Compound Interest (Stage-2)

Q1: Multiple Choice Type:

a. When the present population (P) of a certain locality increases by r% per year, the population in n years will be:

Step 1: Identify that population growth matches the mathematical principle of compounding interest.
Step 2: The starting value is defined as the present population (P).
Step 3: The compound compounding growth factor added each year equals \( \left(1 + \frac{r}{100}\right) \).
Step 4: Apply this factor exponentially across a duration of n years: \[ \text{Population in n years} = P \left(1 + \frac{r}{100}\right)^n \] Answer: iii. \(P\left(1+\frac{\operatorname{r}}{100}\right)^n\)

b. The population of a town decreases by 10% in a particular year and then increases by 15% in the next year. The population at the end of two years is:

Step 1: Note that the rate factor alters across two successive annual timelines.
Step 2: For the initial year, the shrinkage multiplier equals \( \left(1 – \frac{10}{100}\right) \).
Step 3: For the secondary year, the upward growth multiplier equals \( \left(1 + \frac{15}{100}\right) \).
Step 4: Combine the successive year multipliers chronologically to determine the compound scale factor.
Answer: ii. \(\left(1-\frac{10}{100}\right)\left(1+\frac{15}{100}\right)\) times

c. When the cost of a machine decreases by r% per year; the cost of machine in 3 years is:

Step 1: Note that asset value shrinkage represents a negative compound change (depreciation).
Step 2: The annual shrinking compounding factor is written as \( \left(1 – \frac{r}{100}\right) \).
Step 3: Apply the multi-year scale factor exponentially to match a timeframe of 3 years.
Answer: ii. \(\left(1-\frac{\operatorname{r}}{100}\right)^3\) times

d. On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years. Then the amount after 5 years is:

Step 1: Break the multi-year timeline down by individual interest rate rules.
Step 2: For the first 2 years, apply the compounding expression based on x%: \( \left(1 + \frac{x}{100}\right)^2 \).
Step 3: For the next 3 years, apply the compounding expression based on y%: \( \left(1 + \frac{y}{100}\right)^3 \).
Step 4: Multiply the separate timeline components together to form the complete 5-year relationship multiplier.
Answer: iii. \(\left(1+\frac{\operatorname{x}}{100}\right)^2\left(1+\frac{\operatorname{y}}{100}\right)^3\) times

e. The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years. Then cost of machine becomes:

Step 1: Let the initial cost value of the asset be ₹ x.
Step 2: For the first 2 years, scale by the positive growth factor: \( \left(1 + \frac{20}{100}\right)^2 = \left(\frac{120}{100}\right)^2 \).
Step 3: For the subsequent 2 years, scale by the asset depreciation factor: \( \left(1 – \frac{25}{100}\right)^2 = \left(\frac{75}{100}\right)^2 \).
Step 4: Arrange everything chronologically alongside the initial principal parameter expression x.
Answer: iii. ₹ \(x\times\left(\frac{120}{100}\right)^2\times\left(\frac{75}{100}\right)^2\)

Q2: The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value:

i. at the end of 2009:

Step 1: Identify the initial cost of the machine as given at the beginning of 2008.
Step 2: Value at the beginning of 2008 \( (P) = \text{₹ } 44,000 \), Rate of depreciation \( (r) = 12\% \) per annum.
Step 3: Calculate the time duration from the beginning of 2008 to the end of 2009, which spans exactly 2 whole years: Time \( (n) = 2 \) years.
Step 4: Apply the standard formula for asset value depreciation: \[ A = P \left(1 – \frac{r}{100}\right)^n \] Step 5: Substitute the given values into the equation: \[ A = 44000 \left(1 – \frac{12}{100}\right)^2 \] Step 6: Simplify the fraction value within the brackets: \[ A = 44000 \left(\frac{88}{100}\right)^2 \\ A = 44000 \left(\frac{22}{25}\right)^2 \] Step 7: Expand the squared term to perform the calculation: \[ A = 44000 \times \frac{484}{625} \] Step 8: Perform the calculation step-by-step: \[ A = 70.4 \times 484 \\ A = \text{₹ } 34,073.60 \] Answer: Value at the end of 2009 = ₹ 34,073.60

ii. at the beginning of 2007:

