RS Aggarwal Class 8 Mathematics Solutions Chapter 11 Compound Interest

Question 1. Find the compound interest on Rs. 2500 at 10% per annum for 2 years, calculated annually.
Answer: Principal in year 1 is Rs. 2500. Interest for year 1 is Rs. 250 (calculated as 2500 × 10 ÷ 100). Total amount at end of year 1 is Rs. 2750. Principal in year 2 is Rs. 2750. Interest for year 2 is Rs. 275 (calculated as 2750 × 10 ÷ 100). Total amount at end of year 2 is Rs. 3025. Therefore, the compound interest earned is Rs. 525 (3025 - 2500).
In simple words: Year 1: the bank gives you interest on your starting money. Year 2: the bank gives you interest on both your original money and the interest you already earned. This is how compound interest grows faster.

Exam Tip: Always calculate interest for each year separately, using the previous year's amount as the new principal. The compound interest is the final amount minus the original principal.

Question 2. Find the compound interest on Rs. 15625 at 12% per annum for 3 years, calculated annually.
Answer: Principal for year 1 is Rs. 15625. Interest for year 1 is Rs. 1875 (calculated as 15625 × 12 ÷ 100). Total amount at end of year 1 is Rs. 17500. Principal for year 2 is Rs. 17500. Interest for year 2 is Rs. 2100 (calculated as 17500 × 12 ÷ 100). Total amount at end of year 2 is Rs. 19600. Principal for year 3 is Rs. 19600. Interest for year 3 is Rs. 2352 (calculated as 19600 × 12 ÷ 100). Total amount at end of year 3 is Rs. 21952. Therefore, the compound interest earned is Rs. 6327 (21952 - 15625).
In simple words: Each year, you earn interest on a bigger amount because you're earning interest on the interest from the previous year. This causes your money to grow faster than simple interest.

Exam Tip: For multi-year problems, write down the principal and interest for each year separately to avoid calculation errors.

Question 3. A man invests Rs. 5000 at 9% per annum for 2 years. Find the compound interest and compare it with simple interest.
Answer: Principal amount is Rs. 5000. Simple interest would be Rs. 900 (calculated as 5000 × 9 × 2 ÷ 100). Now to calculate compound interest: Principal for year 1 is Rs. 5000. Interest for year 1 is Rs. 450 (calculated as 5000 × 9 ÷ 100). Total amount at end of year 1 is Rs. 5450. Principal for year 2 is Rs. 5450. Interest for year 2 is Rs. 490.5 (calculated as 5450 × 9 ÷ 100). Total amount at end of year 2 is Rs. 5940.5. Compound interest earned is Rs. 940.5 (5940.5 - 5000). The difference between compound interest and simple interest is Rs. 40.5 (940.5 - 900).
In simple words: Simple interest stays the same each year, but compound interest keeps growing. In this case, compound interest earns you Rs. 40.5 more because you earn interest on your interest.

Exam Tip: Always calculate both simple and compound interest separately when the question asks for a comparison. The difference reveals the extra gain from compounding.

Question 4. Ratna borrowed Rs. 25000 from a bank at 8% per annum for 2 years. How much money will she have to pay back?
Answer: Principal in year 1 is Rs. 25000. Interest for year 1 is Rs. 2000 (calculated as 25000 × 8 ÷ 100). Total amount at end of year 1 is Rs. 27000. Principal in year 2 is Rs. 27000. Interest for year 2 is Rs. 2160 (calculated as 27000 × 8 ÷ 100). Total amount at end of year 2 is Rs. 29160. Therefore, Ratna must pay back Rs. 29160 after 2 years to clear her debt.
In simple words: The bank charges interest not just on the borrowed money, but also on the interest already added. So the total amount grows over time.

Exam Tip: In loan problems, the final amount to be paid back is the principal plus the compound interest earned over the entire period.

Question 5. Find the amount on Rs. 20000 at 12% per annum for 2 years, calculated annually. Also calculate how much money the borrower gains after the specified time period.
Answer: Principal amount is Rs. 20000. Simple interest would be Rs. 4800 (calculated as 20000 × 12 × 2 ÷ 100). Compound interest is calculated as follows: Principal for year 1 is Rs. 20000. Interest for year 1 is Rs. 2400 (calculated as 20000 × 12 ÷ 100). Total amount at end of year 1 is Rs. 22400. Principal for year 2 is Rs. 22400. Interest for year 2 is Rs. 2688 (calculated as 22400 × 12 ÷ 100). Total amount at end of year 2 is Rs. 25088. Compound interest earned is Rs. 5088 (25088 - 20000). The gain from compound interest over simple interest is Rs. 288 (5088 - 4800). Therefore, the borrower earns a benefit of Rs. 288 after 2 years due to compounding.
In simple words: When interest is calculated on interest, you get more money compared to simple interest. The extra gain shows the power of compounding over time.

Exam Tip: To find the benefit of compound interest, always subtract simple interest from compound interest. This difference clearly shows the advantage of compounding.

Question 6. Manoj invested Rs. 64000 at 7.5% per annum for 3 years. Calculate the amount he will receive after 3 years.
Answer: Principal in year 1 is Rs. 64000. Interest for year 1 is Rs. 4800 (calculated as 64000 × 7.5 ÷ 100). Total amount at end of year 1 is Rs. 68800. Principal in year 2 is Rs. 68800. Interest for year 2 is Rs. 5160 (calculated as 68800 × 7.5 ÷ 100). Total amount at end of year 2 is Rs. 73960. Principal in year 3 is Rs. 73960. Interest for year 3 is Rs. 5547 (calculated as 73960 × 7.5 ÷ 100). Total amount at end of year 3 is Rs. 79507. Therefore, Manoj will receive Rs. 79507 after 3 years.
In simple words: The amount grows year after year as interest is added to the principal. Each year, the interest becomes larger because it is calculated on a bigger amount.

Exam Tip: Keep track of the amount at the end of each year carefully. This becomes your new principal for the next year in compound interest calculations.

Question 7. Divakaran invested Rs. 6250 at 8% per annum for 1 year, compounded half-yearly. Find the compound interest.
Answer: Principal amount is Rs. 6250. Interest rate is 8% per annum, which equals 4% for half year. Time period is 1 year, which is 2 half years. Principal for the first half year is Rs. 6250. Interest for the first half year is Rs. 250 (calculated as 6250 × 4 ÷ 100). Total amount at end of the first half year is Rs. 6500. Principal for the second half year is Rs. 6500. Interest for the second half year is Rs. 260 (calculated as 6500 × 4 ÷ 100). Total amount at end of the second half year is Rs. 6760. Compound interest earned is Rs. 510 (6760 - 6250). Therefore, Divakaran receives a compound interest of Rs. 510.
In simple words: When interest is calculated half-yearly, the interest rate for each 6-month period is half of the annual rate. The interest is calculated twice per year instead of once.

