Compound Interest

Q1: The compound interest on ₹3,750 for 2 years at 8 % p.a., compounded annually is

i. Calculate the Total Amount

Step 1: Identify the given values from the question statement.
Principal \( (P) = \text{₹}3750 \)
Rate of Interest \( (R) = 8\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Step 2: Substitute the values into the standard compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
\( A = 3750 \times \left(1 + \frac{8}{100}\right)^2 \)
\( A = 3750 \times \left(\frac{108}{100}\right)^2 \)
\( A = 3750 \times \left(\frac{27}{25}\right)^2 \)
Step 3: Solve the squared fraction expression.
\( A = 3750 \times \frac{729}{625} \)
Since \( 3750 \div 625 = 6 \):
\( A = 6 \times 729 \)
\( A = \text{₹}4374 \)

ii. Determine the Compound Interest

Step 1: Subtract the initial principal from the total final amount.
Formula: \( \text{Compound Interest (CI)} = A – P \)
\( \text{CI} = 4374 – 3750 \)
\( \text{CI} = \text{₹}624 \)
Answer:c. ₹624

Q2: A man invests ₹46,875 at 4% p.a. compound interest for 3 years. The interest for the 1st year will be:

Step 1: Identify the given values for the first year calculation.
Principal for the first year \( (P) = \text{₹}46875 \)
Rate of Interest \( (R) = 4\% \text{ per annum} \)
Time interval \( (T) = 1 \text{ year} \)
Step 2: Compute the interest for the first year using the Simple Interest relation (since compound interest and simple interest are identical for the first year).
Formula: \( \text{Interest} = \frac{P \times R \times T}{100} \)
\( \text{Interest} = \frac{46875 \times 4 \times 1}{100} \)
\( \text{Interest} = \frac{187500}{100} \)
\( \text{Interest} = \text{₹}1875 \)
Answer:c. ₹1875

Q3: A man deposits ₹10,000 in a cooperative bank for 3 years at 9% p.a. If interest is compounded annually, then the amount he will get from the bank after 3 years is:

Step 1: Identify the given values from the question statement.
Principal \( (P) = \text{₹}10000 \)
Rate of Interest \( (R) = 9\% \text{ per annum} \)
Time period \( (n) = 3 \text{ years} \)
Step 2: Substitute the values into the standard compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
\( A = 10000 \times \left(1 + \frac{9}{100}\right)^3 \)
\( A = 10000 \times \left(1.09\right)^3 \)
\( A = 10000 \times 1.295029 \)
\( A = \text{₹}12950.29 \)
Answer:a. ₹12,950.29

Q4: ₹16,000 is deposited in a bank for three years. The rates of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes ₹21,384. The rate of interest for the third year will be:

i. Set Up Amount Equation with Successive Rates

Step 1: List all the given values from the problem statement.
Principal \( (P) = \text{₹}16000 \)
Growth Rate for Year 1 \( (R_1) = 8\% \)
Growth Rate for Year 2 \( (R_2) = 12\% \)
Final Amount at the end of 3 years \( (A) = \text{₹}21384 \)
Let the growth rate for the third year be \( R_3\% \).
Step 2: State the compound interest amount formula for successive varying rates.
Formula: \( A = P \times \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right) \)

ii. Solve for the Third Year Rate (R₃)

Step 1: Substitute the known values into the equation framework.
\( 21384 = 16000 \times \left(1 + \frac{8}{100}\right) \times \left(1 + \frac{12}{100}\right) \times \left(1 + \frac{R_3}{100}\right) \)
Step 2: Simplify the fraction components for the first two years.
\( 1 + \frac{8}{100} = \frac{108}{100} = \frac{27}{25} \)
\( 1 + \frac{12}{100} = \frac{112}{100} = \frac{28}{25} \)
Step 3: Calculate the accumulated amount at the end of the second year.
\( 21384 = 16000 \times \frac{27}{25} \times \frac{28}{25} \times \left(1 + \frac{R_3}{100}\right) \)
\( 21384 = \frac{16000 \times 756}{625} \times \left(1 + \frac{R_3}{100}\right) \)
\( 21384 = 19353.6 \times \left(1 + \frac{R_3}{100}\right) \)
Step 4: Isolate the third-year rate term fraction.
\( 1 + \frac{R_3}{100} = \frac{21384}{19353.6} \)
\( 1 + \frac{R_3}{100} \approx 1.10 \)
Step 5: Compute the final percentage rate value of \( R_3 \).
\( \frac{R_3}{100} = 1.10 – 1 \)
\( \frac{R_3}{100} = 0.10 \)
\( R_3 = 0.10 \times 100 \)
\( R_3 = 10\% \)
Answer:b. 10%

