ML Aggarwal Class 9 Maths Solutions Chapter 02 Compound Interest

Question 1. Find the amount and the compound interest on Rs.8000 at 5% per annum for 2 years.
Answer: Start with a principal of Rs.8000 for the first year. Calculate interest by multiplying Rs.8000 by 5% and 1 year, which gives Rs.400. Adding this to the principal yields Rs.8400 after year one. For the second year, the principal now becomes Rs.8400. The interest for year two is Rs.8400 multiplied by 5% and 1 year, equaling Rs.420. After year two, the total amount is Rs.8400 plus Rs.420, which totals Rs.8820. The compound interest over both years is the difference between the final amount and the original principal: Rs.8820 minus Rs.8000 equals Rs.820.
In simple words: The final amount after 2 years is Rs.8820, and the interest earned is Rs.820.

Exam Tip: Always apply the interest from the previous year as part of the new principal for the next year - this is what makes it "compound" interest, not simple interest.

Question 2. A man invested Rs.46875 at 4% per annum compound interest for 3 years. Calculate:
(i) the interest for the first year.
(ii) the amount standing to his credit at the end of the second year.
(iii) the interest for the third year.
Answer:
(i) For the first year, the principal is Rs.46875. Multiply this by 4% and 1 year to get Rs.1875 in interest. So the interest earned in year one is Rs.1875.
(ii) After the first year, the amount becomes Rs.46875 plus Rs.1875, totaling Rs.48750. This becomes the principal for year two. Calculate year two interest: Rs.48750 multiplied by 4% and 1 year gives Rs.1950. After two years, the total amount is Rs.48750 plus Rs.1950, which equals Rs.50700.
(iii) The principal for the third year is now Rs.50700. Interest for year three is Rs.50700 multiplied by 4% and 1 year, equaling Rs.2028. Therefore, the interest for the third year is Rs.2028.
In simple words: The interest earned each year depends on the total amount from the previous year, not the original sum.

Exam Tip: Present all three parts with the final "Hence" statement for each part - examiners expect to see clear identification of what each calculation answers.

Question 3. Calculate the compound interest for the second year on Rs.8000 invested for 3 years at 10% p.a. Also find the sum due at the end of third year.
Answer: The initial principal is Rs.8000. In year one, the interest equals Rs.8000 times 10% times 1, which gives Rs.800. The amount after year one becomes Rs.8000 plus Rs.800, totaling Rs.8800. For year two, use Rs.8800 as the principal. Year two interest is Rs.8800 times 10% times 1, which equals Rs.880. After two years, the amount is Rs.8800 plus Rs.880, equaling Rs.9680. For year three, the principal is Rs.9680. Year three interest is Rs.9680 times 10% times 1, giving Rs.968. The final amount after three years is Rs.9680 plus Rs.968, which totals Rs.10648. The compound interest for the second year is Rs.880, and the sum due at the end of the third year is Rs.10648.
In simple words: The second year interest is Rs.880, and after all three years, the total amount owed is Rs.10648.

Exam Tip: Make sure to distinguish between the compound interest for a specific year and the total compound interest earned - the question asks only for the second year's interest and the final total.

Question 4. Ramesh invested Rs.12800 for three years at the rate of 10% per annum compound interest. Find:
(i) the sum due to Ramesh at the end of the first year.
(ii) the interest he earns for the second year.
(iii) the total amount due to him at the end of three years.
Answer:
(i) The principal for year one is Rs.12800. The interest for year one equals Rs.12800 multiplied by 10% and 1 year, which is Rs.1280. After the first year, the amount standing to Ramesh's credit is Rs.12800 plus Rs.1280, totaling Rs.14080.
(ii) The principal for year two is Rs.14080. The interest earned in year two is Rs.14080 multiplied by 10% and 1 year, which equals Rs.1408. Therefore, Ramesh earns Rs.1408 in interest during the second year.
(iii) After two years, the amount is Rs.14080 plus Rs.1408, equaling Rs.15488. For year three, the principal is Rs.15488. Year three interest is Rs.15488 multiplied by 10% and 1 year, giving Rs.1548.80. The total amount after three years is Rs.15488 plus Rs.1548.80, which totals Rs.17036.80.
In simple words: After one year, Ramesh has Rs.14080. In the second year, he earns Rs.1408 in interest. After three years, his total amount is Rs.17036.80.

Exam Tip: Show each year's calculation separately with clear labeling - this helps examiners quickly verify which part each calculation addresses.

Question 5. The simple interest on a sum of money for 2 years at 12% per annum is Rs.1380. Find:
(i) the sum of money.
(ii) the compound interest on this sum for one year payable half-yearly at the same rate.
Answer:
(i) Use the simple interest formula: S.I. equals P times R times T divided by 100. You know S.I. is Rs.1380, the rate is 12% per annum, and the time is 2 years. Substitute these values: Rs.1380 equals P times 12 times 2 divided by 100. This simplifies to Rs.1380 equals 24P divided by 100, which becomes 138000 equals 24P. Dividing both sides by 24 gives P equals 5750. Therefore, the sum of money is Rs.5750.
(ii) Since the annual rate is 12%, the half-yearly rate becomes 6%. The principal for the first half-year is Rs.5750. Interest for the first half-year is Rs.5750 times 6% times 1, which equals Rs.345. After the first half-year, the amount becomes Rs.5750 plus Rs.345, totaling Rs.6095. This amount serves as the principal for the second half-year. Interest for the second half-year is Rs.6095 times 6% times 1, equaling Rs.365.70. The compound interest for one year payable half-yearly is Rs.345 plus Rs.365.70, which equals Rs.710.70.
In simple words: The original sum is Rs.5750. When interest is added twice in a year at 6% each time, the total compound interest comes to Rs.710.70.

Exam Tip: When the interest is compounded half-yearly, always divide the annual rate by 2 to get the rate for each half-year.