Exam Tip: For half-yearly compounding, divide the annual rate by 2 and count the number of half-year periods. Always double-check that you are using the correct rate for each period.

Question 8. Michael borrowed Rs. 16000 at 10% per annum for 1.5 years. The interest is compounded half-yearly. How much money must he pay back after 1.5 years?
Answer: Principal amount is Rs. 16000. Interest rate is 10% per annum, which equals 5% for half year. Time period is 1.5 years, which is 3 half years. Principal for the first half year is Rs. 16000. Interest for the first half year is Rs. 800 (calculated as 16000 × 5 ÷ 100). Total amount at end of the first half year is Rs. 16800. Principal for the second half year is Rs. 16800. Interest for the second half year is Rs. 840 (calculated as 16800 × 5 ÷ 100). Total amount at end of the second half year is Rs. 17640. Principal for the third half year is Rs. 17640. Interest for the third half year is Rs. 882 (calculated as 17640 × 5 ÷ 100). Total amount at end of the third half year is Rs. 18522. Therefore, Michael must pay Rs. 18522 to the finance company after 1.5 years.
In simple words: Half-yearly compounding means interest is added to your balance 3 times in 1.5 years (every 6 months). This causes the total amount to grow faster.

Exam Tip: When the time is given in years and months, convert it to the number of compounding periods first. For half-yearly, multiply years by 2.

Question 9. Find the amount on Rs. 6000 at 9% per annum for 2 years using the compound interest formula.
Answer: Using the formula \( A = P \left(1 + \frac{R}{100}\right)^n \), where P is Rs. 6000, R is 9%, and n is 2 years: \( A = 6000 \left(1 + \frac{9}{100}\right)^2 = 6000 \left(\frac{109}{100}\right)^2 = 6000 \left(\frac{109}{100}\right)^2 = 6000(1.09 \times 1.09) = 6000(1.1881) = 7128.6 \). The amount including compound interest is Rs. 7128.6. Compound interest earned is Rs. 1128.6 (7128.6 - 6000).
In simple words: The formula lets you calculate the total amount in one step instead of doing it year by year. You multiply the principal by a growth factor raised to the power of the number of years.

Exam Tip: Always use the complete formula and substitute values carefully. Make sure your rate is in percentage form and time is in the correct units.

Question 10. Find the amount on Rs. 10000 at 11% per annum for 2 years, calculated annually.
Answer: Principal amount is Rs. 10000. Interest rate is 11% per annum. Time is 2 years. Using the compound interest formula \( A = P \left(1 + \frac{R}{100}\right)^n \): \( A = 10000 \left(1 + \frac{11}{100}\right)^2 = 10000 \left(\frac{111}{100}\right)^2 = 10000(1.11 \times 1.11) = 10000(1.2321) = 12321 \). The amount including compound interest is Rs. 12321. Compound interest earned is Rs. 2321 (12321 - 10000).
In simple words: By using the formula, you can find the total amount directly without calculating interest year by year. The formula shows how money grows exponentially when interest is compounded.

Exam Tip: The exponent in the formula represents the number of compounding periods. Double-check that n matches your time unit (years, half-years, quarters, etc.).

Question 11. Find the compound interest on Rs. 31250 at 8% per annum for 3 years.
Answer: Principal amount is Rs. 31250. Interest rate is 8% per annum. Time is 3 years. Using the formula \( A = P \left(1 + \frac{R}{100}\right)^n \): \( A = 31250 \left(1 + \frac{8}{100}\right)^3 = 31250 \left(\frac{108}{100}\right)^3 = 31250(1.08 \times 1.08 \times 1.08) = 31250(1.259712) = 39366 \). The amount including compound interest is Rs. 39366. Compound interest earned is Rs. 8116 (39366 - 31250).
In simple words: When you cube the growth factor (raise it to the power of 3), you're essentially compounding the interest over 3 years. The longer the time, the larger the interest earned.

Exam Tip: For calculations with larger exponents, use a calculator to avoid arithmetic errors. Always subtract the principal from the final amount to find just the interest earned.

Question 12. Find the amount on Rs. 10240 at 12.5% per annum for 3 years.
Answer: Principal amount is Rs. 10240. Interest rate is 12.5% per annum, which is \( \frac{25}{2} \)% per annum. Time is 3 years. Using the formula \( A = P \left(1 + \frac{R}{100}\right)^n \): \( A = 10240 \left(1 + \frac{25}{2 \times 100}\right)^3 = 10240 \left(1 + \frac{1}{8}\right)^3 = 10240 \left(\frac{9}{8}\right)^3 = 10240(1.125 \times 1.125 \times 1.125) = 10240(1.423828) = 14580 \). The amount including compound interest is Rs. 14580. Compound interest earned is Rs. 4340 (14580 - 10240).
In simple words: When the rate is expressed as a fraction or decimal, convert it to the same form before using it in the formula. The calculation method remains the same.

Exam Tip: Convert fractional rates to decimals or fractions consistently. Be careful when multiplying decimals to avoid rounding errors.

Question 13. Simple interest on a sum is Rs. 2400 at 8% per annum for 2 years. Find the principal and the compound interest on that principal.
Answer: Using simple interest formula to find principal: Simple Interest = \( \frac{P \times R \times T}{100} \), so \( 2400 = \frac{P \times 8 \times 2}{100} \), which gives \( P = 15000 \). The principal is Rs. 15000. Now to find compound interest using the formula \( A = P \left(1 + \frac{R}{100}\right)^n \): \( A = 15000 \left(1 + \frac{8}{100}\right)^2 = 15000 \left(\frac{108}{100}\right)^2 = 15000(1.08 \times 1.08) = 15000(1.1664) = 17496 \). The amount including compound interest is Rs. 17496. Compound interest earned is Rs. 2496 (17496 - 15000).
In simple words: From the simple interest given, you can work backwards to find the principal. Then use that principal to calculate compound interest, which will be slightly higher than the simple interest.

Exam Tip: Always verify your principal calculation by working backwards with the simple interest formula before proceeding to find compound interest.

Question 14. The difference between compound interest and simple interest on a sum at 10% per annum for 2 years is Rs. 90. Find the principal.
Answer: Let the principal be Rs. P. Simple interest for 2 years at 10% is \( \frac{P \times 10 \times 2}{100} = \frac{20P}{100} = \frac{P}{5} \). Compound interest is found using \( A = P \left(1 + \frac{10}{100}\right)^2 = P \left(\frac{11}{10}\right)^2 = \frac{121P}{100} \), so CI = \( \frac{121P}{100} - P = \frac{121P - 100P}{100} = \frac{21P}{100} \). The difference between CI and SI is \( \frac{21P}{100} - \frac{20P}{100} = \frac{P}{100} = 90 \), which gives \( P = 9000 \). Therefore, the principal is Rs. 9000. We can verify: SI = 1800, CI = 1890, difference = 90.
In simple words: The difference between compound and simple interest increases with the principal. By setting up an equation with the known difference, you can solve for the principal.

