Compound Interest (Stage-2)

Q1: Multiple Choice Type:

a. On ₹ 6,000, the difference between C.I. and S.I. in 2 years and at 10% interest, compounded per year, is:

Step 1: Identify the given values from the statement: Principal \( (P) = \text{₹ } 6,000 \), Time \( (n) = 2 \) years, Rate \( (r) = 10\% \) p.a.
Step 2: Calculate Simple Interest (S.I.): \[ \text{S.I.} = \frac{P \times r \times t}{100} = \frac{6000 \times 10 \times 2}{100} = \text{₹ } 1,200 \] Step 3: Calculate Amount (A) for Compound Interest: \[ A = P \left(1 + \frac{r}{100}\right)^n = 6000 \left(1 + \frac{10}{100}\right)^2 = 6000 \left(\frac{11}{10}\right)^2 \\ A = 6000 \times \frac{121}{100} = 60 \times 121 = \text{₹ } 7,260 \] Step 4: Calculate Compound Interest (C.I.): \[ \text{C.I.} = A – P = 7260 – 6000 = \text{₹ } 1,260 \] Step 5: Find the difference between C.I. and S.I.: \[ \text{Difference} = \text{C.I.} – \text{S.I.} = 1260 – 1200 = \text{₹ } 60 \] Answer: iii. ₹ 60

b. ₹ 10,000 amounts to ₹ 12,500 in one year. The rate of interest per year is:

Step 1: Extract given values: Principal \( (P) = \text{₹ } 10,000 \), Amount \( (A) = \text{₹ } 12,500 \), Time \( (n) = 1 \) year.
Step 2: For a single year period, use the formula: \[ A = P \left(1 + \frac{r}{100}\right) \] Step 3: Insert the values and isolate the rate term: \[ 12500 = 10000 \left(1 + \frac{r}{100}\right) \\ \frac{12500}{10000} = 1 + \frac{r}{100} \\ 1.25 = 1 + \frac{r}{100} \] Step 4: Subtract 1 and multiply by 100: \[ \frac{r}{100} = 0.25 \\ r = 25\% \] Answer: iv. 25%

c. The C.I. on ₹ 16,000 in 2 years at the rate of 20% per annum is:

Step 1: Identify values: Principal \( (P) = \text{₹ } 16,000 \), Time \( (n) = 2 \) years, Rate \( (r) = 20\% \) p.a.
Step 2: Set up the compound amount formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \] Step 3: Substitute values and simplify: \[ A = 16000 \left(1 + \frac{20}{100}\right)^2 = 16000 \left(1 + \frac{1}{5}\right)^2 = 16000 \left(\frac{6}{5}\right)^2 \\ A = 16000 \times \frac{36}{25} = 640 \times 36 = \text{₹ } 23,040 \] Step 4: Determine Compound Interest: \[ \text{C.I.} = A – P = 23040 – 16000 = \text{₹ } 7,040 \] Answer: ii. ₹ 7,040

d. Simple interest, at the same rate for the same period as given above in part (c) is:

Step 1: Use the same parameters: Principal \( (P) = \text{₹ } 16,000 \), Rate \( (r) = 20\% \) p.a., Time \( (t) = 2 \) years.
Step 2: Apply the standard Simple Interest formula: \[ \text{S.I.} = \frac{P \times r \times t}{100} \] Step 3: Calculate the final simple interest amount: \[ \text{S.I.} = \frac{16000 \times 20 \times 2}{100} = 160 \times 40 = \text{₹ } 6,400 \] Answer: ii. ₹ 6,400

e. The difference between C.I. and S.I. in 2 years as given above for parts (c) and (d) is:

Step 1: Fetch values calculated in previous parts: \( \text{C.I.} = \text{₹ } 7,040 \) and \( \text{S.I.} = \text{₹ } 6,400 \).
Step 2: Write down the variation relation formula: \[ \text{Difference} = \text{C.I.} – \text{S.I.} \] Step 3: Perform the subtraction to find the solution: \[ \text{Difference} = 7040 – 6400 = \text{₹ } 640 \] Answer: i. ₹ 640

Q2: The difference between simple interest and compound interest on a certain sum is ₹ 54.40 for 2 years at 8 percent per annum. Find the sum.