Question 6. A person invests Rs.10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to Rs.11200. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of second year.
Answer:
(i) Let the rate of interest be R. You know the amount after one year is Rs.11200. Find the interest earned by subtracting the principal from the amount: Rs.11200 minus Rs.10000 equals Rs.1200. Using the interest formula, Rs.1200 equals Rs.10000 times R times 1 divided by 100. This simplifies to Rs.1200 equals 100R. Dividing both sides by 100 gives R equals 12. Therefore, the rate of interest is 12% per annum.
(ii) After year one, the amount is Rs.11200. This becomes the principal for year two. The interest for year two is Rs.11200 times 12% times 1, which equals Rs.1344. The final amount after two years is Rs.11200 plus Rs.1344, totaling Rs.12544. Therefore, the amount at the end of the second year is Rs.12544.
In simple words: The interest rate is 12% per year. In the second year, starting with Rs.11200, the interest earned is Rs.1344, bringing the total to Rs.12544.

Exam Tip: When finding the rate, use the information from one complete year where both the principal and final amount are given.

Question 7. Mr. Lalit invested Rs.5000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to Rs.5325. Calculate:
(i) the rate of interest.
(ii) the amount at the end of second year, to the nearest rupee.
Answer:
(i) Let the rate of interest be R. The amount after the first year is Rs.5325. Calculate the interest earned: Rs.5325 minus Rs.5000 equals Rs.325. Apply the interest formula: Rs.325 equals Rs.5000 times R times 1 divided by 100. This simplifies to Rs.325 equals 50R. Dividing both sides by 50 gives R equals 6.5. Therefore, the rate of interest is 6.5% per annum.
(ii) After year one, the amount is Rs.5325. For the second year, this becomes the principal. Calculate the interest: Rs.5325 times 6.5% times 1 equals Rs.346.125. The amount after two years is Rs.5325 plus Rs.346.125, which totals Rs.5671.125. Rounded to the nearest rupee, the amount at the end of the second year is Rs.5671.
In simple words: The interest rate is 6.5% per year. After two years, the amount grows to Rs.5671 when rounded.

Exam Tip: Remember to round to the nearest rupee only in the final answer, not during intermediate calculations.

Question 8. A man invests Rs.5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs.5600. Calculate:
(i) the rate of interest per annum.
(ii) the interest accrued in the second year.
(iii) the amount at the end of the third year.
Answer:
(i) Let the rate of interest be R. After the first year, the amount is Rs.5600. The interest earned is Rs.5600 minus Rs.5000, equaling Rs.600. Using the interest formula: Rs.600 equals Rs.5000 times R times 1 divided by 100. This becomes Rs.600 equals 50R. Dividing by 50 gives R equals 12. Therefore, the rate of interest is 12% per annum.
(ii) After year one, the amount is Rs.5600, which becomes the principal for year two. Interest for year two is Rs.5600 times 12% times 1, which equals Rs.672. Therefore, the interest accrued in the second year is Rs.672.
(iii) After two years, the amount is Rs.5600 plus Rs.672, equaling Rs.6272. This becomes the principal for year three. Interest for year three is Rs.6272 times 12% times 1, equaling Rs.752.64. The amount after three years is Rs.6272 plus Rs.752.64, totaling Rs.7024.64. Therefore, the amount at the end of the third year is Rs.7024.64.
In simple words: The rate is 12% per year. In year two, the interest is Rs.672. After all three years, the total amount is Rs.7024.64.

Exam Tip: For part (ii), specifically mention "second year" in your final statement to show you are answering the exact part asked.

Question 9. Find the amount and the compound interest on Rs.2000 at 10% p.a. for 2\(\frac{1}{2}\) years, compounded annually.
Answer: Start with a principal of Rs.2000 for the first year. Interest for year one is Rs.2000 times 10% times 1, which equals Rs.200. After one year, the amount becomes Rs.2000 plus Rs.200, totaling Rs.2200. For year two, use Rs.2200 as the principal. Interest for year two is Rs.2200 times 10% times 1, equaling Rs.220. After two years, the amount is Rs.2200 plus Rs.220, which totals Rs.2420. For the remaining half-year, Rs.2420 becomes the principal. Interest for the half-year is Rs.2420 times 10% times 0.5, equaling Rs.121. The final amount after 2.5 years is Rs.2420 plus Rs.121, totaling Rs.2541. The compound interest is the difference between the final amount and the original principal: Rs.2541 minus Rs.2000 equals Rs.541.
In simple words: After two and a half years, the total amount reaches Rs.2541, and the interest earned is Rs.541.

Exam Tip: When dealing with fractional years, apply the interest formula for the partial year period as well, using the fractional time value.

Question 10. Find the amount and the compound interest on Rs.50000 for 1\(\frac{1}{2}\) years at 8% per annum, the interest being compounded semi-annually.
Answer: When interest compounds semi-annually, the annual rate of 8% is halved to 4% for each half-year. The principal for the first half-year is Rs.50000. Interest for the first half-year is Rs.50000 times 4% times 1, equaling Rs.2000. The amount after the first half-year becomes Rs.50000 plus Rs.2000, totaling Rs.52000. For the second half-year, the principal is now Rs.52000. Interest for the second half-year is Rs.52000 times 4% times 1, equaling Rs.2080. After one year, the amount is Rs.52000 plus Rs.2080, totaling Rs.54080. For the third half-year (the remaining 0.5 years), the principal is Rs.54080. Interest for this half-year is Rs.54080 times 4% times 1, equaling Rs.2163.20. The final amount after 1.5 years is Rs.54080 plus Rs.2163.20, totaling Rs.56243.20. The compound interest is the difference between the final amount and the original principal: Rs.56243.20 minus Rs.50000 equals Rs.6243.20.
In simple words: When interest compounds twice a year, the 8% annual rate becomes 4% per half-year. After 1.5 years, the total amount is Rs.56243.20 and the interest earned is Rs.6243.20.

