Interest

Q1: Multiple Choice Type

i. The compound interest on ₹1,000 at 20% per annum for 2 years is:

Step 1: Use formula for Amount (A): \[ A = P\left(1 + \frac{R}{100}\right)^T = 1000\left(1 + \frac{20}{100}\right)^2 = 1000 \times (1.2)^2 = 1000 \times 1.44 = ₹1440 \]Step 2: Compound Interest = Amount − Principal = ₹1440 − ₹1000 = ₹440
Answer: c. ₹440

ii. A sum of ₹2,000 is put at 10% compound interest. The amount at the end of 2 years will be:

Step 1: Use formula: \[ A = 2000\left(1 + \frac{10}{100}\right)^2 = 2000 \times (1.1)^2 = 2000 \times 1.21 = ₹2,420 \]Answer: c. ₹2,420

iii. The difference between C.I. and S.I. on ₹6,000 at 8% per annum in both the cases and in one year is:

Step 1: For 1 year, C.I. = S.I. (since only one year interest is calculated on same principal)
Answer: b. Nothing

iv. The difference between the C.I. in 1 year and compound interest in 2 years on ₹4,000 at 5% per annum is:

Step 1: Calculate amount for 2 years: \[ A_2 = 4000 \times (1.05)^2 = 4000 \times 1.1025 = ₹4,410 \]Step 2: Calculate amount for 1 year: \[ A_1 = 4000 \times 1.05 = ₹4,200 \]Step 3: Difference in CI = CI(2 years) − CI(1 year) = (4410 − 4000) − (4200 − 4000) = ₹410 − ₹200 = ₹210
Answer: b. ₹210

Q2: A sum of ₹8,000 is invested for 2 years at 10% per annum compound interest. Calculate:

i. Interest for the first year.

Step 1: Principal (P) = ₹8,000
Rate (R) = 10%
Step 2: Interest for first year = \( \frac{P \times R}{100} = \frac{8000 \times 10}{100} = ₹800 \)
Answer: Interest for the first year is ₹800.

ii. Principal for the second year.

Step 1: Principal for second year = Principal + Interest from first year \[ = 8000 + 800 = ₹8,800 \]Answer: Principal for the second year is ₹8,800.

iii. Interest for the second year.

Step 1: Interest for second year = \( \frac{8800 \times 10}{100} = ₹880 \)
Answer: Interest for the second year is ₹880.

iv. Final amount at the end of the second year.

Step 1: Amount after 2 years = Principal for second year + Interest for second year \[ = 8800 + 880 = ₹9680 \]Answer: The final amount at the end of 2 years is ₹9,680.

v. Compound interest earned in 2 years.

Step 1: Compound Interest = Amount − Principal \[ = 9680 – 8000 = ₹1680 \]Answer: The compound interest earned in 2 years is ₹1,680.

Q3: A man borrowed ₹20,000 for 2 years at 8% per annum compound interest. Calculate:

i. Interest for the first year.

Step 1: Principal (P) = ₹20,000
Rate (R) = 8% per annum
Step 2: Interest for first year: \[ SI_1 = \frac{P \times R}{100} = \frac{20000 \times 8}{100} = ₹1,600 \]Answer: Interest for the first year is ₹1,600.

ii. Interest for the second year.

Step 1: Principal for second year = Principal + interest of first year = ₹20,000 + ₹1,600 = ₹21,600
Step 2: Interest for second year: \[ SI_2 = \frac{21600 \times 8}{100} = ₹1,728 \]Answer: Interest for the second year is ₹1,728.

iii. The final amount at the end of the second year.

Step 1: Amount after 2 years: \[ A = 20000 + 1600 + 1728 = ₹23,328 \]Answer: The final amount at the end of 2 years is ₹23,328.

iv. The compound interest for two years.

Step 1: Compound Interest = Amount − Principal \[ CI = 23328 – 20000 = ₹3,328 \]Answer: The compound interest for two years is ₹3,328.

Q4: Calculate the amount and the compound interest on ₹12,000 in 2 years at 10% per year.

Step 1: Given:
Principal (P) = ₹12,000
Rate of interest (R) = 10% per annum
Time (T) = 2 years
Step 2: Use the compound interest formula for amount: \[ A = P \left(1 + \frac{R}{100}\right)^T = 12000 \times \left(1 + \frac{10}{100}\right)^2 = 12000 \times (1.1)^2 = 12000 \times 1.21 = ₹14,520 \]Step 3: Calculate compound interest: \[ CI = A – P = 14520 – 12000 = ₹2,520 \]Answer: The amount after 2 years is ₹14,520 and the compound interest is ₹2,520.