Step 1: Let the initial cost or sum of money be ₹ x.
Step 2: Identify the given parameters from the problem: Rate \( (r) = 8\% \) p.a., Time \( (n \text{ or } t) = 2 \) years, and Difference \( (\text{C.I.} – \text{S.I.}) = \text{₹ } 54.40 \).
Step 3: Write out the formula for Simple Interest (S.I.) in terms of x: \[ \text{S.I.} = \frac{x \times r \times t}{100} \\ \text{S.I.} = \frac{x \times 8 \times 2}{100} = \frac{16x}{100} \] Step 4: Write out the formula for Compound Interest (C.I.) in terms of x: \[ \text{C.I.} = x \left[\left(1 + \frac{r}{100}\right)^n – 1\right] \] Step 5: Substitute the parameters into the compound interest expression: \[ \text{C.I.} = x \left[\left(1 + \frac{8}{100}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\left(\frac{108}{100}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\frac{11664}{10000} – 1\right] \\ \text{C.I.} = x \left[\frac{11664 – 10000}{10000}\right] = \frac{1664x}{10000} \] Step 6: Formulate the difference equation between C.I. and S.I.: \[ \text{C.I.} – \text{S.I.} = 54.40 \\ \frac{1664x}{10000} – \frac{16x}{100} = 54.40 \] Step 7: Make the denominators common to solve the algebraic expression: \[ \frac{1664x – 1600x}{10000} = 54.40 \\ \frac{64x}{10000} = 54.40 \] Step 8: Rearrange the terms to isolate and solve for x: \[ 64x = 54.40 \times 10000 \\ 64x = 544000 \\ x = \frac{544000}{64} \\ x = 8500 \] Answer: The sum is ₹ 8,500.

Q3: Pramod and anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand received ₹ 15 more than Pramod. Find the amount of money lent by each and the interest received.

Step 1: Let the initial sum of money lent by each be ₹ x.
Step 2: Identify the given values from the problem: Rate \( (r) = 5\% \) p.a., Time \( (n \text{ or } t) = 2 \) years.
Step 3: Calculate the Simple Interest (S.I.) received by Pramod in terms of x: \[ \text{S.I.} = \frac{x \times r \times t}{100} \\ \text{S.I.} = \frac{x \times 5 \times 2}{100} = \frac{10x}{100} = \frac{x}{10} \] Step 4: Write out the formula for the Compound Interest (C.I.) received by Anand in terms of x: \[ \text{C.I.} = x \left[\left(1 + \frac{r}{100}\right)^n – 1\right] \] Step 5: Substitute the values into the compound interest expression: \[ \text{C.I.} = x \left[\left(1 + \frac{5}{100}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\left(1 + \frac{1}{20}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\left(\frac{21}{20}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\frac{441}{400} – 1\right] \\ \text{C.I.} = x \left[\frac{441 – 400}{400}\right] = \frac{41x}{400} \] Step 6: Anand received ₹ 15 more than Pramod, so the difference between C.I. and S.I. is ₹ 15: \[ \text{C.I.} – \text{S.I.} = 15 \\ \frac{41x}{400} – \frac{x}{10} = 15 \] Step 7: Equate the denominators to solve the algebraic equation: \[ \frac{41x – 40x}{400} = 15 \\ \frac{x}{400} = 15 \] Step 8: Multiply to find the value of x: \[ x = 15 \times 400 \\ x = 6000 \] Step 9: Calculate the interest received by Pramod (Simple Interest): \[ \text{S.I.} = \frac{6000}{10} = \text{₹ } 600 \] Step 10: Calculate the interest received by Anand (Compound Interest): \[ \text{C.I.} = \frac{41 \times 6000}{400} = 41 \times 15 = \text{₹ } 615 \] Answer: Sum lent by each = ₹ 6,000; Interest received by Pramod = ₹ 600; Interest received by Anand = ₹ 615.

Q4: Simple interest on a sum of money for 2 years at 4% is ₹ 450. Find the compound interest on the same sum and at the same rate for 2 years.

