Compound Interest

Q1: A town has 15625 inhabitants. If the population of this town increases at the rate of 4% per annum, find the number of inhabitants of the town at the end of 3 years.

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Initial population \( (P) = 15625 \)
Rate of growth \( (R) = 4\% \text{ per annum} \)
Time period \( (n) = 3 \text{ years} \)
Step 2: State the population growth formula, which is structurally identical to the compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
Where \( A \) represents the final population after \( n \) years.

ii. Compute the Final Population

Step 1: Substitute the given values into the formula framework.
\( A = 15625 \times \left(1 + \frac{4}{100}\right)^3 \)
Step 2: Simplify the fraction inside the parentheses.
\( 1 + \frac{4}{100} = 1 + \frac{1}{25} = \frac{26}{25} \)
Step 3: Expand and solve the exponential expression.
\( A = 15625 \times \left(\frac{26}{25}\right)^3 \)
\( A = 15625 \times \frac{26 \times 26 \times 26}{25 \times 25 \times 25} \)
Step 4: Perform the division since \( 25 \times 25 \times 25 = 15625 \).
\( A = 15625 \times \frac{17576}{15625} \)
\( A = 17576 \)
Answer:The number of inhabitants of the town at the end of 3 years is 17576

Q2: The population of a town is increasing at the rate of 10% per annum. If its present population is 36300, find:

i. the population after 2 years

Step 1: Identify the given values for the future population calculation.
Present Population \( (P) = 36300 \)
Rate of Increase \( (R) = 10\% \text{ per annum} \)
Time \( (n) = 2 \text{ years} \)
Step 2: Apply the growth formula to find the population after 2 years.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
\( A = 36300 \times \left(1 + \frac{10}{100}\right)^2 \)
\( A = 36300 \times \left(1 + 0.1\right)^2 \)
\( A = 36300 \times \left(1.1\right)^2 \)
\( A = 36300 \times 1.21 \)
\( A = 43923 \)
Answer:The population after 2 years will be 43923

ii. the population 2 years ago

Step 1: Set up the growth equation where the present population is the final value.
Let the population 2 years ago be \( P_{\text{ago}} \).
Formula: \( \text{Present Population} = P_{\text{ago}} \times \left(1 + \frac{R}{100}\right)^n \)
Step 2: Substitute the values and isolate \( P_{\text{ago}} \).
\( 36300 = P_{\text{ago}} \times \left(1 + \frac{10}{100}\right)^2 \)
\( 36300 = P_{\text{ago}} \times \left(1.1\right)^2 \)
\( 36300 = P_{\text{ago}} \times 1.21 \)
\( P_{\text{ago}} = \frac{36300}{1.21} \)
\( P_{\text{ago}} = \frac{3630000}{121} \)
\( P_{\text{ago}} = 30000 \)
Answer:The population 2 years ago was 30000

Q3: The present population of a town is 176400. If the rate of growth in its population is 5% per annum, find:

i. the population 2 years hence

Step 1: Identify the given values for the future population calculation.
Present Population \( (P) = 176400 \)
Rate of Growth \( (R) = 5\% \text{ per annum} \)
Time \( (n) = 2 \text{ years} \)
Step 2: Apply the growth formula to determine the population 2 years hence.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
\( A = 176400 \times \left(1 + \frac{5}{100}\right)^2 \)
\( A = 176400 \times \left(1 + \frac{1}{20}\right)^2 \)
\( A = 176400 \times \left(\frac{21}{20}\right)^2 \)
\( A = 176400 \times \frac{441}{400} \)
\( A = 441 \times 441 \)
\( A = 194481 \)
Answer:The population 2 years hence will be 194481

ii. the population one year ago

Step 1: Set up the growth equation where the present population is the final value.
Let the population 1 year ago be \( P_{\text{ago}} \).
Time \( (n) = 1 \text{ year} \)
Formula: \( \text{Present Population} = P_{\text{ago}} \times \left(1 + \frac{R}{100}\right)^n \)
Step 2: Substitute the values and isolate \( P_{\text{ago}} \).
\( 176400 = P_{\text{ago}} \times \left(1 + \frac{5}{100}\right)^1 \)
\( 176400 = P_{\text{ago}} \times \left(\frac{21}{20}\right) \)
\( P_{\text{ago}} = \frac{176400 \times 20}{21} \)
\( P_{\text{ago}} = 8400 \times 20 \)
\( P_{\text{ago}} = 168000 \)
Answer:The population one year ago was 168000

Q4: Three years ago, the population of a city was 50000. If the annual increase during three successive years be 5%, 8% and 10% respectively, find the present population of the city.

