Compound Interest (Stage 1)

Q1: Multiple Choice Type:

a. The difference between C.I. and S.I. in one year on ₹ 5,000 at the rate of 10% per annum is:

Step 1: Calculate Simple Interest (S.I.) for 1 year.
Formula: \( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{5000 \times 10 \times 1}{100} = ₹ 500 \)
Step 2: Calculate Compound Interest (C.I.) for 1 year.
For the first year, C.I. is calculated on the original principal, same as S.I.
\( \text{C.I.} = P \left( 1 + \frac{R}{100} \right)^T – P \)
\( \text{C.I.} = 5000 \left( 1 + \frac{10}{100} \right)^1 – 5000 = 5500 – 5000 = ₹ 500 \)
Step 3: Find the difference.
Difference = \( \text{C.I.} – \text{S.I.} = 500 – 500 = 0 \)
Answer: i. ₹ 00

b. ₹ 2,000 is saved during the year 2022 and deposited in a bank at the beginning of year 2023 at 8% compound interest. During 2023, ₹ 3,000 more is saved and deposited in the same bank at the beginning of 2024 and at the same rate of interest. The C.I. earned upto the end of 2024 is:

Step 1: Calculate amount of first deposit (₹ 2,000) at the end of 2024 (2 years).
\( A_1 = P(1 + \frac{R}{100})^n = 2000(1 + \frac{8}{100})^2 \)
\( A_1 = 2000(1.08)^2 = 2000 \times 1.1664 = ₹ 2,332.80 \)
Step 2: Calculate amount of second deposit (₹ 3,000) at the end of 2024 (1 year).
\( A_2 = 3000(1 + \frac{8}{100})^1 = 3000 \times 1.08 = ₹ 3,240 \)
Step 3: Calculate Total Amount and Interest.
Total Amount = \( 2332.80 + 3240 = ₹ 5,572.80 \)
Total Principal = \( 2000 + 3000 = ₹ 5,000 \)
C.I. = \( 5572.80 – 5000 = ₹ 572.80 \)
Answer: ii. ₹ 572.80

c. ₹ 1,000 is borrowed at 10% per annum C.I. If ₹ 300 is repaid at the end of each year, the amount of loan outstanding at the end of 2nd year is:

Step 1: Calculate amount after 1st year.
Interest = \( 10\% \text{ of } 1000 = ₹ 100 \)
Amount = \( 1000 + 100 = ₹ 1100 \)
Balance after repayment = \( 1100 – 300 = ₹ 800 \)
Step 2: Calculate amount after 2nd year.
Interest = \( 10\% \text{ of } 800 = ₹ 80 \)
Amount = \( 800 + 80 = ₹ 880 \)
Answer: i. ₹ 880

d. The difference between C.I. and S.I. in 2 years on ₹ 4,000 at 10% per annum is:

Step 1: Calculate S.I. for 2 years.
\( \text{S.I.} = \frac{4000 \times 10 \times 2}{100} = ₹ 800 \)
Step 2: Calculate C.I. for 2 years.
\( A = 4000(1 + \frac{10}{100})^2 = 4000(1.1)^2 = 4000 \times 1.21 = ₹ 4840 \)
\( \text{C.I.} = 4840 – 4000 = ₹ 840 \)
Step 3: Find the difference.
Difference = \( 840 – 800 = ₹ 40 \)
Answer: iv. ₹ 40

e. ₹ 10 is the difference between compound interest and simple interest in 2 years and at 5% per annum. The principal amount is:

Step 1: Assume the Principal and calculate Simple Interest (S.I.).
Let the Principal be \( P \).
Given: Rate \( (R) = 5\% \), Time \( (T) = 2 \) years.
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{P \times 5 \times 2}{100} = \frac{10P}{100} = \frac{P}{10} \)
Step 2: Calculate the Compound Interest (C.I.) in terms of \( P \).
Formula for Amount: \( A = P(1 + \frac{R}{100})^T \)
\( A = P(1 + \frac{5}{100})^2 = P(\frac{21}{20})^2 = \frac{441P}{400} \)
\( \text{C.I.} = A – P \)
\( \text{C.I.} = \frac{441P}{400} – P = \frac{41P}{400} \)
Step 3: Use the given difference to find the Principal.
Given: \( \text{C.I.} – \text{S.I.} = 10 \)
\( \frac{41P}{400} – \frac{P}{10} = 10 \)
To subtract, take L.C.M. of 400 and 10, which is 400:
\( \frac{41P – 40P}{400} = 10 \)
\( \frac{P}{400} = 10 \)
Step 4: Solve for \( P \).
\( P = 10 \times 400 \)
\( P = 4000 \)
Answer: iii. ₹ 4,000

