OP Malhotra Class 9 Maths Solutions Chapter 2 Compound Interest Exercise 2 (A)

Question 1. Find the compound interest in the following: Principal = ₹ 10000, Rate % p.a = 12 %, Number of years = 2.
Answer:
Rate (R) = 12%
Period (T) = 2 years
Interest for the first year = \( \frac { PRT }{ 100 } \)
\( \implies \frac{10000 \times 12 \times 1}{100} \)
\( = \text{₹} 1200 \)
Amount after first year = P + S.I.
\( = \text{₹} 10000 + \text{₹} 1200 = \text{₹} 11200 \)
Principal for the second year = \( \text{₹} 11200 \)
Interest for the second year
\( = \frac{11200 \times 12 \times 1}{100} = \text{₹} 1344 \)
\( \implies \) Interest for 2 years = \( \text{₹} 1200 + \text{₹} 1344 = \text{₹} 2544 \)
Hence compound interest = \( \text{₹} 2544 \)
In simple words: Compound interest is calculated year by year, where the interest from the previous year is added to the principal to form the new principal for the next year.

🎯 Exam Tip: Remember to calculate the amount after each year to use as the principal for the next year's interest calculation in compound interest problems.

Question 2. Principal = ₹50000, Rate % p.a = 10 %, Number of years = 2.
Answer:
Principal (P) = ₹ 50000
Rate (R) = 10% p.a.
Period (T) = 2 years
\( \implies \) Interest after first year = \( \frac { PRT }{ 100 } \)
\( = \frac{50000 \times 10 \times 1}{100} = \text{₹} 5000 \)
\( \implies \) Amount after first year = P + S.I. = \( \text{₹} 50000 + \text{₹} 5000 = \text{₹} 55000 \)
\( \implies \) Principal for the second year = \( \text{₹} 55000 \)
Interest for the second year
\( = \frac{55000 \times 10 \times 1}{100} = \text{₹} 5500 \)
\( \implies \) Interest for 2 years = \( \text{₹} 5000 + \text{₹} 5500 = \text{₹} 10500 \)
Hence compound interest for 2 years = \( \text{₹} 10500 \)
In simple words: The interest earned in the first year is added to the initial principal, and this new, larger amount becomes the principal for calculating interest in the second year.

🎯 Exam Tip: Be careful with the initial principal given in the question; sometimes there's a typo in the provided solution which may vary the final answer slightly.

Question 3. Principal (P) = ₹ 2800, Rate (R) = 10% p.a., Number of years = \( 1\frac { 1 }{ 2 } \).
Answer:
Principal (P) = ₹ 2800
Rate (R) = 10% p.a.
Period (T) = \( 1\frac { 1 }{ 2 } \) years
Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{2800 \times 10 \times 1}{100} \)
\( = \text{₹} 280 \)
\( \implies \) Amount after first year = P + S.I.
\( = \text{₹} 2800 + \text{₹} 280 = \text{₹} 3080 \)
Principal for the second year = \( \text{₹} 3080 \)
Interest for the next 6 months (\( \frac { 1 }{ 2 } \) years)
\( = \frac{3080 \times 10 \times 1}{100 \times 2} = \text{₹} 154 \)
Total interest for \( 1\frac { 1 }{ 2 } \) years = \( \text{₹} 280 + \text{₹} 154 = \text{₹} 434 \)
Hence compound interest for \( 1\frac { 1 }{ 2 } \) years = \( \text{₹} 434 \)
In simple words: When the time period is a mixed fraction like 1.5 years, calculate the interest for the full year first, then for the fractional part using the new principal.

🎯 Exam Tip: For periods involving fractions of a year, calculate interest year-by-year, and then calculate interest for the fractional period using the adjusted principal.

