Compound Interest

Q1: Calculate the amount and the compound interest on ₹10000 for 2 years at 8% p.a., compounded annually.

Step 1: Write the given values.
Principal \( P = 10000 \)
Rate \( R = 8\% \)
Time \( T = 2 \) years
Step 2: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute the values.
\( A = 10000 \left(1 + \frac{8}{100} \right)^2 \)
\( A = 10000 (1.08)^2 \)
Step 4: Solve the power.
\( (1.08)^2 = 1.1664 \)
Step 5: Calculate the amount.
\( A = 10000 \times 1.1664 = 11664 \)
Step 6: Calculate Compound Interest.
\( CI = A – P \)
\( CI = 11664 – 10000 = 1664 \)
Answer: Amount = ₹11664, Compound Interest = ₹1664

Q2: Calculate the amount and the compound interest on ₹64000 for 3 years at \(7\frac{1}{2}\)% per annum compounded annually.

Step 1: Write the given values.
Principal \( P = 64000 \)
Rate \( R = 7\frac{1}{2}\% = 7.5\% \)
Time \( T = 3 \) years
Step 2: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute the values.
\( A = 64000 \left(1 + \frac{7.5}{100} \right)^3 \)
\( A = 64000 (1.075)^3 \)
Step 4: Solve step-by-step.
\( (1.075)^2 = 1.155625 \)
\( (1.075)^3 = 1.155625 \times 1.075 = 1.242296875 \)
Step 5: Calculate amount.
\( A = 64000 \times 1.242296875 \)
\( A = 79507 \) (approx.)
Step 6: Calculate Compound Interest.
\( CI = A – P \)
\( CI = 79507 – 64000 = 15507 \)
Answer: Amount = ₹79507 (approx.), Compound Interest = ₹15507

Q3: How much will ₹12000 amount to in 2 years at compound interest, the rates of interest for the successive years being 10% and 11% respectively?

Step 1: Write the given values.
Principal \( P = 12000 \)
Rate for 1st year \( R_1 = 10\% \)
Rate for 2nd year \( R_2 = 11\% \)
Step 2: Calculate amount after 1st year.
\( A_1 = 12000 \left(1 + \frac{10}{100} \right) \)
\( A_1 = 12000 \times 1.10 \)
\( A_1 = 13200 \)
Step 3: Calculate amount after 2nd year.
\( A_2 = 13200 \left(1 + \frac{11}{100} \right) \)
\( A_2 = 13200 \times 1.11 \)
\( A_2 = 14652 \)
Answer: Amount after 2 years = ₹14652

Q4: Calculate the amount and the compound interest on ₹25000 for 3 years, the rates of interest for the successive years being 8%, 9% and 10%, compounded annually.

Step 1: Write the given values.
Principal \( P = 25000 \)
Rate for 1st year \( R_1 = 8\% \)
Rate for 2nd year \( R_2 = 9\% \)
Rate for 3rd year \( R_3 = 10\% \)
Step 2: Calculate amount after 1st year.
\( A_1 = 25000 \left(1 + \frac{8}{100} \right) \)
\( A_1 = 25000 \times 1.08 \)
\( A_1 = 27000 \)
Step 3: Calculate amount after 2nd year.
\( A_2 = 27000 \left(1 + \frac{9}{100} \right) \)
\( A_2 = 27000 \times 1.09 \)
\( A_2 = 29430 \)
Step 4: Calculate amount after 3rd year.
\( A_3 = 29430 \left(1 + \frac{10}{100} \right) \)
\( A_3 = 29430 \times 1.10 \)
\( A_3 = 32373 \)
Step 5: Calculate Compound Interest.
\( CI = A_3 – P \)
\( CI = 32373 – 25000 \)
\( CI = 7373 \)
Answer: Amount = ₹32373, Compound Interest = ₹7373

Q5: Find the amount and the compound interest on ₹7500 for 2 years 8 months at 10% p.a., compounded annually.

