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

Calculate the amount and the compound interest by using the formula for compound interest :

Question 1. Rate of interest = 4 % p.a., Time (in years) = 2.
Answer:
Principal (P) = ₹ 625
Rate (R) = 4% p.a.
Period (n) = 2 years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 625 \left(1 + \frac{4}{100}\right)^2 \)
= \( ₹ 625 \left(1 + \frac{1}{25}\right)^2 = 625 \left(\frac{26}{25}\right)^2 \)
= \( ₹ 625 \times \frac{26}{25} \times \frac{26}{25} = ₹ 676 \)
And C.I. = A - P = \( ₹ 676 - 625 = ₹ 51 \)
In simple words: Compound interest calculates interest on the initial principal and also on the accumulated interest from previous periods. We find the total amount first and then subtract the principal to get the interest.

🎯 Exam Tip: Remember to express the rate of interest in decimal form or fraction before applying the compound interest formula, and simplify the fraction inside the parentheses for easier calculation.

Question 2. Principal = 8000, Rate of interest = 15 % p.a., Time (in years) = 3.
Answer:
Principal (P) = ₹ 8000
Rate (R) = 15% p.a.
Period (n) = 3 years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 8000 \left(1 + \frac{15}{100}\right)^3 \)
= \( ₹ 8000 \left(\frac{23}{20}\right)^3 \)
= \( ₹ 8000 \times \frac{23}{20} \times \frac{23}{20} \times \frac{23}{20} = ₹ 12167 \)
\( \implies \) C.I. = A - P = \( ₹ 12167 - 8000 = ₹ 4167 \)
In simple words: The compound interest is calculated over three years, where each year's interest is added to the principal for the next year's calculation. The total amount accumulated after 3 years is ₹ 12167, leading to a compound interest of ₹ 4167.

🎯 Exam Tip: When the time period is given in years, ensure the rate is also per annum. If the rate is half-yearly or quarterly, adjust the rate and time accordingly before calculation.

Question 3. Principal = ₹ 1000, Rate of interest = 10 % p.a., Time (in years) = 3.
Answer:
Principal (P) = ₹ 1000
Rate (R) = 10% p.a.
Period (n) = 3 years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 1000 \left(1 + \frac{10}{100}\right)^3 \)
= \( ₹ 1000 \left(\frac{11}{10}\right)^3 \)
= \( ₹ 1000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} = ₹ 1331 \)
\( \implies \) C.I. = A - P = \( ₹ 1331 - 1000 = ₹ 331 \)
In simple words: For a principal of ₹ 1000 at a 10% annual rate compounded for 3 years, the final amount becomes ₹ 1331, meaning ₹ 331 was earned as interest.

🎯 Exam Tip: Always write down the given values for P, R, and n clearly. This helps in correctly substituting them into the formula and avoiding errors.

Question 4. Principal = 8000, Rate of interest = 10 % half yearly, Time (in years) = \( 1\frac{1}{2} \).
Answer:
Principal (P) = ₹ 8000
Rate (R) = 10% p.a. (half-yearly) or 5% half year
Period (n) = \( 1\frac{1}{2} \) years = 3 half years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 8000 \left(1 + \frac{5}{100}\right)^3 \)
= \( ₹ 8000 \times \left(\frac{21}{20}\right)^3 \)
= \( ₹ 8000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} = ₹ 9261 \)
\( \implies \) C.I. = A - P = \( ₹ 9261 - 8000 = ₹ 1261 \)
In simple words: When interest is compounded half-yearly, the annual rate is halved, and the number of years is doubled to get the number of half-year periods. Calculating with these adjusted values gives the correct total amount and compound interest.

🎯 Exam Tip: For half-yearly compounding, remember to divide the annual rate by 2 and multiply the number of years by 2 to get the correct period (n) and rate (R) for the formula.

Question 5. Principal = ₹ 700, Rate of interest = 20 % half yearly, Time (in years) = \( 1\frac{1}{2} \).
Answer:
Principal (P) = ₹ 700
Rate (R) = 20% p.a. (half yearly) or 10% half yearly
Period (n) = \( 1\frac{1}{2} \) years or 3 half years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 700 \left(1 + \frac{10}{100}\right)^3 \)
= \( ₹ 700 \left(\frac{11}{10}\right)^3 \)
= \( ₹ 700 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \)
= \( ₹ \frac{9317}{10} = ₹ 931.70 \)
And C.I. = A - P = \( ₹ 931.70 - 700 = ₹ 231.70 \)
In simple words: We adjust the annual interest rate and time for half-yearly compounding, then calculate the total amount and compound interest. The final interest earned after 1.5 years is ₹ 231.70.

🎯 Exam Tip: Pay close attention to whether the interest is compounded annually, half-yearly, or quarterly, as this affects the values of R and n in the formula.