Q5: A man borrows ₹5,000 at 12% compound interest p.a., interest payable every six months. He pays back ₹1,800 at the end of every six months. The third payment he has to make at the end of 18 months in order to clear the entire loan will be:

i. Adjust Rate and Calculate Balance for the First 6 Months

Step 1: Adjust the annual rate of interest for the half-yearly period.
Annual Rate \( (R) = 12\% \text{ per annum} \)
\( \text{Half-yearly Rate } (R’) = \frac{12\%}{2} = 6\% \text{ per half-year} \)
Initial Principal \( (P_1) = \text{₹}5000 \)
Step 2: Compute the interest and outstanding amount at the end of the first 6 months.
\( \text{Interest for Period 1 } (I_1) = \frac{5000 \times 6 \times 1}{100} = \text{₹}300 \)
\( \text{Amount before repayment } (A_1) = 5000 + 300 = \text{₹}5300 \)
Step 3: Deduct the first repayment of ₹1800 to find the principal for the next period.
\( \text{Remaining Principal for Period 2 } (P_2) = 5300 – 1800 = \text{₹}3500 \)

ii. Calculate Balance for the Second 6 Months (At 12 Months)

Step 1: Compute the interest and outstanding amount for the second half-year period.
\( \text{Interest for Period 2 } (I_2) = \frac{3500 \times 6 \times 1}{100} = \text{₹}210 \)
\( \text{Amount before repayment } (A_2) = 3500 + 210 = \text{₹}3710 \)
Step 2: Deduct the second repayment of ₹1800 to find the principal for the final period.
\( \text{Remaining Principal for Period 3 } (P_3) = 3710 – 1800 = \text{₹}1910 \)

iii. Calculate the Final Clearing Payment (At 18 Months)

Step 1: Compute the interest accrued during the final 6-month period.
\( \text{Interest for Period 3 } (I_3) = \frac{1910 \times 6 \times 1}{100} = \text{₹}114.60 \)
Step 2: Add this final interest to the third period principal to find the total sum required to clear the loan.
\( \text{Final Payment Required } (A_3) = P_3 + I_3 \)
\( A_3 = 1910 + 114.60 = \text{₹}2024.60 \)
Answer:a. ₹2024.60

Q6: The compound interest for the second year on ₹8,000 invested for 3 years at 10% p.a. is:

i. Calculate Principal at the Beginning of the Second Year

Step 1: Identify the given values for the first year.
Initial Principal \( (P_1) = \text{₹}8000 \)
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Time interval \( (T) = 1 \text{ year} \)
Step 2: Calculate the simple interest earned during the first year.
Formula: \( \text{Interest} = \frac{P \times R \times T}{100} \)
\( \text{Interest for Year 1 } (I_1) = \frac{8000 \times 10 \times 1}{100} = \text{₹}800 \)
Step 3: Compute the total amount at the end of Year 1, which becomes the principal for Year 2.
Formula: \( \text{Amount} = \text{Principal} + \text{Interest} \)
\( \text{Principal for Year 2 } (P_2) = 8000 + 800 = \text{₹}8800 \)

ii. Calculate Interest for the Second Year

Step 1: Use the updated principal balance to calculate the interest accrued during the second year.
\( \text{Interest for Year 2 } (I_2) = \frac{8800 \times 10 \times 1}{100} \)
\( I_2 = 88 \times 10 \)
\( I_2 = \text{₹}880 \)
Answer:b. ₹880