Step 1: Let the initial sum of money (principal) be ₹ x.
Step 2: Identify the given values from the problem statement.
Step 3: Simple Interest \( (\text{S.I.}) = \text{₹ } 450 \), Rate \( (r) = 4\% \) p.a., Time \( (t \text{ or } n) = 2 \) years.
Step 4: Write down the simple interest formula to calculate x: \[ \text{S.I.} = \frac{x \times r \times t}{100} \] Step 5: Substitute the known parameters into the equation: \[ 450 = \frac{x \times 4 \times 2}{100} \\ 450 = \frac{8x}{100} \] Step 6: Isolate x to find the initial principal amount: \[ x = \frac{450 \times 100}{8} \\ x = \frac{45000}{8} \\ x = 5625 \] Step 7: Now, apply the standard compound interest formula on this principal: \[ \text{C.I.} = P \left[\left(1 + \frac{r}{100}\right)^n – 1\right] \] Step 8: Substitute \( P = 5625 \), \( r = 4 \), and \( n = 2 \) into the expression: \[ \text{C.I.} = 5625 \left[\left(1 + \frac{4}{100}\right)^2 – 1\right] \] Step 9: Simplify the fractional value inside the brackets: \[ \text{C.I.} = 5625 \left[\left(1 + \frac{1}{25}\right)^2 – 1\right] \\ \text{C.I.} = 5625 \left[\left(\frac{26}{25}\right)^2 – 1\right] \] Step 10: Expand the squared term to execute subtraction: \[ \text{C.I.} = 5625 \left[\frac{676}{625} – 1\right] \\ \text{C.I.} = 5625 \left[\frac{676 – 625}{625}\right] \\ \text{C.I.} = 5625 \left[\frac{51}{625}\right] \] Step 11: Perform the final division and multiplication step where \( 5625 \div 625 = 9 \): \[ \text{C.I.} = 9 \times 51 \\ \text{C.I.} = 459 \] Answer: Compound Interest = ₹ 459

Q5: Compound interest on a certain sum of money at 5% per annum for two years is ₹ 246. Calculate simple interest on the same sum for 3 years at 6% per annum.

Step 1: Let the initial sum of money (principal) be ₹ x.
Step 2: Identify the given values from the first part of the problem statement.
Step 3: Compound Interest \( (\text{C.I.}) = \text{₹ } 246 \), Rate \( (r_1) = 5\% \) p.a., Time \( (n) = 2 \) years.
Step 4: Apply the standard formula connecting compound interest and principal: \[ \text{C.I.} = x \left[\left(1 + \frac{r_1}{100}\right)^n – 1\right] \] Step 5: Substitute the known parameters into the equation: \[ 246 = x \left[\left(1 + \frac{5}{100}\right)^2 – 1\right] \] Step 6: Simplify the fraction within the brackets: \[ 246 = x \left[\left(1 + \frac{1}{20}\right)^2 – 1\right] \\ 246 = x \left[\left(\frac{21}{20}\right)^2 – 1\right] \] Step 7: Expand the squared fraction term: \[ 246 = x \left[\frac{441}{400} – 1\right] \] Step 8: Simplify the subtraction inside the bracket: \[ 246 = x \left[\frac{441 – 400}{400}\right] \\ 246 = x \left[\frac{41}{400}\right] \] Step 9: Rearrange the expression to isolate and solve for x: \[ x = \frac{246 \times 400}{41} \] Step 10: Perform the division step where \( 246 \div 41 = 6 \): \[ x = 6 \times 400 \\ x = 2400 \] Step 11: Now, identify parameters for the second part of the problem statement.
Step 12: Principal \( (P) = \text{₹ } 2,400 \), New Rate \( (r_2) = 6\% \) p.a., New Time \( (t) = 3 \) years.
Step 13: Write down the standard simple interest formula: \[ \text{S.I.} = \frac{P \times r_2 \times t}{100} \] Step 14: Substitute the parameters into the simple interest equation: \[ \text{S.I.} = \frac{2400 \times 6 \times 3}{100} \] Step 15: Simplify the calculation by canceling out common zeros: \[ \text{S.I.} = 24 \times 18 \\ \text{S.I.} = 432 \] Answer: Simple Interest = ₹ 432

Q6: A sum of money, invested at compound interest, amount to ₹ 19,360 in 2 years and to ₹ 23,425.60 in 4 years. Find the rate percent and the original sum of money.

