Compound Interest (Stage 1)

Q1: Multiple Choice Type:

a. Certain sum is lent at 10% compound interest annum. If the interest accrued during the year 2024 was ₹ 1,331 then the interest accrued during the year 2022, was:

Step 1: Understand the relationship between interest of successive years.
Interest for any year is \( 10\% \) more than the interest of the previous year.
Let Interest for 2023 be \( I_{2023} \).
\( I_{2024} = I_{2023} + 10\% \text{ of } I_{2023} = 1.1 \times I_{2023} \)
\( 1,331 = 1.1 \times I_{2023} \)
\( I_{2023} = \frac{1331}{1.1} = ₹ 1,210 \)
Step 2: Calculate interest for the year 2022.
Similarly, \( I_{2023} = 1.1 \times I_{2022} \)
\( 1,210 = 1.1 \times I_{2022} \)
\( I_{2022} = \frac{1210}{1.1} = ₹ 1,100 \)
Answer: 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:

Step 1: Apply the growth formula.
New population = Original population + Increase.
Increase = \( 5\% \) of 45,000.
Population in 2024 = \( 45,000 + 5\% \text{ of } 45,000 \)
Answer: i. 45,000 + 5% of 45,000

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: Use the compound interest formula where Amount is double the Principal.
Let \( P \) be the Principal. Then \( A = 2P \).
\( 2P = P(1 + \frac{10}{100})^n \)
\( 2 = (1.1)^n \)
Step 2: Calculate values of \( (1.1)^n \) by taking n = 5, 8, 10.
\( (1.1)^5 \approx 1.61 \)
\( (1.1)^8 \approx 2.14 \)
\( (1.1)^10 \approx 2.59 \)
None of the specific options are correct.
Answer: iv. none of these

d. The cost of a machine depreciates every year by 10%; the percentage decrease during two years will be:

Step 1: Assume initial cost is 100.
Depreciation after 1st year = \( 10\% \text{ of } 100 = 10 \).
Value after 1st year = \( 100 – 10 = 90 \).
Step 2: Calculate depreciation after 2nd year.
Depreciation after 2nd year = \( 10\% \text{ of } 90 = 9 \).
Value after 2nd year = \( 90 – 9 = 81 \).
Step 3: Find total percentage decrease.
Total decrease = \( 100 – 81 = 19 \).
Percentage decrease = \( 19\% \).
Answer: 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 .

Step 1: Evaluate Assertion (A).
The Compound Interest for a specific year (e.g., 3rd year) is the difference between the total amount accumulated at the end of that year and the total amount accumulated at the end of the preceding year.
\( \text{C.I.}_{3rd\ year} = A_3 – A_2 \). So, A is true.
Step 2: Evaluate Reason (R).
\( A_3 = P + \text{C.I. for 3 years} \)
\( A_2 = P + \text{C.I. for 2 years} \)
\( A_3 – A_2 = (P + \text{C.I. for 3 years}) – (P + \text{C.I. for 2 years}) = \text{C.I. for 3 years} – \text{C.I. for 2 years} \)
The difference between total interest of 3 years and total interest of 2 years is exactly the interest earned in the 3rd year. So, R is true and explains A.
Answer: iii. Both A and R are true and R is the correct reason for A.

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.

Step 1: Evaluate Assertion (A).
The interest for the 5th year specifically is the total interest accumulated over 5 years minus the total interest accumulated over 4 years.
So, A is true.
Step 2: Evaluate Reason (R).
As established in the previous question, the interest for a specific year is the difference between the Amount at the end of that year and the Amount at the end of the previous year.
\( \text{Interest of 5th year} = A_5 – A_4 \). So, R is true.
Step 3: Relationship between A and R.
While both statements are mathematically true, R provides the fundamental definition of how C.I. is calculated from amounts, which explains why the difference in accumulated interests (A) works.
Answer: iii. Both A and R are true and R is the correct reason for A.

g. Statement 1: Rate of C.I. accrued in 3rd year = \( \frac{\text{Amount of 3 years – Amount of 2 years}}{\text{Amount in 2 years}} \times 100\% \)
Statement 2: Amount at the end of 3 years = Amount at the end of 2nd year + Interest on it