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Initial Population 3 years ago \( (P) = 50000 \)
Growth Rate for Year 1 \( (R_1) = 5\% \)
Growth Rate for Year 2 \( (R_2) = 8\% \)
Growth Rate for Year 3 \( (R_3) = 10\% \)
Step 2: State the formula for population growth with successive different rates.
Formula: \( A = P \times \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right) \)
Where \( A \) represents the present population of the city.

ii. Compute the Present Population

Step 1: Substitute the given growth rates into the formula framework.
\( A = 50000 \times \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{8}{100}\right) \times \left(1 + \frac{10}{100}\right) \)
Step 2: Convert the percentage additions into simplified fraction values.
\( 1 + \frac{5}{100} = \frac{105}{100} = \frac{21}{20} \)
\( 1 + \frac{8}{100} = \frac{108}{100} = \frac{27}{25} \)
\( 1 + \frac{10}{100} = \frac{110}{100} = \frac{11}{10} \)
Step 3: Combine and solve the multiplied fraction values.
\( A = 50000 \times \frac{21}{20} \times \frac{27}{25} \times \frac{11}{10} \)
\( A = 50000 \times \frac{21 \times 27 \times 11}{20 \times 25 \times 10} \)
\( A = 50000 \times \frac{6237}{5000} \)
Step 4: Complete the final calculation by dividing the initial population value.
\( A = 10 \times 6237 \)
\( A = 62370 \)
Answer:The present population of the city is 62370

Q5: A farmer has an increase of 12.5% in the output of wheat in his farm every year. This year, he produced 2916 quintals of wheat. What was his annual production of wheat 2 years ago?

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Present Production \( (A) = 2916 \text{ quintals} \)
Rate of Increase \( (R) = 12.5\% = 12\frac{1}{2}\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Let the production 2 years ago be \( P \).
Step 2: State the growth formula used to find the past value.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)

ii. Simplify the Rate Expression

Step 1: Convert the fractional growth rate inside the formula parenthesis into a simplified term.
\( 1 + \frac{12.5}{100} = 1 + \frac{125}{1000} \)
Step 2: Reduce the fraction by dividing both numerator and denominator by 125.
\( 1 + \frac{1}{8} = \frac{9}{8} \)

iii. Calculate Wheat Production 2 Years Ago

Step 1: Substitute the simplified rate term and values into the equation.
\( 2916 = P \times \left(\frac{9}{8}\right)^2 \)
Step 2: Square the fraction expression.
\( 2916 = P \times \frac{81}{64} \)
Step 3: Isolate \( P \) to solve for the original production value.
\( P = \frac{2916 \times 64}{81} \)
Step 4: Perform the division profile since \( 2916 \div 81 = 36 \).
\( P = 36 \times 64 \)
\( P = 2304 \)
Answer:His annual production of wheat 2 years ago was 2304 quintals

Q6: The population of a town is 64000. If the annual birth rate is 11.7% and the annual death rate is 4.2%, calculate the population of the town after 3 years.

i. Calculate Net Annual Growth Rate

Step 1: Identify the annual birth rate and death rate from the statement.
Annual Birth Rate = \( 11.7\% \)
Annual Death Rate = \( 4.2\% \)
Step 2: Find the net annual population growth rate by subtracting the death rate from the birth rate.
Net Growth Rate \( (R) = \text{Birth Rate} – \text{Death Rate} \)
\( R = 11.7\% – 4.2\% \)
\( R = 7.5\% \text{ per annum} \)

ii. Identify Given Values and Formula

Step 1: List all the values needed for the population formula.
Initial Population \( (P) = 64000 \)
Net Growth Rate \( (R) = 7.5\% \text{ per annum} \)
Time period \( (n) = 3 \text{ years} \)
Step 2: State the population growth formula.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
Where \( A \) represents the population after 3 years.

