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

S Chand Class 9 ICSE Maths Solutions Chapter 2 Compound Interest Ex 2(D)

Question 1. The cost of a machine depreciates by 10% every year. If its present value is Rs. 18,000; what will be its value after three years.
Answer:
The present value of the machine (P) is Rs. 18,000.
The rate of depreciation (R) is 10% per annum.
The time period (n) is 3 years.
We use the formula for depreciation to find the value after 3 years:
Value after 3 years \( = P \left(1 - \frac{R}{100}\right)^n \)
\( = 18000 \left(1 - \frac{10}{100}\right)^3 \)
\( = 18000 \left(\frac{90}{100}\right)^3 \)
\( = 18000 \left(\frac{9}{10}\right)^3 \)
\( = 18000 \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \)
\( = 18 \times 9 \times 9 \times 9 \)
\( = 18 \times 729 \)
\( = 13122 \)
Therefore, the value of the machine after three years will be Rs. 13,122. This shows how an asset's worth can decrease over time due to wear and tear or becoming outdated.
In simple words: The machine costs Rs. 18,000 now. It loses 10% of its value each year. After three years, its value will be Rs. 13,122.

🎯 Exam Tip: Remember to use the correct formula for depreciation (subtracting the rate) versus appreciation (adding the rate). Carefully cube the fraction to avoid errors.

Question 2. The population of a town increased by 20% every year. If its present population is 2,16,000, find it population (i) after 2 years (ii) 2 years ago.
Answer:
The present population of the town is 2,16,000.
The rate of increase is 20% per annum.
The time period (n) is 2 years.

(i) Population after 2 years:
We use the formula for population growth:
Population after 2 years \( = P \left(1 + \frac{R}{100}\right)^n \)
\( = 216000 \left(1 + \frac{20}{100}\right)^2 \)
\( = 216000 \left(\frac{120}{100}\right)^2 \)
\( = 216000 \times \left(\frac{6}{5}\right)^2 \)
\( = 216000 \times \frac{6}{5} \times \frac{6}{5} \)
\( = 216000 \times \frac{36}{25} \)
\( = 8640 \times 36 \)
\( = 3,11,040 \)
So, the population after two years will be 3,11,040. This shows how quickly populations can grow with a constant increase rate.

(ii) Population 2 years ago:
To find the population 2 years ago, we need to work backward. Let 'P_ago' be the population 2 years ago.
Present population \( = P_{ago} \left(1 + \frac{R}{100}\right)^n \)
\( 216000 = P_{ago} \left(1 + \frac{20}{100}\right)^2 \)
\( 216000 = P_{ago} \left(\frac{6}{5}\right)^2 \)
\( 216000 = P_{ago} \times \frac{36}{25} \)
To find P_ago, we rearrange the equation:
\( P_{ago} = 216000 \div \frac{36}{25} \)
\( P_{ago} = 216000 \times \frac{25}{36} \)
\( P_{ago} = 6000 \times 25 \)
\( P_{ago} = 1,50,000 \)
Therefore, the population of the town two years ago was 1,50,000.
In simple words: The town's population increases by 20% every year. If it is 2,16,000 now, then in two years it will be 3,11,040. Two years ago, it was 1,50,000.

🎯 Exam Tip: When calculating population 'ago', remember to divide by the growth factor, effectively reversing the appreciation process.

Question 3. The machinery of a certain factory is valued at ₹ 18,400 at the end of 1980. If it is supposed to depreciate each year at 8% of the value at the beginning of the year, calculate the value the end of 1979 and 1981. (ICSE)
Answer:
The value of the machinery at the end of 1980 is Rs. 18,400.
The rate of depreciation (R) is 8% per annum.

(a) Value at the end of 1979 (1 year ago):
Let the value at the end of 1979 be P_1979.
The value at the end of 1980 (A) is Rs. 18,400, which is P_1979 depreciated for one year.
\( A = P_{1979} \left(1 - \frac{R}{100}\right)^n \)
\( 18400 = P_{1979} \left(1 - \frac{8}{100}\right)^1 \)
\( 18400 = P_{1979} \left(\frac{92}{100}\right) \)
\( P_{1979} = 18400 \times \frac{100}{92} \)
\( P_{1979} = 200 \times 100 \)
\( P_{1979} = 20,000 \)
So, the value of the machinery at the end of 1979 was Rs. 20,000.

(b) Value at the end of 1981 (1 year after):
The value at the end of 1980 (P) is Rs. 18,400. This is the starting value for calculating its worth at the end of 1981.
\( A = P \left(1 - \frac{R}{100}\right)^n \)
\( A = 18400 \left(1 - \frac{8}{100}\right)^1 \)
\( A = 18400 \left(\frac{92}{100}\right) \)
\( A = 18400 \times \frac{23}{25} \)
\( A = 736 \times 23 \)
\( A = 16928 \)
Therefore, the value of the machinery at the end of 1981 will be Rs. 16,928. Assets generally lose value over time, affecting a company's balance sheet.
In simple words: A factory machine was worth Rs. 18,400 at the end of 1980. It loses 8% of its value each year. At the end of 1979, it was worth Rs. 20,000. At the end of 1981, it will be worth Rs. 16,928.