Exam Tip: When finding the difference formula, remember that compound interest = A - P and simple interest = (P × R × T)/100. Set their difference equal to the given amount.

Question 15. The difference between compound interest and simple interest on a sum at 10% per annum for 2 years is Rs. 93. Find the principal.
Answer: Let the principal be Rs. P. Simple interest for 2 years at 10% is \( \frac{P \times 10 \times 2}{100} = \frac{20P}{100} = \frac{P}{5} \). Compound interest is \( A = P \left(1 + \frac{10}{100}\right)^2 = P \left(\frac{11}{10}\right)^2 = \frac{121P}{100} \), so CI = \( \frac{121P}{100} - P = \frac{21P}{100} \). The difference is \( \frac{21P}{100} - \frac{20P}{100} = \frac{P}{100} = 93 \), which gives \( P = 9300 \). Therefore, the principal is Rs. 9300. But since Rs. 93 is given, let's verify: if P = 3000, then (CI - SI) = 3000 × (0.1)² = 30, which is not 93. Solving \( \frac{P}{100} = 93 \) gives \( P = 9300 \), but checking with the constraint that the difference should match, we find P = 3000 when verified against the given conditions.
In simple words: Set up an equation based on the difference formula and solve for P. The key is recognizing that the difference depends on the principal squared and the rate squared.

Exam Tip: Always verify your answer by calculating both SI and CI separately and checking that their difference matches the given value.

Question 16. At what rate per annum will Rs. 10240 amount to Rs. 10240 in 2 years if the interest is compounded annually?
Answer: Let P be the principal of Rs. 10240. Rate of interest is \( 6\frac{2}{3} \)% = \( \frac{20}{3} \)% per annum. Time is 2 years. Using the formula \( A = P \left(1 + \frac{R}{100 \times 3}\right)^2 = P \left(1 + \frac{20}{300}\right)^2 = P \left(1 + \frac{20}{300}\right)^2 = P \left(\frac{320}{300}\right)^2 = P \left(\frac{16}{15}\right)^2 = P \times \frac{256}{225} \). So \( 10240 = 10240 \times \frac{256}{225} \), which gives the required amount. Therefore, the principal is Rs. 9000, and the required sum is Rs. 9000.
In simple words: Using the compound interest formula, substitute the known values and solve for the unknown. In this case, we find the principal by working backwards from the amount.

Exam Tip: When finding either rate or time, rearrange the compound interest formula to isolate the unknown variable and solve step by step.

Question 17. At what rate per annum will Rs. 21296 become the compound amount of Rs. 16000 in 3 years?
Answer: Let the rate be R% per annum. Principal P = Rs. 16000. Amount A = Rs. 21296. Time n = 3 years. Using the formula \( A = P \left(1 + \frac{R}{100}\right)^n \): \( 21296 = 16000 \left(1 + \frac{R}{100}\right)^3 \), so \( \frac{21296}{16000} = \left(1 + \frac{R}{100}\right)^3 \), which gives \( 1.331 = \left(1 + \frac{R}{100}\right)^3 \). Taking the cube root: \( 1 + \frac{R}{100} = 1.1 \), so \( \frac{R}{100} = 0.1 \), and \( R = 10 \). Therefore, the required sum is Rs. 16000, and the rate is 10% per annum.
In simple words: Divide the final amount by the principal to get a growth factor. Take the cube root (or appropriate root based on years) to find the annual growth multiplier. Subtract 1 and multiply by 100 to get the rate.

Exam Tip: When solving for rate, carefully take the nth root where n is the number of years. Check your answer by substituting back into the original formula.

Question 18. Let R% per annum be the required rate. A = 4410, P = 4000, n = 2 years. Now, A = P(1 + R/100)^n.
Answer: We have 4410 = 4000(1 + R/100)², which gives 4410/4000 = (1 + R/100)², so 441/400 = (1 + R/100)². Taking square root: 21/20 = 1 + R/100, which gives 21/20 - 1 = R/100, so (21 - 20)/20 = R/100, and 1/20 = R/100. Therefore R = (1 × 100)/20 = 5. Hence, the required rate is 5% per annum.
In simple words: To find the rate, divide the final amount by the principal, then take the square root (for 2 years). Subtract 1 and multiply by 100 to get the percentage rate.

Exam Tip: Always verify your calculated rate by substituting it back into the compound interest formula to confirm the final amount.

Question 19. Let the required rate be R% per annum. A = 774.40, P = 640, n = 2 years. Now, A = P(1 + R/100)^n.
Answer: We have 774.40 = 640(1 + R/100)², which gives 774.40/640 = (1 + R/100)², so 1.21 = (1 + R/100)². Taking square root: (1.1)² = (1 + R/100)², which gives 1.1 = 1 + R/100, so 1.1 - 1 = R/100, and 0.1 = R/100. Therefore R = 0.1 × 100 = 10. Hence, the required rate is 10% per annum.
In simple words: When you have the final amount and principal, divide them to get a ratio. Take the square root to find the annual multiplier. Subtract 1 and multiply by 100 to get the rate.

Exam Tip: Perfect squares and cubes help simplify calculations when taking roots. Learn common values like 1.1² = 1.21, 1.2² = 1.44, 1.08³ to speed up solving.

Question 20. Let the required time be n years. Rate of interest, R = 10%. Principal amount, P = Rs. 1800. Amount with compound interest, A = Rs. 2178. Now, A = P × (1 + R/100)^n.
Answer: We have 2178 = 1800 × (1 + 10/100)^n = 1800 × (1.1)^n, which gives 2178/1800 = (1.1)^n, so 1.21 = (1.1)^n. Since (1.1)² = 1.21, we get n = 2. Therefore, the time is 2 years.
In simple words: Divide the final amount by the principal to get a ratio. Check which power of the growth factor equals this ratio. That power is your time period.

Exam Tip: Memorize common values like 1.1² = 1.21, 1.05² = 1.1025, and 1.08³ = 1.259712 to quickly identify time periods.

Question 21. Let the required time be n years. Rate of interest, R = 8%. Principal amount, P = Rs. 6250. Amount with compound interest, A = Rs. 7290. Now, A = P × (1 + R/100)^n.
Answer: We have 7290 = 6250 × (1 + 8/100)^n = 6250 × (1.08)^n, which gives 7290/6250 = (1.08)^n, so 1.1664 = (1.08)^n. Since (1.08)² = 1.1664, we get n = 2. Therefore, the time is 2 years.
In simple words: Express both the final amount ratio and the growth factor as powers with the same base. Compare the exponents to find the time period.

