Compound Interest (Stage 1)

Q1: Multiple Choice Type

a. For a particular year the simple interest at 10% is ₹800. The compound interest for the next year at the same rate is:

Step 1: Simple Interest for 1 year = ₹800
\( SI = \frac{P \times R \times T}{100} \)
\( 800 = \frac{P \times 10 \times 1}{100} \)
\( P = 800 \times \frac{100}{10} = 8000 \)
Step 2: Amount after 1 year
\( A = P + SI = 8000 + 800 = 8800 \)
Step 3: Compound Interest for next year
\( CI = \frac{8800 \times 10}{100} = 880 \)
Answer: i. ₹880

b. The compound interest on ₹5000 at 10% per annum and in 6 months amounts to:

Step 1: Time = 6 months = \( \frac{1}{2} \) year
Step 2: Simple Interest (since time < 1 year)
\( SI = \frac{5000 \times 10 \times \frac{1}{2}}{100} = 250 \)
Answer: ii. ₹250

c. The compound interest on ₹5000 at 10% per annum and in one year compounded half-yearly is:

Step 1: Rate per half-year = \( \frac{10}{2} = 5\% \)
Number of periods = 2
Step 2: Amount
\( A = 5000 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 5000 (1.05)^2 = 5000 \times 1.1025 = 5512.50 \)
Step 3: Compound Interest
\( CI = 5512.50 – 5000 = 512.50 \)
Answer: 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:

Step 1: Interest for first year
\( CI_1 = \frac{20000 \times 12}{100} = 2400 \)
Step 2: Amount after first year
\( A = 20000 + 2400 = 22400 \)
Step 3: Interest for second year
\( CI_2 = \frac{22400 \times 12}{100} = 2688 \)
Answer: iii. ₹2688

e. During the year 2022, the interest accrued at the rate of 5% is ₹1250. The compound interest accrued at the same rate during the year 2023 is:

Step 1: Find Principal
\( 1250 = \frac{P \times 5}{100} \)
\( P = 1250 \times \frac{100}{5} = 25000 \)
Step 2: Amount after 2022
\( A = 25000 + 1250 = 26250 \)
Step 3: Interest for 2023
\( CI = \frac{26250 \times 5}{100} = 1312.50 \)
Answer: i. ₹1312.50

f. Rates of interest for two consecutive years are 10% and 12% respectively. The percentage increase during these two years is:

Step 1: Let Principal \( = x \)
Step 2: After 1st year (10%)
\( A_1 = x \left(1 + \frac{10}{100}\right) = 1.1x \)
Step 3: After 2nd year (12%)
\( A_2 = 1.1x \left(1 + \frac{12}{100}\right) \)
\( = 1.1x \times 1.12 = 1.232x \)
Step 4: Total increase
\( = 1.232x – x = 0.232x \)
Step 5: Percentage increase
\( = \frac{0.232x}{x} \times 100 = 23.2\% \)
Answer: ii. 23.2%

Q2: ₹ 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: Given Principal
\( P = 16000 \), Rate \( R = 5\% \)
Step 2: For 1st Year (given)
Interest \( = \frac{16000 \times 5}{100} = 800 \)
Final Amount \( = 16000 + 800 = 16800 \)
Step 3: For 2nd Year
Initial Amount \( = 16800 \)
Interest \( = \frac{16800 \times 5}{100} = 840 \)
Final Amount \( = 16800 + 840 = 17640 \)
Step 4: For 3rd Year
Initial Amount \( = 17640 \)
Interest \( = \frac{17640 \times 5}{100} = 882 \)
Final Amount \( = 17640 + 882 = 18522 \)
Step 5: For 4th Year
Initial Amount \( = 18522 \)
Interest \( = \frac{18522 \times 5}{100} = 926.10 \)
Final Amount \( = 18522 + 926.10 = 19448.10 \)
Step 6: Table Completion

Answer: Amount after 4 years = ₹19,448.10

Q3: Calculate the amount and the compound interest on ₹8000 in \(2\frac{1}{2}\) years at 15% per annum.