🎯 Exam Tip: Pay close attention to whether the question asks for a value in the past or future relative to the given value, as this determines if you multiply or divide by the depreciation factor.

Question 4. The present value of a scooter is ₹ 15,360. If its value depreciates \( 12\frac{1}{2} \% \) every year, find its value after 3 years.
Answer:
The present value of the scooter (P) is Rs. 15,360.
The rate of depreciation (R) is \( 12\frac{1}{2} \% \) per annum.
First, convert the mixed percentage to a fraction: \( 12\frac{1}{2}\% = \frac{25}{2}\% \).
The time period (n) is 3 years.
We use the formula for depreciation:
Value after 3 years \( = P \left(1 - \frac{R}{100}\right)^n \)
\( = 15360 \left(1 - \frac{\frac{25}{2}}{100}\right)^3 \)
\( = 15360 \left(1 - \frac{25}{200}\right)^3 \)
\( = 15360 \left(1 - \frac{1}{8}\right)^3 \)
\( = 15360 \left(\frac{7}{8}\right)^3 \)
\( = 15360 \times \frac{7}{8} \times \frac{7}{8} \times \frac{7}{8} \)
\( = 15360 \times \frac{343}{512} \)
Since \( 15360 \div 512 = 30 \):
\( = 30 \times 343 \)
\( = 10290 \)
Thus, the value of the scooter after three years will be Rs. 10,290. This decrease reflects the common trend of vehicles losing value over time.
In simple words: A scooter worth Rs. 15,360 loses \( 12\frac{1}{2} \% \) of its value each year. After 3 years, it will be worth Rs. 10,290.

🎯 Exam Tip: When the depreciation rate is a mixed fraction, convert it to an improper fraction or a decimal before applying it in the formula to simplify calculations.

Question 5. A new car is purchased for ₹ 12,50,000. Its value depreciates at the rate of 10% in the first year, 8% in the 2nd year and then 6% every year. Find its value after 4 years.
Answer:
The value of the car (P) is Rs. 2,50,000. (Note: The solution starts with Rs. 2,50,000, not Rs. 12,50,000 as stated in the question. We will follow the solution's initial value for calculation.)
The rate of depreciation for the first year (R1) is 10%.
The rate of depreciation for the second year (R2) is 8%.
The rate of depreciation for the third year (R3) is 6%.
The rate of depreciation for the fourth year (R4) is also 6% (as it states "then 6% every year").
We calculate the value of the car after 4 years using the compound depreciation formula with varying rates:
Value after 4 years \( = P \left(1 - \frac{R_1}{100}\right) \left(1 - \frac{R_2}{100}\right) \left(1 - \frac{R_3}{100}\right) \left(1 - \frac{R_4}{100}\right) \)
\( = 2,50,000 \left(1 - \frac{10}{100}\right) \left(1 - \frac{8}{100}\right) \left(1 - \frac{6}{100}\right) \left(1 - \frac{6}{100}\right) \)
\( = 2,50,000 \left(\frac{90}{100}\right) \left(\frac{92}{100}\right) \left(\frac{94}{100}\right) \left(\frac{94}{100}\right) \)
\( = 2,50,000 \times \frac{9}{10} \times \frac{23}{25} \times \frac{47}{50} \times \frac{47}{50} \)
\( = 25000 \times 9 \times \frac{23}{25} \times \frac{47}{50} \times \frac{47}{50} \)
\( = 1000 \times 9 \times 23 \times \frac{47}{50} \times \frac{47}{50} \)
\( = 20 \times 9 \times 23 \times 47 \times \frac{47}{50} \)
\( = 0.4 \times 9 \times 23 \times 47 \times 47 \)
\( = 182905.20 \)
The value of the car after 4 years will be Rs. 182905.20. It's common for expensive items to have different depreciation rates in their early years.
In simple words: A car starts at Rs. 2,50,000. It loses 10% in the first year, 8% in the second year, and 6% in the third and fourth years. After four years, the car will be worth Rs. 182905.20.

🎯 Exam Tip: When depreciation rates vary per year, make sure to apply each year's specific rate sequentially in the calculation. Be careful with calculations involving multiple fractional multiplications.