Question 4. Principal = ₹ 2000, Rate % p.a = 20 %, Number of years = 2.
Answer:
Principal (P) = ₹ 2000
Rate (R) = 20% p.a.
Period (T) = 2 years
\( \implies \) Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{2000 \times 20 \times 1}{100} \)
\( = \text{₹} 400 \)
Amount after one year = P + S.I.
\( = \text{₹} 2000 + \text{₹} 400 = \text{₹} 2400 \)
or principal for the second year = \( \text{₹} 2400 \)
Interest for the second year
\( = \frac{2400 \times 20 \times 1}{100} \)
\( = \text{₹} 480 \)
\( \implies \) Total interest for 2 years = \( \text{₹} 400 + \text{₹} 480 = \text{₹} 880 \)
Hence compound interest for 2 years = \( \text{₹} 880 \)
In simple words: The compound interest for two years is the sum of interest from the first year on the initial principal and the interest from the second year on the new principal (initial principal + first year's interest).

🎯 Exam Tip: Clearly show the calculation of principal for each successive year to avoid errors in compound interest problems.

Question 5. Principal = ₹ 20480, Rate % p.a = \( 6\frac { 1 }{ 4 } \)%, Number of years = 2 years 73 days.
Answer:
Principal (P) = ₹ 20480
Rate (R) = \( 6\frac { 1 }{ 4 } \) % = \( \frac { 25 }{ 4 } \) % p.a.
Period (T) = 2 years 73 days
Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{20480 \times 25 \times 1}{100 \times 4} \)
\( = \text{₹} 1280 \)
Amount after one year = P + S.I.
\( = \text{₹} 20480 + \text{₹} 1280 = \text{₹} 21760 \)
or principal for the second year = ₹ 21760
Interest for the second year
\( = \frac{21760 \times 25 \times 1}{4 \times 100} = \text{₹} 1360 \)
\( \implies \) Amount after 2 years = \( \text{₹} 21760 + \text{₹} 1360 = \text{₹} 23120 \)
Principal for the next \( \frac { 1 }{ 5 } \) year (73 days = \( \frac{73}{365} = \frac{1}{5} \) year)
\( = \text{₹} 23120 \)
Interest for \( \frac { 1 }{ 5 } \) year = \( \frac{23120 \times 25 \times 1}{100 \times 4 \times 5} \)
\( = \text{₹} 289 \)
\( \implies \) Total interest for \( 2\frac { 1 }{ 5 } \) years = \( \text{₹} 1280 + \text{₹} 1360 + \text{₹} 289 = \text{₹} 2929 \)
In simple words: When dealing with a period that includes days, convert the days into a fraction of a year (e.g., 73 days = 73/365 year) and calculate the interest for that fractional period separately.

🎯 Exam Tip: Convert days to a fraction of a year carefully (e.g., 73 days / 365 days = 1/5 year) before applying the interest formula for the fractional period.

Question 6. Find the amount and compound interest on a sum of ₹ 15625 at 4% per annum for 3 years compounded annually.
Answer:
Principal (P) = ₹ 15625
Rate (R) = 4% p.a.
Period (T) = 3 years
Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{15625 \times 4 \times 1}{100} = \text{₹} 625 \)
\( \implies \) Amount after one year P + S.I. = \( \text{₹} 15625 + \text{₹} 625 = \text{₹} 16250 \)
or principal for the second year = ₹ 16250
Interest for the second year
\( = \frac{16250 \times 4 \times 1}{100} = \text{₹} 650 \)
Amount after second year = \( \text{₹} 16250 + 650 = \text{₹} 16900 \)
or principal for the third year = \( \text{₹} 16900 \)
Interest for the third year = \( \frac{16900 \times 4 \times 1}{100} \)
\( = \text{₹} 676 \)
Amount after third year = \( \text{₹} 16900 + \text{₹} 676 = \text{₹} 17576 \)
and compound interest = A \( - \) P = \( \text{₹} 17576 - \text{₹} 15625 = \text{₹} 1951 \).
In simple words: To find compound interest over multiple years, calculate the interest for each year, add it to the principal, and use that new amount as the principal for the next year.

🎯 Exam Tip: Always clearly state the principal, rate, and time for each year's calculation, especially when compounding for multiple years.