iii. Compute the Population After 3 Years

Step 1: Substitute the values into the growth formula.
\( A = 64000 \times \left(1 + \frac{7.5}{100}\right)^3 \)
Step 2: Simplify the growth fraction expression inside the parentheses.
\( 1 + \frac{7.5}{100} = 1 + \frac{75}{1000} = 1 + \frac{3}{40} = \frac{43}{40} \)
Step 3: Expand the cubed term.
\( A = 64000 \times \left(\frac{43}{40}\right)^3 \)
\( A = 64000 \times \frac{43 \times 43 \times 43}{40 \times 40 \times 40} \)
Step 4: Complete the final calculation profile since \( 40 \times 40 \times 40 = 64000 \).
\( A = 64000 \times \frac{79507}{64000} \)
\( A = 79507 \)
Answer:The population of the town after 3 years will be 79507

Q7: A mango tree was planted 2 years ago. The rate of its growth is 20% per annum. If at present, the height of the tree is 162 cm, what it was when the tree was planted?

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Present Height \( (A) = 162 \text{ cm} \)
Rate of Growth \( (R) = 20\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Let the height of the tree when it was planted be \( P \).
Step 2: State the growth formula used to calculate the initial height.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)

ii. Calculate Initial Height of the Tree

Step 1: Substitute the given values into the formula framework.
\( 162 = P \times \left(1 + \frac{20}{100}\right)^2 \)
Step 2: Simplify the fraction inside the parentheses.
\( 1 + \frac{20}{100} = 1 + \frac{1}{5} = \frac{6}{5} \)
Step 3: Square the fraction expression.
\( 162 = P \times \left(\frac{6}{5}\right)^2 \)
\( 162 = P \times \frac{36}{25} \)
Step 4: Isolate \( P \) to solve for the original height.
\( P = \frac{162 \times 25}{36} \)
Dividing both 162 and 36 by 18:
\( P = \frac{9 \times 25}{2} \)
\( P = \frac{225}{2} \)
\( P = 112.5 \text{ cm} \)
Answer:The height of the tree when it was planted was 112.5 cm

Q8: Two years ago, the population of a village was 4000. During next year it increased by 6% but due to an epidemic, it decreased by 5% in the following year. What is its population now?

i. Identify Given Values and Formula Framework

Step 1: List all the given values from the problem statement.
Initial population 2 years ago \( (P) = 4000 \)
Growth rate for the first year \( (R_1) = 6\% \text{ (Increase)} \)
Decline rate for the second year \( (R_2) = 5\% \text{ (Decrease)} \)
Step 2: State the combined compound formula for successive variations.
Formula: \( A = P \times \left(1 + \frac{R_1}{100}\right) \times \left(1 – \frac{R_2}{100}\right) \)
Where \( A \) represents the present population of the village.

ii. Compute the Present Population

Step 1: Substitute the given values into the formula layout.
\( A = 4000 \times \left(1 + \frac{6}{100}\right) \times \left(1 – \frac{5}{100}\right) \)
Step 2: Simplify the fraction expressions inside the parentheses.
\( 1 + \frac{6}{100} = \frac{106}{100} = \frac{53}{50} \)
\( 1 – \frac{5}{100} = \frac{95}{100} = \frac{19}{20} \)
Step 3: Combine and solve the multiplied fraction values.
\( A = 4000 \times \frac{53}{50} \times \frac{19}{20} \)
\( A = 4000 \times \frac{53 \times 19}{50 \times 20} \)
\( A = 4000 \times \frac{1007}{1000} \)
Step 4: Complete the final calculation profile.
\( A = 4 \times 1007 \)
\( A = 4028 \)
Answer:The population of the village now is 4028

Q9: The count of bacteria in a culture grows by 10% during first hour, decreases by 8% during second hour and again increases by 12% during third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?

i. Identify Given Values and Formula Framework

Step 1: List all the given values from the problem statement.
Initial Count of Bacteria \( (P) = 13125000 \)
Growth rate for the first hour \( (R_1) = 10\% \text{ (Increase)} \)
Decline rate for the second hour \( (R_2) = 8\% \text{ (Decrease)} \)
Growth rate for the third hour \( (R_3) = 12\% \text{ (Increase)} \)
Step 2: State the combined compound formula for successive growth and decline variations.
Formula: \( A = P \times \left(1 + \frac{R_1}{100}\right) \times \left(1 – \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right) \)
Where \( A \) represents the final count of bacteria after 3 hours.