Step 1: Given
Principal \( P = 8000 \), Rate \( R = 15\% \), Time \( = 2\frac{1}{2} = \frac{5}{2} \) years
Step 2: For first 2 years (Compound Interest)
\( A = 8000 \left(1 + \frac{15}{100}\right)^2 \)
\( A = 8000 (1.15)^2 \)
\( A = 8000 \times 1.3225 = 10580 \)
Step 3: For remaining \( \frac{1}{2} \) year (Simple Interest)
\( SI = \frac{10580 \times 15 \times \frac{1}{2}}{100} \)
\( SI = \frac{10580 \times 15}{200} \)
\( SI = 793.50 \)
Step 4: Final Amount
\( A = 10580 + 793.50 = 11373.50 \)
Step 5: Compound Interest
\( CI = 11373.50 – 8000 = 3373.50 \)
Answer: Amount = ₹11,373.50 and CI = ₹3,373.50

Q4: Calculate the amount and the compound interest on: ₹4600 in 2 years when the rates of interest of successive years are 10% and 12% respectively.

Step 1: Given
Principal \( P = 4600 \)
Rate for 1st year \( = 10\% \)
Rate for 2nd year \( = 12\% \)
Step 2: Interest for 1st year
\( CI_1 = \frac{4600 \times 10}{100} = 460 \)
Step 3: Amount after 1st year
\( A_1 = 4600 + 460 = 5060 \)
Step 4: Interest for 2nd year
\( CI_2 = \frac{5060 \times 12}{100} = 607.20 \)
Step 5: Final Amount
\( A = 5060 + 607.20 = 5667.20 \)
Step 6: Total Compound Interest
\( CI = 5667.20 – 4600 = 1067.20 \)
Answer: Amount = ₹5,667.20 and CI = ₹1,067.20

Q5: 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: Given
Principal \( P = 75000 \)
Rate for 1st year \( = 15\% \)
Rate for 2nd year \( = 15\% \)
Rate for 3rd year \( = 16\% \)
Step 2: After 1st year
\( CI_1 = \frac{75000 \times 15}{100} = 11250 \)
\( A_1 = 75000 + 11250 = 86250 \)
Step 3: After 2nd year
\( CI_2 = \frac{86250 \times 15}{100} = 12937.50 \)
\( A_2 = 86250 + 12937.50 = 99187.50 \)
Step 4: After 3rd year
\( CI_3 = \frac{99187.50 \times 16}{100} = 15870 \)
\( A_3 = 99187.50 + 15870 = 115057.50 \)
Answer: Amount = ₹1,15,057.50

Q6: Calculate the amount and the compound interest on ₹16,000 in 3 years, when the rates of interest for successive years are 10%, 14% and 15% respectively.

Step 1: Given
Principal \( P = 16000 \)
Rates: \(10\%, 14\%, 15\%\)
Step 2: After 1st year
\( CI_1 = \frac{16000 \times 10}{100} = 1600 \)
\( A_1 = 16000 + 1600 = 17600 \)
Step 3: After 2nd year
\( CI_2 = \frac{17600 \times 14}{100} = 2464 \)
\( A_2 = 17600 + 2464 = 20064 \)
Step 4: After 3rd year
\( CI_3 = \frac{20064 \times 15}{100} = 3009.60 \)
\( A_3 = 20064 + 3009.60 = 23073.60 \)
Step 5: Total Compound Interest
\( CI = 23073.60 – 16000 = 7073.60 \)
Answer: Amount = ₹23,073.60 and CI = ₹7,073.60

Q7: Calculate the compound interest for the second year on ₹8000 invested for 3 years at 10% per annum.

Step 1: Given
Principal \( P = 8000 \), Rate \( R = 10\% \)
Step 2: Interest for 1st year
\( CI_1 = \frac{8000 \times 10}{100} = 800 \)
Amount after 1st year:
\( A_1 = 8000 + 800 = 8800 \)
Step 3: Interest for 2nd year (on new principal)
\( CI_2 = \frac{8800 \times 10}{100} = 880 \)
Answer: Compound Interest for 2nd year = ₹880

Q8: Find the compound interest, correct to the nearest rupee, on ₹2400 for \(2\frac{1}{2}\) years at 5 percent per annum.