Q5: Calculate the amount and the compound interest on ₹10,000 in 3 years at 8% per annum.

Step 1: Given:
Principal (P) = ₹10,000
Rate of interest (R) = 8% per annum
Time (T) = 3 years
Step 2: Use the compound interest formula for amount: \[ A = P \left(1 + \frac{R}{100}\right)^T = 10000 \times \left(1 + \frac{8}{100}\right)^3 = 10000 \times (1.08)^3 \]Calculate \( (1.08)^3 \): \[ 1.08 \times 1.08 = 1.1664 \quad \Rightarrow \quad 1.1664 \times 1.08 = 1.259712 \]So, \[ A = 10000 \times 1.259712 = ₹12,597.12 \]Step 3: Calculate compound interest: \[ CI = A – P = 12597.12 – 10000 = ₹2,597.12 \]Answer: The amount after 3 years is ₹12,597.12 and the compound interest is ₹2,597.12.

Q6: Calculate the compound interest on ₹5,000 in 2 years, if the rates of interest for successive years are 10% and 12% respectively.

Step 1: Principal (P) = ₹5,000
Step 2: After 1st year at 10%: \[ A_1 = P \times \left(1 + \frac{10}{100}\right) = 5000 \times 1.10 = ₹5,500 \]Step 3: After 2nd year at 12%: \[ A_2 = A_1 \times \left(1 + \frac{12}{100}\right) = 5500 \times 1.12 = ₹6,160 \]Step 4: Compound Interest (CI) = Amount after 2 years − Principal \[ CI = 6160 – 5000 = ₹1,160 \]Answer: The compound interest on ₹5,000 in 2 years is ₹1,160.

Q7: Calculate the compound interest on ₹15,000 in 3 years; if the rates of interest for successive years are 6%, 8% and 10% respectively.

Step 1: Principal (P) = ₹15,000
Step 2: After 1st year at 6%: \[ A_1 = 15000 \times \left(1 + \frac{6}{100}\right) = 15000 \times 1.06 = ₹15,900 \]Step 3: After 2nd year at 8%: \[ A_2 = 15900 \times \left(1 + \frac{8}{100}\right) = 15900 \times 1.08 = ₹17,172 \]Step 4: After 3rd year at 10%: \[ A_3 = 17172 \times \left(1 + \frac{10}{100}\right) = 17172 \times 1.10 = ₹18,889.20 \]Step 5: Compound Interest = Amount after 3 years − Principal \[ CI = 18889.20 – 15000 = ₹3,889.20 \]Answer: The compound interest on ₹15,000 for 3 years is ₹3,889.20.

Q8: Mohan borrowed ₹16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan would have to pay at the end of 3 years.

Step 1: Given:
Principal (P) = ₹16,000
Rate of interest (R) = 5% per annum
Time (T) = 3 years
Step 2: Use compound interest formula for amount: \[ A = P \left(1 + \frac{R}{100}\right)^T = 16000 \times \left(1 + \frac{5}{100}\right)^3 = 16000 \times (1.05)^3 \]Calculate \( (1.05)^3 \): \[ 1.05 \times 1.05 \times 1.05 = 1.157625 \]So, \[ A = 16000 \times 1.157625 = ₹18,522 \]Answer: Mohan has to pay ₹18,522 at the end of 3 years.

Q9: Rekha borrowed ₹40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.

Step 1: Principal (P) = ₹40,000
Rate (R) = 10% per annum
Step 2: Amount after 1st year: \[ A_1 = P \times \left(1 + \frac{R}{100}\right) = 40000 \times 1.10 = ₹44,000 \]Step 3: Amount after 2nd year: \[ A_2 = A_1 \times \left(1 + \frac{R}{100}\right) = 44000 \times 1.10 = ₹48,400 \]Step 4: Interest for second year = Amount after 2nd year − Amount after 1st year \[ = 48400 – 44000 = ₹4,400 \]Answer: The interest paid by Rekha for the second year is ₹4,400.

Q10: Calculate the compound interest for the second year on ₹15,000 invested for 5 years at 6% per annum.

Step 1: Principal (P) = ₹15,000
Rate (R) = 6% per annum
Step 2: Amount after 1st year: \[ A_1 = P \times \left(1 + \frac{R}{100}\right) = 15000 \times 1.06 = ₹15,900 \]Step 3: Amount after 2nd year: \[ A_2 = A_1 \times \left(1 + \frac{R}{100}\right) = 15900 \times 1.06 = ₹16,854 \]Step 4: Compound interest for second year = Amount after 2nd year − Amount after 1st year \[ = 16854 – 15900 = ₹954 \]Answer: The compound interest for the second year is ₹954.