Question 7. To renovate his shop, Anurag obtained a loan of ₹ 8000 from a bank. If the rate of interest at 5% per annum is compounded annually, calculate the compound interest that Anurag will have to pay after 3 years.
Answer:
Amount of loan (Principal) (P) = ₹ 8000
Rate of interest (R) = 5% p.a.
Period (T) = 3 years
\( \implies \) Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{8000 \times 5 \times 1}{100} = \text{₹} 400 \)
Amount after one year = P + S.I.
\( = \text{₹} 8000 + \text{₹} 400 = \text{₹} 8400 \)
or principal for the second year = Rs. 8400
Interest for the second year
\( = \frac{8400 \times 5 \times 1}{100} \)
\( = \text{₹} 420 \)
Amount after two years = \( \text{₹} 8400 + \text{₹} 420 = \text{₹} 8820 \)
or principal for the third year = ₹ 8820
Interest for the third year = \( \frac{8820 \times 5 \times 1}{100} \)
\( = \text{₹} 441 \)
\( \implies \) Amount after 3 years = \( \text{₹} 8820 + \text{₹} 441 = \text{₹} 9261 \)
\( \implies \) C.I. = A \( - \) P = \( 9261 - 8000 = \text{₹} 1261 \)
In simple words: Anurag's compound interest is calculated by adding the interest earned each year to the principal, causing the interest to grow faster over time.

🎯 Exam Tip: For word problems, correctly identify the principal, rate, and time before starting calculations to avoid errors.

Question 8. Maria's investment (P) = ₹ 93750, Rate of interest (R) = 9.6%, Period (T) = 3 years, compounded annually. Calculate.
(i) the amount standing to her credit at the end of the second year.
(ii) the interest for the 3rd year.
Answer:
Maria's investment (P) = ₹ 93750
Rate of interest (R) = 9.6%
Period (T) = 3 years
Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{93750 \times 9.6 \times 1}{100} \)
\( = \frac{93750 \times 96 \times 1}{100 \times 10} \)
\( = \text{₹} 9000 \)
Amount after one year = P + S.I.
\( = \text{₹} 93750 + 9000 = \text{₹} 102750 \)
or principal for the second year = ₹ 102750
Interest for the second year
\( = \frac{102750 \times 9.6 \times 1}{100} \)
\( = \frac{102750 \times 96 \times 1}{100 \times 10} \)
\( = \text{₹} 9864 \)
(i) \( \implies \) Amount after 2 years = \( \text{₹} 102750 + \text{₹} 9864 = \text{₹} 112614 \)
or Principal for the third year = ₹ 112614
(ii) \( \implies \) Interest for the third year = \( \frac{112614 \times 9.6 \times 1}{100} \)
\( = \frac{112614 \times 96 \times 1}{100 \times 10} = \text{₹} 10810.94 \)
In simple words: This problem requires calculating the amount and interest year by year, using the previous year's total as the new principal, and remembering to convert decimal rates for easier calculation.

🎯 Exam Tip: Pay attention to decimal rates (e.g., 9.6%); convert them to fractions for easier calculation if needed (9.6 = 96/10) to avoid errors.

Question 9. A sum of ₹ 9,600 is invested for 3 years at 10% p.a. compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year ?
(iii) Find the compound interest earned in the first 2 years.
(iv) Find the compound interest at the end of 3 years.
Answer:
Rate of interest (R) = 10% p.a.
Period (T) = 3 years
Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{9600 \times 10 \times 1}{100} = \text{₹} 960 \)
(i) \( \implies \) Amount after the first year = P + S.I. = \( \text{₹} 9600 + \text{₹} 960 = \text{₹} 10560 \)
(ii) Principal for the second year = ₹ 10560
Interest for the second year = \( \frac{10560 \times 10 \times 1}{100} \)
\( = \text{₹} 1056 \)
Amount after the second year = \( \text{₹} 10560 + \text{₹} 1056 = \text{₹} 11616 \)
(iii) Compound interest for two years = \( \text{₹} 960 + \text{₹} 1056 = \text{₹} 2016 \)
Principal for the third year = ₹ 11616
Interest for the third year = \( \frac{11616 \times 10 \times 1}{100} \)
\( = \text{₹} 1161.60 \)
(iv) Compound interest after the end of third year = \( \text{₹} 2016 + \text{₹} 1161.60 = \text{₹} 3177.60 \)
In simple words: This question requires step-by-step calculation of amount and interest for each year, showing how the interest builds up over three years.

🎯 Exam Tip: Break down multi-part compound interest questions into annual calculations; this helps to systematically find the amount due and total interest for each requested period.