Q7: A person took a loan of ₹6,000 from a bank and agreed to pay back the amount along with interest in 2 years. If the rate of compound interest for the first year is 10% and second year is 12 %, the amount he had to pay after 2 years will be:

i. Calculate Interest and Amount for the First Year

Step 1: Identify given values for the first year.
Initial Principal \( (P_1) = \text{₹}6000 \)
Rate of Interest for Year 1 \( (R_1) = 10\% \text{ per annum} \)
Time \( (T) = 1 \text{ year} \)
Step 2: Calculate the Simple Interest for the first year.
Formula: \( \text{Interest} = \frac{P \times R \times T}{100} \)
\( \text{Interest for Year 1 } (I_1) = \frac{6000 \times 10 \times 1}{100} = \text{₹}600 \)
Step 3: Find the Amount at the end of the first year.
\( \text{Amount after Year 1 } (A_1) = 6000 + 600 = \text{₹}6600 \)
This amount becomes the principal for the second year.

ii. Calculate Interest and Final Amount for the Second Year

Step 1: Identify values for the second year.
Principal for Year 2 \( (P_2) = \text{₹}6600 \)
Rate of Interest for Year 2 \( (R_2) = 12\% \text{ per annum} \)
Step 2: Calculate the Simple Interest for the second year.
\( \text{Interest for Year 2 } (I_2) = \frac{6600 \times 12 \times 1}{100} \)
\( I_2 = 66 \times 12 = \text{₹}792 \)
Step 3: Calculate the total final amount to be paid after 2 years.
\( \text{Final Amount } (A_2) = P_2 + I_2 \)
\( A_2 = 6600 + 792 = \text{₹}7392 \)
Answer:d. ₹7392

Q8: Nikita invests ₹6,000 for two years at a certain rate of interest compounded annually. At the end of the first year, it amounts to ₹6,720. The rate of interest p.a. is:

i. Calculate Interest Earned in the First Year

Step 1: Identify the given values for the first year.
Principal \( (P) = \text{₹}6000 \)
Amount at the end of Year 1 \( (A_1) = \text{₹}6720 \)
Time interval \( (T) = 1 \text{ year} \)
Step 2: Find the interest accrued during the first year.
Formula: \( \text{Interest for Year 1 } (I_1) = A_1 – P \)
\( I_1 = 6720 – 6000 \)
\( I_1 = \text{₹}720 \)

ii. Determine the Rate of Interest

Step 1: Apply the Simple Interest formula for the first year to find the rate \( (R) \).
Formula: \( \text{Interest} = \frac{P \times R \times T}{100} \)
\( 720 = \frac{6000 \times R \times 1}{100} \)
\( 720 = 60 \times R \)
Step 2: Isolate and calculate the value of \( R \).
\( R = \frac{720}{60} \)
\( R = 12\% \)
Answer:c. 12%

Q9: The compound interest on ₹8,640 for 3 years at 8% p.a. is:

i. Calculate the Total Amount

Step 1: Identify the given values from the question statement.
Principal \( (P) = \text{₹}8640 \)
Rate of Interest \( (R) = 8\% \text{ per annum} \)
Time period \( (n) = 3 \text{ years} \)
Step 2: Substitute the values into the standard compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
\( A = 8640 \times \left(1 + \frac{8}{100}\right)^3 \)
\( A = 8640 \times \left(\frac{108}{100}\right)^3 \)
\( A = 8640 \times \left(\frac{27}{25}\right)^3 \)
Step 3: Expand the cubed fraction expression profile.
\( A = 8640 \times \frac{19683}{15625} \)
\( A = \frac{170061120}{15625} \)
\( A = \text{₹}10883.91 \)

ii. Determine the Compound Interest

Step 1: Subtract the initial principal from the total final amount.
Formula: \( \text{Compound Interest (CI)} = A – P \)
\( \text{CI} = 10883.91 – 8640 \)
\( \text{CI} = \text{₹}2243.91 \)
Answer: 2243.91