Step 1: Given
Principal \( P = 2400 \), Rate \( R = 5\% \), Time \( = 2\frac{1}{2} = \frac{5}{2} \) years
Step 2: For first 2 years (Compound Interest)
\( A = 2400 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 2400 (1.05)^2 \)
\( A = 2400 \times 1.1025 = 2646 \)
Step 3: For remaining \( \frac{1}{2} \) year (Simple Interest)
\( SI = \frac{2646 \times 5 \times \frac{1}{2}}{100} \)
\( SI = \frac{2646 \times 5}{200} = 66.15 \)
Step 4: Final Amount
\( A = 2646 + 66.15 = 2712.15 \)
Step 5: Compound Interest
\( CI = 2712.15 – 2400 = 312.15 \)
Step 6: Correct to nearest rupee
\( CI \approx 312 \)
Answer: Compound Interest = ₹312

Q9: A borrowed ₹2500 from B at 12% per annum compound interest. After 2 years, A gave ₹2936 and a watch to B to clear the account. Find the cost of the watch.

Step 1: Given
Principal \( P = 2500 \), Rate \( R = 12\% \), Time \( T = 2 \) years
Step 2: Amount after 2 years (Compound Interest)
\( A = 2500 \left(1 + \frac{12}{100}\right)^2 \)
\( A = 2500 (1.12)^2 \)
\( A = 2500 \times 1.2544 = 3136 \)
Step 3: Total amount to be paid = ₹3136
Step 4: Amount paid in cash = ₹2936
Step 5: Cost of watch
\( \text{Cost of watch} = 3136 – 2936 = 200 \)
Answer: Cost of the watch = ₹200

Q10: 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: Given
Principal \( P = 50000 \)
Rates: \(6\%, 8\%, 10\%\)
Step 2: After 1st year
\( CI_1 = \frac{50000 \times 6}{100} = 3000 \)
\( A_1 = 50000 + 3000 = 53000 \)
Step 3: After 2nd year
\( CI_2 = \frac{53000 \times 8}{100} = 4240 \)
\( A_2 = 53000 + 4240 = 57240 \)
Step 4: After 3rd year
\( CI_3 = \frac{57240 \times 10}{100} = 5724 \)
\( A_3 = 57240 + 5724 = 62964 \)
Answer: Amount = ₹62,964

Q11: Govind borrows ₹18,000 at 10% simple interest. He immediately invests the money at 10% compound interest compounded half-yearly. How much money does Govind gain in one year?

Step 1: Simple Interest (SI) on borrowed money
\( SI = \frac{18000 \times 10 \times 1}{100} = 1800 \)
Step 2: Amount to be returned after 1 year
\( = 18000 + 1800 = 19800 \)
Step 3: Compound Interest (half-yearly)
Rate per half-year \( = \frac{10}{2} = 5\% \)
Number of periods = 2
Step 4: Amount after 1 year
\( A = 18000 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 18000 (1.05)^2 \)
\( A = 18000 \times 1.1025 = 19845 \)
Step 5: Gain
\( \text{Gain} = 19845 – 19800 = 45 \)
Answer: Gain = ₹45

Q12: Find the compound interest on ₹4000 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: Given
Principal \( P = 4000 \)
Rates: \(8\%, 10\%, 10\%\)
Step 2: After 1st year
\( CI_1 = \frac{4000 \times 8}{100} = 320 \)
\( A_1 = 4000 + 320 = 4320 \)
Step 3: After 2nd year
\( CI_2 = \frac{4320 \times 10}{100} = 432 \)
\( A_2 = 4320 + 432 = 4752 \)
Step 4: After 3rd year
\( CI_3 = \frac{4752 \times 10}{100} = 475.20 \)
\( A_3 = 4752 + 475.20 = 5227.20 \)
Step 5: Total Compound Interest
\( CI = 5227.20 – 4000 = 1227.20 \)
Answer: Compound Interest = ₹1,227.20