Step 1: Let the original sum of money (principal) be ₹ x.
Step 2: Identify the given values from the problem statement.
Step 3: Amount after 2 years \( (A_2) = \text{₹ } 19,360 \) and Amount after 4 years \( (A_4) = \text{₹ } 23,425.60 \).
Step 4: Let the rate of interest per annum be \( r \% \).
Step 5: Write down the compound interest amount formulas for both periods: \[ A_2 = x \left(1 + \frac{r}{100}\right)^2 \\ A_4 = x \left(1 + \frac{r}{100}\right)^4 \] Step 6: Substitute the known values into the respective equations: \[ 19360 = x \left(1 + \frac{r}{100}\right)^2 \quad \text{— (Equation 1)} \\ 23425.60 = x \left(1 + \frac{r}{100}\right)^4 \quad \text{— (Equation 2)} \] Step 7: Divide Equation 2 by Equation 1 to eliminate the variable x: \[ \frac{23425.60}{19360} = \frac{x \left(1 + \frac{r}{100}\right)^4}{x \left(1 + \frac{r}{100}\right)^2} \\ 1.21 = \left(1 + \frac{r}{100}\right)^{4 – 2} \\ 1.21 = \left(1 + \frac{r}{100}\right)^2 \] Step 8: Express 1.21 as a fraction to simplify taking the square root: \[ \frac{121}{100} = \left(1 + \frac{r}{100}\right)^2 \] Step 9: Take the square root on both sides of the expression: \[ \sqrt{\frac{121}{100}} = 1 + \frac{r}{100} \\ \frac{11}{10} = 1 + \frac{r}{100} \\ 1.1 = 1 + \frac{r}{100} \] Step 10: Subtract 1 from both sides to isolate the rate fraction term: \[ \frac{r}{100} = 1.1 – 1 \\ \frac{r}{100} = 0.1 \] Step 11: Multiply by 100 to solve for the interest rate percentage value r: \[ r = 0.1 \times 100 \\ r = 10\% \] Step 12: Substitute the value of \( r = 10 \) back into Equation 1 to find the original sum x: \[ 19360 = x \left(1 + \frac{10}{100}\right)^2 \] Step 13: Simplify the terms inside the brackets: \[ 19360 = x \left(1 + \frac{1}{10}\right)^2 \\ 19360 = x \left(\frac{11}{10}\right)^2 \] Step 14: Expand the squared fractional term: \[ 19360 = x \left(\frac{121}{100}\right) \] Step 15: Rearrange the expression to solve directly for original sum x: \[ x = \frac{19360 \times 100}{121} \] Step 16: Perform the division step where \( 19360 \div 121 = 160 \): \[ x = 160 \times 100 \\ x = 16000 \] Answer: Rate of interest = 10% per annum and Original sum of money = ₹ 16,000

Q7: A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

Step 1: Let the initial sum of money (principal) be ₹ x.
Step 2: According to the first condition, the money becomes 3 times itself: Amount \( (A_1) = 3x \) in Time \( (n_1) = 8 \) years.
Step 3: Let the rate of interest per annum be \( r \% \).
Step 4: Write down the standard compound interest amount formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \] Step 5: Substitute the parameters of the first condition into the formula: \[ 3x = x \left(1 + \frac{r}{100}\right)^8 \] Step 6: Divide both sides by x to eliminate the variable: \[ 3 = \left(1 + \frac{r}{100}\right)^8 \quad \text{— (Equation 1)} \] Step 7: According to the second condition, we want the money to become 27 times itself: Amount \( (A_2) = 27x \).
Step 8: Let the required time period for this growth be \( n_2 \) years.
Step 9: Set up the equation for the second condition: \[ 27x = x \left(1 + \frac{r}{100}\right)^{n_2} \] Step 10: Divide both sides by x to simplify the expression: \[ 27 = \left(1 + \frac{r}{100}\right)^{n_2} \quad \text{— (Equation 2)} \] Step 11: Express the number 27 as a base of 3 to link it to Equation 1: \[ 3^3 = \left(1 + \frac{r}{100}\right)^{n_2} \] Step 12: Substitute the value of 3 from Equation 1 into this expression: \[ \left[\left(1 + \frac{r}{100}\right)^8\right]^3 = \left(1 + \frac{r}{100}\right)^{n_2} \] Step 13: Use exponent rules to multiply the powers on the left-hand side: \[ \left(1 + \frac{r}{100}\right)^{8 \times 3} = \left(1 + \frac{r}{100}\right)^{n_2} \\ \left(1 + \frac{r}{100}\right)^{24} = \left(1 + \frac{r}{100}\right)^{n_2} \] Step 14: Compare the exponents on both sides since the geometric bases match exactly: \[ n_2 = 24 \] Answer: Time period = 24 years

Q8: On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on ₹ 9,430 for 10 years, both at the rate of 5 percent per annum?