Question 6. ₹ 40960 to Amar to purchase a shop at 12.5% per annum. If the interest is compounded semi-annually, find the interest paid by Amar after \( 1\frac{1}{2} \) years.
Answer:
Amount of loan (P) = ₹ 40960
Rate (R) = 12.5% = \( 12\frac{1}{2}\% = \frac{25}{2}\% \) p.a.
= \( \frac{25}{4}\% \) semi annually
Period (n) = \( 1\frac{1}{2} \) years or 3 half years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 40960 \left(1 + \frac{25}{4 \times 100}\right)^3 \)
= \( ₹ 40960 \left(\frac{17}{16}\right)^3 \)
= \( ₹ 40960 \times \frac{17}{16} \times \frac{17}{16} \times \frac{17}{16} \)
= \( ₹ 49130 \)
\( \implies \) C. Interest = A - P
= \( ₹ 49130 - 40960 = ₹ 8170 \)
In simple words: Amar borrowed ₹ 40960, and with semi-annual compounding at a rate of 12.5% per annum for 1.5 years, the total amount to be paid back is ₹ 49130, meaning the interest paid is ₹ 8170.

🎯 Exam Tip: Converting mixed fractions and percentages into proper fractions or decimals first can prevent calculation errors, especially when dealing with semi-annual compounding.

Question 7. Sudhir lent ₹ 2000 at compound interest at 10% payable yearly, while Prashant lent ₹ 2000 at compound interest at 10% payable half-yearly. Find the difference in the interest received by Sudhir and Prashant at the end of one year.
Answer:
Incase of Sudhir,
Amount lent (P) = ₹ 2000
Rate (R) = 10% p.a.
Period = 1 year
\( \implies \) Interest = \( \frac{PRT}{100} \)
= \( ₹ \frac{2000 \times 10 \times 1}{100} = ₹ 200 \)
Incase of Prashant,
Principal (P) = ₹ 2000
Rate (R) = 10% p.a. or 5% half-yearly
Period (n) = 1 year or 2 half years
\( \implies \) Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( ₹ 2000 \left(1 + \frac{5}{100}\right)^2 \)
= \( ₹ 2000 \times \frac{21}{20} \times \frac{21}{20} = ₹ 2205 \)
\( \implies \) C. Interest = A - P = \( ₹ 2205 - 2000 = ₹ 205 \)
Difference in their interest paid = \( ₹ 205 - 200 = ₹ 5 \)
In simple words: Sudhir's interest is calculated annually, while Prashant's is calculated semi-annually, resulting in a slightly higher interest for Prashant. The difference between their earnings is ₹ 5.

🎯 Exam Tip: For problems involving different compounding periods, calculate the interest for each case separately and then find the difference. Always adjust the rate and time for the specific compounding frequency.

Question 8. How much will ₹25000 amount to in 2 years at compound interest, if the rates for the successive years be 4 and 5 per cent per year ?
Answer:
Principal (P) = ₹ 25000
Period (n) = 2 years
Rate for the first year (\( R_1 \)) = 4% p.a.
And for the second year (\( R_2 \)) = 5%
\( \implies \) Amount = \( P \left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right) \)
= \( ₹ 25000 \left(1 + \frac{4}{100}\right)\left(1 + \frac{5}{100}\right) \)
= \( ₹ 25000 \times \frac{26}{25} \times \frac{21}{20} = ₹ 27300 \)
In simple words: When interest rates are different for successive years, we apply each year's rate to the principal plus accumulated interest from the previous year. The investment grows to ₹ 27300 after two years.

🎯 Exam Tip: When different rates are given for successive years, calculate the amount for each year individually, using the previous year's amount as the new principal, or use the formula for varying rates directly.

Question 9. Umesh set up a small factory by investing ₹ 40,000. During the first three successive years his profits were 5%, 10% and 15% respectively. If each year the profit was on previous year's capital, calculate his total profit.
Answer:
Investment of Umesh (P) = ₹ 40,000
Period (n) = 3 years
Rate of profit for the first year (\( r_1 \)) = 5%
For second year (\( r_2 \)) = 10%
And for third year (\( r_3 \)) = 15%
\( \implies \) Total amount after 3 years
= \( P \left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right)\left(1 + \frac{r_3}{100}\right) \)
= \( ₹ 40000 \left(1 + \frac{5}{100}\right)\left(1 + \frac{10}{100}\right)\left(1 + \frac{15}{100}\right) \)
= \( ₹ 40000 \times \frac{21}{20} \times \frac{11}{10} \times \frac{23}{20} = ₹ 53130 \)
Profit = A - P = \( ₹ 53130 - 40000 \)
= \( ₹ 13130 \)
In simple words: Umesh's investment of ₹ 40,000 generated profits at increasing rates over three years, with each year's profit added to the capital for the next year. His total capital grew to ₹ 53130, making his total profit ₹ 13130.

🎯 Exam Tip: When dealing with successive profits or depreciations, use the compound interest formula with positive rates for profit and negative rates for depreciation, adjusting for each period's specific rate.