Question 10. Shanker takes a loan of ₹ 10,000 at a compound interest rate of 10% per annum (p.a.)
(i) Find the interest after one year.
(ii) Find the compound interest for 2 years.
(iii) Find the sum of money required to clear the debt at the end of 2 years.
(iv) Find the difference between the compound interest and the simple interest at the same rate for 2 years.
Answer:
Amount of loan taken by Shanker (P) = ₹ 10000
Rate of interest (R) = 10% p.a.
Period (T) = 2 years
(i) Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{10000 \times 10 \times 1}{100} = \text{₹} 1000 \)
Amount after one year = P + S.I. = ₹ 11000
Principal for the second year = ₹ 11000
Interest for the second year
\( = \frac{11000 \times 10 \times 1}{100} = \text{₹} 1100 \)
(ii) \( \implies \) C.I. for 2 years = \( \text{₹} 1000 + \text{₹} 1100 = \text{₹} 2100 \)
(iii) Amount after 2 years = \( \text{₹} 11000 + \text{₹} 1100 = \text{₹} 12100 \)
(iv) Simple interest for 2 years
\( = \frac{10000 \times 10 \times 2}{100} = \text{₹} 2000 \)
Now difference between C.I. and S.I. for 2 years
\( = 2100 - \text{₹} 2000 = \text{₹} 100 \)
In simple words: This problem compares compound interest, where interest grows on previous interest, with simple interest, where interest is only on the original principal, showing their difference over time.

🎯 Exam Tip: When comparing CI and SI for the same period, calculate both separately, remembering that SI is always based on the original principal.

Question 11. Find compound interest on ₹ 5000 at 12% p.a. for 1 year, compounded half yearly.
Answer:
Principal (P) = ₹ 5000
Rate (R) = 12% p.a. or 6% half yearly
Period (T) = 1 year or 2 half years
Interest for the first half year = \( \frac { PRT }{ 100 } \)
\( = \frac{5000 \times 6 \times 1}{100} = \text{₹} 300 \)
Amount after one half year = P + S.I.
\( = \text{₹} 5000 + 300 = \text{₹} 5300 \)
Principal for the second half year = ₹ 5300
Interest for the second half year
\( = \frac{5300 \times 6 \times 1}{100} = \text{₹} 318 \)
\( \implies \) Amount after 2 half years = \( \text{₹} 5300 + \text{₹} 318 = \text{₹} 5618 \)
and C.I. for 2 half years = Rs. A \( - \) P = \( \text{₹} 5618 - \text{₹} 5000 = \text{₹} 618 \)
In simple words: When interest is compounded half-yearly, you divide the annual rate by two and multiply the number of years by two to find the effective rate and periods.

🎯 Exam Tip: For half-yearly compounding, halve the annual rate and double the number of years before applying the simple interest formula for each period.

Question 12. Find the amount and the compound interest on ₹ 16000 for \( 1\frac { 1 }{ 2 } \) years at 10% p.a. the interest being compounded half-yearly.
Answer:
Rate (R) = 10% or 5% half yearly
Period (T) = \( 1\frac { 1 }{ 2 } \) years or 3 half years
Interest for the first half year = \( \frac { PRT }{ 100 } \)
\( = \frac{16000 \times 5 \times 1}{100} = \text{₹} 800 \)
Amount after first half year = P + S.I. = \( \text{₹} 16000 + \text{₹} 800 = \text{₹} 16800 \)
or Principal for the second half year = ₹ 16800
Interest for the second half year
\( = \frac{16800 \times 5 \times 1}{100} = \text{₹} 840 \)
Amount after 2 half years = \( \text{₹} 16800 + \text{₹} 840 = \text{₹} 17640 \)
or Principal for the third half year = ₹ 17640
Interest for the third half year
\( = \frac{17640 \times 5 \times 1}{100} = \text{₹} 882 \)
Amount after third half year = \( \text{₹} 17640 + \text{₹} 882 = \text{₹} 18522 \)
\( \implies \) C.I. for 3 half years = A \( - \) P = \( \text{₹} 18522 - 16000 = \text{₹} 2522 \)
In simple words: For half-yearly compounding over a mixed time period, convert everything to half-year terms: the rate becomes half the annual rate, and the time becomes double the number of years.

