Compound Interest (Stage 1)

Q1: Multiple Choice Type:

a. ₹ 300 and ₹ 360 are the compound interest for two consecutive years. The rate of interest is:

Step 1: Understand that the difference between C.I. of two consecutive years is the interest on the first year’s C.I.
Interest on ₹ 300 for 1 year = \( 360 – 300 = ₹ 60 \)
Step 2: Calculate the Rate of Interest \( (R) \).
\( R = \frac{\text{Interest} \times 100}{P \times T} \)
\( R = \frac{60 \times 100}{300 \times 1} = 20\% \)
Answer: 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:

Step 1: The amount at the end of the 5th year acts as the Principal for the 6th year.
Principal \( (P) = ₹ 5,000 \), Amount \( (A) = ₹ 6,000 \)
Interest for 1 year = \( 6,000 – 5,000 = ₹ 1,000 \)
Step 2: Calculate the Rate of Interest \( (R) \).
\( R = \frac{1000 \times 100}{5000 \times 1} = 20\% \)
Answer: 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:

Step 1: Let the C.I. at the end of 2019 be \( x \).
The C.I. for the next year (2020) is the C.I. of the previous year plus interest on it.
\( 3850 = x + (10\% \text{ of } x) \)
Step 2: Solve for \( x \).
\( 3850 = x + 0.1x = 1.1x \)
\( x = \frac{3850}{1.1} = ₹ 3,500 \)
Answer: 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:

Step 1: The amount at the end of 6 years becomes the principal for the 7th year.
Principal for 7th year \( (P) = ₹ 4,500 \)
Step 2: Calculate interest for the 7th year.
Interest = \( 12\% \text{ of } 4,500 = \frac{12 \times 4500}{100} = ₹ 540 \)
Step 3: Find the total amount at the end of 7 years.
Amount = \( 4,500 + 540 = ₹ 5,040 \)
Answer: 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:

Step 1: Calculate the difference in interest.
Difference = \( 690 – 600 = ₹ 90 \)
Step 2: Calculate the Rate of Interest \( (R) \).
\( R = \frac{90 \times 100}{600 \times 1} = 15\% \)
Answer: 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:

Step 1: Calculate the interest earned in the second year.
Interest = \( 2,760 – 2,400 = ₹ 360 \)
Step 2: Calculate the Rate of Interest \( (R) \) using the first amount as principal.
\( R = \frac{360 \times 100}{2400 \times 1} = 15\% \)
Answer: ii. 15%

Q2: A certain sum amounts to ₹ 5,292 in two years and ₹ 5,556.60 in three years, interest compounded annually, Find:

i. the rate of interest

Step 1: Identify the Principal for the third year.
Since interest is compounded annually, the amount at the end of the 2nd year becomes the Principal for the 3rd year.
Principal for 3rd year \( (P_3) = ₹ 5,292 \)
Amount at the end of 3rd year \( (A_3) = ₹ 5,556.60 \)
Step 2: Calculate interest earned in the 3rd year.
Interest = \( A_3 – P_3 \)
Interest = \( 5,556.60 – 5,292 = ₹ 264.60 \)
Step 3: Calculate the Rate of Interest \( (R) \).
Using formula: \( R = \frac{\text{Interest} \times 100}{P \times T} \)
\( R = \frac{264.60 \times 100}{5,292 \times 1} \)
\( R = \frac{26460}{5292} = 5\% \)
Answer: Rate of Interest = 5%

ii. the original sum.

Step 1: Use the Amount formula for 2 years to find the original Principal \( (P) \).
Formula: \( A = P \left( 1 + \frac{R}{100} \right)^n \)
Given: \( A = 5,292 \), \( R = 5 \), \( n = 2 \)
\( 5,292 = P \left( 1 + \frac{5}{100} \right)^2 \)
Step 2: Simplify the equation.
\( 5,292 = P \left( \frac{21}{20} \right)^2 \)
\( 5,292 = P \left( \frac{441}{400} \right) \)
Step 3: Solve for \( P \).
\( P = \frac{5,292 \times 400}{441} \)
\( P = 12 \times 400 \)
\( P = 4,800 \)
Answer: Original Sum = ₹ 4,800

Q3: 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.