Step 1: Evaluate Statement 1.
Rate of interest is calculated as \( \frac{\text{Interest}}{\text{Principal}} \times 100 \). For the 3rd year, the interest is \( (A_3 – A_2) \) and the principal is \( A_2 \).
Formula: \( R = \frac{A_3 – A_2}{A_2} \times 100\% \). So, Statement 1 is true.
Step 2: Evaluate Statement 2.
In compound interest, the Amount at the end of any period is the Principal (Amount of previous period) plus the interest earned on it.
\( A_3 = A_2 + \text{Interest on } A_2 \). So, Statement 2 is true.
Answer: i. Both the statements are true.

h. Statement 1: If P is the sum invested for 2 years at 20% rate of interest. then Amount in 2 years = ₹ \( P \times \frac{20}{100} \times \frac{20}{100} \)
Statement 2: Interest accrued in 2 years = ₹ \( (P \times \frac{120}{100} \times \frac{120}{100} – P) \).

Step 1: Evaluate Statement 1.
Amount formula: \( A = P(1 + \frac{R}{100})^n \). For \( R = 20\% \) and \( n = 2 \):
\( A = P(1 + \frac{20}{100})^2 = P(\frac{120}{100})^2 \).
The expression \( P \times \frac{20}{100} \times \frac{20}{100} \) only calculates interest on interest for the 2nd year. So, Statement 1 is false.
Step 2: Evaluate Statement 2.
Interest \( (I) = \text{Amount} – \text{Principal} \).
\( I = [P(1 + \frac{20}{100})^2] – P = [P \times (\frac{120}{100}) \times (\frac{120}{100})] – P \).
So, Statement 2 is true.
Answer: iv. Statement 1 is false, and statement 2 is true.

Q2: 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: Identify the given values and formula for different rates.
When rates are different for successive years, the amount formula is:
\( A = P \left( 1 + \frac{R_1}{100} \right) \left( 1 + \frac{R_2}{100} \right) \)
Given:
Amount \( (A) = ₹ 6,593.40 \)
Rate for 1st year \( (R_1) = 10\% \)
Rate for 2nd year \( (R_2) = 11\% \)
Step 2: Substitute the values into the formula.
\( 6,593.40 = P \left( 1 + \frac{10}{100} \right) \left( 1 + \frac{11}{100} \right) \)
\( 6,593.40 = P \left( \frac{110}{100} \right) \left( \frac{111}{100} \right) \)
Step 3: Simplify the equation to find \( P \).
\( 6,593.40 = P \times 1.1 \times 1.11 \)
\( 6,593.40 = P \times 1.221 \)
Step 4: Calculate the Principal sum.
\( P = \frac{6,593.40}{1.221} \)
\( P = 5,400 \)
Answer: ₹ 5,400

Q3: 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: Assume the initial cost and calculate the value after the first year.
Let the initial cost of the machine be \( ₹ x \).
Depreciation rate for the 1st year = \( 10\% \).
Value after 1st year = \( x – (10\% \text{ of } x) = x – 0.1x = 0.9x \)
Step 2: Calculate the value after the second year.
Depreciation rate for the 2nd year = \( 10\% \).
Value after 2nd year = \( 0.9x – (10\% \text{ of } 0.9x) = 0.9x – 0.09x = 0.81x \)
Step 3: Calculate the value after the third year.
Depreciation rate for the 3rd year = \( 15\% \).
Value after 3rd year = \( 0.81x – (15\% \text{ of } 0.81x) \)
Value after 3rd year = \( 0.81x – 0.1215x = 0.6885x \)
Step 4: Calculate the total depreciation amount.
Total Depreciation = \( \text{Initial Cost} – \text{Final Value} \)
Total Depreciation = \( x – 0.6885x = 0.3115x \)
Step 5: Find the total depreciation percentage.
Total Depreciation % = \( \frac{\text{Total Depreciation}}{\text{Initial Cost}} \times 100 \)
Total Depreciation % = \( \frac{0.3115x}{x} \times 100 = 31.15\% \)
Answer: 31.15%