Q11: A man invests ₹9,600 at 10% per annum compound interest for 3 years. Calculate:

i. The interest for the first year.

Step 1: Principal (P) = ₹9,600
Rate (R) = 10%
Step 2: Interest for first year: \[ SI_1 = \frac{9600 \times 10}{100} = ₹960 \]Answer: The interest for the first year is ₹960.

ii. The amount at the end of the first year.

Step 1: Amount after first year: \[ A_1 = P + SI_1 = 9600 + 960 = ₹10,560 \]Answer: The amount at the end of the first year is ₹10,560.

iii. The interest for the second year.

Step 1: Principal for second year = Amount after first year = ₹10,560
Step 2: Interest for second year: \[ SI_2 = \frac{10560 \times 10}{100} = ₹1,056 \]Answer: The interest for the second year is ₹1,056.

iv. The interest for the third year.

Step 1: Principal for third year = Amount after second year = Amount after first year + Interest for second year = ₹10,560 + ₹1,056 = ₹11,616
Step 2: Interest for third year: \[ SI_3 = \frac{11616 \times 10}{100} = ₹1,161.60 \]Answer: The interest for the third year is ₹1,161.60.

Q12: A person invests ₹5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to ₹5,600. Calculate:

i. The rate of interest per annum.

Step 1: Given:
Principal (P) = ₹5,000
Amount after 1 year (A) = ₹5,600
Step 2: Use compound interest formula: \[ A = P \left(1 + \frac{R}{100}\right)^1 = P \left(1 + \frac{R}{100}\right) \] Substitute values: \[ 5600 = 5000 \times \left(1 + \frac{R}{100}\right) \\ \Rightarrow 1 + \frac{R}{100} = \frac{5600}{5000} = 1.12 \]Step 3: Solve for rate: \[ \frac{R}{100} = 1.12 – 1 = 0.12 \\ \Rightarrow R = 12\% \]Answer: Rate of interest = 12% per annum.

ii. The amount at the end of the second year.

Step 1: Use the amount formula for 2 years: \[ A_2 = P \left(1 + \frac{R}{100}\right)^2 = 5000 \times (1.12)^2 = 5000 \times 1.2544 = ₹6,272 \]Answer: The amount at the end of the second year is ₹6,272.

Q13: Calculate the difference between the compound interest and the simple interest on ₹7,500 in two years and at 8% per annum.

Step 1: Given:
Principal (P) = ₹7,500
Rate (R) = 8% per annum
Time (T) = 2 years
Step 2: Calculate Simple Interest (SI): \[ SI = \frac{P \times R \times T}{100} = \frac{7500 \times 8 \times 2}{100} = ₹1,200 \]Step 3: Calculate Amount for Compound Interest (CI): \[ A = P \times \left(1 + \frac{R}{100}\right)^T = 7500 \times (1.08)^2 = 7500 \times 1.1664 = ₹8,748 \]Step 4: Calculate Compound Interest: \[ CI = A – P = 8748 – 7500 = ₹1,248 \]Step 5: Difference between CI and SI: \[ CI – SI = 1248 – 1200 = ₹48 \]Answer: The difference between compound interest and simple interest is ₹48.

Q14: Calculate the difference between the compound interest and the simple interest on ₹8,000 in three years at 10% per annum.

Step 1: Given:
Principal (P) = ₹8,000
Rate (R) = 10% per annum
Time (T) = 3 years
Step 2: Calculate Simple Interest (SI): \[ SI = \frac{P \times R \times T}{100} = \frac{8000 \times 10 \times 3}{100} = ₹2,400 \]Step 3: Calculate Amount for Compound Interest (CI): \[ A = P \times \left(1 + \frac{R}{100}\right)^T = 8000 \times (1.10)^3 = 8000 \times 1.331 = ₹10,648 \]Step 4: Calculate Compound Interest: \[ CI = A – P = 10648 – 8000 = ₹2,648 \]Step 5: Difference between CI and SI: \[ CI – SI = 2648 – 2400 = ₹248 \]Answer: The difference between compound interest and simple interest is ₹248.

Q15: Rohit borrowed ₹40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gives less interest and by how much?