Step 1: Calculate the interest earned in the first year.
Principal \( (P) = ₹ 8,000 \)
Amount after 1 year \( (A_1) = ₹ 9,440 \)
Interest for 1st year = \( A_1 – P = 9,440 – 8,000 = ₹ 1,440 \)
Step 2: Find the Rate of Interest \( (R) \).
Using the formula: \( R = \frac{\text{Interest} \times 100}{P \times T} \)
\( R = \frac{1440 \times 100}{8000 \times 1} \)
\( R = \frac{144}{8} = 18\% \)
Answer: Rate of Interest = 18% p.a.

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

Step 1: The amount at the end of the 1st year becomes the Principal for the 2nd year.
Principal for 2nd year \( (P_2) = ₹ 9,440 \)
Step 2: Calculate interest for the 2nd year.
Interest = \( 18\% \text{ of } 9,440 = \frac{18 \times 9440}{100} = ₹ 1,699.20 \)
Step 3: Find the Amount at the end of the 2nd year.
\( A_2 = P_2 + \text{Interest} = 9,440 + 1,699.20 = ₹ 11,139.20 \)
Answer: Amount at end of 2nd year = ₹ 11,139.20

iii. the interest accrued in the third year.

Step 1: The amount at the end of the 2nd year becomes the Principal for the 3rd year.
Principal for 3rd year \( (P_3) = ₹ 11,139.20 \)
Step 2: Calculate interest for the 3rd year only.
Interest = \( \frac{P_3 \times R \times T}{100} \)
Interest = \( \frac{11139.20 \times 18 \times 1}{100} \)
Interest = \( \frac{200505.6}{100} = ₹ 2,005.056 \)
Rounding to two decimal places: \( ₹ 2,005.06 \)
Answer: Interest in 3rd year = ₹ 2,005.06

Q4: 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.

i. Calculate the rate of interest per annum.

Step 1: Find the extra interest earned in the third year.
The difference between the C.I. of two consecutive years is the interest on the C.I. of the preceding year.
Difference in Interest = \( 1,197.90 – 1,089 = ₹ 108.90 \)
Step 2: Calculate the Rate of Interest \( (R) \).
This \( ₹ 108.90 \) is the interest on \( ₹ 1,089 \) for one year.
\( R = \frac{\text{Difference} \times 100}{\text{C.I. of 2nd year} \times 1} \)
\( R = \frac{108.90 \times 100}{1089} \)
\( R = \frac{10890}{1089} = 10\% \)
Answer: Rate of Interest = 10% p.a.

ii. Calculate the sum of money (Principal).

Step 1: Find the C.I. for the first year.
Let the C.I. for the 1st year be \( x \).
The C.I. for the 2nd year is the C.I. of the 1st year plus interest on it.
\( 1,089 = x + (10\% \text{ of } x) \)
\( 1,089 = 1.1x \)
\( x = \frac{1089}{1.1} = ₹ 990 \)
Step 2: Calculate the Principal \( (P) \).
For the first year, C.I. is equal to S.I.
\( \text{S.I. for 1st year} = ₹ 990 \)
\( 990 = \frac{P \times 10 \times 1}{100} \)
\( 990 = \frac{P}{10} \)
\( P = 990 \times 10 = ₹ 9,900 \)
Answer: Sum of money = ₹ 9,900

Q5: 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: Find the difference between the interests of two successive years.
In compound interest, the interest for any year is the interest on the amount at the beginning of that year. Therefore, the difference in interest between two successive years is the interest on the interest of the preceding year.
Difference in Interest = \( 7,410 – 5,700 = ₹ 1,710 \)
Step 2: Calculate the Rate of Interest \( (R) \).
This difference of \( ₹ 1,710 \) is the interest on \( ₹ 5,700 \) for one year.
Using the formula: \( R = \frac{\text{Interest} \times 100}{P \times T} \)
\( R = \frac{1,710 \times 100}{5,700 \times 1} \)
Step 3: Simplify the calculation.
\( R = \frac{1710}{57} \)
\( R = 30\% \)
Answer: Rate of Interest = 30% p.a.