Q4: Rachna borrow ₹ 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: Calculate the balance for the first six months.
Principal \( (P_1) = ₹ 12,000 \)
Semi-annual Rate \( (r) = \frac{10\%}{2} = 5\% \)
Interest for the 1st six months = \( 5\% \text{ of } 12,000 = \frac{5 \times 12,000}{100} = ₹ 600 \)
Amount = \( 12,000 + 600 = ₹ 12,600 \)
Repayment at the end of 6 months = \( ₹ 4,000 \)
Balance at the end of 1st six months = \( 12,600 – 4,000 = ₹ 8,600 \)
Step 2: Calculate the balance for the second six months (at 12 months).
Principal for the 2nd period \( (P_2) = ₹ 8,600 \)
Interest for the 2nd six months = \( 5\% \text{ of } 8,600 = \frac{5 \times 8,600}{100} = ₹ 430 \)
Amount = \( 8,600 + 430 = ₹ 9,030 \)
Repayment at the end of 12 months = \( ₹ 4,000 \)
Balance at the end of 2nd six months = \( 9,030 – 4,000 = ₹ 5,030 \)
Step 3: Calculate the final payment at the end of 18 months.
Principal for the 3rd period \( (P_3) = ₹ 5,030 \)
Interest for the 3rd six months = \( 5\% \text{ of } 5,030 = \frac{5 \times 5,030}{100} = ₹ 251.50 \)
Total amount due at the end of 18 months = \( 5,030 + 251.50 = ₹ 5,281.50 \)
Step 4: Determine the final payment.
To clear the loan entirely at the end of 18 months, the third payment must be equal to the total outstanding amount.
Third payment = \( ₹ 5,281.50 \)
Answer: ₹ 5,281.50

Q5: 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: Assume the Principal and calculate interest for the first year.
Let the Principal \( (P) \) be \( ₹ x \).
Rate \( (R) = 10\% \) per annum.
Interest for 1st year \( (I_1) = \frac{x \times 10 \times 1}{100} = ₹ 0.1x \)
Step 2: Calculate the amount at the end of the second year.
Amount after 1st year \( (A_1) = x + 0.1x = 1.1x \)
Interest for 2nd year \( (I_2) = 10\% \text{ of } 1.1x = 0.11x \)
Amount after 2nd year \( (A_2) = 1.1x + 0.11x = 1.21x \)
Step 3: Calculate the interest for the third year.
The amount after the 2nd year acts as the Principal for the 3rd year.
Interest for 3rd year \( (I_3) = 10\% \text{ of } 1.21x = 0.121x \)
Step 4: Use the given condition to find the sum \( (x) \).
Given: \( I_1 + I_3 = 2,652 \)
\( 0.1x + 0.121x = 2,652 \)
\( 0.221x = 2,652 \)
Step 5: Solve for \( x \).
\( x = \frac{2652}{0.221} \)
\( x = \frac{2652000}{221} \)
\( x = 12,000 \)
Answer: ₹ 12,000

Q6: During every financial year, the value of a machine depreciates by 12%. Find the cost of a machine which depreciates by ₹ 2,640 during the second financial year of its purchase.

Step 1: Assume the initial cost and calculate depreciation for the first year.
Let the initial cost of the machine be \( ₹ x \).
Rate of depreciation \( (R) = 12\% \) per annum.
Depreciation for the 1st year = \( 12\% \text{ of } x = 0.12x \)
Step 2: Calculate the value of the machine at the beginning of the second year.
Value after 1st year = \( \text{Initial Cost} – \text{Depreciation of 1st year} \)
Value at beginning of 2nd year = \( x – 0.12x = 0.88x \)
Step 3: Calculate the depreciation during the second year.
Depreciation for the 2nd year = \( 12\% \text{ of the value at the beginning of 2nd year} \)
Depreciation for 2nd year = \( 0.12 \times 0.88x = 0.1056x \)
Step 4: Equate the calculated depreciation to the given value.
Given depreciation in 2nd year = \( ₹ 2,640 \)
\( 0.1056x = 2,640 \)
Step 5: Solve for the initial cost \( (x) \).
\( x = \frac{2,640}{0.1056} \)
\( x = \frac{26400000}{1056} \)
\( x = 25,000 \)
Answer: ₹ 25,000

Q7: Find the sum on which the difference between the simple interest and the compound interest at the rate of 8% per annum compounded annually is ₹ 64 in 2 years.