Exam Tip: If the ratio doesn't match a common value, you may need to use logarithms or trial-and-error with different time periods to find n.

Question 22. The population of a town is 125000. The rate of increase is 2%. Find the population after 3 years.
Answer: Initial population P = 125000. Rate of increase R = 2% per annum. Time n = 3 years. Using the growth formula \( \text{Population} = P \left(1 + \frac{R}{100}\right)^n \): \( \text{Population} = 125000 \left(1 + \frac{2}{100}\right)^3 = 125000 \left(1.02\right)^3 = 125000 \left(\frac{102}{100}\right)^3 = 125000(1.02 \times 1.02 \times 1.02) = 125000 \times \left(\frac{51}{50}\right)^3 = 125000 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} = (51 \times 51 \times 51) = 132651 \). Therefore, the population of the town after 3 years is 132651.
In simple words: Population growth follows the same compound interest formula. Each year, the population increases by a percentage of the current population, not the original population.

Exam Tip: For population or growth problems, use the compound interest formula with growth rate instead of interest rate. Always express the rate as a decimal in calculations.

Question 23. The population of a town is 50000. The rate of increase for the first year is 5%, for the second year is 4%, and for the third year is 3%. Find the population after 3 years.
Answer: Initial population = 50000. Rate for year 1: p = 5%. Rate for year 2: q = 4%. Rate for year 3: r = 3%. Time = 3 years. When rates are different for different years: \( \text{Population} = P \times \left(1 + \frac{p}{100}\right) \times \left(1 + \frac{q}{100}\right) \times \left(1 + \frac{r}{100}\right) = 50000 \times \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{4}{100}\right) \times \left(1 + \frac{3}{100}\right) = 50000 \times \left(\frac{105}{100}\right) \times \left(\frac{104}{100}\right) \times \left(\frac{103}{100}\right) = 50000 \times \frac{21}{20} \times \frac{26}{25} \times \frac{103}{100} = (21 \times 26 \times 103) = 56238 \). Therefore, the population of the town after 3 years is 56238.
In simple words: When growth rates change each year, multiply by the growth factors for each year separately. Do not average the rates or add them together.

Exam Tip: For variable rate problems, write separate multiplication factors for each year. The final population equals the initial population times all these growth factors multiplied together.

Question 24. Population of the city in 2009 is 120000. Rate of increase is 6%. Find the population of the city in the year 2010. Also find the population in 2011 with a rate of decrease of 5%.
Answer: Initial population in 2009, P = 120000. Rate of increase for 2010, R = 6%. Using the growth formula, population in 2010 is: \( \text{Population} = 120000 \left(1 + \frac{6}{100}\right)^1 = 120000 \times 1.06 = 127200 \). Now for year 2011, population in 2010 becomes P = 127200. Rate of decrease, R = 5%. Using the depreciation formula \( V = V_0 \left(1 - \frac{r}{100}\right)^n \): \( \text{Population} = 127200 \left(1 - \frac{5}{100}\right)^1 = 127200 \times 0.95 = 120840 \). Therefore, the population in 2010 is 127200, and the population in 2011 is 120840.
In simple words: Use the growth formula when population increases and the depreciation formula when it decreases. Treat each year's ending population as the starting point for the next year.

Exam Tip: For increase use (1 + r/100) and for decrease use (1 - r/100). Never mix these formulas or forget to update your principal for the next period.

Question 25. The initial count of bacteria is 500000. The rate of increase is 2%. Find the count of bacteria at the end of 2 hours.
Answer: Initial count of bacteria, P = 500000. Rate of increase, R = 2%. Time, n = 2 hours. Using the growth formula \( \text{Count} = P \left(1 + \frac{R}{100}\right)^n \): \( \text{Count} = 500000 \left(1 + \frac{2}{100}\right)^2 = 500000 \left(1.02\right)^2 = 500000 \left(\frac{102}{100}\right)^2 = 500000(1.02 \times 1.02) = 500000 \times \left(\frac{51}{50}\right)^2 = 500000 \times \frac{51}{50} \times \frac{51}{50} = (500 \times 51 \times 51) = 520200 \). Therefore, the count of bacteria at the end of 2 hours is 520200.
In simple words: Bacteria, money, and populations all grow using the same exponential formula. The rate tells you the percentage increase per time period.

Exam Tip: In growth problems, always identify the initial amount, the rate, and the time period clearly before substituting into the formula.

Question 27. A sum of money triples itself in 16 years at simple interest. Find the rate of interest per annum.
Answer: Let the principal be P. According to the problem, the amount after 16 years is 3P, which means the simple interest earned is 3P - P = 2P. Using the formula SI = (P × R × T) / 100, we have 2P = (P × R × 16) / 100. Simplifying: 2P = (16PR) / 100, so 2 = (16R) / 100, giving us 200 = 16R, and therefore R = 12.5% per annum.
In simple words: If your money becomes three times bigger in 16 years, the interest rate must be 12.5% every year. You earn enough interest each year so that after 16 years, the interest equals twice your starting amount.

Exam Tip: Always identify the principal and final amount first, then calculate the interest earned by subtraction. Use SI = (P × R × T) / 100 to find the rate.

Question 28. A sum of money becomes Rs. 1,080 in 2 years and Rs. 1,188 in 3 years at compound interest. Find the principal and the rate of interest.
Answer: Let the principal be P and the annual rate of interest be R%. The compound interest formula gives us A = P(1 + R/100)^n. After 2 years: 1080 = P(1 + R/100)^2. After 3 years: 1188 = P(1 + R/100)^3. Dividing the second equation by the first: 1188 / 1080 = (1 + R/100). So (1 + R/100) = 11/10 = 1.1, which gives us R = 10%. Now substituting back: 1080 = P(1.1)^2 = P × 1.21, so P = 1080 / 1.21 = 1000. Therefore, the principal is Rs. 1,000 and the rate of interest is 10% per annum.
In simple words: When you divide the amount in year 3 by the amount in year 2, you get the growth factor for one year. This tells you the interest rate directly. Once you know the rate, you can work backward to find the starting amount.

Exam Tip: Dividing consecutive year amounts is a clever shortcut - it instantly reveals the annual growth multiplier without needing to expand the full formula.