🎯 Exam Tip: Always adjust both the rate and the time period correctly when interest is compounded more frequently than annually (e.g., half-yearly, quarterly).

Question 13. Calculate the amount due and the compound interest on ₹ 40000 for 2 years when the rate of interest successive years is 7% and 8% respectively.
Answer:
Principal (P) = ₹ 40000
Period (T) = 2 years
Rate of interest for the first year (\( R_1 \)) = 7%
and for the second year (\( R_2 \)) = 8%
\( \implies \) Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{40000 \times 7 \times 1}{100} = \text{₹} 2800 \)
Amount after first year = P + S.I.
\( = \text{₹} 40000 + \text{₹} 2800 = \text{₹} 42800 \)
Principal for the second year = ₹ 42800
Interest for the second year = \( \frac{42800 \times 8 \times 1}{100} \)
\( = \text{₹} 3424 \)
\( \implies \) Amount after 2 years = \( \text{₹} 42800 + \text{₹} 3424 = \text{₹} 46224 \)
and compound interest for 2 years = A \( - \) P = \( \text{₹} 46224 - 40000 = \text{₹} 6224 \)
In simple words: When interest rates change each year, calculate the interest for each year separately using the specific rate for that year and the updated principal.

🎯 Exam Tip: If the interest rate varies for successive years, ensure you use the correct rate for each period's calculation, always with the updated principal.

Question 14. If the simple interest on a sum of money for 2 years at 5% per annum is ₹ 50, what will be the compound interest on the same sum at the same rate for the same time.
Answer:
S. Interest for 2 years = ₹ 50
Period (T) = 2 years
Rate (R) = 5% p.a.
We know that S.I. and C.I. is same for the first year.
\( \implies \) S. Interest for the first year = \( \frac { 50 }{ 2 } = \text{₹} 25 \)
and S.I. for the second year = ₹ 25
\( \implies \) Principal (P) = \( \frac{\text{S.I.} \times 100}{R \times T} \)
\( = \frac{50 \times 100}{5 \times 2} = \text{₹} 500 \)
Now amount after first year = P + S.I.
\( = \text{₹} 500 + 25 = \text{₹} 525 \)
Principal for the second year = ₹ 525
\( \implies \) Interest for the second year
\( = \frac{525 \times 5 \times 1}{100}=\frac{2625}{100} = \text{₹} 26.25 \)
C.I. for 2 years = \( \text{₹} 25 + 26.25 = \text{₹} 51.25 \)
In simple words: First, use the simple interest to find the original amount (principal), then use that principal to calculate the compound interest for the same period.

🎯 Exam Tip: Always remember the formula for calculating principal from simple interest (\( P = \frac{SI \times 100}{R \times T} \)) when it's not directly given.

Question 15. A man invests ₹46,875 at 4% per annum compound interest for 3 years. Calculate :
(i) the interest for the 1st year;
(ii) the amount standing to his credit at the end of the 2nd year;
(iii) the interest for the 3rd year.
Answer:
Investment (P) = ₹ 46875
Rate (R) = 4% p.a.
Period (T) = 3 years
(i) \( \implies \) Interest for the first year = \( \frac { PRT }{ 100 } \)
\( = \frac{46875 \times 4 \times 1}{100} = \text{₹} 1875 \)
Amount after one year = P + S.I.
\( = \text{₹} 46875 + \text{₹} 1875 = \text{₹} 48750 \)
Principal for the second year = ₹ 48750
Interest for the second year = \( \frac{48750 \times 4 \times 1}{100} \)
\( = \text{₹} 1950 \)
(ii) \( \implies \) Amount after second year = \( \text{₹} 48750 + \text{₹} 1950 = \text{₹} 50700 \)
(iii) Principal for the third year = ₹ 50700
Interest for the third year = \( \frac{50700 \times 4 \times 1}{100} \)
\( = \text{₹} 2028 \)
In simple words: This problem involves a multi-step calculation, finding the interest for each year and then the amount accumulated, especially at the end of the second year.

🎯 Exam Tip: Clearly label each step (e.g., "Interest for 1st year," "Amount after 2nd year") to make your solution easy to follow and ensure all sub-parts of the question are addressed.