Step 1: Identify the given values for calculating the Simple Interest (S.I.).
Step 2: Principal for simple interest \( (P_1) = \text{₹ } 9,430 \), Rate \( (r) = 5\% \) p.a., and Time \( (t) = 10 \) years.
Step 3: Calculate the Simple Interest using the formula: \[ \text{S.I.} = \frac{P_1 \times r \times t}{100} \\ \text{S.I.} = \frac{9430 \times 5 \times 10}{100} \\ \text{S.I.} = 943 \times 5 = \text{₹ } 4,715 \] Step 4: According to the problem statement, this interest is equal to the compound interest: Compound Interest \( (\text{C.I.}) = \text{₹ } 4,715 \).
Step 5: Let the required sum of money for compound interest be ₹ x.
Step 6: Identify the parameters for the compound interest part: Rate \( (r) = 5\% \) p.a., Time \( (n) = 2 \) years.
Step 7: Write down the compound interest formula in terms of principal: \[ \text{C.I.} = x \left[\left(1 + \frac{r}{100}\right)^n – 1\right] \] Step 8: Substitute the known parameters into the equation: \[ 4715 = x \left[\left(1 + \frac{5}{100}\right)^2 – 1\right] \] Step 9: Simplify the fractional value inside the bracket term: \[ 4715 = x \left[\left(1 + \frac{1}{20}\right)^2 – 1\right] \\ 4715 = x \left[\left(\frac{21}{20}\right)^2 – 1\right] \] Step 10: Expand the squared fraction expression: \[ 4715 = x \left[\frac{441}{400} – 1\right] \] Step 11: Perform the subtraction step inside the brackets: \[ 4715 = x \left[\frac{441 – 400}{400}\right] \\ 4715 = x \left[\frac{41}{400}\right] \] Step 12: Rearrange the expression terms to isolate and solve for x: \[ x = \frac{4715 \times 400}{41} \] Step 13: Perform the division step where \( 4715 \div 41 = 115 \): \[ x = 115 \times 400 \\ x = 46000 \] Answer: The required sum of money is ₹ 46,000.

Q9: Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228. Find the sum.

Step 1: Let the initial cost or sum of money be ₹ x.
Step 2: Identify the parameters for the simple interest calculation: Time \( (t) = 4 \) years, Rate \( (r_1) = 4\% \) p.a.
Step 3: Formulate the Simple Interest (S.I.) equation using the standard formula: \[ \text{S.I.} = \frac{x \times r_1 \times t}{100} \\ \text{S.I.} = \frac{x \times 4 \times 4}{100} = \frac{16x}{100} \] Step 4: Identify the parameters for the compound interest calculation: Time \( (n) = 3 \) years, Rate \( (r_2) = 5\% \) p.a.
Step 5: Write out the formula for Compound Interest (C.I.) in terms of x: \[ \text{C.I.} = x \left[\left(1 + \frac{r_2}{100}\right)^n – 1\right] \] Step 6: Substitute the known compound interest values into the equation: \[ \text{C.I.} = x \left[\left(1 + \frac{5}{100}\right)^3 – 1\right] \\ \text{C.I.} = x \left[\left(1 + \frac{1}{20}\right)^3 – 1\right] \\ \text{C.I.} = x \left[\left(\frac{21}{20}\right)^3 – 1\right] \] Step 7: Expand the cubed fraction expression inside the bracket: \[ \text{C.I.} = x \left[\frac{9261}{8000} – 1\right] \\ \text{C.I.} = x \left[\frac{9261 – 8000}{8000}\right] = \frac{1261x}{8000} \] Step 8: According to the problem statement, simple interest exceeds compound interest by ₹ 228: \[ \text{S.I.} – \text{C.I.} = 228 \\ \frac{16x}{100} – \frac{1261x}{8000} = 228 \] Step 9: Make the denominators common by multiplying the first fraction by \( \frac{80}{80} \): \[ \frac{16x \times 80}{100 \times 80} – \frac{1261x}{8000} = 228 \\ \frac{1280x}{8000} – \frac{1261x}{8000} = 228 \\ \frac{1280x – 1261x}{8000} = 228 \\ \frac{19x}{8000} = 228 \] Step 10: Rearrange the terms to isolate and solve for sum x: \[ 19x = 228 \times 8000 \\ x = \frac{228 \times 8000}{19} \] Step 11: Perform the final division step where \( 228 \div 19 = 12 \): \[ x = 12 \times 8000 \\ x = 96000 \] Answer: The sum of money is ₹ 96,000.

