Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Set 14.2 Solutions

Compound Interest class 8 practice set 14.2

Question 1. On the construction work of a flyover bridge there were 320 workers initially.
The number of workers was increased by 25% every year. Find the number of workers after 2 years.
Solution: Here, P = Initial number of workers = 320 R = Increase in the number of workers per year = 25% N = 2 years A = Number of workers after 2 years \[A = P \left[1 + \frac{R}{100}\right]^N\] \[= 320 \left[1 + \frac{25}{100}\right]^2\] \[= 320 \left[\frac{100+25}{100}\right]^2\] \[= 320 \left[\frac{125}{100}\right]^2\] \[= 320 \left[\frac{25 \times 5}{25 \times 4}\right]^2\] \[= 320 \left[\frac{5}{4}\right]^2\] \[= 320 \left[\frac{25}{16}\right]\] \[= 20 \times 25\] \[= 500\] ∴ The number of workers after 2 years would be 500.
Answer: 500 workers
In simple words: This problem calculates the future number of workers using the compound interest formula, where the initial number of workers increases by a certain percentage each year. The calculation shows a steady growth over two years.

🎯 Exam Tip: Remember to correctly identify the principal (P), rate (R), and number of years (N) before applying the compound interest formula for growth-related problems.

Question 2. A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
Solution: Here, P = Present number of sheeps = 200 R = Increase in number of sheeps per year = 8% N = 3 years A = Number of sheeps after 3 years \[A = P \left[1 + \frac{R}{100}\right]^N\] \[= 200 \left[1 + \frac{8}{100}\right]^3\] \[= 200 \left[\frac{100+8}{100}\right]^3\] \[= 200 \left[\frac{108}{100}\right]^3\] \[= 200 \left[\frac{27 \times 4}{25 \times 4}\right]^3\] \[= 200 \left[\frac{27}{25}\right]^3\] \[= 200 \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25}\] \[= 8 \times \frac{27}{25} \times 27 \times 27\] \[= \frac{0.32}{25} \times 27 \times 27 \times 27\] \[= 0.0128 \times 27 \times 27 \times 27\] \[= 251.9424\] ∴ The number of sheeps with the shepherd after 2 years would be 252 (approx).
Answer: 252 sheep (approximately)
In simple words: This problem calculates the future number of sheep by applying the compound interest formula, considering an annual increase of 8%. The calculation extends over three years, showing the cumulative growth.

🎯 Exam Tip: When dealing with real-world scenarios like population or animal count, it's appropriate to round the final answer to the nearest whole number if the context demands it.

Question 3. In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Solution: Here, P = Present number of trees in the forest = 40,000 R = Increase in the number of trees per year = 5% N = 3 years A = Number of trees after 3 years \[A = P \left[1 + \frac{R}{100}\right]^N\] \[= 40000 \left[1 + \frac{5}{100}\right]^3\] \[= 40000 \left[\frac{100+5}{100}\right]^3\] \[= 40000 \left[\frac{105}{100}\right]^3\] \[= 40000 \left[\frac{21 \times 5}{20 \times 5}\right]^3\] \[= 40000 \left[\frac{21}{20}\right]^3\] \[= 40000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\] \[= 5 \times 21 \times 21 \times 21\] \[= 5 \times 9261\] \[= 46,305\] ∴ The expected number of trees in the forest after 3 years is 46,305.
Answer: 46,305 trees
In simple words: This problem estimates the future tree count in a forest by applying a 5% annual growth rate over three years using the compound interest formula. The final number reflects this consistent increase.

🎯 Exam Tip: Pay close attention to the units (e.g., trees, workers, sheep) and ensure your final answer is expressed clearly with the correct unit.

Question 4. The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.
Solution: Here, P = Cost price of machine = Rs 2,50,000 R = Rate of depreciation per year = 10% N = 2 years A = Depreciated price of the machine after 2 years \[A = P \left[1 - \frac{R}{100}\right]^N\] \[= 2,50,000 \left[1 + \frac{(-10)}{100}\right]^2\] \[= 2,50,000 \left[1 - \frac{10}{100}\right]^2\] \[= 2,50,000 \left[\frac{100-10}{100}\right]^2\] \[= 2,50,000 \left[\frac{90}{100}\right]^2\] \[= 2,50,000 \left[\frac{9}{10}\right]^2\] \[= 2,50,000 \left[\frac{81}{100}\right]\] \[= 2,500 \times 81\] \[= \text{Rs } 2,02,500\] Depreciation in price = Cost price (P) - Depreciated price (A) \[= 2,50,000 - 2,02,500\] \[= \text{Rs } 47,500\] ∴ The depreciation in price of the machine after 2 years would be Rs 47,500.
Answer: Rs 47,500
In simple words: This problem calculates the depreciation of a machine's value over two years at a 10% annual rate. The depreciation is found by subtracting the final value from the initial cost.