Exam Tip: Always divide the annual rate by 2 when compounding semi-annually, and count the total number of half-year periods in the given time.

Question 11. Calculate the amount and the compound interest on Rs.5000 in 2 years when the rate of interest for successive years is 6% and 8% respectively.
Answer: The principal for the first year is Rs.5000. Since the rate for year one is 6%, the interest is Rs.5000 times 6% times 1, which equals Rs.300. The amount after year one is Rs.5000 plus Rs.300, totaling Rs.5300. For year two, use Rs.5300 as the principal, but the interest rate changes to 8%. Interest for year two is Rs.5300 times 8% times 1, equaling Rs.424. The amount after two years is Rs.5300 plus Rs.424, totaling Rs.5724. The compound interest is calculated by subtracting the original principal from the final amount: Rs.5724 minus Rs.5000 equals Rs.724.
In simple words: When the interest rate changes each year, apply each rate to that year's principal. The final amount after two years is Rs.5724, and the total interest earned is Rs.724.

Exam Tip: When rates vary by year, clearly label which rate applies to which year in your working to avoid confusion.

Question 12. A sum of Rs 9600 is invested for 3 years at 10% per annum at compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in 2 years.
(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.
(v) Hence, write down the compound interest for the third year.
Answer:
(i) The starting amount for year one is Rs 9600. We calculate the interest for the first year using the formula:
\[ I = \frac{9600 \times 10 \times 1}{100} = \frac{96000}{100} = Rs 960 \]
The amount at the end of the first year becomes:
\[ Rs 9600 + Rs 960 = Rs 10560 \]
Therefore, the sum due at the end of the first year is Rs 10560.
(ii) For the second year, the principal now becomes Rs 10560. The interest earned in the second year is:
\[ I = \frac{10560 \times 10 \times 1}{100} = \frac{105600}{100} = Rs 1056 \]
The amount at the end of the second year is:
\[ Rs 10560 + Rs 1056 = Rs 11616 \]
Therefore, the sum due at the end of the second year is Rs 11616.
(iii) The compound interest over 2 years is found by subtracting the original principal from the final amount:
\[ \text{C.I.} = Rs 11616 - Rs 9600 = Rs 2016 \]
Hence, the compound interest earned in 2 years is Rs 2016.
(iv) The difference between the amounts in parts (ii) and (i) is:
\[ Rs 11616 - Rs 10560 = Rs 1056 \]
The interest on this sum for one year at 10% is:
\[ I = \frac{1056 \times 10 \times 1}{100} = \frac{10560}{100} = Rs 105.60 \]
Therefore, the difference is Rs 1056 and the interest earned on it is Rs 105.60.
(v) For the third year, the principal becomes Rs 11616. The interest for the third year is:
\[ I = \frac{11616 \times 10 \times 1}{100} = \frac{116160}{100} = Rs 1161.60 \]
Hence, the compound interest for the third year is Rs 1161.60.
In simple words: Every year, interest is calculated on the total amount (principal plus earlier interest). So each year, you earn interest on a bigger amount than the previous year, which makes the interest grow faster.

Exam Tip: Always remember that in compound interest, each year's principal includes the previous year's interest. This is the key difference from simple interest.

Question 13. The simple interest on a certain sum of money for 2 years at 10% per annum is Rs 1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.
Answer:
Let the principal be Rs x. Given that the simple interest is Rs 1600, we use the simple interest formula:
\[ 1600 = \frac{x \times 10 \times 2}{100} \]
\[ 1600 = \frac{20x}{100} \]
\[ 160000 = 20x \]
\[ x = Rs 8000 \]
Now we calculate the compound interest for 3 years on this principal.
Year 1: Principal = Rs 8000. Interest earned:
\[ I = \frac{8000 \times 10 \times 1}{100} = \frac{80000}{100} = Rs 800 \]
Amount after year 1 = Rs 8000 + Rs 800 = Rs 8800
Year 2: Principal = Rs 8800. Interest earned:
\[ I = \frac{8800 \times 10 \times 1}{100} = \frac{88000}{100} = Rs 880 \]
Amount after year 2 = Rs 8800 + Rs 880 = Rs 9680
Year 3: Principal = Rs 9680. Interest earned:
\[ I = \frac{9680 \times 10 \times 1}{100} = \frac{96800}{100} = Rs 968 \]
Amount after year 3 = Rs 9680 + Rs 968 = Rs 10648
The compound interest is the final amount minus the original principal:
\[ \text{C.I.} = Rs 10648 - Rs 8000 = Rs 2648 \]
Therefore, the amount due is Rs 10648 and the compound interest is Rs 2648 respectively.
In simple words: First, find the original sum using the simple interest formula. Then calculate compound interest by finding the interest each year on the growing amount.

Exam Tip: This question tests your ability to work backwards using the simple interest formula to find the principal, then switch to compound interest calculations.

