10% p.a.

8. In what time will ₹ 1,500 yield ₹ 496.50 as compound interest at 20% per year compounded half-yearly ?
Step 1: P = 1500, C.I. = 496.50, A = 1996.50, r = 10% per half-yr.
Step 2: 1996.50 = 1500(1.1)n.
Step 3: (1.1)n = 1996.50 / 1500 = 1.331.
Step 4: (1.1)n = (1.1)3 ⇒ n = 3 half-years.
Answer: 1 1/2 years

9. Calculate the C.I. on ₹ 3,500 at 6% per annum for 3 years, the interest being compounded half-yearly.
Do not use mathematical tables. Use the necessary information from the following :
(1.06)3 = 1.191016; (1.03)3 = 1.092727
(1.06)6 = 1.418519; (1.03)6 = 1.194052
Step 1: P = 3500, r = 3% per half-yr, n = 6 half-yrs.
Step 2: A = 3500(1.03)6.
Step 3: Using given value: A = 3500 × 1.194052 = 4179.182.
Step 4: C.I. = 4179.182 - 3500 = 679.182.
Answer: Compound Interest = ₹ 679.18

10. Find the difference between compound interest and simple interest on ₹ 12,000 and in 1 1/2 years at 10% p.a. compounded yearly.
Step 1: S.I. = 12000 × 10 × 1.5 / 100 = 1800.
Step 2: C.I. Amount = 12000(1.1)(1.05) = 13860.
Step 3: C.I. = 13860 - 12000 = 1860.
Step 4: Difference = 1860 - 1800 = 60.
Answer: ₹ 60

11. Find the difference between compound interest and simple interest on ₹ 12,000 and in 1 1/2 years at 10% compounded half-yearly.
Step 1: S.I. = 1800.
Step 2: C.I. half-yearly Amount = 12000(1.05)3 = 12000 × 1.157625 = 13891.50.
Step 3: C.I. = 13891.50 - 12000 = 1891.50.
Step 4: Difference = 1891.50 - 1800 = 91.50.
Answer: ₹ 91.50

EXERCISE 3 (D)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) When the present population (P) of a certain locality increases by r% per year, the population in n years will be :
(i) P(1+r/100)n - P (ii) P(1+r/100)n + P (iii) P(1+r/100)n (iv) P(1+r/100n)
Step 1: The standard growth formula represents the future amount entirely.
Answer: (iii) P(1+r/100)n

(b) The population of a town decreases by 10% in a particular year and then increases by 15% in the next year. The population at the end of two years is :
(i) (1+10/100)3(1+15/100) times (ii) (1-10/100)(1+15/100) times
(iii) (1-10/100)(1-15/100) times (iv) (1+10/100)(1-15/100) times
Step 1: Decrease implies minus, increase implies plus.
Answer: (ii) (1-10/100)(1+15/100) times

(c) When the cost of a machine decreases by r% per year; the cost of machine in 3 years is :
(i) (1+r/100)3 times (ii) (1-r/100)3 times (iii) (1-r/100)2 times (iv) (1+r/100)2 times
Step 1: Decrease means subtracting the rate over 3 periods.
Answer: (ii) (1-r/100)3 times

(d) On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years. Then the amount after 5 years is :
(i) (1+x/100)(1+y/100) times (ii) (1+x/100)2(1+y/100)3 times
(iii) (1+x/100)2(1+y/100)3 times (iv) (1+x/100)2(1-y/100)3 times
Step 1: Combine the separate compound growth factors based on their durations.
Answer: (ii) (1+x/100)2(1+y/100)3 times

(e) The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years. Then cost of machine becomes :
(i) ₹ x × (120/100)(75/100) (ii) ₹ x × (120/100)2 × (75/100)
(iii) ₹ x × (120/100)2 × (75/100)2 (iv) ₹ x × (80/100)2 × (75/100)2
Step 1: Increase 20% ⇒ 120/100 factor for 2 yrs. Decrease 25% ⇒ 75/100 factor for 2 yrs.
Answer: (iii) ₹ x × (120/100)2 × (75/100)2