Q6: 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.

Step 1: Find the reduction in depreciation.
The difference between the depreciation of two consecutive years is the depreciation on the first year’s depreciation value.
Reduction in depreciation = \( 4,000 – 3,600 = ₹ 400 \)
Step 2: Calculate the Rate of Depreciation \( (R) \).
The reduction of \( ₹ 400 \) is the depreciation on the 1st year’s depreciation of \( ₹ 4,000 \).
\( R = \frac{\text{Reduction} \times 100}{\text{Depreciation of 1st year} \times 1} \)
\( R = \frac{400 \times 100}{4000} = 10\% \)
Answer: Rate of Depreciation = 10% p.a.

ii. the original cost of the machine.

Step 1: Use the first year’s depreciation to find the original cost \( (P) \).
Depreciation for the 1st year = \( \frac{P \times R \times 1}{100} \)
\( 4,000 = \frac{P \times 10}{100} \)
Step 2: Solve for \( P \).
\( P = 4,000 \times 10 = ₹ 40,000 \)
Answer: Original Cost = ₹ 40,000

iii. its cost at the end of the third year.

Step 1: Find the cost at the end of the second year.
Cost after 2nd year = \( \text{Original Cost} – (\text{Dep. 1st year} + \text{Dep. 2nd year}) \)
Cost after 2nd year = \( 40,000 – (4,000 + 3,600) = ₹ 32,400 \)
Step 2: Calculate depreciation for the third year.
Depreciation of 3rd year = \( 10\% \text{ of } 32,400 \)
Depreciation of 3rd year = \( \frac{10}{100} \times 32,400 = ₹ 3,240 \)
Step 3: Calculate the final cost.
Cost at the end of 3rd year = \( 32,400 – 3,240 = ₹ 29,160 \)
Answer: Cost at the end of 3rd year = ₹ 29,160

Q7: 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.

Step 1: Calculate the interest for the first year.
Principal \( (P_1) = ₹ 12,800 \)
Rate \( (R) = 10\% \)
Interest for 1st year = \( \frac{12,800 \times 10 \times 1}{100} = ₹ 1,280 \)
Step 2: Find the amount (sum due) at the end of the first year.
Amount \( (A_1) = P_1 + \text{Interest} \)
\( A_1 = 12,800 + 1,280 = ₹ 14,080 \)
Answer: Sum due at the end of 1st year = ₹ 14,080

ii. the interest he earns for the second year.

Step 1: Identify the Principal for the second year.
The amount at the end of the 1st year becomes the Principal for the 2nd year.
Principal \( (P_2) = ₹ 14,080 \)
Step 2: Calculate interest for the second year.
Interest for 2nd year = \( \frac{14,080 \times 10 \times 1}{100} = ₹ 1,408 \)
Answer: Interest for 2nd year = ₹ 1,408

iii. the total amount due to him at the end of the third year.

Step 1: Find the Amount at the end of the second year.
Amount \( (A_2) = P_2 + \text{Interest of 2nd year} \)
\( A_2 = 14,080 + 1,408 = ₹ 15,488 \)
Step 2: Calculate the Interest for the third year.
Principal \( (P_3) = ₹ 15,488 \)
Interest for 3rd year = \( \frac{15,488 \times 10 \times 1}{100} = ₹ 1,548.80 \)
Step 3: Calculate the final Amount at the end of the third year.
Amount \( (A_3) = P_3 + \text{Interest of 3rd year} \)
\( A_3 = 15,488 + 1,548.80 = ₹ 17,036.80 \)
Answer: Total amount due at the end of 3rd year = ₹ 17,036.80