Question 14. Vikram borrowed Rs 20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after 2\(\frac{1}{2}\) years.
Answer:
Vikram pays simple interest to the bank. Using the simple interest formula:
\[ S.I. = \frac{P \times R \times T}{100} = \frac{20000 \times 10 \times 2.5}{100} = \frac{500000}{100} = Rs 5000 \]
Now we calculate what Vikram receives from Venkat using compound interest.
Year 1: Principal = Rs 20000. Interest:
\[ I = \frac{20000 \times 10 \times 1}{100} = \frac{200000}{100} = Rs 2000 \]
Amount after year 1 = Rs 20000 + Rs 2000 = Rs 22000
Year 2: Principal = Rs 22000. Interest:
\[ I = \frac{22000 \times 10 \times 1}{100} = \frac{220000}{100} = Rs 2200 \]
Amount after year 2 = Rs 22000 + Rs 2200 = Rs 24200
For the remaining half year: Principal = Rs 24200. Interest for 6 months:
\[ I = \frac{24200 \times 10 \times 0.5}{100} = \frac{242000}{200} = Rs 1210 \]
Amount after 2.5 years = Rs 24200 + Rs 1210 = Rs 25410
The compound interest Vikram receives from Venkat:
\[ C.I. = Rs 25410 - Rs 20000 = Rs 5410 \]
Vikram's gain is the difference between what he receives and what he pays:
\[ \text{Gain} = Rs 5410 - Rs 5000 = Rs 410 \]
Therefore, Vikram gains Rs 410 after 2.5 years.
In simple words: Vikram pays simple interest (lower amount) to the bank but receives compound interest (higher amount) from his friend. The difference is his profit.

Exam Tip: When comparing simple and compound interest, always calculate both separately and then find the difference. The gain comes from the extra interest earned through compounding.

Question 15. A man borrows Rs 6000 at 5% compound interest. If he repays Rs 1200 at the end of each year, find the amount outstanding at the beginning of the third year.
Answer:
Year 1: Principal for first year = Rs 6000, rate = 5%. Interest earned:
\[ I = \frac{6000 \times 5 \times 1}{100} = \frac{30000}{100} = Rs 300 \]
Amount at end of year 1 = Rs 6000 + Rs 300 = Rs 6300
After repaying Rs 1200:
Principal for year 2 = Rs 6300 - Rs 1200 = Rs 5100
Year 2: Principal = Rs 5100, rate = 5%. Interest earned:
\[ I = \frac{5100 \times 5 \times 1}{100} = \frac{25500}{100} = Rs 255 \]
Amount at end of year 2 = Rs 5100 + Rs 255 = Rs 5355
After repaying Rs 1200:
Principal for year 3 = Rs 5355 - Rs 1200 = Rs 4155
Therefore, the amount outstanding at the beginning of the third year is Rs 4155.
In simple words: Each year, interest is added to what's left. Then a payment is made. This leaves a smaller amount for the next year's interest calculation.

Exam Tip: Track the principal carefully - it changes after each repayment. The outstanding amount at the start of year 3 is calculated after the year 2 repayment.

Question 16. Mr. Raina deposits Rs 1,600 in a bank every year in the beginning of the year, at 5% per annum compound interest. Calculate the amount due to him at the end of 2 years. Also find his gain in two years.
Answer:
First year: Principal = Rs 1,600, Rate = 5%, Time = 1 year. Using the interest formula:
\[ I = \frac{1,600 \times 5 \times 1}{100} = \frac{8000}{100} = Rs 80 \]
Amount after year 1 = Rs 1,600 + Rs 80 = Rs 1,680
Second year: At the start of year 2, he deposits another Rs 1,600, so the principal becomes:
P = Rs 1,680 + Rs 1,600 = Rs 3,280
Rate = 5%, Time = 1 year
\[ I = \frac{3,280 \times 5 \times 1}{100} = \frac{16,400}{100} = Rs 164 \]
Amount after year 2 = Rs 3,280 + Rs 164 = Rs 3,444
The total gain is the final amount minus the total money invested:
\[ \text{Gain} = Rs 3,444 - (Rs 1,600 \times 2) = Rs 3,444 - Rs 3,200 = Rs 244 \]
Therefore, at the end of 2 years the amount is Rs 3,444 and the gain is Rs 244.
In simple words: He puts money in at the start of each year. The interest on the first year's deposit helps earn more interest in the second year. The gain is what the interest adds beyond his total deposits.

Exam Tip: Remember that deposits are made at the beginning of each year. The first year's deposit earns interest for two full years (partially), so it contributes more to the gain.

Question 17. Mr. Dubey borrows Rs 100000 from State Bank of India at 11% per annum compound interest. He repays Rs 41000 at the end of first year and Rs 47700 at the end of second year. Find the amount outstanding at the beginning of the third year.
Answer:
Year 1: Principal = Rs 100000, rate = 11%. Interest earned:
\[ I = \frac{100000 \times 11 \times 1}{100} = \frac{1100000}{100} = Rs 11000 \]
Amount at end of year 1 = Rs 100000 + Rs 11000 = Rs 111000
After repayment of Rs 41000:
Principal for year 2 = Rs 111000 - Rs 41000 = Rs 70000
Year 2: Principal = Rs 70000, rate = 11%. Interest earned:
\[ I = \frac{70000 \times 11 \times 1}{100} = \frac{770000}{100} = Rs 7700 \]
Amount at end of year 2 = Rs 70000 + Rs 7700 = Rs 77700
After repayment of Rs 47700:
Principal for year 3 = Rs 77700 - Rs 47700 = Rs 30000
Therefore, the amount outstanding at the beginning of the third year is Rs 30000.
In simple words: Interest is added each year. Then a payment reduces what's owed. The remaining balance becomes the principal for next year's interest.

Exam Tip: Always subtract the repayment from the amount (principal plus interest) to find the new principal for the next year.

Question 18. Jaya borrowed Rs 50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays Rs 33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.
Answer:
Year 1: Principal = Rs 50000, rate = 12%. Interest earned:
\[ I = \frac{50000 \times 12 \times 1}{100} = \frac{600000}{100} = Rs 6000 \]
Amount at end of year 1 = Rs 50000 + Rs 6000 = Rs 56000
After repayment of Rs 33000:
Principal for year 2 = Rs 56000 - Rs 33000 = Rs 23000
Year 2: Principal = Rs 23000, rate = 15%. Interest earned:
\[ I = \frac{23000 \times 15 \times 1}{100} = \frac{345000}{100} = Rs 3450 \]
Amount at end of year 2 = Rs 23000 + Rs 3450 = Rs 26450
Therefore, Jaya must pay Rs 26450 at the end of the second year to clear her debt.
In simple words: In year 1, interest is added at 12%. After she pays, the leftover debt becomes the principal for year 2. Then interest at 15% is added to find what's left to pay.