🎯 Exam Tip: For depreciation problems, remember to use a minus sign in the compound interest formula \( \left[1 - \frac{R}{100}\right]^N \) to represent a decrease in value.

Question 5. Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.
Solution: Here, A = Rs 4036.80, R = 16 p.c.p.a. and N = 2 years
(i) Find Principal (P): \[A = P \left[1 + \frac{R}{100}\right]^N\] \[\text{∴ } 4036.80 = P \left[1 + \frac{16}{100}\right]^2\] \[\text{∴ } 4036.80 = P \left[\frac{100+16}{100}\right]^2\] \[\text{∴ } 4036.80 = P \left[\frac{116}{100}\right]^2\] \[\text{∴ } 4036.80 = P \left[\frac{29 \times 4}{25 \times 4}\right]^2\] \[\text{∴ } 4036.80 = P \left[\frac{29}{25}\right]^2\] \[\text{∴ } 4036.80 = P \times \frac{29}{25} \times \frac{29}{25}\] \[\text{∴ } P = \frac{4036.80 \times 25 \times 25}{29 \times 29}\] \[\text{∴ } P = \frac{2523000}{841}\] \[\text{∴ } P = 3000\] \[\text{∴ } P = \text{Rs } 3000\]
(ii) Interest = Amount (A) - Principal (P) \[= 4036.80 - 3000\] \[= \text{Rs } 1036.80\] ∴ The compound interest after 2 years would be Rs 1036.80.
Answer: Rs 1036.80
In simple words: This problem first requires finding the initial principal amount, given the final amount, rate, and time. Once the principal is determined, the compound interest is calculated by subtracting the principal from the total amount.

🎯 Exam Tip: When the principal is unknown, rearrange the compound interest formula to solve for P. Ensure all calculations are precise, especially with decimal values.

Question 6. A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Solution: Here, P = Rs 15,000, R = 12 p.c.p.a, and N = 3 years \[A = P \left[1 + \frac{R}{100}\right]^N\] \[\text{∴ } A = 15000 \left[1 + \frac{12}{100}\right]^3\] \[= 15000 \left[\frac{100+12}{100}\right]^3\] \[= 15000 \left[\frac{112}{100}\right]^3\] \[= 15000 \left[\frac{28 \times 4}{25 \times 4}\right]^3\] \[= 15000 \left[\frac{28}{25}\right]^3\] \[= 15000 \times \frac{28}{25} \times \frac{28}{25} \times \frac{28}{25}\] \[= \frac{24 \times 28 \times 28 \times 28}{25}\] \[= 0.96 \times 21952\] \[= \text{Rs } 21,073.92\] ∴ The amount required to settle the loan after 3 years is Rs 21,073.92.
Answer: Rs 21,073.92
In simple words: This problem calculates the total amount to be repaid for a loan after three years, considering compound interest at an annual rate of 12%. The calculation uses the standard compound interest formula for future value.

🎯 Exam Tip: Double-check your arithmetic, especially when dealing with higher powers (N=3 in this case), to avoid calculation errors.

Question 7. A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Solution: Here, A = Rs 13,924, R = 18 p.c.p.a., and N = 2 years \[A = P \left[1 + \frac{R}{100}\right]^N\] \[\text{∴ } 13924 = P \left[1 + \frac{18}{100}\right]^2\] \[\text{∴ } 13924 = P \left[\frac{100+18}{100}\right]^2\] \[\text{∴ } 13924 = P \left[\frac{118}{100}\right]^2\] \[\text{∴ } 13924 = P \left[\frac{59 \times 2}{50 \times 2}\right]^2\] \[\text{∴ } 13924 = P \left[\frac{59}{50}\right]^2\] \[\text{∴ } 13924 = P \times \frac{59}{50} \times \frac{59}{50}\] \[\text{∴ } P = \frac{13924 \times 50 \times 50}{59 \times 59}\] \[\text{∴ } P = \frac{236 \times 50 \times 50}{59 \times 1}\] \[\text{∴ } P = 4 \times 50 \times 50\] \[\text{∴ } P = \text{Rs } 10,000\] The principal is Rs 10,000.
Answer: Rs 10,000
In simple words: This problem involves working backward from a given final amount, rate, and time to find the original principal amount. It demonstrates the application of the compound interest formula in reverse.

🎯 Exam Tip: When finding the principal (P), ensure you correctly isolate P by multiplying by the reciprocal of the compounded factor. Simplification of fractions before squaring can make calculations easier.