Question 29. The value of a scooter depreciates at the rate of 10% per annum. Its present value is Rs. 56,000. What was its value 3 years ago?
Answer: We use the depreciation formula: V = V₀(1 - R/100)^n, where V is the present value, V₀ is the value n years ago, R is the depreciation rate, and n is the number of years. Here, V = 56,000, R = 10%, and n = 3. So 56,000 = V₀(1 - 10/100)^3 = V₀(0.9)^3 = V₀ × 0.729. Therefore, V₀ = 56,000 / 0.729 = 56,000 / (729/1000) = (56,000 × 1000) / 729 ≈ Rs. 76,820 (approximately). More precisely, using the fraction form: V₀ = 56,000 / (9/10)^3 = 56,000 × (10/9)^3 = 56,000 × 1000 / 729 ≈ Rs. 76,820.
In simple words: If something loses value at 10% each year, it retains 90% of its value yearly. Going backward 3 years means multiplying by (10/9) three times to reverse the depreciation.

Exam Tip: For depreciation problems asking about the past, rearrange the formula to solve for V₀ - divide the current value by the depreciation factor raised to the power of years.

Question 30. A machine's value depreciates at 10% per annum. Its current value is Rs. 291,600. Find its value 3 years ago.
Answer: Using the depreciation formula V = V₀(1 - R/100)^n in the form where V is present and V₀ is past value: 291,600 = V₀(1 - 10/100)^3. This simplifies to 291,600 = V₀(9/10)^3 = V₀ × (729/1000). Solving for V₀: V₀ = 291,600 × (1000/729) = (291,600 × 1000) / 729. Performing the division: V₀ = 291,600,000 / 729 = 400,000. Therefore, the machine's value 3 years ago was Rs. 400,000. We can verify: 400,000 × (9/10)^3 = 400,000 × (729/1000) = 291,600. ✓
In simple words: Start with today's value and divide by the depreciation factor three times (or divide by its cube). This reverses the three years of value loss.

Exam Tip: Always verify your answer by applying depreciation forward - if you calculated the past value correctly, applying the formula should give you back the present value.

Question 1. Find the compound interest on Rs. 8,000 for 1 year at 5% per annum, compounded half-yearly.
Answer: Principal P = Rs. 8,000, Time n = 1 year = 2 half-years, Annual rate = 10%, Half-yearly rate = 5%. Using the compound interest formula with half-yearly compounding: A = P(1 + R/100)^n = 8,000 × (1 + 5/100)^2 = 8,000 × (105/100)^2 = 8,000 × (21/20)^2 = 8,000 × (21/20) × (21/20) = 8,000 × (441/400) = (8,000 × 441) / 400 = 20 × 441 = Rs. 8,820. Therefore, the compound interest earned = 8,820 - 8,000 = Rs. 820.
In simple words: When interest is compounded every six months instead of yearly, the money grows faster because you earn interest on your interest more frequently. This particular investment grows from Rs. 8,000 to Rs. 8,820, giving you Rs. 820 extra.

Exam Tip: Always adjust the rate and time period together - if compounding is half-yearly, halve the rate and double the time in years to match the compounding frequency.

Question 3. Find the compound interest on Rs. 12,800 for 1 year at 7.5% per annum, compounded half-yearly.
Answer: Principal P = Rs. 12,800, Annual rate = 7.5% = 15/2 %, Half-yearly rate = (1/2) × (15/2) = 15/4 %, Time = 1 year = 2 half-years. The amount after 1 year is A = P × (1 + R/100)^n = 12,800 × (1 + 15/400)^2 = 12,800 × (1 + 15/(4×100))^2 = 12,800 × ((400 + 15)/400)^2 = 12,800 × (415/400)^2 = 12,800 × (83/80)^2 = 12,800 × (83/80) × (83/80) = (2 × 83 × 83) = 2 × 6,889 = Rs. 13,778. Compound interest = 13,778 - 12,800 = Rs. 978.
In simple words: With a rate of 7.5% per year paid every six months, each half-year gets 3.75% interest. The first period's interest itself earns interest in the second period, producing the extra Rs. 978.

Exam Tip: Convert fractional rates carefully - halving a rate like 7.5% means calculating 3.75% per half-year, and working with fractions (83/80) often simplifies calculations better than decimals.

Question 4. Find the compound interest on Rs. 160,000 for 2 years at 10% per annum, compounded half-yearly.
Answer: Principal P = Rs. 160,000, Annual rate = 10%, Half-yearly rate = 5%, Time = 2 years = 4 half-years. Using the compound interest formula: A = P(1 + R/100)^n = 160,000 × (1 + 5/100)^4 = 160,000 × (105/100)^4 = 160,000 × (21/20)^4 = 160,000 × (21/20) × (21/20) × (21/20) × (21/20) = 160,000 × (21 × 21 × 21 × 21) / (20 × 20 × 20 × 20) = (160,000 × (21/20)^4) = Rs. 194,481. Compound interest = 194,481 - 160,000 = Rs. 34,481.
In simple words: With interest compounded four times over two years (every six months), your money multiplies by 1.05 four times. This creates a snowball effect where each compounding period builds on the previous one's growth.

Exam Tip: For multi-period compounding, calculate (1 + r/100)^n step by step or recognize patterns - here (21/20)^4 can be computed as ((21/20)^2)^2 to reduce arithmetic.

Question 6. A sum of Rs. 40,960 invested at some rate of interest becomes Rs. 49,130 after 1.5 years, when the interest is compounded half-yearly. Find the rate of interest per annum.
Answer: Principal P = Rs. 40,960, Amount A = Rs. 49,130, Time = 1.5 years = 3 half-years. Using A = P(1 + r)^n where r is the half-yearly rate: 49,130 = 40,960 × (1 + r)^3. So (1 + r)^3 = 49,130 / 40,960 = (49,130 / 40,960). Simplifying the fraction: GCD analysis gives us (1 + r)^3 = (1331/1000) = (11/10)^3. Taking the cube root: 1 + r = 11/10, so r = 1/10 = 0.1 = 10%. Since this is the half-yearly rate, the annual rate R = 2 × 10% = 20% per annum.
In simple words: Divide the final amount by the starting amount and take the cube root (since there are three half-year periods). This gives you the growth multiplier for each half-year. Subtract 1 to get the half-yearly interest rate, then multiply by 2 to convert to the annual rate.

Exam Tip: Recognizing perfect cubes like 1331/1000 = (11/10)^3 saves time on root calculations - always try to express growth factors as neat powers when possible.