Exam Tip: When the interest rate changes each year, calculate each year separately with its own rate. Always subtract repayments before finding the next year's principal.

Exercise 2.2

Question 1. Find the amount and the compound interest on Rs 5000 for 2 years at 6% per annum, interest payable yearly.
Answer:
Using the compound interest formula:
\[ A = P\left(1 + \frac{r}{100}\right)^n \]
Substituting values, P = Rs 5000, r = 6%, n = 2:
\[ A = Rs 5000\left(1 + \frac{6}{100}\right)^2 = Rs 5000 \times \left(\frac{106}{100}\right)^2 \]
\[ = Rs 5000 \times \left(\frac{53}{50}\right)^2 = Rs 5000 \times \frac{53}{50} \times \frac{53}{50} \]
\[ = Rs \frac{14045000}{2500} = Rs 5618 \]
The compound interest is:
\[ C.I. = A - P = Rs 5618 - Rs 5000 = Rs 618 \]
Therefore, the amount is Rs 5618 and the compound interest is Rs 618 respectively.
In simple words: Apply the compound interest formula with the rate and time given. The difference between the final amount and the starting amount gives you the interest earned.

Exam Tip: Always simplify fractions before squaring them - it makes calculations much faster and reduces errors.

Question 2. Find the amount and the compound interest on Rs 8000 for 4 years at 10% per annum, interest reckoned yearly.
Answer:
Using the compound interest formula with P = Rs 8000, r = 10%, n = 4:
\[ A = Rs 8000\left(1 + \frac{10}{100}\right)^4 = Rs 8000 \times \left(\frac{110}{100}\right)^4 \]
\[ = Rs 8000 \times \left(\frac{11}{10}\right)^4 = Rs 8000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \]
\[ = Rs \frac{117128000}{10000} = Rs 11712.80 \]
The compound interest is:
\[ C.I. = A - P = Rs 11712.80 - Rs 8000 = Rs 3712.80 \]
Therefore, the amount is Rs 11712.80 and the compound interest is Rs 3712.80 respectively.
In simple words: When using the formula for 4 years, multiply the rate fraction four times. The interest grows significantly because interest earns interest each year.

Exam Tip: For longer time periods, use the formula method rather than year-by-year calculation - it saves time and reduces arithmetic errors.

Question 3. If the interest is compounded half-yearly, calculate the amount when the principal is Rs 7400, the rate of interest is 5% and the duration is one year.
Answer:
When interest is compounded half-yearly, we divide the annual rate by 2:
\[ \text{Rate per half-year} = \frac{5}{2} = 2.5\% \]
Since the duration is 1 year, the number of conversion periods is:
\[ n = 2 \text{ (half-yearly periods)} \]
Using the formula with P = Rs 7400, r = 2.5%, n = 2:
\[ A = Rs 7400\left(1 + \frac{2.5}{100}\right)^2 = Rs 7400 \times \left(\frac{102.5}{100}\right)^2 \]
\[ = Rs 7400 \times \left(\frac{1025}{1000}\right)^2 = Rs 7400 \times \left(\frac{41}{40}\right)^2 \]
\[ = Rs 7400 \times \frac{41}{40} \times \frac{41}{40} = Rs \frac{12439400}{1600} = Rs 7774.625 \]
Therefore, the amount is Rs 7774.625.
In simple words: When compounding happens twice a year, divide the yearly rate by 2 and count 2 periods instead of 1. This gives more interest than yearly compounding.

Exam Tip: Always adjust the rate and number of periods when compounding frequency changes - half-yearly means divide the rate by 2 and multiply periods by 2.

Question 4. Find the amount and the compound interest on Rs 5000 at 10% p.a. for 1\(\frac{1}{2}\) years, compound interest reckoned semi-annually.
Answer:
Since interest is compounded half-yearly, the rate per conversion period is:
\[ \text{Rate} = \frac{10}{2} = 5\% \]
The number of half-yearly periods in 1.5 years is:
\[ n = 3 \text{ half-years} \]
Using the formula with P = Rs 5000, r = 5%, n = 3:
\[ A = Rs 5000\left(1 + \frac{5}{100}\right)^3 = Rs 5000 \times \left(\frac{105}{100}\right)^3 \]
\[ = Rs 5000 \times \left(\frac{21}{20}\right)^3 = Rs 5000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \]
\[ = Rs 5000 \times \frac{9261}{8000} = Rs 5788.125 \]
The compound interest is:
\[ C.I. = A - P = Rs 5788.125 - Rs 5000 = Rs 788.125 \]
Therefore, the amount is Rs 5788.125 and the compound interest is Rs 788.125 respectively.
In simple words: For 1.5 years with half-yearly compounding, there are 3 periods. Divide the 10% rate by 2 to get 5% per period, then apply the formula.

Exam Tip: Always count the exact number of compounding periods - if compounding happens twice yearly, 1.5 years means 3 periods, not 1.5 periods.

Question 5. Find the amount and the compound interest on Rs.100000 compounded quarterly for 9 months at the rate of 4% p.a.
Answer: The yearly interest rate is 4%, so the quarterly rate becomes 1%. Since the money stays invested for 9 months, there are 3 conversion periods.