Q8: 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: Calculate the difference between the interests of two successive half-years.
In compound interest, the difference between the interest of two consecutive periods is the interest earned on the interest of the preceding period.
Difference in Interest = \( 760.50 – 650 = ₹ 110.50 \)
Step 2: Calculate the rate of interest per half-year \( (r) \).
The difference of \( ₹ 110.50 \) is the interest on \( ₹ 650 \) for one half-year period.
\( r = \frac{\text{Difference} \times 100}{\text{Interest of 1st half-year} \times 1} \)
\( r = \frac{110.50 \times 100}{650} \)
\( r = \frac{11050}{650} = 17\% \) per half-year.
Step 3: Calculate the annual rate of interest \( (R) \).
Since the interest is compounded half-yearly, the annual rate is double the half-yearly rate.
\( R = 2 \times r \)
\( R = 2 \times 17\% = 34\% \)
Answer: Rate of Interest = 34% p.a.

Q9: 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.

Step 1: Calculate the interest for the first six months.
Principal \( (P) = ₹ 15,000 \)
Amount after 6 months \( (A_1) = ₹ 15,600 \)
Interest for the first 6 months = \( A_1 – P = 15,600 – 15,000 = ₹ 600 \)
Step 2: Find the semi-annual (half-yearly) rate of interest \( (r) \).
Using the formula: \( r = \frac{\text{Interest} \times 100}{P \times T} \)
Since the period is one semi-annual term, \( T = 1 \):
\( r = \frac{600 \times 100}{15,000 \times 1} = 4\% \) per half-year.
Step 3: Calculate the annual rate of interest \( (R) \).
Annual Rate = \( 2 \times \text{Semi-annual Rate} \)
\( R = 2 \times 4\% = 8\% \)
Answer: Rate of Interest = 8% p.a.

ii. the total amount of money that Geeta must pay at the end of 18 months in order to clear the account.

Step 1: Identify the number of semi-annual conversion periods.
Time = 18 months = 3 semi-annual periods.
Semi-annual rate \( (r) = 4\% \)
Step 2: Calculate the amount at the end of the second 6 months (12 months).
Principal for 2nd period = \( ₹ 15,600 \)
Interest = \( 4\% \text{ of } 15,600 = \frac{4 \times 15,600}{100} = ₹ 624 \)
Amount after 12 months \( (A_2) = 15,600 + 624 = ₹ 16,224 \)
Step 3: Calculate the amount at the end of the third 6 months (18 months).
Principal for 3rd period = \( ₹ 16,224 \)
Interest = \( 4\% \text{ of } 16,224 = \frac{4 \times 16,224}{100} = ₹ 648.96 \)
Amount after 18 months \( (A_3) = 16,224 + 648.96 = ₹ 16,872.96 \)
Answer: Total amount to pay = ₹ 16,872.96

Q10: ₹ 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.

Step 1: Calculate the interest for the first year.
Principal \( (P_1) = ₹ 8,000 \)
Rate \( (R) = 7\% \)
Interest for 1st year = \( \frac{8,000 \times 7 \times 1}{100} = ₹ 560 \)
Step 2: Find the amount at the end of the first year and the balance after repayment.
Amount after 1st year = \( 8,000 + 560 = ₹ 8,560 \)
Repayment at the end of 1st year = \( ₹ 3,560 \)
Balance Principal for 2nd year \( (P_2) = 8,560 – 3,560 = ₹ 5,000 \)
Step 3: Calculate the interest for the second year.
Interest for 2nd year = \( \frac{5,000 \times 7 \times 1}{100} = ₹ 350 \)
Answer: Interest for 2nd year = ₹ 350

ii. the total interest paid in two years

Step 1: Add the interest from the first year and the second year.
Total Interest = \( \text{Interest of 1st year} + \text{Interest of 2nd year} \)
Total Interest = \( 560 + 350 = ₹ 910 \)
Answer: Total Interest = ₹ 910

iii. the total amount of money paid in two years to clear the debt.