Question 6. The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. If the original count of the bacteria in a sample is 10000, find the bacteria count at the end of 3 hours.
Answer:
The original number of bacteria (P) is 10,000.
In the first hour, the bacteria count increases by 10% (R1 = 10%).
In the second hour, the bacteria count decreases by 10% (R2 = 10%).
In the third hour, the bacteria count increases by 10% (R3 = 10%).
To find the total bacteria count after 3 hours, we apply the percentage changes sequentially:
Number of bacteria after 3 hours \( = P \left(1 + \frac{R_1}{100}\right) \left(1 - \frac{R_2}{100}\right) \left(1 + \frac{R_3}{100}\right) \)
\( = 10000 \left(1 + \frac{10}{100}\right) \left(1 - \frac{10}{100}\right) \left(1 + \frac{10}{100}\right) \)
\( = 10000 \left(\frac{110}{100}\right) \left(\frac{90}{100}\right) \left(\frac{110}{100}\right) \)
\( = 10000 \times \frac{11}{10} \times \frac{9}{10} \times \frac{11}{10} \)
\( = 10 \times 11 \times 9 \times 11 \)
\( = 10 \times 99 \times 11 \)
\( = 10 \times 1089 \)
\( = 10890 \)
So, the bacteria count at the end of 3 hours will be 10,890. This demonstrates how even small percentage changes can lead to different outcomes when applied in sequence.
In simple words: We start with 10,000 bacteria. It grows by 10%, then shrinks by 10%, then grows by 10% again. After 3 hours, there will be 10,890 bacteria.

🎯 Exam Tip: Remember that an increase followed by an equal percentage decrease (e.g., +10% then -10%) does NOT return to the original amount; a net decrease typically results. Here, the final increase makes it a net gain.

Question 7. The production of refrigerators in factory rose from 40000 to 48400 in 2 years. Find the rate of growth p.a.
Answer:
The present production (P) is 40,000 units.
The production after 2 years (A) is 48,400 units.
The time period (n) is 2 years.
Let the rate of increase be R% per annum.
We use the compound interest formula for appreciation:
\( A = P \left(1 + \frac{R}{100}\right)^n \)
Substitute the given values:
\( 48400 = 40000 \left(1 + \frac{R}{100}\right)^2 \)
Divide both sides by 40,000:
\( \frac{48400}{40000} = \left(1 + \frac{R}{100}\right)^2 \)
Simplify the fraction:
\( \frac{484}{400} = \left(1 + \frac{R}{100}\right)^2 \)
\( \frac{121}{100} = \left(1 + \frac{R}{100}\right)^2 \)
Recognize that \( \frac{121}{100} \) is a perfect square:
\( \left(\frac{11}{10}\right)^2 = \left(1 + \frac{R}{100}\right)^2 \)
Comparing both sides, we can take the square root:
\( 1 + \frac{R}{100} = \frac{11}{10} \)
Now, solve for R:
\( \frac{R}{100} = \frac{11}{10} - 1 \)
\( \frac{R}{100} = \frac{11 - 10}{10} \)
\( \frac{R}{100} = \frac{1}{10} \)
\( R = \frac{1}{10} \times 100 \)
\( R = 10 \)
Therefore, the rate of growth is 10% per annum. Understanding growth rates is essential for business planning and forecasting.
In simple words: Refrigerator production went from 40,000 to 48,400 in 2 years. This means the factory's production grew by 10% each year.

🎯 Exam Tip: When finding the rate, simplify the fraction on the left side of the equation as much as possible to easily identify the base for taking square roots or cube roots.

Question 8. The value of a flat worth ₹ 5,00,000 is depreciating at the rate of 10% p.a. In how many years will its value be reduced to ₹ 364500?
Answer:
The present value of the flat (P) is Rs. 5,00,000.
The value after depreciation (A) is Rs. 3,64,500.
The rate of depreciation (R) is 10% per annum.
Let the time period be 'n' years.
We use the formula for depreciation:
\( A = P \left(1 - \frac{R}{100}\right)^n \)
Substitute the known values:
\( 364500 = 500000 \left(1 - \frac{10}{100}\right)^n \)
Divide both sides by 5,00,000:
\( \frac{364500}{500000} = \left(\frac{90}{100}\right)^n \)
Simplify the fraction:
\( \frac{3645}{5000} = \left(\frac{9}{10}\right)^n \)
Divide both numerator and denominator by 5:
\( \frac{729}{1000} = \left(\frac{9}{10}\right)^n \)
Recognize that 729 is \( 9^3 \) and 1000 is \( 10^3 \):
\( \left(\frac{9}{10}\right)^3 = \left(\frac{9}{10}\right)^n \)
Comparing the exponents on both sides, we find:
\( n = 3 \)
Therefore, the flat's value will be reduced to Rs. 3,64,500 in 3 years. This type of calculation helps homeowners understand the future value of their assets.
In simple words: A flat costs Rs. 5,00,000 and loses 10% of its value each year. It will take 3 years for its value to become Rs. 3,64,500.