\( A = P\left(1 + \frac{r}{100}\right)^n = \text{Rs.}100000 \times \left(1 + \frac{1}{100}\right)^3 = \text{Rs.}100000 \times \left(\frac{101}{100}\right)^3 = \text{Rs.}100000 \times \frac{101}{100} \times \frac{101}{100} \times \frac{101}{100} = \text{Rs.}\frac{100000 \times 101 \times 101 \times 101}{1000000} = \text{Rs.}\frac{1030301}{10} = \text{Rs.}103030.10 \)

Compound interest = Final amount - Principal = Rs.103030.10 - Rs.100000 = Rs.3030.10

Hence, the amount is Rs.103030.10 and the compound interest is Rs.3030.10.
In simple words: To find how much money grows in compound interest, use the formula with the quarterly rate and number of quarters. The interest earned is what's left after subtracting the starting amount from the final amount.

Exam Tip: Always convert the annual rate to the conversion period rate (divide by the number of periods in a year) and count the total number of conversion periods correctly - this is where most mistakes happen.

Question 6. Find the difference between C.I. and S.I. on sum of Rs.4800 for 2 years at 5% per annum payable yearly.
Answer: For simple interest:

\( S.I. = \frac{P \times R \times T}{100} = \frac{4800 \times 5 \times 2}{100} = \frac{48000}{100} = \text{Rs.}480 \)

For compound interest:

\( C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] = \text{Rs.}4800 \times \left[\left(1 + \frac{5}{100}\right)^2 - 1\right] = \text{Rs.}4800 \times \left[\left(\frac{105}{100}\right)^2 - 1\right] = \text{Rs.}4800 \times \left[\left(\frac{21}{20}\right)^2 - 1\right] = \text{Rs.}4800 \times \left[\frac{441}{400} - 1\right] = \text{Rs.}4800 \times \left[\frac{441 - 400}{400}\right] = \text{Rs.}4800 \times \frac{41}{400} = \text{Rs.}\frac{196800}{400} = \text{Rs.}492 \)

Difference = C.I. - S.I. = Rs.492 - Rs.480 = Rs.12

Hence, the difference between C.I. and S.I. is Rs.12.
In simple words: Simple interest stays the same each year, but compound interest grows faster because you earn interest on your interest. The difference shows how much extra compound interest gives you.

Exam Tip: Always calculate both S.I. and C.I. separately using their own formulas - don't try to combine or simplify. The final subtraction gives the difference directly.

Question 7. Find the difference between the simple interest and compound interest on Rs.2500 for 2 years at 4% per annum, compound interest being reckoned semi-annually.
Answer: Since the interest is calculated half-yearly, the rate per half-year is \( \frac{4}{2} = 2\% \). In 2 years, there are 4 half-year periods.

For simple interest (using yearly rate):

\( S.I. = \frac{P \times R \times T}{100} = \frac{\text{Rs.}2500 \times 2 \times 4}{100} = \frac{\text{Rs.}20000}{100} = \text{Rs.}200 \)

For compound interest (with semi-annual compounding):

\( C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] = \text{Rs.}2500 \times \left[\left(1 + \frac{2}{100}\right)^4 - 1\right] = \text{Rs.}2500 \times \left[\left(\frac{102}{100}\right)^4 - 1\right] = \text{Rs.}2500 \times \left[\left(\frac{51}{50}\right)^4 - 1\right] = \text{Rs.}2500 \times \left[\frac{6765201}{6250000} - 1\right] = \text{Rs.}2500 \times \left[\frac{6765201 - 6250000}{6250000}\right] = \text{Rs.}2500 \times \frac{515201}{6250000} = \text{Rs.}\frac{515201}{2500} = \text{Rs.}206.084 \)

Difference = C.I. - S.I. = Rs.206.084 - Rs.200 = Rs.6.084

Hence, the difference between C.I. and S.I. is Rs.6.084.
In simple words: When interest compounds twice a year instead of once, it grows faster and the gap between compound and simple interest becomes larger.

Exam Tip: When compounding is semi-annual, divide the annual rate by 2 and double the number of periods. This is crucial - forgetting to adjust the rate or periods is a common error.

Question 8. Find the amount and the compound interest on Rs.2000 in 2 years if the rate is 4% for the first year and 3% for the second year.
Answer: When the rate changes each year, calculate the amount after each year separately, using the previous year's amount as the new principal.

For the first year with P = Rs.2000 and rate = 4%:

\( A = P\left(1 + \frac{r}{100}\right)^n = \text{Rs.}2000 \times \left(1 + \frac{4}{100}\right)^1 = \text{Rs.}2000 \times \frac{104}{100} = \text{Rs.}2000 \times \frac{26}{25} = \text{Rs.}80 \times 26 = \text{Rs.}2080 \)

For the second year with P = Rs.2080 and rate = 3%:

\( A = \text{Rs.}2080 \times \left(1 + \frac{3}{100}\right)^1 = \text{Rs.}2080 \times \frac{103}{100} = \text{Rs.}\frac{214240}{100} = \text{Rs.}2142.40 \)

Compound interest = Final amount - Principal = Rs.2142.40 - Rs.2000 = Rs.142.40

Hence, the amount is Rs.2142.40 and the compound interest is Rs.142.40.
In simple words: When rates change year to year, work through each year one at a time. Each year's ending amount becomes the starting amount for the next year.

Exam Tip: Keep track of the principal for each year carefully - the amount at the end of one year becomes the principal for the next year. Write it down clearly to avoid mixing up numbers.

Question 9. Find the compound interest on Rs.3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.
Answer: Apply each year's rate in sequence, using each year's final amount as the next year's principal.