Q10: If the interest is compounded half-yearly, then, C.I. when the principal is ₹7,400, the rate of interest is 5% p.a. and the duration is one year, is:

i. Adjust Rate and Time for Half-Yearly Compounding

Step 1: Convert the annual interest rate and duration into half-yearly parameters.
Principal \( (P) = \text{₹}7400 \)
Annual Rate \( (R) = 5\% \text{ per annum} \)
\( \text{Half-yearly Rate } (R’) = \frac{5\%}{2} = 2.5\% \text{ per half-year} \)
Time duration = \( 1 \text{ year} = 2 \text{ half-yearly periods } (n = 2) \)

ii. Calculate Total Amount and Compound Interest

Step 1: Substitute the adjusted values into the standard compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R’}{100}\right)^n \)
\( A = 7400 \times \left(1 + \frac{2.5}{100}\right)^2 \)
\( A = 7400 \times \left(1 + \frac{25}{1000}\right)^2 \)
\( A = 7400 \times \left(1 + \frac{1}{40}\right)^2 \)
\( A = 7400 \times \left(\frac{41}{40}\right)^2 \)
Step 2: Expand and evaluate the squared fraction to find the final total amount.
\( A = 7400 \times \frac{1681}{1600} \)
\( A = 74 \times \frac{1681}{16} \)
\( A = \frac{124394}{16} \)
\( A = \text{₹}7774.625 \approx \text{₹}7774.63 \)
Step 3: Deduct the initial principal from the total final amount to get the compound interest.
Formula: \( \text{Compound Interest (CI)} = A – P \)
\( \text{CI} = 7774.63 – 7400 \)
\( \text{CI} = \text{₹}374.63 \)
Answer:b. ₹374.63

Q11: The simple interest on a sum of money for 2 years at 4% per annum is ₹340. The compound interest on this sum for one year payable half-yearly at the same rate is:

i. Find the Sum of Money (Principal)

Step 1: Identify the given values from the simple interest part.
Simple Interest \( (\text{SI}) = \text{₹}340 \)
Rate of Interest \( (R) = 4\% \text{ per annum} \)
Time \( (T) = 2 \text{ years} \)
Step 2: Apply the Simple Interest formula to compute the Principal \( (P) \).
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( 340 = \frac{P \times 4 \times 2}{100} \)
\( 340 = \frac{8P}{100} \)
\( P = \frac{340 \times 100}{8} \)
\( P = \text{₹}4250 \)

ii. Adjust Parameters for Half-Yearly Compounding

Step 1: Convert the annual values into half-yearly values for a duration of 1 year.
Principal \( (P) = \text{₹}4250 \)
\( \text{Half-yearly Rate } (R’) = \frac{4\%}{2} = 2\% \text{ per half-year} \)
Time duration = \( 1 \text{ year} = 2 \text{ half-yearly periods } (n = 2) \)

iii. Compute the Compound Interest

Step 1: Substitute the updated parameters into the compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R’}{100}\right)^n \)
\( A = 4250 \times \left(1 + \frac{2}{100}\right)^2 \)
\( A = 4250 \times \left(1.02\right)^2 \)
\( A = 4250 \times 1.0404 \)
\( A = \text{₹}4421.70 \)
Step 2: Calculate the final Compound Interest by subtracting the principal.
Formula: \( \text{Compound Interest (CI)} = A – P \)
\( \text{CI} = 4421.70 – 4250 \)
\( \text{CI} = \text{₹}171.70 \)
Answer:c. ₹171.70

Q12: The compound interest on a certain sum of money at 5% p.a. for two years is ₹246. The simple interest on the same sum for three years at 6% p.a. will be:

i. Find the Sum of Money (Principal)

Step 1: Identify the given variables from the Compound Interest portion.
Compound Interest \( (\text{CI}) = \text{₹}246 \)
Rate of Interest \( (R) = 5\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Step 2: Set up the equation using the standard Compound Interest formula.
Formula: \( \text{CI} = P \times \left[\left(1 + \frac{R}{100}\right)^n – 1\right] \)
\( 246 = P \times \left[\left(1 + \frac{5}{100}\right)^2 – 1\right] \)
\( 246 = P \times \left[\left(\frac{21}{20}\right)^2 – 1\right] \)
Step 3: Expand the fractional expression and solve for Principal \( (P) \).
\( 246 = P \times \left[\frac{441}{400} – 1\right] \)
\( 246 = P \times \left[\frac{441 – 400}{400}\right] \)
\( 246 = P \times \frac{41}{400} \)
\( P = \frac{246 \times 400}{41} \)
Since \( 246 \div 41 = 6 \):
\( P = 6 \times 400 = \text{₹}2400 \)
The initial sum of money is ₹2400.