Step 9: Let the initial cost of the machine at the beginning of 2007 be ₹ x.
Step 10: The time duration from the beginning of 2007 to the beginning of 2008 is exactly \( 1 \) year.
Step 11: Set up the depreciation relation for a 1-year timeline: \[ \text{Value at the beginning of 2008} = x \left(1 – \frac{12}{100}\right)^1 \] Step 12: Substitute the known value at the beginning of 2008 into our formula expression: \[ 44000 = x \left(\frac{88}{100}\right) \] Step 13: Isolate the variable x to find the previous year’s value: \[ x = \frac{44000 \times 100}{88} \] Step 14: Perform the final division step where \( 44000 \div 88 = 500 \): \[ x = 500 \times 100 \\ x = 50000 \] Answer: Value at the beginning of 2007 = ₹ 50,000

Q3: The value of an article decreased for two years at the rate of 10% per year and then in the third year it increased by 10%. Find the original value of the article, if its value at the end of 3 years is ₹ 40,095.

Step 1: Let the initial cost or original value of the article be ₹ x.
Step 2: Identify the compounding parameters across the three distinct years from the statement.
Step 3: Rate of depreciation for the first 2 years \( (r_1) = 10\% \) per year.
Step 4: Rate of appreciation for the third year \( (r_2) = 10\% \).
Step 5: Given the final value at the end of 3 years \( (A) = \text{₹ } 40,095 \).
Step 6: Formulate the chronological multi-year value equation: \[ A = x \left(1 – \frac{r_1}{100}\right)^2 \left(1 + \frac{r_2}{100}\right) \] Step 7: Substitute the given parameter numerical values into the equation: \[ 40095 = x \left(1 – \frac{10}{100}\right)^2 \left(1 + \frac{10}{100}\right) \] Step 8: Simplify the fraction values inside each bracket: \[ 40095 = x \left(1 – \frac{1}{10}\right)^2 \left(1 + \frac{1}{10}\right) \\ 40095 = x \left(\frac{9}{10}\right)^2 \left(\frac{11}{10}\right) \] Step 9: Expand the squared term fraction block component: \[ 40095 = x \left(\frac{81}{100}\right) \left(\frac{11}{10}\right) \\ 40095 = x \left(\frac{891}{1000}\right) \] Step 10: Rearrange the terms to isolate and solve for the original value x: \[ x = \frac{40095 \times 1000}{891} \] Step 11: Perform the division step where \( 40095 \div 891 = 45 \): \[ x = 45 \times 1000 \\ x = 45000 \] Answer: The original value of the article was ₹ 45,000.

Q4: According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74,088?

Step 1: Identify the given values from the problem statement.
Step 2: Initial Population \( (P) = 64,000 \), Final Population \( (A) = 74,088 \), and Growth Rate \( (r) = 5\% \) per annum.
Step 3: Let the required time period after the year 2009 be \( n \) years.
Step 4: Write down the compound growth formula for population expansion: \[ A = P \left(1 + \frac{r}{100}\right)^n \] Step 5: Substitute the known parameters into the equation: \[ 74088 = 64000 \left(1 + \frac{5}{100}\right)^n \] Step 6: Isolate the exponential component term by dividing both sides by 64,000: \[ \frac{74088}{64000} = \left(1 + \frac{1}{20}\right)^n \\ \frac{74088}{64000} = \left(\frac{21}{20}\right)^n \] Step 7: Reduce the fraction on the left-hand side by dividing the numerator and denominator by 8: \[ \frac{9261}{8000} = \left(\frac{21}{20}\right)^n \] Step 8: Express the simplified fractional fraction block as a perfect power of base \( \frac{21}{20} \): \[ \left(\frac{21}{20}\right)^3 = \left(\frac{21}{20}\right)^n \] Step 9: Compare indices on both sides since the geometric bases match exactly: \[ n = 3 \] Answer: Time period = 3 years

Q5: The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 2,85,120.