Q2: Calculate the difference between the simple interest and the compound interest on ₹4000 in 2 years at 8% per annum compounded yearly.

Step 1: Given
Principal \( P = 4000 \), Rate \( R = 8\% \), Time \( T = 2 \) years
Step 2: Simple Interest
\( SI = \frac{P \times R \times T}{100} \)
\( SI = \frac{4000 \times 8 \times 2}{100} = 640 \)
Step 3: Compound Interest
\( A = 4000 \left(1 + \frac{8}{100}\right)^2 \)
\( A = 4000 (1.08)^2 \)
\( A = 4000 \times 1.1664 = 4665.60 \)
\( CI = 4665.60 – 4000 = 665.60 \)
Step 4: Difference
\( CI – SI = 665.60 – 640 = 25.60 \)
Answer: Difference = ₹25.60

Q3: A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by ₹96, find the sum of money.

Step 1: Let Principal \( = P \)
Rate \( R = 8\% \)
Step 2: Interest for 1st year
\( CI_1 = \frac{P \times 8}{100} = 0.08P \)
Step 3: Amount after 1st year
\( A_1 = P + 0.08P = 1.08P \)
Step 4: Interest for 2nd year
\( CI_2 = \frac{1.08P \times 8}{100} = 0.0864P \)
Step 5: Given difference
\( CI_2 – CI_1 = 96 \)
\( 0.0864P – 0.08P = 96 \)
\( 0.0064P = 96 \)
Step 6: Solve for \(P\)
\( P = \frac{96}{0.0064} = 15000 \)
Answer: Principal = ₹15,000

Q4: A man invests ₹5600 at 14% per annum compound interest for 2 years. Calculate:

i. the interest for the first year.

Step 1: Interest for 1st year
\( CI_1 = \frac{5600 \times 14}{100} = 784 \)
Answer: ₹784

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

Step 2: Amount after 1st year
\( A_1 = 5600 + 784 = 6384 \)
Answer: ₹6,384

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

Step 3: Interest for 2nd year
\( CI_2 = \frac{6384 \times 14}{100} = 893.76 \)
Step 4: Rounded value
\( CI_2 \approx 894 \)
Answer: ₹894

Q5: A man saves ₹ 3,000 every year and invests it at the end of the year at 10% compound interest. Calculate the total amount of his savings at the end of the third year.

Step 1: Amount at the end of the 1st year.
Since the man saves ₹ 3,000 at the end of the year, no interest is earned for the 1st year.
Savings at the end of Year 1 = \( ₹ 3,000 \)
Step 2: Amount at the end of the 2nd year.
The amount from the 1st year earns interest for the 2nd year, plus a new saving of ₹ 3,000 is added at the end.
Interest for 2nd year = \( 10\% \text{ of } 3,000 = \frac{10 \times 3000}{100} = ₹ 300 \)
Total at end of 2nd year = \( (\text{Principal} + \text{Interest}) + \text{New Saving} \)
Total = \( (3,000 + 300) + 3,000 = ₹ 6,300 \)
Step 3: Amount at the end of the 3rd year.
The balance of ₹ 6,300 earns interest for the 3rd year, plus a new saving of ₹ 3,000 is added at the end.
Interest for 3rd year = \( 10\% \text{ of } 6,300 = \frac{10 \times 6300}{100} = ₹ 630 \)
Total at end of 3rd year = \( (\text{Principal} + \text{Interest}) + \text{New Saving} \)
Total = \( (6,300 + 630) + 3,000 = ₹ 9,930 \)
Answer: ₹ 9,930

Q6: A man lends ₹12,500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. of the first year and the compound interest for the third year.