Question 7. Swati borrowed Rs. 40,960 from a bank to buy a scooter. The bank charged interest at the rate of 12.5% per annum, compounded half-yearly. What amount will she pay after 1.5 years?
Answer: Principal P = Rs. 40,960, Annual rate = 12.5% = 25/2 %, Half-yearly rate = (25/4) % = 6.25%, Time = 1.5 years = 3 half-years. The amount is A = P(1 + R/100)^n = 40,960 × (1 + 25/(4×100))^3 = 40,960 × (1 + 25/400)^3 = 40,960 × ((400 + 25)/400)^3 = 40,960 × (425/400)^3 = 40,960 × (17/16)^3 = 40,960 × (17/16) × (17/16) × (17/16) = (10 × 17 × 17 × 17) = 10 × 4,913 = Rs. 49,130. Therefore, Swati will pay Rs. 49,130 after 1.5 years.
In simple words: At 12.5% annual interest paid every six months, each period's interest is 6.25%. Over 1.5 years (three half-year periods), the principal grows from Rs. 40,960 to Rs. 49,130 because the interest itself earns interest each time it's compounded.

Exam Tip: Always convert the time period and rate to match the compounding frequency. If compounding is half-yearly, express both rate and time in half-year units to use the formula correctly.

Question 8. Aslam has to pay Rs. 125,000 to the bank for 1.5 years at a rate of 12% per annum, compounded half-yearly. What is the interest he pays?
Answer: Principal P = Rs. 125,000, Annual rate = 12%, Half-yearly rate = 6%, Time = 1.5 years = 3 half-years. Using the compound interest formula: A = P(1 + R/100)^n = 125,000 × (1 + 6/100)^3 = 125,000 × (106/100)^3 = 125,000 × (53/50)^3 = 125,000 × (53/50) × (53/50) × (53/50) = Rs. 148,877. Compound interest = A - P = 148,877 - 125,000 = Rs. 23,877.
In simple words: Every six months, 6% interest is added to the amount. By the end of three periods (1.5 years), the combined effect of three interest calculations raises the total owed from Rs. 125,000 to Rs. 148,877, making the interest Rs. 23,877.

Exam Tip: Compare compound interest (paid when compounded half-yearly) with simple interest (Rs. 22,500 for the same principal, rate, and time) - notice the difference is Rs. 1,377, which shows the value of compounding multiple times per year.

Question 9. A bank offers 25% per annum compounded half-yearly. Kamla invests Rs. 65,536 with this bank. How much interest will she earn after 2 years?
Answer: Principal P = Rs. 65,536, Annual rate = 25%, Half-yearly rate = 12.5%, Time = 2 years = 4 half-years. Amount = P(1 + R/100)^n = 65,536 × (1 + 12.5/100)^4 = 65,536 × (1 + 1/8)^4 = 65,536 × (9/8)^4. Computing (9/8)^4 = ((9/8)^2)^2 = (81/64)^2 = 6,561/4,096. So A = 65,536 × (6,561/4,096) = 16 × 6,561 = Rs. 104,976. Alternatively, 65,536 × (9/8)^4 = 65,536 × (9/8) × (9/8) × (9/8) × (9/8) = Rs. 104,976. Compound interest = 104,976 - 65,536 = Rs. 39,440. Additionally, for simple interest comparison: SI = (P × R × T) / 100 = (65,536 × 25 × 2) / 100 = Rs. 32,768. The difference between compound and simple interest = 39,440 - 32,768 = Rs. 6,672, showing how much extra earning comes from half-yearly compounding.
In simple words: With 12.5% added every six months over two years (four times), the principal multiplies by (9/8) four times in total. This produces Rs. 39,440 in interest, which is significantly more than the Rs. 32,768 that simple interest would give for the same terms.

Exam Tip: To see the power of compounding, always calculate what simple interest would earn and compare - the difference highlights how frequently compounding adds extra earnings beyond the basic annual rate.

Question 10. Sudershan invests Rs. 32,000 at 5% per annum for 6 months, compounded quarterly. How much will he receive?
Answer: Principal P = Rs. 32,000, Annual rate = 5%, Quarterly rate = (5/4) %, Time = 6 months = 2 quarter years. Using the compound interest formula: A = P(1 + R/100)^n = 32,000 × (1 + (5/4)/100)^2 = 32,000 × (1 + 5/400)^2 = 32,000 × ((400 + 5)/400)^2 = 32,000 × (405/400)^2 = 32,000 × (81/80)^2 = 32,000 × (81/80) × (81/80) = (5 × 81 × 81) = 5 × 6,561 = Rs. 32,805. Therefore, Sudershan will receive Rs. 32,805 after 6 months.
In simple words: At 5% annual interest paid quarterly, each three-month period earns 1.25%. Over 6 months (two quarters), your Rs. 32,000 grows to Rs. 32,805 because the first quarter's interest also earns interest in the second quarter.

Exam Tip: Match the compounding period to the time given - for 6 months with quarterly compounding, you have exactly 2 periods, making the exponent 2 instead of a fraction.

Question 10. Arun borrows Rs. 390,625 at 16% per annum for 1 year, compounded quarterly. What amount must he pay after 1 year?
Answer: Principal P = Rs. 390,625, Annual rate = 16%, Quarterly rate = (16/4) % = 4%, Time = 1 year = 4 quarter years. Using the compound interest formula: A = P(1 + R/100)^n = 390,625 × (1 + 4/100)^4 = 390,625 × (104/100)^4 = 390,625 × (26/25)^4 = 390,625 × (26/25) × (26/25) × (26/25) × (26/25) = Rs. 456,976. Therefore, Arun must pay Rs. 456,976 after 1 year.
In simple words: With interest compounded every three months at 4% per quarter, the amount is multiplied by 1.04 four times over the year. This produces a final amount of Rs. 456,976 that Arun must repay.

Exam Tip: For quarterly compounding, divide the annual rate by 4 and multiply the years by 4 to get the exponent - 1 year becomes 4 compounding periods.

Question 1. Find the compound interest on Rs. 5,000 at 8% per annum for 2 years, compounded half-yearly.
Answer: (c) Rs. 832

Principal P = Rs. 5,000, Annual rate = 8%, Half-yearly rate = 4%, Time = 2 years = 4 half-years. The amount is A = P(1 + R/100)^n = 5,000 × (1 + 4/100)^2. Wait, let me recalculate: A = 5,000 × (1 + 4/100)^2. Actually, for 2 years with half-yearly compounding, n = 4 (not 2). So A = 5,000 × (1 + 4/100)^2 would be for 1 year. For 2 years: A = 5,000 × (1.04)^4 = 5,000 × (104/100)^2 = 5,000 × (27/25)^2 = 5,000 × (27/25) × (27/25) = Rs. 5,832. Compound interest = 5,832 - 5,000 = Rs. 832.
In simple words: Over two years with interest added every six months, your money grows by four compounding steps of 4% each. The final amount reaches Rs. 5,832, so you earn Rs. 832 in interest.

Exam Tip: Count compounding periods carefully - two years of half-yearly compounding means 4 periods total, not 2. Always multiply years by the compounding frequency.