Step 1: Given:
Principal (P) = ₹40,000
Time (T) = 2 years
Rohit’s rate (Compound Interest) = 10% per annum
Manish’s rate (Simple Interest) = 10.5% per annum
Step 2: Amount for Rohit using compound interest formula: \[ A = P \left(1 + \frac{R}{100}\right)^T = 40000 \times \left(1 + \frac{10}{100}\right)^2 = 40000 \times (1.10)^2 = 40000 \times 1.21 = ₹48,400 \]Step 3: Compound Interest (CI) for Rohit: \[ CI = A – P = 48400 – 40000 = ₹8,400 \]Step 4: Simple Interest (SI) formula: \[ SI = \frac{P \times R \times T}{100} = \frac{40000 \times 10.5 \times 2}{100} = ₹8,400 \]Step 5: Compound Interest for Rohit = ₹8,400
Simple Interest for Manish = ₹8,400
Answer: Both Rohit and Manish pay the same amount of interest, ₹8,400. Hence, neither gives less interest.

Q16: Mr. Sharma lends ₹24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.

Step 1: Given:
Simple Interest Principal (P₁) = ₹24,000
Simple Interest Rate (R₁) = 13% per annum
Time (T) = 2 years
Compound Interest Principal (P₂) = ₹24,000
Compound Interest Rate (R₂) = 12% per annum
Step 2: Simple Interest formula: \[ SI = \frac{P_1 \times R_1 \times T}{100} = \frac{24000 \times 13 \times 2}{100} = ₹6,240 \]Step 3: Calculate amount after 2 years using compound interest formula: \[ A = P_2 \times \left(1 + \frac{R_2}{100}\right)^T = 24000 \times \left(1 + \frac{12}{100}\right)^2 = 24000 \times (1.12)^2 \]Calculate \( (1.12)^2 \): \[ 1.12 \times 1.12 = 1.2544 \]So, \[ A = 24000 \times 1.2544 = ₹30,105.60 \]Step 4: Compound Interest: \[ CI = A – P_2 = 30105.60 – 24000 = ₹6,105.60 \]Step 5: Total interest = Simple Interest + Compound Interest \[ = 6240 + 6105.60 = ₹12,345.60 \]Answer: Mr. Sharma earns a total interest of ₹12,345.60 in 2 years.

Q17: Peter borrows ₹12,000 for 2 years at 10% p.a. compound interest. He repays ₹8,000 at the end of first year. Find:

i. The amount at the end of first year, before making the repayment.

Step 1: Principal (P) = ₹12,000
Rate (R) = 10% per annum
Time = 1 year
Step 2: Calculate amount after 1 year: \[ A_1 = P \times \left(1 + \frac{R}{100}\right)^1 = 12000 \times 1.10 = ₹13,200 \]Answer: The amount at the end of first year before repayment is ₹13,200.

ii. The amount at the end of first year, after making the repayment.

Step 3: Amount after repayment: \[ 13,200 – 8,000 = ₹5,200 \]Answer: The amount at the end of first year after repayment is ₹5,200.

iii. The principal for second year.

Step 4: Principal for second year is the remaining amount: \[ P_2 = ₹5,200 \]Answer: The principal for the second year is ₹5,200.

iv. The amount to be paid at the end of the second year to clear the account.

Step 5: Calculate amount at end of second year: \[ A_2 = P_2 \times \left(1 + \frac{R}{100}\right) = 5200 \times 1.10 = ₹5,720 \]Answer: The amount to be paid at the end of the second year to clear the account is ₹5,720.

Q18: Gautam takes a loan of ₹16,000 for 2 years at 15% p.a. compound interest. He repays ₹9,000 at the end of first year. How much must he pay at the end of second year to clear the debt?

Step 1: Given:
Principal (P) = ₹16,000
Rate of interest (R) = 15% per annum
Time = 2 years
Step 2: Calculate amount after 1st year: \[ A_1 = P \times \left(1 + \frac{R}{100}\right) = 16000 \times 1.15 = ₹18,400 \]Step 3: Amount remaining after repayment of ₹9,000 at end of 1st year: \[ Remaining\ Principal = 18400 – 9000 = ₹9,400 \]Step 4: Calculate amount to be paid at end of 2nd year: \[ A_2 = Remaining\ Principal \times \left(1 + \frac{R}{100}\right) = 9400 \times 1.15 = ₹10,810 \]Answer: Gautam must pay ₹10,810 at the end of the second year to clear the debt.

Q19: A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹12,000. Find:

i. The sum of money.