Q10: A certain sum of money amounts to ₹ 23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest.

Step 1: Let the initial sum of money (principal) be ₹ x.
Step 2: Identify the given values from the simple interest part: Amount \( (A_{\text{S.I.}}) = \text{₹ } 23,400 \), Time \( (t) = 3 \) years, and Rate \( (r) = 10\% \) p.a.
Step 3: Write down the simple interest amount formula: \[ A_{\text{S.I.}} = x + \text{S.I.} = x + \frac{x \times r \times t}{100} \\ A_{\text{S.I.}} = x \left(1 + \frac{r \times t}{100}\right) \] Step 4: Substitute the known simple interest parameters into the equation: \[ 23400 = x \left(1 + \frac{10 \times 3}{100}\right) \\ 23400 = x \left(1 + \frac{30}{100}\right) \\ 23400 = x \left(\frac{130}{100}\right) \] Step 5: Rearrange the terms to solve for the principal sum x: \[ x = \frac{23400 \times 100}{130} \\ x = \frac{234000}{13} \\ x = 18000 \] Step 6: Now, identify parameters for the second part of the problem to calculate the compound interest amount.
Step 7: Principal \( (P) = \text{₹ } 18,000 \), Rate \( (r) = 10\% \) p.a., and Time \( (n) = 2 \) years.
Step 8: Apply the standard compound interest amount formula: \[ A_{\text{C.I.}} = P \left(1 + \frac{r}{100}\right)^n \] Step 9: Substitute the values into the compound amount formula: \[ A_{\text{C.I.}} = 18000 \left(1 + \frac{10}{100}\right)^2 \] Step 10: Simplify the fraction value within the brackets: \[ A_{\text{C.I.}} = 18000 \left(1 + \frac{1}{10}\right)^2 \\ A_{\text{C.I.}} = 18000 \left(\frac{11}{10}\right)^2 \] Step 11: Expand the squared fraction expression: \[ A_{\text{C.I.}} = 18000 \times \frac{121}{100} \] Step 12: Simplify by canceling out the zeros from the denominator: \[ A_{\text{C.I.}} = 180 \times 121 \\ A_{\text{C.I.}} = 21780 \] Answer: The required amount under compound interest is ₹ 21,780.

Q11: Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12,600 at the end of the first year and ₹ 17,640 at the end of the second year. Find the sum borrowed.

Step 1: Let the initial sum borrowed by Mohit be ₹ x.
Step 2: Identify the given values from the problem statement: Rate of interest \( (r) = 5\% \) per annum.
Step 3: Calculate the total accumulated amount due at the end of the first year before payment: \[ \text{Amount after 1 year} = x \left(1 + \frac{5}{100}\right) = x \left(1 + \frac{1}{20}\right) = \frac{21x}{20} \] Step 4: Subtract the first year repayment of ₹ 12,600 to find the remaining principal balance: \[ \text{Remaining Principal balance} = \frac{21x}{20} – 12600 \] Step 5: This balance acts as the principal for the second year. Compute the final amount due at the end of the second year: \[ \text{Amount after 2 years} = \left(\frac{21x}{20} – 12600\right) \left(1 + \frac{5}{100}\right) \\ \text{Amount after 2 years} = \left(\frac{21x}{20} – 12600\right) \left(\frac{21}{20}\right) \] Step 6: According to the problem statement, this final amount is completely cleared by paying ₹ 17,640: \[ \left(\frac{21x}{20} – 12600\right) \left(\frac{21}{20}\right) = 17640 \] Step 7: Isolate the term inside the bracket by multiplying by \( \frac{20}{21} \) on the right side: \[ \frac{21x}{20} – 12600 = \frac{17640 \times 20}{21} \] Step 8: Perform the division step where \( 17640 \div 21 = 840 \): \[ \frac{21x}{20} – 12600 = 840 \times 20 \\ \frac{21x}{20} – 12600 = 16800 \] Step 9: Transpose the numerical constant term to isolate the variable expression: \[ \frac{21x}{20} = 16800 + 12600 \\ \frac{21x}{20} = 29400 \] Step 10: Rearrange the remaining terms to find the initial sum borrowed x: \[ x = \frac{29400 \times 20}{21} \] Step 11: Perform the final division and multiplication step where \( 29400 \div 21 = 1400 \): \[ x = 1400 \times 20 \\ x = 28000 \] Answer: The sum borrowed is ₹ 28,000.