Question 8. The population of a suburb is 16,000. Find the rate of increase in the population after two years is 17,640.
Solution: Here, P = Population of a suburb = 16,000 N = 2 years A = Increase in the population after 2 years = 17,640 R = Rate of increase in population \[A = P \left[1 + \frac{R}{100}\right]^N\] \[\text{∴ } 17640 = 16000 \left[1 + \frac{R}{100}\right]^2\] \[\text{∴ } \frac{17640}{16000} = \left[1 + \frac{R}{100}\right]^2\] \[\text{∴ } \frac{1764}{1600} = \left[1 + \frac{R}{100}\right]^2\] \[\text{∴ } \frac{441 \times 4}{400 \times 4} = \left[1 + \frac{R}{100}\right]^2\] \[\text{∴ } \frac{441}{400} = \left[1 + \frac{R}{100}\right]^2\] \[\text{∴ } \left[\frac{21}{20}\right]^2 = \left[1 + \frac{R}{100}\right]^2\] \[\text{∴ } \frac{21}{20} = 1 + \frac{R}{100} \text{ [Taking square root on both sides]}\] \[\text{∴ } \frac{21}{20} - 1 = \frac{R}{100}\] \[\text{∴ } \frac{21-20}{20} = \frac{R}{100}\] \[\text{∴ } \frac{1}{20} = \frac{R}{100}\] \[\text{∴ } R = \frac{100}{20}\] \[\text{∴ } R = 5\] i.e., R = 5% ∴ The rate of increase in the population is 5 p.c.p.a.
Answer: 5 p.c.p.a.
In simple words: This problem determines the annual rate of population increase given the initial population, the final population after two years, and the time period. It involves solving the compound interest formula for the rate R.

🎯 Exam Tip: When solving for the rate (R), remember to take the square root (or appropriate root) of both sides of the equation. Also, express the final answer as a percentage.

Question 9. In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.
Solution: Here, P = Rs 700, R = 10 p.c.p.a., A = Rs 847 \[A = P \left[1 + \frac{R}{100}\right]^N\] \[\text{∴ } 847 = 700 \left[1 + \frac{10}{100}\right]^N\] \[\text{∴ } 847 = 700 \left[\frac{100+10}{100}\right]^N\] \[\text{∴ } 847 = 700 \left[\frac{110}{100}\right]^N\] \[\text{∴ } 847 = 700 \left[\frac{11}{10}\right]^N\] \[\text{∴ } \frac{847}{700} = \left[\frac{11}{10}\right]^N\] \[\text{∴ } \frac{121 \times 7}{100 \times 7} = \left[\frac{11}{10}\right]^N\] \[\text{∴ } \frac{121}{100} = \left[\frac{11}{10}\right]^N\] \[\text{∴ } \left[\frac{11}{10}\right]^2 = \left[\frac{11}{10}\right]^N \text{ [If } a^N = a^m, \text{ then } N=m\text{]}\] \[\text{∴ } N = 2\] ∴ Rs 700 will amount to Rs 847 in 2 years.
Answer: 2 years
In simple words: This problem asks for the time period (N) it takes for a principal amount to reach a certain final amount at a given compound interest rate. It requires solving the compound interest formula for N, often by comparing powers.

🎯 Exam Tip: When solving for N, simplify the fractions to their lowest common base. Recognizing perfect squares or cubes will help in equating the powers to find N.

Question 10. Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a. Solution: Here, P = Rs 20,000, R = 8 p.c.p.a., N = 2 years
(i) Simple interest (I) \[I = \frac{PNR}{100}\] \[\text{∴ } I = \frac{20,000 \times 2 \times 8}{100} = \text{Rs } 3200\]
(ii) Compound Interest (I): \[A = P \left[1 + \frac{R}{100}\right]^N\] \[= 20000 \left[1 + \frac{8}{100}\right]^2\] \[= 20000 \left[\frac{100+8}{100}\right]^2\] \[= 20000 \left[\frac{108}{100}\right]^2\] \[= 20000 \left[\frac{27 \times 4}{25 \times 4}\right]^2\] \[= 20000 \left[\frac{27}{25}\right]^2\] \[= 20000 \times \frac{27}{25} \times \frac{27}{25}\] \[= 32 \times 27 \times 27\] \[= \text{Rs } 23,328\] Compound interest (I) = Amount (A) - Principal (P) \[= 23,328 - 20,000\] \[= \text{Rs } 3328 \text{...(ii)}\]
(iii) Difference = Compound interest - Simple interest \[= 3328 - 3200 \text{ [Form (i) and (ii)]}\] \[= \text{Rs } 128\] ∴ The difference between compound interest and simple interest is Rs 128. [Note: The question is modified as per the answer given in the textbook.]
Answer: Rs 128
In simple words: This problem calculates both the simple interest and compound interest for the same principal, rate, and time, then finds the difference between the two. It highlights how compound interest yields more interest than simple interest over time.

🎯 Exam Tip: Clearly separate the calculations for simple interest and compound interest before finding their difference. This minimizes confusion and helps in identifying any calculation errors.

Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Practice Set 14.2 Intext Questions and Activities

Question 1. Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class. (Textbook pg. no. 90) Solution: (Students should attempt this activity at their own.)
Answer: Students should attempt this activity at their own.
In simple words: This is an activity requiring students to research real-world banking schemes and interest rates and present their findings visually.

🎯 Exam Tip: For activity-based questions, focus on thorough research and clear presentation. A well-organized chart showing different schemes and their interest rates will fetch good evaluation.