2. The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value :
(i) at the end of 2009.
(ii) at the beginning of 2007.
Step 1: (i) Value at end of 2009 means 2 years later. A = 44000(1 - 12/100)2.
Step 2: A = 44000(0.88)2 = 44000 × 0.7744 = 34073.60.
Step 3: (ii) Value 1 year before (beginning of 2007). Let it be P. P(0.88) = 44000.
Step 4: P = 44000 / 0.88 = 50000.
Answer: (i) ₹ 34,073.60 (ii) ₹ 50,000

TEST YOURSELF
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The amount of ₹ 1,000 in 2 years and at 20% compound interest compounded per year is :
(i) ₹ 1,200 (ii) ₹ 1,400 (iii) ₹ 800 (iv) ₹ 1,440
Step 1: A = 1000(1.20)2 = 1000 × 1.44 = 1440.
Answer: (iv) ₹ 1,440

(b) The difference between C.I. and S.I. at 10% in 2 years on ₹ 100 is :
(i) ₹ 1 (ii) ₹ 41 (iii) ₹ 00 (iv) none of these
Step 1: Diff = P(r/100)2 = 100(0.1)2 = 1.
Answer: (i) ₹ 1

Case-Study Based Question
1. The compound interest formula is : A = P(1 + r/100)n ...
(a) Calculate the amount when the interest is compounded quarterly.
(b) What do you observe from the two cases discussed above ?
Step 1: (a) For quarterly, P = 4000, r = 6% p.a., n = 1 yr. Rate per quarter = 6/4 = 1.5%.
Step 2: A = 4000(1 + 1.5/100)4 = 4000(1.015)4.
Step 3: 1.0154 ≈ 1.06136.
Step 4: A = 4000 × 1.06136 = 4245.45.
Step 5: (b) As the compounding frequency increases, the accumulated amount increases.
Answer: (a) ₹ 4,245.45 (b) The more frequently interest is compounded, the higher the total amount.

2. Based on the above information, answer the following :
(i) If the number of blood donors increases by 15% every year, then what will be the number of units to be collected by IRCS in the year 2026-27.
(ii) If 2024-25 is taken as a base year and it is predicted that the number of units to be collected by IRCS in 3 years will be 39,930, then what will be rate of increase of donors?
Step 1: (i) Base units in 2024-25 = 1.46 crore. Time to 2026-27 = 2 years.
Step 2: Units = 1.46 × (1 + 15/100)2 = 1.46 × (1.15)2 = 1.46 × 1.3225 = 1.93085 crore.
Step 3: (ii) Base yearly typical = 30000 units. 39930 = 30000(1 + r/100)3.
Step 4: (1 + r/100)3 = 39930 / 30000 = 1.331.
Step 5: 1 + r/100 = 1.1 ⇒ r = 10%.
Answer: (i) 1.93085 crore units, (ii) Rate of increase = 10%