ii. Compute the Final Bacteria Count

Step 1: Substitute the given values into the formula layout.
\( A = 13125000 \times \left(1 + \frac{10}{100}\right) \times \left(1 – \frac{8}{100}\right) \times \left(1 + \frac{12}{100}\right) \)
Step 2: Simplify the fraction expressions inside the parentheses.
\( 1 + \frac{10}{100} = \frac{110}{100} = \frac{11}{10} \)
\( 1 – \frac{8}{100} = \frac{92}{100} = \frac{23}{25} \)
\( 1 + \frac{12}{100} = \frac{112}{100} = \frac{28}{25} \)
Step 3: Combine and solve the multiplied fraction values.
\( A = 13125000 \times \frac{11}{10} \times \frac{23}{25} \times \frac{28}{25} \)
\( A = 13125000 \times \frac{11 \times 23 \times 28}{10 \times 25 \times 25} \)
\( A = 13125000 \times \frac{7084}{6250} \)
Step 4: Complete the final division and multiplication profile.
Since \( \frac{13125000}{6250} = 2100 \):
\( A = 2100 \times 7084 \)
\( A = 14876400 \)
Answer:The count of bacteria after 3 hours will be 14876400

Q10: In a factory, the production of scooters was 40000 per year, which rose to 57600 in 2 years. Find the rate of growth per annum.

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Initial Production \( (P) = 40000 \)
Final Production \( (A) = 57600 \)
Time period \( (n) = 2 \text{ years} \)
Let the rate of growth per annum be \( R\% \).
Step 2: State the standard formula for uniform exponential growth.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)

ii. Solve for the Rate of Growth (R)

Step 1: Substitute the given values into the formula layout.
\( 57600 = 40000 \times \left(1 + \frac{R}{100}\right)^2 \)
Step 2: Isolate the squared expression by dividing both sides by 40000.
\( \frac{57600}{40000} = \left(1 + \frac{R}{100}\right)^2 \)
\( \frac{576}{400} = \left(1 + \frac{R}{100}\right)^2 \)
Step 3: Express the numerical fraction on the left as a perfect square.
We know that \( 576 = 24^2 \) and \( 400 = 20^2 \).
\( \left(\frac{24}{20}\right)^2 = \left(1 + \frac{R}{100}\right)^2 \)
Step 4: Take the square root on both sides to clear the exponents.
\( \frac{24}{20} = 1 + \frac{R}{100} \)
Reduce the fraction on the left by dividing by 4:
\( \frac{6}{5} = 1 + \frac{R}{100} \)
Step 5: Isolate and compute the final value of R.
\( \frac{R}{100} = \frac{6}{5} – 1 \)
\( \frac{R}{100} = \frac{1}{5} \)
\( R = \frac{100}{5} \)
\( R = 20\% \)
Answer:The rate of growth per annum is 20%

Q11: Amit started a shop by investing ₹500000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.

i. Identify Given Values and Formula Framework

Step 1: List all the given values from the problem statement.
Initial Investment \( (P) = \text{₹}500000 \)
Loss rate for the first year \( (R_1) = 5\% \text{ (Decrease)} \)
Profit rate for the second year \( (R_2) = 10\% \text{ (Increase)} \)
Profit rate for the third year \( (R_3) = 12\% \text{ (Increase)} \)
Step 2: State the combined compound formula for successive variations (loss and profits).
Formula: \( A = P \times \left(1 – \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right) \)
Where \( A \) represents the final amount at the end of 3 years.

ii. Compute the Final Amount

Step 1: Substitute the given values into the formula layout.
\( A = 500000 \times \left(1 – \frac{5}{100}\right) \times \left(1 + \frac{10}{100}\right) \times \left(1 + \frac{12}{100}\right) \)
Step 2: Simplify the fraction expressions inside the parentheses.
\( 1 – \frac{5}{100} = \frac{95}{100} = \frac{19}{20} \)
\( 1 + \frac{10}{100} = \frac{110}{100} = \frac{11}{10} \)
\( 1 + \frac{12}{100} = \frac{112}{100} = \frac{28}{25} \)
Step 3: Combine and multiply the fraction values.
\( A = 500000 \times \frac{19}{20} \times \frac{11}{10} \times \frac{28}{25} \)
\( A = 500000 \times \frac{19 \times 11 \times 28}{20 \times 10 \times 25} \)
\( A = 500000 \times \frac{5852}{5000} \)
Step 4: Complete the multiplication profile by dividing out the common base.
\( A = 100 \times 5852 \)
\( A = \text{₹}585200 \)
The final amount after 3 years is ₹585200.