Step 1: Let the population of the town at the beginning of 1998 be \( x \).
Step 2: Identify the given values from the problem statement.
Step 3: Rate of decrease during 1998 \( (r_1) = 12\% \), Rate of increase during 1999 \( (r_2) = 8\% \).
Step 4: Given the final population at the end of 1999 \( (A) = 2,85,120 \).
Step 5: Formulate the chronological successive change equation: \[ A = x \left(1 – \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right) \] Step 6: Substitute the given parameter values into the equation: \[ 285120 = x \left(1 – \frac{12}{100}\right)\left(1 + \frac{8}{100}\right) \] Step 7: Simplify the fraction values inside each bracket term: \[ 285120 = x \left(\frac{88}{100}\right)\left(\frac{108}{100}\right) \\ 285120 = x \left(\frac{22}{25}\right)\left(\frac{27}{25}\right) \] Step 8: Multiply the fractional values together: \[ 285120 = x \left(\frac{594}{625}\right) \] Step 9: Rearrange the terms to isolate and solve for the initial population \( x \): \[ x = \frac{285120 \times 625}{594} \] Step 10: Perform the division step where \( 285120 \div 594 = 480 \): \[ x = 480 \times 625 \\ x = 300000 \] Answer: The population of the town at the beginning of 1998 was 3,00,000.

Q6: A sum of money, invested at compound interest, amounts to ₹ 16,500 in 1 year and to ₹ 19,965 in 3 years. Find the rate per cent and the original sum of money invested.

Step 1: Let the original sum of money (principal) invested be ₹ x.
Step 2: Identify the given values from the problem statement.
Step 3: Amount after 1 year \( (A_1) = \text{₹ } 16,500 \) and Amount after 3 years \( (A_3) = \text{₹ } 19,965 \ bag\).
Step 4: Let the required rate of interest per annum be \( r \% \).
Step 5: Write down the compound interest amount formulas for both periods: \[ A_1 = x \left(1 + \frac{r}{100}\right)^1 \\ A_3 = x \left(1 + \frac{r}{100}\right)^3 \] Step 6: Substitute the known values into the respective equations: \[ 16500 = x \left(1 + \frac{r}{100}\right)^1 \quad \text{— (Equation 1)} \\ 19965 = x \left(1 + \frac{r}{100}\right)^3 \quad \text{— (Equation 2)} \] Step 7: Divide Equation 2 by Equation 1 to eliminate the variable x: \[ \frac{19965 = x \left(1 + \frac{r}{100}\right)^3}{16500 = x \left(1 + \frac{r}{100}\right)^1} \\ \frac{19965}{16500} = \left(1 + \frac{r}{100}\right)^{3 – 1} \\ 1.21 = \left(1 + \frac{r}{100}\right)^2 \] Step 8: Express 1.21 as a fractional fraction block to simplify taking the 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: \[ 16500 = x \left(1 + \frac{10}{100}\right)^1 \] Step 13: Simplify the terms inside the brackets: \[ 16500 = x \left(1 + \frac{1}{10}\right) \\ 16500 = x \left(\frac{11}{10}\right) \] Step 14: Rearrange the expression to solve directly for original sum x: \[ x = \frac{16500 \times 10}{11} \] Step 15: Perform the division step where \( 16500 \div 11 = 1500 \): \[ x = 1500 \times 10 \\ x = 15000 \] Answer: Rate per cent = 10% per annum and Original sum of money invested = ₹ 15,000

Q7: The difference between C.I. and S.I. on ₹ 7,500 for two years is ₹ 12 at the same rate of interest per annum. Find the rate of interest.

Step 1: Identify the given values from the problem statement.
Step 2: Principal \( (P) = \text{₹ } 7,500 \), Time \( (t \text{ or } n) = 2 \) years, and Difference \( (\text{C.I.} – \text{S.I.}) = \text{₹ } 12 \).
Step 3: Let the required rate of interest per annum be \( r \% \).
Step 4: Calculate the Simple Interest (S.I.) in terms of r: \[ \text{S.I.} = \frac{P \times r \times t}{100} \\ \text{S.I.} = \frac{7500 \times r \times 2}{100} \\ \text{S.I.} = 75 \times 2r \\ \text{S.I.} = 150r \quad \text{— (Equation 1)} \] Step 5: Formulate the Compound Interest (C.I.) expression in terms of r: \[ \text{C.I.} = P \left[\left(1 + \frac{r}{100}\right)^n – 1\right] \] Step 6: Substitute the given values into the compound interest formula: \[ \text{C.I.} = 7500 \left[\left(1 + \frac{r}{100}\right)^2 – 1\right] \] Step 7: Expand the squared term using the algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \): \[ \text{C.I.} = 7500 \left[1 + \frac{2r}{100} + \frac{r^2}{10000} – 1\right] \\ \text{C.I.} = 7500 \left[\frac{2r}{100} + \frac{r^2}{10000}\right] \] Step 8: Multiply 7,500 across the bracket terms to finalize the expression for C.I.: \[ \text{C.I.} = \frac{7500 \times 2r}{100} + \frac{7500 \times r^2}{10000} \\ \text{C.I.} = 150r + \frac{3r^2}{4} \quad \text{— (Equation 2)} \] Step 9: According to the problem statement, the difference between C.I. and S.I. is ₹ 12: \[ \text{C.I.} – \text{S.I.} = 12 \] Step 10: Substitute Equation 1 and Equation 2 into this relationship: \[ \left(150r + \frac{3r^2}{4}\right) – 150r = 12 \] Step 11: Cancel out the matching linear terms to isolate the quadratic component: \[ \frac{3r^2}{4} = 12 \] Step 12: Rearrange the expression to isolate and solve for \( r^2 \): \[ 3r^2 = 12 \times 4 \\ 3r^2 = 48 \\ r^2 = \frac{48}{3} \\ r^2 = 16 \] Step 13: Take the square root on both sides of the expression to find r: \[ r = \sqrt{16} \\ r = 4 \] Answer: Rate of interest = 4% per annum