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Quick Review Flashcards - Click to flip and test your knowledge!
Question
In the context of compound interest, what is the 'amount' (A)?
Answer
The sum of the principal and the interest accumulated over a specific period.
Question
How is compound interest (C.I.) fundamentally defined as a process?
Answer
As a repeated simple interest computation where the principal grows in each conversion period.
Question
What makes manual computation of compound interest 'quite tedious' as the number of conversion periods increases?
Answer
The requirement to calculate simple interest repeatedly for every year or half-year.
Question
What is the standard formula for the amount (A) when interest is compounded yearly?
Answer
A = P(1 + \frac{r}{100})^n
Question
In the compound interest formula, what does the variable P represent?
Answer
The principal, which is the initial sum of money invested or borrowed.
Question
In the compound interest formula, what does the variable r represent?
Answer
The rate of interest compounded yearly.
Question
In the compound interest formula, what does the variable n represent?
Answer
The number of years (or conversion periods).
Question
Formula: Calculate Compound Interest (C.I.) using Amount (A) and Principal (P).
Answer
C.I. = A - P
Question
What is the direct formula to calculate Compound Interest (C.I.) without finding the amount first?
Answer
C.I. = P[(1 + \frac{r}{100})^n - 1]
Question
Formula: What is the amount (A) when rates for successive years are different (r_1, r_2, r_3)?
Answer
A = P(1 + \frac{r_1}{100})(1 + \frac{r_2}{100})(1 + \frac{r_3}{100})...
Question
How is the principal (P) calculated if the amount (A), rate (r), and time (n) are known?
Answer
By rearranging the formula to P = \frac{A}{(1 + \frac{r}{100})^n}.
Question
To find the rate percent (r) when A and P are known, what is the first algebraic step?
Answer
Divide the amount by the principal to isolate the term (1 + \frac{r}{100})^n.
Question
If \frac{A}{P} = (\frac{21}{20})^3 and the time is 3 years, what is the value of (1 + \frac{r}{100})?
Answer
\frac{21}{20}
Question
When finding the number of years (n), if (\frac{11}{10})^n = (\frac{11}{10})^3, what is the value of n?
Answer
3 years.
Question
When interest is compounded half-yearly, how is the annual rate (r) adjusted in the formula?
Answer
The rate percent is divided by 2 (i.e., \frac{r}{2}).
Question
When interest is compounded half-yearly, how is the number of years (n) adjusted in the formula?
Answer
The number of years is multiplied by 2 (i.e., n \times 2).
Question
What is the formula for Amount (A) when interest is compounded half-yearly?
Answer
A = P(1 + \frac{r}{2 \times 100})^{n \times 2}
Question
Although not in the I.C.S.E. syllabus, what is the formula for Amount (A) when interest is compounded quarterly?
Answer
A = P(1 + \frac{r}{4 \times 100})^{n \times 4}
Question
How is the amount calculated when the time is not an exact number of years (e.g., 2 \frac{1}{2} years) and interest is compounded yearly?
Answer
Calculate the amount for the full years, then use that as the principal for the remaining fractional year.
Question
Formula: Amount (A) for 2 \frac{1}{2} years at 10\% interest compounded yearly.
Answer
A = P(1 + \frac{10}{100})^2 \times (1 + \frac{10}{2 \times 100})^1
Question
What is the conversion period for interest compounded half-yearly?
Answer
Six months.
Question
In growth problems, what formula is used to find production after n years given an initial production and growth rate r?
Answer
Production after n years = Initial production \times (1 + \frac{r}{100})^n
Question
Concept: Depreciation
Answer
Definition: The reduction in the value of an asset (like machinery) over time due to wear and tear or age.
Question
What is the formula for the value of a machine after n years if it depreciates at r\% every year?
Answer
Value after n years = Present value \times (1 - \frac{r}{100})^n
Question
How is the present value of a machine calculated if its value n years ago and depreciation rate are known?
Answer
Present value = Value n years ago \times (1 - \frac{r}{100})^n
Question
What is 'scrap value' in the context of depreciation?
Answer
The reduced value of an asset at the time it is sold or discarded after several years of use.
Question
Formula: Population after n years given present population (P) and growth rate (r).
Answer
Population after n years = P(1 + \frac{r}{100})^n
Question
How do you find the population n years ago if the present population and growth rate r are given?
Answer
Present population = Population n years ago \times (1 + \frac{r}{100})^n
Question
In population problems, which variable in the standard A = P(1 + \frac{r}{100})^n formula represents the population at the earlier point in time?