iii. Calculate the Net Profit

Step 1: Subtract the initial investment principal from the final total amount.
Formula: \( \text{Net Profit} = A – P \)
\( \text{Net Profit} = 585200 – 500000 \)
\( \text{Net Profit} = \text{₹}85200 \)
Answer:His net profit for the entire period of three years is ₹85200

Q12: The value of a machine depreciates 10% annually. Its present value is ₹64800. Find:

i. its value after 2 years

Step 1: Identify the given values for the future depreciation calculation.
Present Value \( (P) = \text{₹}64800 \)
Rate of Depreciation \( (R) = 10\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Step 2: State the depreciation formula used to find the future value.
Formula: \( A = P \times \left(1 – \frac{R}{100}\right)^n \)
Where \( A \) represents the value of the machine after \( n \) years.
Step 3: Substitute the values into the formula framework and simplify.
\( A = 64800 \times \left(1 – \frac{10}{100}\right)^2 \)
\( A = 64800 \times \left(1 – \frac{1}{10}\right)^2 \)
\( A = 64800 \times \left(\frac{9}{10}\right)^2 \)
\( A = 64800 \times \frac{81}{100} \)
Step 4: Perform the final multiplication step.
\( A = 648 \times 81 \)
\( A = \text{₹}52488 \)
Answer:The value of the machine after 2 years will be ₹52488

ii. its value 2 years ago.

Step 1: Set up the depreciation equation where the present value is the final result.
Let the value of the machine 2 years ago be \( P_{\text{ago}} \).
Formula: \( \text{Present Value} = P_{\text{ago}} \times \left(1 – \frac{R}{100}\right)^n \)
Step 2: Substitute the known values into the depreciation framework.
\( 64800 = P_{\text{ago}} \times \left(1 – \frac{10}{100}\right)^2 \)
\( 64800 = P_{\text{ago}} \times \left(\frac{9}{10}\right)^2 \)
\( 64800 = P_{\text{ago}} \times \frac{81}{100} \)
Step 3: Isolate \( P_{\text{ago}} \) to solve for the original purchase price value.
\( P_{\text{ago}} = \frac{64800 \times 100}{81} \)
Since \( 64800 \div 81 = 800 \):
\( P_{\text{ago}} = 800 \times 100 \)
\( P_{\text{ago}} = \text{₹}80000 \)
Answer:The value of the machine 2 years ago was ₹80000

Q13: A refrigerator was purchased one year ago for ₹20000. Its value depreciates at the rate of 15% per annum. Find:

i. its present value,

Step 1: Identify given values regarding its purchase one year ago.
Purchase Value 1 year ago \( (P) = \text{₹}20000 \)
Rate of Depreciation \( (R) = 15\% \text{ per annum} \)
Time passed \( (n) = 1 \text{ year} \)
Step 2: Apply the depreciation formula to determine its present value.
Formula: \( \text{Present Value} = P \times \left(1 – \frac{R}{100}\right)^n \)
\( \text{Present Value} = 20000 \times \left(1 – \frac{15}{100}\right)^1 \)
\( \text{Present Value} = 20000 \times \left(\frac{85}{100}\right) \)
\( \text{Present Value} = 200 \times 85 \)
\( \text{Present Value} = \text{₹}17000 \)
Answer:The present value of the refrigerator is ₹17000

ii. its value after 1 year.