Step 1: Assume the Principal and express Simple Interest (S.I.).
Let the sum (Principal) be \( ₹ P \).
Given: Rate \( (R) = 8\% \), Time \( (T) = 2 \) years.
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{P \times 8 \times 2}{100} = \frac{16P}{100} = 0.16P \)
Step 2: Express Compound Interest (C.I.) in terms of \( P \).
Formula for Amount: \( A = P(1 + \frac{R}{100})^T \)
\( A = P(1 + \frac{8}{100})^2 = P(1.08)^2 = 1.1664P \)
\( \text{C.I.} = A – P \)
\( \text{C.I.} = 1.1664P – P = 0.1664P \)
Step 3: Use the given difference to find the Principal.
Given: \( \text{C.I.} – \text{S.I.} = 64 \)
\( 0.1664P – 0.16P = 64 \)
\( 0.0064P = 64 \)
Step 4: Solve for \( P \).
\( P = \frac{64}{0.0064} \)
\( P = \frac{640000}{64} \)
\( P = 10,000 \)
Answer: ₹ 10,000

Q8: A sum of ₹ 13,500 is invested at 16% per annum compound interest for 5 years. Calculate:

i. the interest for the first year.

Step 1: Identify the Principal and Rate for the first year.
Principal \( (P_1) = ₹ 13,500 \)
Rate \( (R) = 16\% \)
Step 2: Calculate interest using the Simple Interest formula for 1 year.
\( \text{Interest for 1st year} = \frac{P_1 \times R \times T}{100} \)
\( \text{Interest for 1st year} = \frac{13500 \times 16 \times 1}{100} \)
\( \text{Interest for 1st year} = 135 \times 16 = ₹ 2,160 \)
Answer: ₹ 2,160

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

Step 1: Add the interest to the Principal.
\( \text{Amount } (A_1) = P_1 + \text{Interest for 1st year} \)
\( A_1 = 13,500 + 2,160 = ₹ 15,660 \)
Answer: ₹ 15,660

iii. the interest for the second year, correct to the nearest rupee.

Step 1: Identify the Principal for the second year.
In compound interest, the amount at the end of the 1st year becomes the principal for the 2nd year.
Principal for 2nd year \( (P_2) = ₹ 15,660 \)
Step 2: Calculate interest for the second year.
\( \text{Interest for 2nd year} = \frac{P_2 \times R \times T}{100} \)
\( \text{Interest for 2nd year} = \frac{15660 \times 16 \times 1}{100} \)
\( \text{Interest for 2nd year} = \frac{250560}{100} = ₹ 2,505.60 \)
Step 3: Round the interest to the nearest rupee.
Interest correct to the nearest rupee = \( ₹ 2,506 \)
Answer: ₹ 2,506

Q9: Saurabh invests \( ₹ 48,000 \) for 7 years at \( 10\% \) per annum compound interest. Calculate:

i. the interest for the first year.

Step 1: Identify the Principal and Rate for the first year.
Principal \( (P_1) = ₹ 48,000 \)
Rate \( (R) = 10\% \)
Step 2: Calculate interest for the 1st year.
\( \text{Interest for 1st year} = \frac{P_1 \times R \times T}{100} \)
\( \text{Interest for 1st year} = \frac{48,000 \times 10 \times 1}{100} = ₹ 4,800 \)
Answer: ₹ 4,800

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

Step 1: Find the Amount at the end of the 1st year (Principal for 2nd year).
\( A_1 = P_1 + \text{Interest}_1 = 48,000 + 4,800 = ₹ 52,800 \)
Step 2: Calculate interest for the 2nd year.
\( \text{Interest for 2nd year} = \frac{52,800 \times 10 \times 1}{100} = ₹ 5,280 \)
Step 3: Calculate the total Amount at the end of the 2nd year.
\( A_2 = \text{Principal for 2nd year} + \text{Interest for 2nd year} \)
\( A_2 = 52,800 + 5,280 = ₹ 58,080 \)
Answer: ₹ 58,080

iii. the interest for the third year.