ii. Calculate the Simple Interest

Step 1: Identify the variables for the Simple Interest transaction.
Principal \( (P) = \text{₹}2400 \)
New Rate of Interest \( (R_{\text{new}}) = 6\% \text{ per annum} \)
New Time Period \( (T) = 3 \text{ years} \)
Step 2: Compute the Simple Interest value profile using the standard interest equation.
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( \text{SI} = \frac{2400 \times 6 \times 3}{100} \)
\( \text{SI} = 24 \times 18 \)
\( \text{SI} = \text{₹}432 \)
Answer:a. ₹432

Q13: Ramesh wants to get ₹6,050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is:

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Desired Final Amount \( (A) = \text{₹}6050 \)
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Let the initial principal amount he needs to invest be \( P \).
Step 2: State the standard compound interest amount formula.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)

ii. Solve for the Principal Amount (P)

Step 1: Substitute the given values into the formula framework.
\( 6050 = P \times \left(1 + \frac{10}{100}\right)^2 \)
Step 2: Simplify the expression inside the parentheses.
\( 1 + \frac{10}{100} = 1 + 0.1 = 1.1 \)
Step 3: Evaluate the squared term.
\( 6050 = P \times (1.1)^2 \)
\( 6050 = P \times 1.21 \)
Step 4: Isolate \( P \) to determine the initial investment money required.
\( P = \frac{6050}{1.21} \)
Multiply the numerator and denominator by 100 to eliminate decimals:
\( P = \frac{605000}{121} \)
Since \( 605 \div 121 = 5 \):
\( P = 5000 \)
Answer:b. ₹5000

Q14: The difference between the compound and simple interest on a certain sum deposited for 2 years at 5% p.a. is ₹12. The sum will be:

i. Express Simple Interest and Compound Interest using Formula

Step 1: Identify the given values and set up the variable profile.
Let the Principal sum of money be \( P \).
Rate of Interest \( (R) = 5\% \text{ per annum} \)
Time period \( (n \text{ or } T) = 2 \text{ years} \)
Step 2: State the dedicated formula for the difference between Compound Interest (CI) and Simple Interest (SI) for a 2-year period.
Formula: \( \text{CI} – \text{SI} = P \times \left(\frac{R}{100}\right)^2 \)

ii. Solve for the Principal Sum (P)

Step 1: Substitute the given values into the difference formula configuration.
\( 12 = P \times \left(\frac{5}{100}\right)^2 \)
Step 2: Simplify the fraction inside the parentheses.
\( \frac{5}{100} = \frac{1}{20} \)
\( 12 = P \times \left(\frac{1}{20}\right)^2 \)
Step 3: Square the fraction and solve for \( P \).
\( 12 = P \times \frac{1}{400} \)
\( P = 12 \times 400 \)
\( P = 4800 \)
Answer:c. ₹4800

Q15: At what rate of compound interest p.a. will 20,000 amount to ₹26,620 in 3 years?

i. Set Up the Compound Interest Amount Formula

Step 1: Identify the given values from the problem statement.
Principal \( (P) = \text{₹}20000 \)
Amount \( (A) = \text{₹}26620 \)
Time \( (n) = 3 \text{ years} \)
Let the required rate of interest be \( R\% \) per annum.
Step 2: State the standard formula for uniform compound amount.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)

ii. Solve for the Rate of Interest (R)

Step 1: Substitute the given values into the formula framework.
\( 26620 = 20000 \times \left(1 + \frac{R}{100}\right)^3 \)
Step 2: Isolate the exponential term by dividing both sides by the principal.
\( \frac{26620}{20000} = \left(1 + \frac{R}{100}\right)^3 \)
Cancel out the zero from both parts and reduce the remaining fraction by dividing by 2:
\( \frac{2662}{2000} = \left(1 + \frac{R}{100}\right)^3 \)
\( \frac{1331}{1000} = \left(1 + \frac{R}{100}\right)^3 \)
Step 3: Express the numerical fraction on the left as a perfect cube.
We know that \( 1331 = 11^3 \) and \( 1000 = 10^3 \).
\( \left(\frac{11}{10}\right)^3 = \left(1 + \frac{R}{100}\right)^3 \)
Step 4: Take the cube root on both sides to eliminate the exponents.
\( \frac{11}{10} = 1 + \frac{R}{100} \)
\( \frac{R}{100} = \frac{11}{10} – 1 \)
\( \frac{R}{100} = \frac{11 – 10}{10} \)
\( \frac{R}{100} = \frac{1}{10} \)
Step 5: Solve for the final value of R.
\( R = \frac{1 \times 100}{10} \)
\( R = 10\% \)
Answer:d. 10%