Q8: A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 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 lent out be ₹ x.
Step 2: According to the first condition, the money becomes 3 times itself: Amount \( (A_1) = 3x \) in Time \( (n_1) = 10 \) 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: \[ 3x = x \left(1 + \frac{r}{100}\right)^{10} \] Step 6: Divide both sides by x to simplify the expression: \[ 3 = \left(1 + \frac{r}{100}\right)^{10} \quad \text{— (Equation 1)} \] 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 expansion 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 eliminate the variable: \[ 27 = \left(1 + \frac{r}{100}\right)^{n_2} \quad \text{— (Equation 2)} \] Step 11: Express the number 27 as a power with base 3 to connect 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: \[ \left[\left(1 + \frac{r}{100}\right)^{10}\right]^3 = \left(1 + \frac{r}{100}\right)^{n_2} \] Step 13: Use laws of exponents to multiply the powers on the left-hand side: \[ \left(1 + \frac{r}{100}\right)^{10 \times 3} = \left(1 + \frac{r}{100}\right)^{n_2} \\ \left(1 + \frac{r}{100}\right)^{30} = \left(1 + \frac{r}{100}\right)^{n_2} \] Step 14: Compare exponents on both sides since the geometric bases match exactly: \[ n_2 = 30 \] Answer: Time period = 30 years

Q9: Mr. Sharma borrowed a certain sum of money at 10% per annum annually. If by paying ₹ 19,360 at the end of the second year and ₹ 31,944 at the end of the third year he clears the debt; find the sum borrowed by him.

Step 1: Let the initial sum borrowed by Mr. Sharma be ₹ x.
Step 2: Identify the given parameter values: Rate of interest \( (r) = 10\% \) per annum compounded annually.
Step 3: Formulate the accumulated amount due at the end of the first year: \[ \text{Amount after 1 year} = x \left(1 + \frac{10}{100}\right) = x \left(\frac{11}{10}\right) = 1.1x \] Step 4: This amount becomes the principal for the second year. Formulate the amount accumulated at the end of the second year before any payment: \[ \text{Amount before payment at end of 2nd year} = 1.1x \left(1 + \frac{10}{100}\right) = 1.1x \times 1.1 = 1.21x \] Step 5: Subtract the payment of ₹ 19,360 made at the end of the second year to find the remaining balance: \[ \text{Remaining Principal balance after 2nd year} = 1.21x – 19360 \] Step 6: This remaining balance acts as the principal for the third year. Formulate the total amount due at the end of the third year: \[ \text{Amount at end of 3rd year} = (1.21x – 19360) \left(1 + \frac{10}{100}\right) \\ \text{Amount at end of 3rd year} = (1.21x – 19360) \left(1.1\right) \] Step 7: Equate this third-year amount to the final clearing payment of ₹ 31,944 as given in the problem statement: \[ (1.21x – 19360) \times 1.1 = 31944 \] Step 8: Divide both sides by 1.1 to simplify and isolate the bracket term: \[ 1.21x – 19360 = \frac{31944}{1.1} \\ 1.21x – 19360 = 29040 \] Step 9: Transpose the numerical constant value to combine it with the right-hand side term: \[ 1.21x = 29040 + 19360 \\ 1.21x = 48400 \] Step 10: Divide by 1.21 to solve for the initial sum borrowed x: \[ x = \frac{48400}{1.21} \\ x = \frac{4840000}{121} \] Step 11: Perform the final division step where \( 484 \div 121 = 4 \): \[ x = 40000 \] Answer: The sum borrowed by him was ₹ 40,000.