Step 1: Write the given values.
Principal \( P = 7500 \)
Rate \( R = 10\% \)
Time \( = 2 \text{ years } 8 \text{ months} \)
Step 2: Convert time into years.
\( 8 \text{ months} = \frac{8}{12} = \frac{2}{3} \text{ year} \)
Total time \( = 2 + \frac{2}{3} \)
Step 3: Calculate amount for 2 full years.
\( A = 7500 \left(1 + \frac{10}{100} \right)^2 \)
\( A = 7500 (1.1)^2 \)
\( A = 7500 \times 1.21 \)
\( A = 9075 \)
Step 4: For remaining \( \frac{2}{3} \) year, use simple interest on ₹9075.
\( SI = \frac{P \times R \times T}{100} \)
\( SI = \frac{9075 \times 10 \times \frac{2}{3}}{100} \)
\( SI = \frac{9075 \times 20}{300} \)
\( SI = 605 \)
Step 5: Calculate final amount.
\( A = 9075 + 605 = 9680 \)
Step 6: Calculate Compound Interest.
\( CI = A – P \)
\( CI = 9680 – 7500 = 2180 \)
Answer: Amount = ₹9680, Compound Interest = ₹2180

Q6: If the simple interest on a sum of money for 3 years at 8% per annum is ₹7500, find the compound interest on the same sum, for the same period at the same rate.

Step 1: Use simple interest formula to find principal.
\( SI = \frac{P \times R \times T}{100} \)
Step 2: Substitute the values.
\( 7500 = \frac{P \times 8 \times 3}{100} \)
Step 3: Simplify.
\( 7500 = \frac{24P}{100} \)
\( 7500 = \frac{6P}{25} \)
Step 4: Solve for \( P \).
\( P = \frac{7500 \times 25}{6} \)
\( P = 31250 \)
Step 5: Calculate compound amount.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
\( A = 31250 (1.08)^3 \)
Step 6: Solve step-by-step.
\( (1.08)^2 = 1.1664 \)
\( (1.08)^3 = 1.1664 \times 1.08 = 1.259712 \)
Step 7: Calculate amount.
\( A = 31250 \times 1.259712 \)
\( A = 39366 \) (approx.)
Step 8: Calculate Compound Interest.
\( CI = A – P \)
\( CI = 39366 – 31250 = 8116 \)
Answer: Compound Interest = ₹8116 (approx.)

Q7: Calculate the amount and the compound interest on ₹16000 for 1 year at 15% per annum, compounded half-yearly.

Step 1: Write the given values.
Principal \( P = 16000 \)
Annual Rate \( R = 15\% \)
Time \( T = 1 \) year
Step 2: Convert into half-yearly terms.
Rate per half-year \( = \frac{15}{2} = 7.5\% \)
Number of periods \( = 2 \)
Step 3: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^n \)
Step 4: Substitute values.
\( A = 16000 \left(1 + \frac{7.5}{100} \right)^2 \)
\( A = 16000 (1.075)^2 \)
Step 5: Solve.
\( (1.075)^2 = 1.155625 \)
\( A = 16000 \times 1.155625 = 18490 \)
Step 6: Calculate Compound Interest.
\( CI = A – P \)
\( CI = 18490 – 16000 = 2490 \)
Answer: Amount = ₹18490, Compound Interest = ₹2490

Q8: Find the amount and the compound interest on ₹125000 for \(1\frac{1}{2}\) years at 12% per annum, compounded half-yearly.

Step 1: Write the given values.
Principal \( P = 125000 \)
Annual Rate \( R = 12\% \)
Time \( T = 1\frac{1}{2} = 1.5 \text{ years} \)
Step 2: Convert into half-yearly terms.
Rate per half-year \( = \frac{12}{2} = 6\% \)
Number of periods \( = 1.5 \times 2 = 3 \)
Step 3: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^n \)
Step 4: Substitute values.
\( A = 125000 \left(1 + \frac{6}{100} \right)^3 \)
\( A = 125000 (1.06)^3 \)
Step 5: Solve step-by-step.
\( (1.06)^2 = 1.1236 \)
\( (1.06)^3 = 1.1236 \times 1.06 = 1.191016 \)
Step 6: Calculate amount.
\( A = 125000 \times 1.191016 \)
\( A = 148877 \) (approx.)
Step 7: Calculate Compound Interest.
\( CI = A – P \)
\( CI = 148877 – 125000 = 23877 \)
Answer: Amount = ₹148877 (approx.), Compound Interest = ₹23877