Q16: In what time will ₹5000 amount to ₹5832 at 8% rate of compound interest p.a.?

i. Set Up the Compound Interest Amount Formula

Step 1: Identify the given values from the problem statement.
Principal \( (P) = \text{₹}5000 \)
Amount \( (A) = \text{₹}5832 \)
Rate of Interest \( (R) = 8\% \text{ per annum} \)
Let the required time period be \( n \) years.
Step 2: State the standard formula for total compound amount.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)

ii. Solve for the Time Period (n)

Step 1: Substitute the given values into the formula framework.
\( 5832 = 5000 \times \left(1 + \frac{8}{100}\right)^n \)
Step 2: Isolate the exponential term by dividing both sides by the principal.
\( \frac{5832}{5000} = \left(1 + \frac{8}{100}\right)^n \)
Step 3: Simplify the fractions on both sides to their lowest terms.
Dividing both 5832 and 5000 by 8:
\( \frac{729}{625} = \left(1 + \frac{2}{25}\right)^n \)
\( \frac{729}{625} = \left(\frac{27}{25}\right)^n \)
Step 4: Express the fraction on the left side as a power with base \( \frac{27}{25} \).
We know that \( 729 = 27^2 \) and \( 625 = 25^2 \).
\( \left(\frac{27}{25}\right)^2 = \left(\frac{27}{25}\right)^n \)
Step 5: Equate the exponents since the bases are identical.
\( n = 2 \text{ years} \)
Answer:a. 2 years

Q17: A machine depreciates at the rate of 10% of its value at the beginning of a year. If the present value of a machine is ₹8000, its value after 3 years will be:

i. Identify Given Values and Formula

Step 1: List all the given values from the problem statement.
Present Value \( (P) = \text{₹}8000 \)
Rate of Depreciation \( (R) = 10\% \text{ per annum} \)
Time period \( (n) = 3 \text{ years} \)
Step 2: State the standard formula for uniform annual depreciation.
Formula: \( A = P \times \left(1 – \frac{R}{100}\right)^n \)
Where \( A \) represents the value of the machine after 3 years.

ii. Compute the Value After 3 Years

Step 1: Substitute the given values into the depreciation formula.
\( A = 8000 \times \left(1 – \frac{10}{100}\right)^3 \)
Step 2: Simplify the fraction expression inside the parentheses.
\( 1 – \frac{10}{100} = 1 – \frac{1}{10} = \frac{9}{10} \)
Step 3: Expand the cubed fraction expression profile.
\( A = 8000 \times \left(\frac{9}{10}\right)^3 \)
\( A = 8000 \times \frac{9 \times 9 \times 9}{10 \times 10 \times 10} \)
\( A = 8000 \times \frac{729}{1000} \)
Step 4: Complete the final multiplication calculation by clearing out the zeros.
\( A = 8 \times 729 \)
\( A = \text{₹}5832 \)
Answer:b. ₹5832

Q18: The present population of a town is 2,00,000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be:

i. Identify Given Values and Formula Framework

Step 1: List all the given values from the problem statement.
Present Population \( (P) = 200000 \)
Growth rate for the first year \( (R_1) = 10\% \)
Growth rate for the second year \( (R_2) = 15\% \)
Step 2: State the compound formula for population growth over successive varying rates.
Formula: \( A = P \times \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \)
Where \( A \) represents the population of the town after two years.