Question 2. Find the compound interest on Rs. 10,000 at 10% per annum for 3 years, compounded annually.
Answer: (b) Rs. 3,310

Principal P = Rs. 10,000, Annual rate = 10%, Time = 3 years. Using the compound interest formula: A = P(1 + R/100)^n = 10,000 × (1 + 10/100)^3 = 10,000 × (110/100)^3 = 10,000 × (11/10)^3 = 10,000 × (11/10) × (11/10) × (11/10) = 10,000 × (1,331/1,000) = 10 × 1,331 = Rs. 13,310. Compound interest = 13,310 - 10,000 = Rs. 3,310.
In simple words: Each year, the 10% interest is calculated on the growing balance, not just the original amount. After three years of this compounding effect, the interest accumulated totals Rs. 3,310.

Exam Tip: Use (11/10)^3 rather than expanding decimals - recognizing (1.1)^3 = 1.331 or (11/10)^3 = 1331/1000 makes calculations cleaner and reduces rounding errors.

Question 3. A principal amount of Rs. 10,000 is invested for 1 year at different rates: 12% for the first quarter, 10% for the second quarter, and then another rate. If the interest for two different compounding methods is Rs. 1,872 and Rs. 961 respectively, what is the principal for the second scenario?
Answer: (a) Rs. 1,872

Here, A = P × (1 + p/100) × (1 + q/100). We have Rs. 10,000 × (1 + 12/100) × (1 + 10/100) = Rs. [10,000 × (100 + 12)/100 × (100 + 20)/100] = Rs. [10,000 × (112/100) × (120/100)] = Rs. [10,000 × (28/25) × (6/5)] = Rs. (8 × 28 × 6) = Rs. 1,872. Compound interest = Rs. (11,872 - 10,000) = Rs. 1,872.
In simple words: When different rates apply to different periods, multiply the individual growth factors together. Over the two periods shown, your Rs. 10,000 grows to Rs. 11,872, generating Rs. 1,872 in total interest.

Exam Tip: For varying rates across periods, multiply (1 + r₁/100), (1 + r₂/100), and so on - don't add the rates together, and don't calculate simple interest separately for each period.

Question 4. A sum of Rs. 4,000 is invested at 10% per annum for 1 year and 3 months, where interest for the first year is compounded annually and then simple interest is charged for the remaining 3 months. Find the compound interest.
Answer: (c) Rs. 961

Here, A = P × (1 + R/100)² × (1 + (r × R)/100). We have Rs. 4,000 × (1 + 10/100)² × (1 + ((1/4) × 10)/100) = Rs. 4,000 × (110/100)² × ((100 + 2.5)/100) = Rs. 4,000 × (11/10)² × (41/40) = Rs. 4,000 × (121/100) × (41/40) = Rs. (11 × 11 × 41) = Rs. 4,961. Compound interest = Rs. (4,961 - 4,000) = Rs. 961.
In simple words: First, compound the interest annually for the full 1 year using (1.1)² conceptually, then apply simple interest for the fractional quarter-year on that new amount. The two-step process yields Rs. 961 in total interest.

Exam Tip: For fractional-year problems, apply compound interest for complete years first, then simple interest on the enlarged principal for the remaining fraction of a year.

Question 5. A principal amount is invested such that compound interest with rates p%, q%, and r% for 1st, 2nd, and 3rd year respectively yields a total of Rs. 30,051 with principal Rs. 25,000. Find the compound interest.
Answer: (b) Rs. 5,051

Here, A = P × (1 + p/100) × (1 + q/100) × (1 + r/100). We have Rs. 25,000 × (1 + 5/100) × (1 + 6/100) × (1 + 8/100) = Rs. 25,000 × (105/100) × (106/100) × (108/100) = Rs. 25,000 × (21/20) × (53/50) × (27/25) = Rs. (21 × 53 × 27) = Rs. 30,051. Compound interest = Rs. (30,051 - 25,000) = Rs. 5,051.
In simple words: When the interest rate changes each year, calculate the growth factor for each year separately and multiply them together. Each year's new balance becomes the base for the next year's interest calculation.

Exam Tip: For multi-year varying rates, express each rate as a fraction (105/100, 106/100, etc.) and multiply all fractions before simplifying - this avoids repeated decimal calculations and reduces error.

Question 6. A sum of Rs. 6,250 is invested at 8% per annum for 1 year, compounded half-yearly. Find the compound interest earned.
Answer: Principal P = Rs. 6,250, Annual rate = 8%, Half-yearly rate = 4%, Time = 1 year = 2 half-years. Using the formula: A = P(1 + R/100)^n = 6,250 × (1 + 4/100)^2 = 6,250 × (104/100)^2 = 6,250 × (26/25)^2 = 6,250 × (26/25) × (26/25) = Rs. 6,760. Compound interest = 6,760 - 6,250 = Rs. 510.
In simple words: At 8% annual interest paid half-yearly, each six-month period earns 4%. Over 1 year (two periods), your Rs. 6,250 grows to Rs. 6,760 because the first period's interest also earns interest in the second period.

Exam Tip: For half-yearly compounding, always halve the annual rate and double the time period in years. Then use the compound interest formula with these adjusted values.

Question 7. A sum of Rs. 40,000 is invested at some rate of interest per annum, compounded quarterly, for 6 months. If the compound interest earned is Rs. 1,209, find the annual rate of interest.
Answer: (a) Rs. 1,209

Time = 6 months = 2 quarter years, Rate compounded quarterly = (r/4)% per quarter. Now, A = P(1 + R/100)^n = Rs. 40,000 × (1 + (r/(4×100)))^2 = Rs. 40,000 × (1 + r/400)^2 = Rs. 40,000 × ((400 + r)/400)^2 = Rs. 40,000 × (203/200)^2. This gives A = Rs. 40,000 × (203/200) × (203/200) = Rs. (203 × 203) = Rs. 41,209. Compound interest = Rs. (41,209 - 40,000) = Rs. 1,209. Therefore, the rate is 12% per annum (since (203/200) corresponds to a 1.5% quarterly rate, or 6% semi-annually, or 12% annually).
In simple words: Over six months with quarterly compounding, the principal grows such that two compounding steps occur. By working backward from the interest earned, you can determine that the annual rate producing Rs. 1,209 interest is 12%.

Exam Tip: Use the compound interest formula to set up an equation when the rate is unknown - isolate the growth factor and solve by taking square roots or recognizing perfect squares like (203/200).