For the first year with P = Rs.3125 and rate = 4%:

\( A = \text{Rs.}3125 \times \left(1 + \frac{4}{100}\right)^1 = \text{Rs.}3125 \times \frac{104}{100} = \text{Rs.}3125 \times \frac{26}{25} = \text{Rs.}\frac{81250}{25} = \text{Rs.}3250 \)

For the second year with P = Rs.3250 and rate = 5%:

\( A = \text{Rs.}3250 \times \left(1 + \frac{5}{100}\right)^1 = \text{Rs.}3250 \times \frac{105}{100} = \text{Rs.}3250 \times \frac{21}{20} = \text{Rs.}\frac{68250}{20} = \text{Rs.}3412.50 \)

For the third year with P = Rs.3412.50 and rate = 6%:

\( A = \text{Rs.}3412.50 \times \left(1 + \frac{6}{100}\right)^1 = \text{Rs.}3412.50 \times \frac{106}{100} = \text{Rs.}3412.50 \times \frac{53}{50} = \text{Rs.}\frac{180862.5}{50} = \text{Rs.}3617.25 \)

Compound interest = Final amount - Principal = Rs.3617.25 - Rs.3125 = Rs.492.25

Hence, the compound interest is Rs.492.25.
In simple words: When rates change every year, multiply step by step through each year. The answer from one year becomes the starting point for the next.

Exam Tip: For multi-year problems with changing rates, always write down each year's calculation separately and clearly label which year you're working on. This prevents errors and makes checking easier.

Question 10. What sum of money will amount to Rs.9261 in 3 years at 5% per annum compound interest?
Answer: We need to find the principal P when the final amount is known. Use the compound interest formula and solve for P.

Given: A = Rs.9261, rate = 5%, n = 3

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( 9261 = P\left(1 + \frac{5}{100}\right)^3 = P\left(\frac{105}{100}\right)^3 = P\left(\frac{21}{20}\right)^3 = P \times \frac{9261}{8000} \)

\( P = \frac{9261 \times 8000}{9261} = \text{Rs.}8000 \)

Hence, Rs.8000 will grow to Rs.9261 in 3 years at 5% per annum compound interest.
In simple words: If you know the final amount and the rate, divide the final amount by the growth factor to find how much you started with.

Exam Tip: Always rearrange the formula carefully to isolate P. Double-check your simplified fraction - here \( \frac{21}{20} \) gives a nice cube of \( \frac{9261}{8000} \), which means the original numbers were chosen to work out neatly.

Question 11. What sum invested at 4% per annum compounded semi-annually amounts to Rs.7803 at the end of one year?
Answer: Since interest compounds semi-annually, the rate per half-year is \( \frac{4}{2} = 2\% \). In one year, there are 2 conversion periods.

Given: A = Rs.7803, rate (semi-annual) = 2%, n = 2

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( 7803 = P\left(1 + \frac{2}{100}\right)^2 = P\left(\frac{102}{100}\right)^2 = P\left(\frac{51}{50}\right)^2 = P \times \frac{2601}{2500} \)

\( P = \frac{7803 \times 2500}{2601} = \text{Rs.}7500 \)

Hence, Rs.7500 will grow to Rs.7803 in 1 year at 4% per annum compounded semi-annually.
In simple words: When money compounds twice yearly, fewer periods means faster calculation. Rearrange the formula the same way to find the starting amount.

Exam Tip: In semi-annual compounding problems, always divide the annual rate by 2 and count the number of six-month periods. Here n = 2 because one year has two six-month periods.

Question 12. What sum invested for \( 1\frac{1}{2} \) years compounded half-yearly at the rate of 4% p.a. will amount to Rs.132651?
Answer: Since interest is compounded half-yearly, the rate per half-year is \( \frac{4}{2} = 2\% \). In \( 1\frac{1}{2} \) years, there are 3 conversion periods (since each period is 6 months).

Given: A = Rs.132651, rate (semi-annual) = 2%, n = 3

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( 132651 = P\left(1 + \frac{2}{100}\right)^3 = P\left(\frac{102}{100}\right)^3 = P\left(\frac{51}{50}\right)^3 = P \times \frac{132651}{125000} \)

\( P = \frac{132651 \times 125000}{132651} = \text{Rs.}125000 \)

Hence, Rs.125000 will grow to Rs.132651 in \( 1\frac{1}{2} \) years at 4% per annum compounded semi-annually.
In simple words: To count conversion periods with semi-annual compounding, multiply the years by 2. Here, 1.5 years becomes 3 half-year periods.

Exam Tip: Mixed numbers like \( 1\frac{1}{2} \) need careful handling. Convert to 1.5 years first, then multiply by 2 to get the number of half-yearly periods - this step is easy to skip and causes mistakes.

Question 13. On what sum will the compound interest for 2 years at 4% per annum be Rs.5712?
Answer: We know the compound interest amount but need to find the principal. Use the formula relating compound interest to principal.

\( C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)

\( 5712 = P\left[\left(1 + \frac{4}{100}\right)^2 - 1\right] = P\left[\left(\frac{104}{100}\right)^2 - 1\right] = P\left[\left(\frac{26}{25}\right)^2 - 1\right] = P\left[\frac{676}{625} - 1\right] = P\left[\frac{676 - 625}{625}\right] = P \times \frac{51}{625} \)

\( P = \frac{5712 \times 625}{51} = 112 \times 625 = \text{Rs.}70000 \)

Hence, the principal is Rs.70000.
In simple words: When you know only the interest earned (not the final amount), use the C.I. formula directly. Divide the given interest by the bracketed expression to find the principal.

Exam Tip: Notice that 51 and 625 appear in the simplified form - these numbers often have common factors with the given C.I. Make sure to simplify fractions fully before the final multiplication.

Question 14. A man invests Rs.1200 for two years at compound interest. After one year the money amounts to Rs.1275. Find the interest for the second year correct to the nearest rupee.
Answer: First, find the rate of interest from the first year's data. Then use that rate to calculate interest for the second year.

Let the rate of interest be r% per annum. After one year, Rs.1200 becomes Rs.1275.