Step 1: Given
Principal \( P = 12500 \)
Rates: \(12\%, 15\%, 18\%\)
Step 2: Interest for 1st year
\( CI_1 = \frac{12500 \times 12}{100} = 1500 \)
Step 3: Amount after 1st year
\( A_1 = 12500 + 1500 = 14000 \)
Step 4: After 2nd year
\( CI_2 = \frac{14000 \times 15}{100} = 2100 \)
\( A_2 = 14000 + 2100 = 16100 \)
Step 5: Interest for 3rd year
\( CI_3 = \frac{16100 \times 18}{100} = 2898 \)
Step 6: Required difference
\( = CI_3 – CI_1 = 2898 – 1500 = 1398 \)
Answer: Difference = ₹1,398

Q7: A man borrows ₹ 6,000 at 5 percent C.I. per annum. If he repays ₹ 1,200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Step 1: Calculate the interest and outstanding balance for the first year.
Principal for the 1st year \( (P_1) = ₹ 6,000 \)
Interest for the 1st year = \( 5\% \text{ of } 6,000 = \frac{5 \times 6000}{100} = ₹ 300 \)
Amount at the end of the 1st year = \( 6,000 + 300 = ₹ 6,300 \)
Repayment at the end of the 1st year = \( ₹ 1,200 \)
Balance at the end of the 1st year = \( 6,300 – 1,200 = ₹ 5,100 \)
Step 2: Calculate the interest and outstanding balance for the second year.
Principal for the 2nd year \( (P_2) = ₹ 5,100 \)
Interest for the 2nd year = \( 5\% \text{ of } 5,100 = \frac{5 \times 5100}{100} = ₹ 255 \)
Amount at the end of the 2nd year = \( 5,100 + 255 = ₹ 5,355 \)
Repayment at the end of the 2nd year = \( ₹ 1,200 \)
Balance at the end of the 2nd year = \( 5,355 – 1,200 = ₹ 4,155 \)
Step 3: Determine the amount at the beginning of the third year.
The amount outstanding at the end of the 2nd year becomes the principal at the beginning of the 3rd year.
Outstanding loan = \( ₹ 4,155 \)
Answer: ₹ 4,155

Q8: A man borrows ₹ 5,000 at 12 percent compound interest payable every six months. He repays ₹ 1,800 at the end of every six months. Calculate the third payment he 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 \( (P) = ₹ 5,000 \)
Rate for six months = \( \frac{12\%}{2} = 6\% \)
Interest for the 1st six months = \( 6\% \text{ of } 5,000 = \frac{6 \times 5000}{100} = ₹ 300 \)
Amount = \( 5,000 + 300 = ₹ 5,300 \)
Repayment at the end of 6 months = \( ₹ 1,800 \)
Balance at the end of 1st six months = \( 5,300 – 1,800 = ₹ 3,500 \)
Step 2: Calculate the balance for the second six months (at 12 months).
Principal for the 2nd period = \( ₹ 3,500 \)
Interest for the 2nd six months = \( 6\% \text{ of } 3,500 = \frac{6 \times 3500}{100} = ₹ 210 \)
Amount = \( 3,500 + 210 = ₹ 3,710 \)
Repayment at the end of 12 months = \( ₹ 1,800 \)
Balance at the end of 2nd six months = \( 3,710 – 1,800 = ₹ 1,910 \)
Step 3: Calculate the final payment at the end of 18 months.
Principal for the 3rd period = \( ₹ 1,910 \)
Interest for the 3rd six months = \( 6\% \text{ of } 1,910 = \frac{6 \times 1910}{100} = ₹ 114.60 \)
Total amount due at the end of 18 months = \( 1,910 + 114.60 = ₹ 2,024.60 \)
Step 4: Determine the final payment.
To clear the loan, the third payment must equal the total outstanding amount.
Third payment = \( ₹ 2,024.60 \)
Answer: ₹ 2,024.60

Q9: On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is ₹ 180/-. Find the sum lent out, if the rate of interest in both the cases is 10% per annum.