Q10: The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is ₹ 15. Find the sum of money lent out.

Step 1: Let the initial sum of money lent out be ₹ x.
Step 2: Identify the parameters for the simple interest calculation: Rate \( (r) = 10\% \) p.a., Time \( (t) = 1 \) year.
Step 3: Formulate the Simple Interest (S.I.) equation using the standard formula: \[ \text{S.I.} = \frac{x \times r \times t}{100} \\ \text{S.I.} = \frac{x \times 10 \times 1}{100} = \frac{10x}{100} = \frac{x}{10} \] Step 4: Identify the parameters for the compound interest calculation: Time \( = 1 \) year, Rate \( (r) = 10\% \) p.a. compounded half-yearly.
Step 5: Adjust parameters for semi-annual cycles: Half-yearly rate \( (R) = \frac{10}{2}\% = 5\% \), conversion periods \( (n) = 1 \times 2 = 2 \) half-years.
Step 6: 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 7: Substitute the adjusted values into the compound interest expression: \[ \text{C.I.} = x \left[\left(1 + \frac{5}{100}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\left(1 + \frac{1}{20}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\left(\frac{21}{20}\right)^2 – 1\right] \\ \text{C.I.} = x \left[\frac{441}{400} – 1\right] = \frac{41x}{400} \] Step 8: According to the problem statement, 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 9: Make the denominators common to solve the algebraic equation: \[ \frac{41x – 40x}{400} = 15 \\ \frac{x}{400} = 15 \] Step 10: Rearrange the terms to isolate and solve for sum x: \[ x = 15 \times 400 \\ x = 6000 \] Answer: The sum of money lent out is ₹ 6,000.

Q11: The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?

Step 1: Calculate the investment time duration for each person up to the target age of 25 years.
Step 2: Time period for Pramod \( (n_1) = 25 – 16 = 9 \) years.
Step 3: Time period for Rohit \( (n_2) = 25 – 18 = 7 \) years.
Step 4: Let Pramod’s investment principal be \( P_1 \) and Rohit’s investment principal be \( P_2 \).
Step 5: Identify the given values from the problem statement: Rate of interest \( (r) = 5\% \) p.a.
Step 6: Write down the standard compound interest amount formula: \[ A = P \left(1 + \frac{r}{100}\right)^n \] Step 7: Formulate the equation for Pramod’s final accumulated amount after 9 years: \[ \text{Pramod’s Amount} = P_1 \left(1 + \frac{5}{100}\right)^9 \] Step 8: Formulate the equation for Rohit’s final accumulated amount after 7 years: \[ \text{Rohit’s Amount} = P_2 \left(1 + \frac{5}{100}\right)^7 \] Step 9: Since both get the same sum on attaining the age of 25 years, equate both expressions: \[ P_1 \left(1 + \frac{5}{100}\right)^9 = P_2 \left(1 + \frac{5}{100}\right)^7 \] Step 10: Rearrange the terms to isolate the ratio of their investments \( \frac{P_1}{P_2} \): \[ \frac{P_1}{P_2} = \frac{\left(1 + \frac{5}{100}\right)^7}{\left(1 + \frac{5}{100}\right)^9} \] Step 11: Apply exponent division rules to simplify the powers: \[ \frac{P_1}{P_2} = \frac{1}{\left(1 + \frac{5}{100}\right)^{9 – 7}} \\ \frac{P_1}{P_2} = \frac{1}{\left(1 + \frac{5}{100}\right)^2} \] Step 12: Simplify the fraction value inside the bracket term: \[ \frac{P_1}{P_2} = \frac{1}{\left(1 + \frac{1}{20}\right)^2} \\ \frac{P_1}{P_2} = \frac{1}{\left(\frac{21}{20}\right)^2} \] Step 13: Expand the squared terms to evaluate the fractional inverse: \[ \frac{P_1}{P_2} = \frac{1}{\frac{441}{400}} \\ \frac{P_1}{P_2} = \frac{400}{441} \] Answer: They must invest their money in the ratio of 400 : 441.