Question 8. A sum of Rs. 24,000 is invested at 5% per annum for 2 years, compounded annually. Find the compound interest.
Answer: (b) Rs. 26,460

Principal P = Rs. 24,000, Annual rate = 5%, Time = 2 years. Using the compound interest formula: A = P(1 + R/100)^n = 24,000 × (1 + 5/100)^2 = 24,000 × (105/100)^2 = 24,000 × (21/20)^2 = 24,000 × (21/20) × (21/20) = Rs. (60 × 21 × 21) = Rs. 26,460. Therefore, the compound interest = Rs. 26,460 (or the interest earned is Rs. 26,460 - 24,000 = Rs. 2,460).
In simple words: At 5% annual interest, your Rs. 24,000 grows as the interest from the first year (Rs. 1,200) also earns 5% interest in the second year, adding an extra Rs. 60. The total amount becomes Rs. 26,460.

Exam Tip: Always distinguish between the total amount and the interest earned - the amount is what you have at the end, while interest is the amount minus the principal.

Question 9. A principal sum depreciates at 10% per annum. Its present value is Rs. 60,000. Find its value 3 years ago.
Answer: (c) Rs. 43,740

Let V₀ be the value 3 years ago. Using the depreciation formula: V = V₀(1 - R/100)^n, where V is the present value, R is the depreciation rate, and n is the number of years. We have 60,000 = V₀(1 - 10/100)^3 = V₀(90/100)^3 = V₀(9/10)^3. So V₀ = 60,000 / (9/10)^3 = 60,000 × (10/9)^3 = 60,000 × (1,000/729) = Rs. (60 × 9 × 9 × 9) = Rs. 43,740.
In simple words: If something loses 10% of its value each year for 3 years, you reverse the depreciation by dividing the current value by (0.9)³ or multiplying by (10/9)³ to find the starting value.

Exam Tip: For depreciation problems asking "how much was it before," rearrange the formula to V₀ = V / (1 - R/100)^n - this inverts the depreciation to reveal the past value.

Question 11. A quantity grows such that its population is 33,275 after 3 years. If the rate of growth is 10% per annum, what was the population 3 years ago?
Answer: (a) 25,000

Let P be the population 3 years ago. Using the growth formula: V = P(1 + r/100)^n, where V is the present population, r is the growth rate, and n is the number of years. We have 33,275 = P(1 + 10/100)^3 = P(110/100)^3 = P(11/10)^3. So P = 33,275 / (11/10)^3 = 33,275 × (10/11)^3 = (33,275 × 1,000) / (1,331) = 33,275,000 / 1,331 = 25,000.
In simple words: If a population grows by 10% each year for 3 years, it multiplies by (1.1)³ or (11/10)³. Working backward from 33,275, you divide by this growth factor to find the original population was 25,000.

Exam Tip: Recognize the perfect cube 1331 = 11³ - if the present value is divisible by 1331 or a related number, the original value calculation simplifies dramatically.

Question 12. A principal is invested at 5% per annum simple interest. After 3 years, the amount is Rs. 9,261. If the same principal is invested at 5% per annum compound interest, what is the compound interest earned?
Answer: (d) Rs. 1,261

From simple interest: SI = (P × R × T) / 100 = (P × 5 × 3) / 100. The amount after 3 years is A = P + SI = P + (15P/100) = (100P + 15P) / 100 = 115P / 100. Given that A = 9,261, we have 115P / 100 = 9,261, so P = (9,261 × 100) / 115 = Rs. 8,000. Now, for compound interest at 5% per annum for 3 years: A = 8,000 × (1 + 5/100)^3 = 8,000 × (105/100)^3 = 8,000 × (21/20)^3 = 8,000 × (21/20) × (21/20) × (21/20) = Rs. 9,261. Compound interest = 9,261 - 8,000 = Rs. 1,261.
In simple words: First, use the simple interest formula to find the principal from the given amount. Then recalculate the amount using compound interest, which produces more interest because of the compounding effect over three years.

Exam Tip: Problems comparing simple and compound interest often require you to find the principal first from one scenario, then apply it to the other - always identify what is given and solve systematically.

Question 13. A sum of money doubles in a certain period at a fixed rate of compound interest. Find the compound interest if the period is 3 years and the initial principal is Rs. 1,920.
Answer: (d) Rs. 480

We have 510 = {P × (1 + 25/(100 × 2))^2} - P = {P × (1 + 25/200)^2} - P = {P × (9/8)^2} - P = {P × (81/64)} - P = P × {81/64 - 1} = P × {(81 - 64)/64} = P × {17/64}. So 510 = 17P / 64, giving P = (510 × 64) / 17 = Rs. 1,920. Now, SI = (P × R × T) / 100 = (1,920 × 2 × 25) / (100 × 2) = Rs. 480. Therefore, the answer is Rs. 480.
In simple words: From the compound interest earned (Rs. 510) and the compound interest formula, you can work backward to find the principal. Once you have the principal, the simple interest can be calculated for the given rate and time period.

Exam Tip: Problems that mix compound and simple interest calculations require careful tracking of which formula applies to which part of the problem - solve the given scenario first, then apply the learned principal to the new scenario.

Question 14. A principal becomes Rs. 4,913 after 3 years at a certain rate of compound interest, compounded annually. Find the principal.
Answer: (d) Rs. 4,096

We have Rs. 4,913 = {P × (1 + 25/(100 × 4))^3} = {P × (16 + 1)/16)^3} = {P × (17/16)^3}. So P = Rs. (4,913 × (16/17)^3) = Rs. (4,913 × 4,096) / 4,913 = Rs. 4,096.
In simple words: Given the final amount and the time period, use the compound interest formula rearranged to solve for P. Recognize that 4,913 may factor as 17³, and this helps identify the growth factor and calculate the original principal.

Exam Tip: Always check if the given amount is a perfect power - if 4,913 = 17³, then recognizing this structure helps you quickly identify the growth factor and solve for the principal.

Question 15. A principal is invested at a certain rate of compound interest. If the amount after 2 years is Rs. 7,500 and after another year (3 years total) it is Rs. 8,427, find the rate of interest.
Answer: (c) 6%

Here, A = P(1 + R/100). We have Rs. 7,500 × (1 + R/100) = Rs. 8,427. So (1 + R/100) = 8,427 / 7,500 = (8,427 / 7,500). Simplifying: (8,427 / 7,500) = (53/50) (after dividing both by 159). So 1 + R/100 = 53/50, giving R/100 = 53/50 - 1 = 3/50, so R = (3/50) × 100 = 6%. Therefore, the rate of interest is **6% per annum**.
In simple words: Divide the amount in year 3 by the amount in year 2 to find the single-year growth multiplier. This directly gives you (1 + R/100), from which you subtract 1 and multiply by 100 to get the rate as a percentage.

Exam Tip: The ratio of consecutive year amounts in compound interest directly reveals the annual growth factor - this is a much faster method than setting up the full compound interest formula when you have amounts for two successive years.