Step 1: Assume the Principal and calculate Simple Interest (S.I.) for 1 year.
Let the sum (Principal) be \( P \).
Rate \( (R) = 10\% \) per annum, Time \( (T) = 1 \) year.
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{P \times 10 \times 1}{100} = \frac{P}{10} \)
Step 2: Calculate Compound Interest (C.I.) payable half-yearly.
For half-yearly, Rate \( (r) = \frac{10\%}{2} = 5\% \) per half-year.
Time \( (n) = 1 \text{ year} = 2 \text{ half-years} \).
Formula for Amount: \( A = P(1 + \frac{r}{100})^n \)
\( A = P(1 + \frac{5}{100})^2 = P(\frac{21}{20})^2 = \frac{441P}{400} \)
\( \text{C.I.} = A – P = \frac{441P}{400} – P = \frac{41P}{400} \)
Step 3: Equate the difference between C.I. and S.I. to ₹ 180.
Given: \( \text{C.I.} – \text{S.I.} = 180 \)
\( \frac{41P}{400} – \frac{P}{10} = 180 \)
Take L.C.M. of 400 and 10, which is 400:
\( \frac{41P – 40P}{400} = 180 \)
\( \frac{P}{400} = 180 \)
Step 4: Find the Principal \( (P) \).
\( P = 180 \times 400 \)
\( P = 72,000 \)
Answer: ₹ 72,000

Q10: A manufacturer estimates that his machine depreciates by 15% of its value at the beginning of the year. Find the original value (cost) of the machine, if it depreciates by ₹ 5,355 during the second year.

Step 1: Assume the original value of the machine.
Let the original cost of the machine be \( P \).
Rate of depreciation \( (R) = 15\% \) per annum.
Step 2: Calculate the value of the machine at the beginning of the second year.
Value at the end of 1st year = \( P – (15\% \text{ of } P) \)
Value = \( P – 0.15P = 0.85P \)
This value \( (0.85P) \) is the principal for the 2nd year.
Step 3: Calculate the depreciation during the second year.
Depreciation in 2nd year = \( 15\% \text{ of the value at the beginning of 2nd year} \)
Depreciation = \( \frac{15}{100} \times 0.85P \)
Depreciation = \( 0.15 \times 0.85P = 0.1275P \)
Step 4: Equate the calculated depreciation to the given amount.
Given depreciation in 2nd year = \( ₹ 5,355 \)
\( 0.1275P = 5,355 \)
\( P = \frac{5,355}{0.1275} \)
Step 5: Solve for \( P \).
\( P = \frac{53550000}{1275} \)
\( P = 42,000 \)
Answer: ₹ 42,000

Q11: A man borrows ₹ 10,000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year 42% of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt?

Step 1: Calculate the balance at the end of the first year.
Principal \( (P_1) = ₹ 10,000 \)
Interest for the 1st year = \( 5\% \text{ of } 10,000 = \frac{5 \times 10000}{100} = ₹ 500 \)
Amount at the end of the 1st year = \( 10,000 + 500 = ₹ 10,500 \)
Repayment at the end of the 1st year = \( 35\% \text{ of } 10,000 = \frac{35 \times 10000}{100} = ₹ 3,500 \)
Balance at the end of the 1st year = \( 10,500 – 3,500 = ₹ 7,000 \)
Step 2: Calculate the balance at the end of the second year.
Principal for the 2nd year \( (P_2) = ₹ 7,000 \)
Interest for the 2nd year = \( 5\% \text{ of } 7,000 = \frac{5 \times 7000}{100} = ₹ 350 \)
Amount at the end of the 2nd year = \( 7,000 + 350 = ₹ 7,350 \)
Repayment at the end of the 2nd year = \( 42\% \text{ of } 10,000 = \frac{42 \times 10000}{100} = ₹ 4,200 \)
Balance at the end of the 2nd year = \( 7,350 – 4,200 = ₹ 3,150 \)
Step 3: Calculate the final payment at the end of the third year.
Principal for the 3rd year \( (P_3) = ₹ 3,150 \)
Interest for the 3rd year = \( 5\% \text{ of } 3,150 = \frac{5 \times 3150}{100} = ₹ 157.50 \)
Total amount to be paid at the end of the 3rd year = \( 3,150 + 157.50 = ₹ 3,307.50 \)
Answer: ₹ 3,307.50