Q9: A sum of ₹12,500 is deposited for \(1\frac{1}{2}\) years, compounded half yearly. It amounts to ₹13,000 at the end of first half year. Find:

i. The rate of interest

Step 1: Write the given values.
Principal \( P = 12500 \)
Amount after first half-year \( A = 13000 \)
Step 2: Use compound formula for one half-year.
\( A = P \left(1 + \frac{R}{100} \right) \)
Step 3: Substitute values.
\( 13000 = 12500 \left(1 + \frac{R}{100} \right) \)
Step 4: Solve.
\( \frac{13000}{12500} = 1 + \frac{R}{100} \)
\( 1.04 = 1 + \frac{R}{100} \)
\( 0.04 = \frac{R}{100} \)
\( R = 4\% \) (per half-year)
Step 5: Convert to annual rate.
Annual rate \( = 4\% \times 2 = 8\% \)
Answer: Rate = 8% per annum

ii. the final amount. Give your answer correct to the nearest rupee.

Step 1: Total time \( = 1.5 \) years \( = 3 \) half-years
Step 2: Use compound formula for one half-year.
\( A = P \left(1 + \frac{R}{100} \right) \)
Step 3: Substitute values.
\( A = 12500 \left(1 + \frac{4}{100} \right)^3 \)
\( A = 12500 \left(1 + 0.04 \right)^3 \)
\( A = 12500 \left(1.04 \right)^3 \)
Step 4: Solve step-by-step.
\( (1.04)^2 = 1.0816 \)
\( (1.04)^3 = 1.0816 \times 1.04 = 1.124864 \)
Step 5: Final amount.
\( A = 12500 \times 1.124864 = 14060.8 \)
Nearest rupee \( = 14061 \)
Answer: Final Amount = ₹14061

Q10: The simple interest on a sum of money at 12% per annum for 1 year is ₹900. Find:

i. the sum of money

Step 1: Use simple interest formula.
\( SI = \frac{P \times R \times T}{100} \)
Step 2: Substitute given values.
\( 900 = \frac{P \times 12 \times 1}{100} \)
Step 3: Simplify.
\( 900 = \frac{12P}{100} \)
\( 900 = \frac{3P}{25} \)
Step 4: Solve for \( P \).
\( P = \frac{900 \times 25}{3} \)
\( P = 7500 \)
Answer: Sum of money = ₹7500

ii. the compound interest on this sum for 1 year, payable half-yearly at the same rate

Step 1: Given values.
\( P = 7500 \), \( R = 12\% \), Time \( = 1 \) year
Step 2: Convert to half-yearly terms.
Rate per half-year \( = \frac{12}{2} = 6\% \)
Number of periods \( = 2 \)
Step 3: Calculate amount.
\( A = 7500 \left(1 + \frac{6}{100} \right)^2 \)
\( A = 7500 (1.06)^2 \)
Step 4: Solve.
\( (1.06)^2 = 1.1236 \)
\( A = 7500 \times 1.1236 = 8427 \)
Step 5: Calculate compound interest.
\( CI = A – P \)
\( CI = 8427 – 7500 = 927 \)
Answer: Compound Interest = ₹927

Q11: What sum of money will amount to ₹18150 in 2 years at 10% per annum, compounded annually?

Step 1: Write the given values.
Amount \( A = 18150 \)
Rate \( R = 10\% \)
Time \( T = 2 \) years
Step 2: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute values.
\( 18150 = P (1.1)^2 \)
\( 18150 = P \times 1.21 \)
Step 4: Solve for \( P \).
\( P = \frac{18150}{1.21} \)
\( P = 15000 \)
Answer: Required sum = ₹15000

Q12: What sum of money will amount to ₹93170 in 3 years at 10% per annum, compounded annually?

Step 1: Write the given values.
Amount \( A = 93170 \)
Rate \( R = 10\% \)
Time \( T = 3 \) years
Step 2: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute values.
\( 93170 = P (1.1)^3 \)
\( (1.1)^3 = 1.331 \)
So, \( 93170 = P \times 1.331 \)
Step 4: Solve for \( P \).
\( P = \frac{93170}{1.331} \)
\( P = 70000 \)
Answer: Required sum = ₹70000

Q13: On what sum of money will the compound interest for 2 years at 8% per annum be ₹7488?