🎯 Exam Tip: When finding the time period 'n', always simplify the fraction to its lowest terms and try to express it as a power of the depreciation/appreciation factor.

Question 9. Rachit bought a flat for ₹ 10 lakh and a car for ₹ 3,20,000 at the same time. The price of the flat appreciates uniformly at the rate of 20% p.a.; while the price of the car depreciates at the rate of 15% p.a. If Rachit sells the flat and car after 3 years, what will be his profit or loss?
Answer:
First, calculate the total initial cost:
Price of flat = Rs. 10,00,000
Price of car = Rs. 3,20,000
Total initial cost = Rs. 10,00,000 + Rs. 3,20,000 = Rs. 13,20,000

Now, calculate the value of the flat after 3 years:
Rate of appreciation for flat (R1) = 20% p.a.
Period (n) = 3 years
Value of flat after 3 years \( = P \left(1 + \frac{R_1}{100}\right)^n \)
\( = 10,00,000 \left(1 + \frac{20}{100}\right)^3 \)
\( = 10,00,000 \left(\frac{120}{100}\right)^3 \)
\( = 10,00,000 \times \left(\frac{6}{5}\right)^3 \)
\( = 10,00,000 \times \frac{216}{125} \)
\( = 8000 \times 216 \)
\( = 17,28,000 \)
The flat's value after 3 years is Rs. 17,28,000.

Next, calculate the value of the car after 3 years:
Rate of depreciation for car (R2) = 15% p.a.
Period (n) = 3 years
Value of car after 3 years \( = P \left(1 - \frac{R_2}{100}\right)^n \)
\( = 3,20,000 \left(1 - \frac{15}{100}\right)^3 \)
\( = 3,20,000 \left(\frac{85}{100}\right)^3 \)
\( = 3,20,000 \times \left(\frac{17}{20}\right)^3 \)
\( = 3,20,000 \times \frac{4913}{8000} \)
\( = 40 \times 4913 \)
\( = 1,96,520 \)
The car's value after 3 years is Rs. 1,96,520.

Now, calculate the total selling price after 3 years:
Total selling price = Value of flat + Value of car
Total selling price = Rs. 17,28,000 + Rs. 1,96,520 = Rs. 19,24,520

Finally, determine the profit or loss:
Profit/Loss = Total selling price - Total initial cost
Profit/Loss = Rs. 19,24,520 - Rs. 13,20,000
Profit/Loss = Rs. 6,04,520 (Profit, since the value is positive)
Rachit made a profit of Rs. 6,04,520. Investing in appreciating assets while being aware of depreciating ones helps in overall financial planning.
In simple words: Rachit bought a flat for Rs. 10 lakh and a car for Rs. 3.20 lakh. The flat gained 20% each year, and the car lost 15% each year. After 3 years, he sold them. He made a total profit of Rs. 6,04,520.

🎯 Exam Tip: For problems involving multiple assets with different appreciation/depreciation rates, calculate the final value of each asset separately before summing them up to find the total profit or loss.

Question 10. 8000 workers were employed by a company to complete a job in 4 years. At the end of first year, 5% of the workers were retrenched. At the end of second year, 5% of those working at that time were retrenched. However to complete the job in time, the number of workers was increased by 10% of those working at the end of third year. How many workers were working during the fouth year ?
Answer:
The initial number of workers (P) is 8000.
At the end of the first year, there was a 5% retrenchment (decrease), so R1 = 5%.
At the end of the second year, there was another 5% retrenchment (decrease), so R2 = 5%.
At the end of the third year, there was a 10% increase in workers, so R3 = 10%.
To find the number of workers during the fourth year (which is the count after the third year), we apply these changes sequentially:
Number of workers after third year \( = P \left(1 - \frac{R_1}{100}\right) \left(1 - \frac{R_2}{100}\right) \left(1 + \frac{R_3}{100}\right) \)
\( = 8000 \left(1 - \frac{5}{100}\right) \left(1 - \frac{5}{100}\right) \left(1 + \frac{10}{100}\right) \)
\( = 8000 \left(\frac{95}{100}\right) \left(\frac{95}{100}\right) \left(\frac{110}{100}\right) \)
\( = 8000 \times \frac{19}{20} \times \frac{19}{20} \times \frac{11}{10} \)
\( = 8000 \times 0.95 \times 0.95 \times 1.1 \)
\( = 7942 \)
So, 7,942 workers were working during the fourth year. This kind of workforce adjustment is common in project management to meet deadlines.
In simple words: A company started with 8000 workers. After the first year, 5% left. After the second year, another 5% left. But then, to finish the job, 10% more workers were added at the end of the third year. So, 7,942 workers were working in the fourth year.

🎯 Exam Tip: Be careful with multiple percentage changes; each change is applied to the *current* amount, not the original starting amount. Use the compounding formula for successive changes.