Step 1: Find the amount due at the end of the second year.
Amount to be paid at the end of 2nd year = \( P_2 + \text{Interest of 2nd year} \)
Amount = \( 5,000 + 350 = ₹ 5,350 \)
Step 2: Calculate the total money paid over two years.
Total Money Paid = \( \text{First Repayment} + \text{Final Payment} \)
Total Money Paid = \( 3,560 + 5,350 = ₹ 8,910 \)
Answer: Total amount paid = ₹ 8,910

Q11: 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: 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 = \( x + 0.1x = 1.1x \)
Interest for 2nd year = \( 10\% \text{ of } 1.1x = 0.11x \)
Amount after 2nd year = \( 1.1x + 0.11x = 1.21x \)
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 \( (I_3) = 10\% \text{ of } 1.21x \)
\( I_3 = \frac{10}{100} \times 1.21x = ₹ 0.121x \)
Step 4: Use the given condition to find \( x \).
Given: \( I_3 – I_1 = 252 \)
\( 0.121x – 0.1x = 252 \)
\( 0.021x = 252 \)
Step 5: Solve for the Principal.
\( x = \frac{252}{0.021} \)
\( x = \frac{252000}{21} = 12,000 \)
Answer: Sum Invested = ₹ 12,000

Q12: 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: Calculate the balance at the end of the first year.
Principal \( (P_1) = ₹ 10,000 \)
Interest for the 1st year = \( 10\% \text{ of } 10,000 = \frac{10 \times 10,000}{100} = ₹ 1,000 \)
Amount at the end of 1st year = \( 10,000 + 1,000 = ₹ 11,000 \)
Repayment at the end of 1st year = \( 30\% \text{ of sum borrowed } (10,000) = ₹ 3,000 \)
Balance at the end of 1st year = \( 11,000 – 3,000 = ₹ 8,000 \)
Step 2: Calculate the balance at the end of the second year.
Principal for the 2nd year \( (P_2) = ₹ 8,000 \)
Interest for the 2nd year = \( 10\% \text{ of } 8,000 = \frac{10 \times 8,000}{100} = ₹ 800 \)
Amount at the end of 2nd year = \( 8,000 + 800 = ₹ 8,800 \)
Repayment at the end of 2nd year = \( 30\% \text{ of sum borrowed } (10,000) = ₹ 3,000 \)
Balance at the end of 2nd year = \( 8,800 – 3,000 = ₹ 5,800 \)
Answer: Amount left unpaid = ₹ 5,800

Q13: 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: Calculate the balance after the first year repayment.
Principal \( (P_1) = ₹ 10,000 \)
Interest for the 1st year = \( 10\% \text{ of } 10,000 = ₹ 1,000 \)
Amount at the end of 1st year \( (A_1) = 10,000 + 1,000 = ₹ 11,000 \)
Repayment at the end of 1st year = \( 20\% \text{ of } 11,000 = \frac{20}{100} \times 11,000 = ₹ 2,200 \)
Balance at the end of 1st year = \( 11,000 – 2,200 = ₹ 8,800 \)
Step 2: Calculate the balance after the second year repayment.
Principal for the 2nd year \( (P_2) = ₹ 8,800 \)
Interest for the 2nd year = \( 10\% \text{ of } 8,800 = ₹ 880 \)
Amount at the end of 2nd year \( (A_2) = 8,800 + 880 = ₹ 9,680 \)
Repayment at the end of 2nd year = \( 20\% \text{ of } 9,680 = \frac{20}{100} \times 9,680 = ₹ 1,936 \)
Balance at the end of 2nd year = \( 9,680 – 1,936 = ₹ 7,744 \)
Answer: Amount left unpaid = ₹ 7,744