ii. Compute the Population After 2 Years

Step 1: Substitute the given growth rates into the formula framework.
\( A = 200000 \times \left(1 + \frac{10}{100}\right) \times \left(1 + \frac{15}{100}\right) \)
Step 2: Simplify the fraction expressions inside the parentheses.
\( 1 + \frac{10}{100} = \frac{110}{100} = \frac{11}{10} \)
\( 1 + \frac{15}{100} = \frac{115}{100} = \frac{23}{20} \)
Step 3: Combine and multiply the fraction values.
\( A = 200000 \times \frac{11}{10} \times \frac{23}{20} \)
\( A = 200000 \times \frac{11 \times 23}{200} \)
\( A = 200000 \times \frac{253}{200} \)
Step 4: Complete the final calculation profile by dividing the base.
\( A = 1000 \times 253 \)
\( A = 253000 \)
Answer:a. 2,53,000

Q19: A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for ₹10,000 and is sold for ₹7,744. The number of years, that the machine was used is:

i. Set Up the Depreciation Formula

Step 1: List all the given values from the problem statement.
Purchase Price \( (P) = \text{₹}10000 \)
Selling Price \( (A) = \text{₹}7744 \)
Rate of Depreciation \( (R) = 12\% \text{ per annum} \)
Let the number of years the machine was used be \( n \).
Step 2: State the standard annual depreciation formula framework.
Formula: \( A = P \times \left(1 – \frac{R}{100}\right)^n \)

ii. Solve for the Number of Years (n)

Step 1: Substitute the known values into the equation.
\( 7744 = 10000 \times \left(1 – \frac{12}{100}\right)^n \)
Step 2: Isolate the exponential term by dividing both sides by 10000.
\( \frac{7744}{10000} = \left(1 – 0.12\right)^n \)
\( 0.7744 = \left(0.88\right)^n \)
Step 3: Convert both sides into fractional base form or powers to compare bases.
\( \frac{7744}{10000} = \left(\frac{88}{100}\right)^n \)
Step 4: Check if the left-side fraction can be expressed as a perfect square of the right-side base.
We know that \( 88 \times 88 = 7744 \) and \( 100 \times 100 = 10000 \).
\( \left(\frac{88}{100}\right)^2 = \left(\frac{88}{100}\right)^n \)
Step 5: Equate the exponents since the bases match exactly.
\( n = 2 \text{ years} \)
Answer:a. 2

Q20: The value of a machine depreciates every year at a constant rate. If the values of the machine in 2006 and 2008 are ₹25,000 and ₹19,360 respectively, then the annual rate of depreciation is:

i. Identify Given Values and Setup Formula

Step 1: Identify the time interval and baseline values from the question statement.
Initial value in 2006 \( (P) = \text{₹}25000 \)
Final value in 2008 \( (A) = \text{₹}19360 \)
Time duration \( (n) = 2008 – 2006 = 2 \text{ years} \)
Let the required annual rate of depreciation be \( R\% \).
Step 2: State the standard formula for constant annual depreciation.
Formula: \( A = P \times \left(1 – \frac{R}{100}\right)^n \)

ii. Solve for the Rate of Depreciation (R)

Step 1: Substitute the known values into the equation profile.
\( 19360 = 25000 \times \left(1 – \frac{R}{100}\right)^2 \)
Step 2: Isolate the exponential expression by dividing both sides by 25000.
\( \frac{19360}{25000} = \left(1 – \frac{R}{100}\right)^2 \)
Cancel out the trailing zero from both the numerator and denominator:
\( \frac{1936}{2500} = \left(1 – \frac{R}{100}\right)^2 \)
Step 3: Express the fraction on the left side as a perfect square.
We know that \( 1936 = 44^2 \) and \( 2500 = 50^2 \).
\( \left(\frac{44}{50}\right)^2 = \left(1 – \frac{R}{100}\right)^2 \)
Step 4: Take the square root on both sides to clear the exponents.
\( \frac{44}{50} = 1 – \frac{R}{100} \)
Step 5: Isolate and solve for the rate value \( R \).
\( \frac{R}{100} = 1 – \frac{44}{50} \)
\( \frac{R}{100} = \frac{50 – 44}{50} \)
\( \frac{R}{100} = \frac{6}{50} \)
\( R = \frac{6 \times 100}{50} \)
\( R = 6 \times 2 \)
\( R = 12\% \)
Answer:c. 12%