Step 1: Write the given values.
Compound Interest \( CI = 7488 \)
Rate \( R = 8\% \)
Time \( T = 2 \) years
Step 2: Use relation:
\( CI = A – P \)
and \( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute in formula.
\( CI = P \left(1 + \frac{R}{100} \right)^T – P \)
\( 7488 = P (1.08)^2 – P \)
Step 4: Simplify.
\( (1.08)^2 = 1.1664 \)
\( 7488 = P (1.1664 – 1) \)
\( 7488 = P (0.1664) \)
Step 5: Solve for \( P \).
\( P = \frac{7488}{0.1664} \)
\( P = 45000 \)
Answer: Required sum = ₹45000

Q14: The difference between the simple interest and the compound interest on a sum of money for 2 years at 12% per annum is ₹216. Find the sum.

i. Express Simple Interest (SI) using Formula

Step 1: Identify given values and set up the Simple Interest equation profile.
Let the Principal sum of money be \( P \).
Rate of Interest \( (R) = 12\% \text{ per annum} \)
Time \( (T) = 2 \text{ years} \)
Step 2: Compute the expression for total Simple Interest over 2 years.
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( \text{SI} = \frac{P \times 12 \times 2}{100} \)
\( \text{SI} = \frac{24P}{100} \)

ii. Express Compound Interest (CI) using Formula

Step 1: Write down the compound interest formula in terms of Principal (P).
Formula: \( \text{CI} = P \times \left[\left(1 + \frac{R}{100}\right)^n – 1\right] \)
Step 2: Substitute the given values into the formula framework.
\( \text{CI} = P \times \left[\left(1 + \frac{12}{100}\right)^2 – 1\right] \)
\( \text{CI} = P \times \left[\left(\frac{112}{100}\right)^2 – 1\right] \)
\( \text{CI} = P \times \left[\frac{12544}{10000} – 1\right] \)
\( \text{CI} = P \times \left[\frac{12544 – 10000}{10000}\right] \)
\( \text{CI} = \frac{2544P}{10000} \)

iii. Solve for Principal (P) using the Difference Relation

Step 1: Set up the equation for the difference between CI and SI.
Given: \( \text{CI} – \text{SI} = 216 \)
\( \frac{2544P}{10000} – \frac{24P}{100} = 216 \)
Step 2: Match denominators to simplify the fractions.
\( \frac{2544P}{10000} – \frac{2400P}{10000} = 216 \)
\( \frac{144P}{10000} = 216 \)
Step 3: Isolate and solve for the value of P.
\( 144P = 216 \times 10000 \)
\( P = \frac{2160000}{144} \)
\( P = 15000 \)
Answer:The sum is ₹15000

Q15: The difference between the simple interest and the compound interest on a sum of money for 3 years at 10% per annum is ₹558. Find the sum.

i. Express Simple Interest (SI) using Formula

Step 1: Identify the given values and set up the Simple Interest expression in terms of Principal (P).
Let the Principal sum of money be \( P \).
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Time \( (T) = 3 \text{ years} \)
Step 2: Compute the expression for total Simple Interest over 3 years.
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( \text{SI} = \frac{P \times 10 \times 3}{100} \)
\( \text{SI} = \frac{30P}{100} = \frac{3P}{10} \)

ii. Express Compound Interest (CI) using Formula

Step 1: Write down the compound interest formula in terms of Principal (P).
Formula: \( \text{CI} = P \times \left[\left(1 + \frac{R}{100}\right)^n – 1\right] \)
Step 2: Substitute the given rate and time period into the formula layout.
\( \text{CI} = P \times \left[\left(1 + \frac{10}{100}\right)^3 – 1\right] \)
\( \text{CI} = P \times \left[\left(\frac{110}{100}\right)^3 – 1\right] \)
\( \text{CI} = P \times \left[\left(\frac{11}{10}\right)^3 – 1\right] \)
\( \text{CI} = P \times \left[\frac{1331}{1000} – 1\right] \)
\( \text{CI} = P \times \left[\frac{1331 – 1000}{1000}\right] \)
\( \text{CI} = \frac{331P}{1000} \)