Step 1: Identify the Principal for the third year.
The amount at the end of the 2nd year becomes the Principal for the 3rd year.
Principal for 3rd year \( (P_3) = ₹ 58,080 \)
Step 2: Calculate interest for the 3rd year only.
\( \text{Interest for 3rd year} = \frac{58,080 \times 10 \times 1}{100} = ₹ 5,808 \)
Answer: ₹ 5,808

Q10: 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 was \( ₹ 920 \), find:

i. the rate of interest charged

Step 1: Assume the rate of interest and calculate interest for the first year.
Let the rate of interest be \( R\% \) per annum.
Principal \( (P_1) = ₹ 12,000 \)
Interest for 1st year = \( \frac{12,000 \times R \times 1}{100} = 120R \)
Step 2: Find the Principal for the second year after repayment.
Amount after 1st year = \( 12,000 + 120R \)
Repayment = \( ₹ 4,000 \)
Principal for 2nd year \( (P_2) = (12,000 + 120R) – 4,000 = 8,000 + 120R \)
Step 3: Use the given interest for the second year to find \( R \).
Interest for 2nd year = \( \frac{P_2 \times R \times 1}{100} \)
\( 920 = \frac{(8,000 + 120R) \times R}{100} \)
\( 92,000 = 8,000R + 120R^2 \)
Dividing by 40:
\( 3R^2 + 200R – 2,300 = 0 \)
Step 4: Solve the quadratic equation.
\( 3R^2 + 230R – 30R – 2,300 = 0 \)
\( R(3R + 230) – 10(3R + 230) = 0 \)
\( (R – 10)(3R + 230) = 0 \)
Since rate cannot be negative, \( R = 10 \).
Answer: Rate of interest = 10% p.a.

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

Step 1: Calculate the Principal for the second year using \( R = 10 \).
\( P_2 = 8,000 + 120(10) = 8,000 + 1,200 = ₹ 9,200 \)
Step 2: Calculate the amount at the end of the second year.
Amount = \( P_2 + \text{Interest of 2nd year} \)
Amount = \( 9,200 + 920 = ₹ 10,120 \)
Answer: Debt at the end of 2nd year = ₹ 10,120

Q11: 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

Step 1: Understand the relationship between interest of consecutive years.
In compound interest, the interest for the second year is the interest on the first year’s interest plus the first year’s interest itself. The difference between them is the interest earned on the first year’s interest.
\( \text{Increase in interest} = 1,725 – 1,500 = ₹ 225 \)
Step 2: Calculate the rate of interest for the second year.
This \( ₹ 225 \) is the interest on \( ₹ 1,500 \) for one year.
\( R_2 = \frac{\text{Increase} \times 100}{\text{Interest of 1st year} \times 1} \)
\( R_2 = \frac{225 \times 100}{1500} \)
\( R_2 = \frac{225}{15} = 15\% \)
Answer: Rate of interest for 2nd year = 15% p.a.

ii. third year.

Step 1: Find the increase in interest between the second and third year.
The difference between the interest of the third year and the second year is the interest earned on the second year’s interest.
\( \text{Increase in interest} = 2,070 – 1,725 = ₹ 345 \)
Step 2: Calculate the rate of interest for the third year.
This \( ₹ 345 \) is the interest on \( ₹ 1,725 \) for one year.
\( R_3 = \frac{\text{Increase} \times 100}{\text{Interest of 2nd year} \times 1} \)
\( R_3 = \frac{345 \times 100}{1725} \)
\( R_3 = \frac{34500}{1725} = 20\% \)
Answer: Rate of interest for 3rd year = 20% p.a.