Question 10. Himachal Pradesh State Electricity Board issued in July 1988, 20 year bonds worth ₹ 6.25 crore. The issue price of each bond is ₹ 100 and it carries an annual interest of 11.5%, compounded half-yearly. Jasbir invested ₹ 5000 in these bonds. Find the amount that he gets on maturity of the bonds in 2008. [Given that (1.0575)40 = 9.35869]
Answer:
Jasbir's investment = ₹ 5000
Period (n) = 2008 - 1988 = 20 years = 40 half years
Rate of interest (r) = 11.5% p.a. (half-yearly) or 5.75% half-yearly
\( \implies \) Amount of the bonds after 40 half years
= \( P \left(1 + \frac{r}{100}\right)^n = ₹ 5000 \left(1 + \frac{5.75}{100}\right)^{40} \)
= \( ₹ 5000 (1 + 0.0575)^{40} \)
= \( ₹ 5000 \times (1.0575)^{40} \)
Given that \( (1.0575)^{40} = 9.35869 \)
= \( ₹ 5000 \times 9.35869 = ₹ 46793.45 \)
\( \implies \) Maturity value of the bonds = \( ₹ 46793.45 \)
In simple words: Jasbir invested in bonds that compound interest half-yearly. By calculating the total number of compounding periods and the half-yearly interest rate, we find his investment matures to ₹ 46793.45 after 20 years.

🎯 Exam Tip: For problems involving maturity value with given powers, ensure you correctly identify the rate per period and the total number of periods (n) to match the given power, then substitute the provided value directly.

Question 11. A district contains 64000 inhabitants. If the population increases at the rate of \( 2\frac{1}{2} \) % per annum, find the number of inhabitants at the end of 3 years.
Answer:
Population of a district (P) = 64000
Rate of increase (R) = \( 2\frac{1}{2}\% \) p.a.
Period (n) = 3 years
\( \implies \) Population after 3 years = \( P \left(1 + \frac{R}{100}\right)^n \)
= \( 64000 \left(1 + \frac{2.5}{100}\right)^3 \)
= \( 64000 \left(1 + \frac{5}{2 \times 100}\right)^3 \)
= \( 64000 \times \left(\frac{41}{40}\right)^3 \)
= \( 64000 \times \frac{41}{40} \times \frac{41}{40} \times \frac{41}{40} \)
= \( 64000 \times \frac{68921}{64000} = 68921 \)
In simple words: The population growth is calculated using a compound interest-like formula. Starting with 64,000 inhabitants, a 2.5% annual increase leads to 68,921 inhabitants after 3 years.

🎯 Exam Tip: Population growth problems are similar to compound interest. Convert the percentage rate to a fraction or decimal, and be careful with calculations involving powers for accurate results.

Question 12. The Nagar Palika of a certain city started campaign to kill stray dogs which numbered 1250 in the city. As a result, the population of stray dogs started decreasing at the rate of 20% per month. Calculate the number of stray dogs in the city three months after the campaign started.
Answer:
In a city no. of stray dogs in the beginning (P) = 1250
Rate of decrease (R) = 20% per month
Period (n) = 3 months
\( \implies \) Number of stray dogs after 3 months
= \( P \left(1 - \frac{R}{100}\right)^n = 1250 \left(1 - \frac{20}{100}\right)^3 \)
= \( 1250 \left(\frac{4}{5}\right)^3 \)
= \( 1250 \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = 640 \)
In simple words: When a quantity decreases at a constant rate over time, we use a formula similar to compound interest but with subtraction instead of addition for the rate. After three months, the number of stray dogs reduces from 1250 to 640.

🎯 Exam Tip: For problems involving depreciation or decrease, use a negative sign for the rate (1 - R/100) in the formula. Ensure the rate and period units (e.g., months) match.

Question 13. 8000 blood donors were registered with a charitable hospital. Some student organizations started mobilizing people for this noble cause. As a result, the number of donors registered, increased at the rate of 20% per half year. Find the total number of new registrants during \( 1\frac{1}{2} \) years.
Answer:
No. of blood donors in the beginning (P) = 8000
Rate of Increase = 20% per half year
Period (n) = \( 1\frac{1}{2} \) years or 3 half year
\( \implies \) Increased donors after 3 half years
= \( P \left(1 + \frac{R}{100}\right)^n \)
= \( 8000 \left(1 + \frac{20}{100}\right)^3 \)
= \( 8000 \left(\frac{6}{5}\right)^3 \)
= \( 8000 \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} = 13824 \)
\( \implies \) Net increase in donors = A - P = \( 13824 - 8000 = 5824 \)
In simple words: The number of blood donors grew by 20% every half-year. Over \( 1\frac{1}{2} \) years (three half-year periods), the total number of donors increased from 8000 to 13824, meaning 5824 new donors registered.

🎯 Exam Tip: Similar to compound interest, when growth occurs half-yearly, the rate is applied for each half-year period. Ensure the total number of periods (n) correctly reflects the given time in half-years.