iii. Solve for Principal (P) using the Difference Relation

Step 1: Set up the equation for the difference between Compound Interest and Simple Interest.
Given: \( \text{CI} – \text{SI} = 558 \)
\( \frac{331P}{1000} – \frac{3P}{10} = 558 \)
Step 2: Balance denominators to simplify the fractions.
\( \frac{331P}{1000} – \frac{300P}{1000} = 558 \)
\( \frac{31P}{1000} = 558 \)
Step 3: Isolate and solve for the value of P.
\( 31P = 558 \times 1000 \)
\( P = \frac{558000}{31} \)
\( P = 18000 \)
Answer:The sum is ₹18000

Q16: The difference between the compound interest for 1 year, compounded half-yearly and the simple interest for 1 year on a certain sum of money at 10% per annum is ₹360. Find the sum.

i. Express Simple Interest (SI) using Formula

Step 1: Identify the given values and set up the Simple Interest expression in terms of Principal (P).
Let the Principal sum of money be \( P \).
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Time \( (T) = 1 \text{ year} \)
Step 2: Compute the expression for total Simple Interest over 1 year.
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( \text{SI} = \frac{P \times 10 \times 1}{100} \)
\( \text{SI} = \frac{10P}{100} \)

ii. Express Compound Interest (CI) using Formula

Step 1: Adjust the rate and time period for half-yearly compounding.
Annual Rate \( = 10\% \)
\( \text{Half-yearly Rate } (R’) = \frac{10\%}{2} = 5\% \text{ per half-year} \)
Time \( = 1 \text{ year} = 2 \text{ half-yearly periods } (n = 2) \)
Step 2: Write down the compound interest formula for half-yearly compounding.
Formula: \( \text{CI} = P \times \left[\left(1 + \frac{R’}{100}\right)^n – 1\right] \)
\( \text{CI} = P \times \left[\left(1 + \frac{5}{100}\right)^2 – 1\right] \)
\( \text{CI} = P \times \left[\left(\frac{105}{100}\right)^2 – 1\right] \)
\( \text{CI} = P \times \left[\left(\frac{21}{20}\right)^2 – 1\right] \)
\( \text{CI} = P \times \left[\frac{441}{400} – 1\right] \)
\( \text{CI} = P \times \left[\frac{441 – 400}{400}\right] \)
\( \text{CI} = \frac{41P}{400} \)

iii. Solve for Principal (P) using the Difference Relation

Step 1: Set up the equation for the difference between Compound Interest and Simple Interest.
Given: \( \text{CI} – \text{SI} = 360 \)
\( \frac{41P}{400} – \frac{10P}{100} = 360 \)
Step 2: Balance denominators to simplify the fractions.
\( \frac{41P}{400} – \frac{40P}{400} = 360 \)
\( \frac{P}{400} = 360 \)
Step 3: Isolate and solve for the value of P.
\( P = 360 \times 400 \)
\( P = 144000 \)
Answer:The sum is ₹144000

Q17: At what rate per cent per annum compound interest will ₹6250 amount to ₹7290 in 2 years?

i. Set Up the Compound Interest Amount Formula

Step 1: Identify the given values from the problem statement.
Principal \( (P) = \text{₹}6250 \)
Amount \( (A) = \text{₹}7290 \)
Time \( (n) = 2 \text{ years} \)
Let the required rate of interest be \( R\% \) per annum.
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 Rate of Interest (R)