Answer
The principal (P).
Question
In population problems, which variable in the standard A = P(1 + \frac{r}{100})^n formula represents the population at the later point in time?
Answer
The amount (A).
Question
When calculating the difference between C.I. and S.I. for 2 years, what is the standard formula for Simple Interest (S.I.)?
Answer
S.I. = \frac{P \times r \times 2}{100}
Question
Concept: Conversion Periods
Answer
Definition: The fixed intervals of time (e.g., year, half-year) after which interest is added to the principal.
Question
Under what condition does a sum of money double itself in n years in a compound interest scenario?
Answer
When (1 + \frac{r}{100})^n = 2.
Question
If the rate of interest is 20\% per annum, what is the rate used for half-yearly compounding calculations?
Answer
10\% per half-year.
Question
In a 2 \frac{1}{2} year period, how many half-yearly conversion periods are there?
Answer
5
Question
If a sum of money is borrowed and repaid in two yearly instalments, how is the total sum borrowed calculated?
Answer
By finding the present value (Principal) for each instalment separately and adding them together.
Question
Why is the interest for the second year in compound interest higher than the interest for the first year?
Answer
Because the interest from the first year is added to the principal, increasing the sum on which interest is calculated.
Question
Formula: Find the rate (r) when the amount in 1 year and amount in 2 years are given (yearly compounding).
Answer
r = \frac{\text{Amount in 2nd year} - \text{Amount in 1st year}}{\text{Amount in 1st year}} \times 100
Question
In growth problems, which formula applies to 'the rapid growth of plants' or 'inflation'?
Answer
The standard compound interest formula: A = P(1 + \frac{r}{100})^n.
Question
If the question asks for 'the rate of interest per annum' and the calculation for a half-year yields 5\%, what is the final answer?
Answer
10\% per annum (5\% \times 2).
Question
True or False: The principal remains constant throughout the entire duration in compound interest.
Answer
False; the principal increases after every conversion period.
Question
When comparing yearly and half-yearly compounding for the same rate and time, which yields a higher amount?
Answer
Half-yearly compounding.
Question
What is the value of (1 + \frac{10}{100})^2 expressed as a decimal?
Answer
1.21
Question
If (1 + \frac{r}{100})^n = 1.44 for n=2, what is the value of 1 + \frac{r}{100}?
Answer
1.2 (since \sqrt{1.44} = 1.2).
Question
In the formula for depreciation, why is a minus sign used inside the bracket?
Answer
To represent the decrease in value over time.
Question
When n = 1 \frac{1}{2} years and interest is compounded half-yearly, what value of n is used in the power of the formula?
Answer
3 (conversion periods).
Question
How do you calculate the C.I. earned in the 3^{rd} year only?
Answer
Subtract the Amount at the end of 2 years from the Amount at the end of 3 years (A_3 - A_2).
Question
If an asset depreciates by 10\% in year 1 and 12\% in year 2, what is the formula for final value V?
Answer
V = P(1 - \frac{10}{100})(1 - \frac{12}{100})
Question
In the equation \frac{9261}{8000} = (1 + \frac{r}{100})^3, what is the cubical root of \frac{9261}{8000}?
Answer
\frac{21}{20}
Question
If a sum of money doubles in 5 years, how many years will it take to become eight times itself (2^3) at the same rate?
Answer
15 years (5 \times 3).
Question
What is the relationship between Amount (A) and Principal (P) for the first year if interest is compounded annually?
Answer
They are the same as in simple interest for one year.
Question
In the 'Direct Method' formula for C.I., what does the term (1 + \frac{r}{100})^n - 1 represent?
Answer
The total interest earned on a principal of 1 unit.
Question
How is the 'rate of growth' usually expressed in population or industrial problems?
Answer
As a percentage (r\%) per year.
Question
If 18,000 amounts to 23,805 in 2 years, what is the total compound interest earned?
Answer
5,805 (23,805 - 18,000).
Question
Cloze: In compound interest, the interest for the first year becomes _____ for the second year.
Answer
Part of the principal
Question
Formula: Calculate simple interest (I) to find principal (P) when I, R, and T are known.
Answer
P = \frac{I \times 100}{R \times T}
Question
If interest is reckoned half-yearly, what is the amount formula for n=1 year?
Answer
A = P(1 + \frac{r}{200})^2
Question
What is the result of 1,600(1 + \frac{20}{100})^2?
Answer
2,304
Question
To find the rate r from (1 + \frac{r}{100}) = \frac{11}{10}, what is the final percentage?
Answer
10\%
Question
When solving for n in (\frac{21}{20})^n = \frac{9261}{8000}, what power is 9261 of 21?
Answer
The 3^{rd} power (cube).