Step 1: Given:
Simple Interest (SI) = ₹12,000
Rate (R) = 8% per annum
Time (T) = 5 years
Step 2: Use Simple Interest formula: \[ SI = \frac{P \times R \times T}{100} \] Substitute the values: \[ 12000 = \frac{P \times 8 \times 5}{100} \]Step 3: Solve for \(P\): \[ 12000 = \frac{40P}{100} \\ 12000 = 0.4P \\ P = \frac{12000}{0.4} = ₹30,000 \]Answer: The sum of money is ₹30,000.

ii. the compound interest earned by this money in two years. at 10% p.a. compound interest.

Step 1: Given:
Principal (P) = ₹30,000
Rate (R) = 10% per annum
Time (T) = 2 years
Step 2: Calculate amount \(A\) using compound interest formula: \[ A = P \times \left(1 + \frac{R}{100}\right)^T = 30000 \times (1.10)^2 = 30000 \times 1.21 = ₹36,300 \]Step 3: Calculate compound interest (CI): \[ CI = A – P = 36300 – 30000 = ₹6,300 \]Answer: The compound interest earned in 2 years at 10% p.a. is ₹6,300.

Q20: Find the amount and the C.I. on ₹12,000 in one year per annum compounded half-yearly.

Step 1: Given:
Principal (P) = ₹12,000
Rate of interest (R) = (assumed 10% per annum for example, please specify if different)
Time (T) = 1 year
Number of compounding periods per year (n) = 2 (half-yearly)
Step 2: Calculate rate per half-year: \[ r = \frac{R}{n} = \frac{10}{2} = 5\% \]Step 3: Total number of compounding periods: \[ N = n \times T = 2 \times 1 = 2 \]Step 4: Use compound interest formula: \[ A = P \times \left(1 + \frac{r}{100}\right)^N = 12000 \times \left(1 + \frac{5}{100}\right)^2 = 12000 \times (1.05)^2 \]Calculate \( (1.05)^2 \): \[ 1.05 \times 1.05 = 1.1025 \]So, \[ A = 12000 \times 1.1025 = ₹13,230 \]Step 5: Calculate compound interest: \[ CI = A – P = 13230 – 12000 = ₹1,230 \]Answer: The amount after one year is ₹13,230 and the compound interest earned is ₹1,230.

Q21: Find the amount and the C.I. on ₹8,000 in \(1\frac{1}{2}\) years at 20% per year compounded half-yearly.

Step 1: Given:
Principal (P) = ₹8,000
Rate of interest (R) = 20% per annum
Time (T) = \(1\frac{1}{2} = \frac{3}{2}\) years
Compounding frequency (n) = 2 (half-yearly)
Step 2: Calculate rate per half-year: \[ r = \frac{R}{n} = \frac{20}{2} = 10\% \]Step 3: Calculate total number of compounding periods: \[ N = n \times T = 2 \times \frac{3}{2} = 3 \]Step 4: Use compound interest formula: \[ A = P \times \left(1 + \frac{r}{100}\right)^N = 8000 \times (1 + 0.10)^3 = 8000 \times (1.10)^3 \]Calculate \( (1.10)^3 \): \[ 1.10 \times 1.10 = 1.21, \quad 1.21 \times 1.10 = 1.331 \]So, \[ A = 8000 \times 1.331 = ₹10,648 \]Step 5: Calculate compound interest: \[ CI = A – P = 10648 – 8000 = ₹2,648 \]Answer: The amount after \(1\frac{1}{2}\) years is ₹10,648 and the compound interest earned is ₹2,648.

Q22: Find the amount and the compound interest on ₹24,000, 2 years at 10% per annum compounded yearly.

Step 1: Given:
Principal (P) = ₹24,000
Rate of interest (R) = 10% per annum
Time (T) = 2 years
Compounding frequency (n) = 1 (yearly)
Step 2: Calculate rate per compounding period: \[ r = \frac{R}{n} = \frac{10}{1} = 10\% \]Step 3: Total number of compounding periods: \[ N = n \times T = 1 \times 2 = 2 \]Step 4: Use compound interest formula: \[ A = P \times \left(1 + \frac{r}{100}\right)^N = 24000 \times (1 + 0.10)^2 = 24000 \times (1.10)^2 \]Calculate \( (1.10)^2 \): \[ 1.10 \times 1.10 = 1.21 \]So, \[ A = 24000 \times 1.21 = ₹29,040 \]Step 5: Calculate compound interest: \[ CI = A – P = 29040 – 24000 = ₹5,040 \]Answer: The amount after 2 years is ₹29,040 and the compound interest earned is ₹5,040.