Step 1: Substitute the given values into the formula framework.
\( 7290 = 6250 \times \left(1 + \frac{R}{100}\right)^2 \)
Step 2: Isolate the exponential term by dividing both sides by the principal.
\( \frac{7290}{6250} = \left(1 + \frac{R}{100}\right)^2 \)
\( \frac{729}{625} = \left(1 + \frac{R}{100}\right)^2 \)
Step 3: Express the fraction on the left side as a perfect square.
We know that \( 729 = 27^2 \) and \( 625 = 25^2 \).
\( \left(\frac{27}{25}\right)^2 = \left(1 + \frac{R}{100}\right)^2 \)
Step 4: Take the square root on both sides to eliminate the exponents.
\( \frac{27}{25} = 1 + \frac{R}{100} \)
\( \frac{R}{100} = \frac{27}{25} – 1 \)
\( \frac{R}{100} = \frac{27 – 25}{25} \)
\( \frac{R}{100} = \frac{2}{25} \)
Step 5: Solve for the final value of R.
\( R = \frac{2 \times 100}{25} \)
\( R = 2 \times 4 \)
\( R = 8\% \)
Answer:The rate of interest is 8% per annum

Q18: At what rate per cent per annum will ₹3000 amount to ₹3993 in 3 years, the interest being compounded annually?

Step 1: Write the given values.
Principal \( P = 3000 \)
Amount \( A = 3993 \)
Time \( T = 3 \) years
Step 2: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute the values.
\( 3993 = 3000 \left(1 + \frac{R}{100} \right)^3 \)
Step 4: Simplify.
\( \frac{3993}{3000} = \left(1 + \frac{R}{100} \right)^3 \)
\( 1.331 = \left(1 + \frac{R}{100} \right)^3 \)
Step 5: Take cube root.
\( 1 + \frac{R}{100} = \sqrt[3]{1.331} = 1.1 \)
Step 6: Solve for \( R \).
\( \frac{R}{100} = 1.1 – 1 = 0.1 \)
\( R = 10\% \)
Answer: Rate = 10% per annum

Q19: In what time will ₹5120 amount to ₹7290 at \(12\frac{1}{2}\)% per annum, compounded annually?

Step 1: Write the given values.
Principal \( P = 5120 \)
Amount \( A = 7290 \)
Rate \( R = 12\frac{1}{2}\% = 12.5\% \)
Step 2: Use compound interest formula.
\( A = P \left(1 + \frac{R}{100} \right)^T \)
Step 3: Substitute values.
\( 7290 = 5120 \left(1 + \frac{12.5}{100} \right)^T \)
\( 7290 = 5120 \left(1 + \frac{1}{8} \right)^T \)
Step 4: Simplify.
\( \frac{7290}{5120} = \left( \frac{9}{8} \right)^T \)
\( \frac{729}{512} = \left( \frac{9}{8} \right)^T \)
\( \left( \frac{9}{8} \right)^3 = \left( \frac{9}{8} \right)^T \)
Step 5: Find value of \( T \).
Check powers:
\( T = 3 \) years
Answer: Time = 3 years

Q20: A certain sum of money amounts to ₹7260 in 2 years and to ₹7986 in 3 years, interest being compounded annually Find the rate per cent per annum.

i. Set Up Amount Equations for Different Years

Step 1: Identify the variables and write the formula.
Let the principal sum be \( P \) and the rate of interest be \( R\% \) per annum.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
Step 2: Create Equation 1 for the amount at the end of 2 years.
\( 7260 = P \times \left(1 + \frac{R}{100}\right)^2 \) —- (Equation 1)
Step 3: Create Equation 2 for the amount at the end of 3 years.
\( 7986 = P \times \left(1 + \frac{R}{100}\right)^3 \) —- (Equation 2)

ii. Solve for the Rate of Interest (R)

Step 1: Divide Equation 2 by Equation 1 to isolate the terms with R.
\( \frac{7986}{7260} = \frac{P \times \left(1 + \frac{R}{100}\right)^3}{P \times \left(1 + \frac{R}{100}\right)^2} \)
Step 2: Simplify the algebraic fraction expression.
The terms \( P \) cancel out, and the power reduces to 1.
\( \frac{7986}{7260} = 1 + \frac{R}{100} \)
Step 3: Reduce the numerical fraction by dividing both parts by 726.
\( \frac{11}{10} = 1 + \frac{R}{100} \)
\( 1.1 = 1 + \frac{R}{100} \)
Step 4: Isolate and compute the final value of R.
\( \frac{R}{100} = 1.1 – 1 \)
\( \frac{R}{100} = 0.1 \)
\( R = 0.1 \times 100 \)
\( R = 10\% \)
Answer:The rate of interest is 10% per annum