Step 1: Use the newly calculated present value as the new base principal.
New Principal Base \( (P_{\text{new}}) = \text{₹}17000 \)
Time period ahead \( (n) = 1 \text{ year} \)
Step 2: Apply the depreciation formula to find its value 1 year into the future.
Formula: \( \text{Value after 1 year} = P_{\text{new}} \times \left(1 – \frac{R}{100}\right)^n \)
\( \text{Value after 1 year} = 17000 \times \left(1 – \frac{15}{100}\right)^1 \)
\( \text{Value after 1 year} = 17000 \times \left(\frac{85}{100}\right) \)
\( \text{Value after 1 year} = 170 \times 85 \)
\( \text{Value after 1 year} = \text{₹}14450 \)
Answer:The value of the refrigerator after 1 year will be ₹14450

Q14: A machine depreciates each year at 8% of its value in the beginning of the year. If its value be ₹57500 at the end of the year 2015, find:

i. its value at the end of the year 2014,

Step 1: Set up the depreciation relation between the year 2014 and the year 2015.
Let the value at the end of 2014 be \( P_{2014} \).
Value at the end of 2015 \( (A) = \text{₹}57500 \)
Rate of Depreciation \( (R) = 8\% \text{ per annum} \)
Time interval \( (n) = 1 \text{ year} \)
Step 2: Formulate the equation where the value in 2015 is the depreciated outcome of 2014.
Formula: \( A = P_{2014} \times \left(1 – \frac{R}{100}\right)^n \)
\( 57500 = P_{2014} \times \left(1 – \frac{8}{100}\right)^1 \)
\( 57500 = P_{2014} \times \left(\frac{92}{100}\right) \)
Step 3: Isolate \( P_{2014} \) to determine the initial value.
\( P_{2014} = \frac{57500 \times 100}{92} \)
Dividing both 57500 and 92 by 23:
\( P_{2014} = \frac{2500 \times 100}{4} \)
\( P_{2014} = 2500 \times 25 \)
\( P_{2014} = \text{₹}62500 \)
Answer:The value of the machine at the end of the year 2014 was ₹62500

ii. its value at the end of the year 2016.

Step 1: Use the known value at the end of 2015 as the new baseline principal.
Principal value at the beginning of 2016 \( (P_{2015}) = \text{₹}57500 \)
Time period ahead \( (n) = 1 \text{ year} \)
Step 2: Apply the depreciation formula to determine the value at the end of 2016.
Formula: \( \text{Value}_{2016} = P_{2015} \times \left(1 – \frac{R}{100}\right)^n \)
\( \text{Value}_{2016} = 57500 \times \left(1 – \frac{8}{100}\right)^1 \)
\( \text{Value}_{2016} = 57500 \times \left(\frac{92}{100}\right) \)
\( \text{Value}_{2016} = 575 \times 92 \)
\( \text{Value}_{2016} = \text{₹}52900 \)
Answer:The value of the machine at the end of the year 2016 will be ₹52900

Q15: The value of a machine depreciates at the rate of \(16\frac{2}{3}\) % per annum. It was purchased 3 years ago. If its present value is ₹62500, find its purchase price.

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Present Value \( (A) = \text{₹}62500 \)
Rate of Depreciation \( (R) = 16\frac{2}{3}\% = \frac{50}{3}\% \text{ per annum} \)
Time period \( (n) = 3 \text{ years} \)
Let the purchase price of the machine 3 years ago be \( P \).
Step 2: State the uniform depreciation formula used to calculate the past value.
Formula: \( A = P \times \left(1 – \frac{R}{100}\right)^n \)

ii. Simplify the Depreciation Rate Fraction

Step 1: Substitute the fractional rate into the formula parenthesis expression.
\( 1 – \frac{R}{100} = 1 – \frac{\frac{50}{3}}{100} \)
Step 2: Simplify the compound fraction term.
\( 1 – \frac{50}{3 \times 100} = 1 – \frac{50}{300} \)
Step 3: Reduce to the lowest terms.
\( 1 – \frac{1}{6} = \frac{5}{6} \)

iii. Compute the Original Purchase Price

Step 1: Substitute the values and the simplified fraction back into the main equation.
\( 62500 = P \times \left(\frac{5}{6}\right)^3 \)
Step 2: Expand the cubed fraction expression profile.
\( 62500 = P \times \frac{5 \times 5 \times 5}{6 \times 6 \times 6} \)
\( 62500 = P \times \frac{125}{216} \)
Step 3: Isolate \( P \) to calculate the original machine price.
\( P = \frac{62500 \times 216}{125} \)
Step 4: Perform the division profile since \( 62500 \div 125 = 500 \).
\( P = 500 \times 216 \)
\( P = \text{₹}108000 \)
Answer:The purchase price of the machine was ₹108000