\( A = P\left(1 + \frac{r}{100}\right)^n \)

\( 1275 = 1200\left(1 + \frac{r}{100}\right) \implies \frac{1275}{1200} = 1 + \frac{r}{100} \implies \frac{1275}{1200} - 1 = \frac{r}{100} \implies \frac{1275 - 1200}{1200} = \frac{r}{100} \implies \frac{75}{1200} = \frac{r}{100} \)

\( r = \frac{7500}{1200} = \frac{25}{4} = 6\frac{1}{4} \) %

The principal for the second year is Rs.1275 (the amount at the end of year 1). The interest for the second year is:

\( \text{Interest} = \frac{P \times R \times T}{100} = \frac{\text{Rs.}1275 \times \frac{25}{4} \times 1}{100} = \frac{\text{Rs.}1275 \times 25}{400} = \frac{\text{Rs.}31875}{400} = \text{Rs.}79.6875 \approx \text{Rs.}80 \)

Hence, the interest for the second year is Rs.80.
In simple words: First work out what rate was used (by comparing start and end of year 1). Then apply that rate to the new principal (year 1's ending amount) to find year 2's interest.

Exam Tip: Always round at the very end, not during intermediate steps. Here, Rs.79.6875 rounds to Rs.80 - rounding earlier could give a slightly different answer.

Question 15. At what rate percent per annum compound interest will Rs.2304 amount to Rs.2500 in 2 years?
Answer: We must find the rate when principal, amount, and time are known. Rearrange the compound interest formula to solve for r.

\( A = P\left(1 + \frac{r}{100}\right)^n \)

Given: A = Rs.2500, P = Rs.2304, n = 2

\( 2500 = 2304\left(1 + \frac{r}{100}\right)^2 \implies \left(1 + \frac{r}{100}\right)^2 = \frac{2500}{2304} \implies 1 + \frac{r}{100} = \sqrt{\frac{2500}{2304}} = \frac{50}{48} = \frac{25}{24} \)

\( \frac{r}{100} = \frac{25}{24} - 1 = \frac{25 - 24}{24} = \frac{1}{24} \)

\( r = \frac{100}{24} = \frac{25}{6} = 4\frac{1}{6} \) % per annum

Hence, the rate is \( 4\frac{1}{6} \) % per annum.
In simple words: Divide the final amount by the starting amount to get the growth factor. Take the square root (since n = 2) to find the yearly multiplier. Subtract 1 and multiply by 100 to get the percentage rate.

Exam Tip: Always take the nth root when solving for rate with compound interest - here the square root because n = 2. If n were 3, you'd take the cube root instead. Check that your final rate is reasonable by substituting back into the original formula.

Question 16. A sum compounded annually becomes \( \frac{16}{25} \) times of itself in two years. Determine the rate of interest per annum.
Answer: Suppose the principal is P. Because the amount grows to \( \frac{16}{25}P \) in two years, we have
\( \Rightarrow \frac{16}{25}P = P\left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \frac{16}{25} = \left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \left(\frac{4}{5}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \frac{4}{5} = 1 + \frac{r}{100} \)
\( \Rightarrow \frac{r}{100} = \frac{4}{5} - 1 \)
\( \Rightarrow \frac{r}{100} = \frac{4 - 5}{5} \)
\( \Rightarrow \frac{r}{100} = \frac{-1}{5} \)
\( \Rightarrow r = -20 \)
Since the rate cannot be negative, there is an error in the problem statement (the sum should grow, not shrink). Assuming the sum becomes \( \frac{25}{16} \) times itself instead:
\( \frac{25}{16} = \left(1 + \frac{r}{100}\right)^2 \)
\( \left(\frac{5}{4}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \)
\( \frac{5}{4} = 1 + \frac{r}{100} \)
\( \frac{r}{100} = \frac{5}{4} - 1 = \frac{1}{4} \)
\( r = 25 \)
In simple words: When a sum grows to a larger multiple of itself over two years, we use the compound interest formula to work backwards and find the interest rate. The rate comes out to be 25% per annum.

Exam Tip: Always take the square root of both sides when the exponent is 2, and verify that your rate is positive - a negative rate would mean the money is decreasing, which is impossible in this context.

Question 17. At what rate percent will Rs.2000 amount to Rs.2315.25 in 3 years at compound interest?
Answer: We use the compound interest formula to find the rate. Given that the principal is Rs.2000 and it grows to Rs.2315.25 in 3 years:
\( \Rightarrow 2315.25 = 2000\left(1 + \frac{r}{100}\right)^3 \)
\( \Rightarrow \frac{2315.25}{2000} = \left(1 + \frac{r}{100}\right)^3 \)
\( \Rightarrow \frac{9261}{8000} = \left(1 + \frac{r}{100}\right)^3 \)
\( \Rightarrow \left(\frac{21}{20}\right)^3 = \left(1 + \frac{r}{100}\right)^3 \)
Taking the cube root of both sides:
\( \Rightarrow \frac{21}{20} = 1 + \frac{r}{100} \)
\( \Rightarrow \frac{r}{100} = \frac{21}{20} - 1 \)
\( \Rightarrow \frac{r}{100} = \frac{21 - 20}{20} \)
\( \Rightarrow \frac{r}{100} = \frac{1}{20} \)
\( \Rightarrow r = 5 \)
In simple words: To discover what interest rate turns Rs.2000 into Rs.2315.25 over three years, we divide the final amount by the starting amount and take the cube root. The interest rate is 5% per year.

Exam Tip: When the time period is 3 years, remember to take the cube root (not the square root) when isolating the rate expression - this is a common mistake.

Question 18. If Rs.40000 amounts to Rs.48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum.
Answer: When interest is compounded half-yearly, the rate per half-year becomes \( \frac{r}{2} \) and the number of half-year periods in 2 years is 4. Using the compound interest formula:
\( \Rightarrow 48620.25 = 40000\left(1 + \frac{r}{200}\right)^4 \)
\( \Rightarrow \frac{48620.25}{40000} = \left(1 + \frac{r}{200}\right)^4 \)
\( \Rightarrow \frac{194481}{160000} = \left(1 + \frac{r}{200}\right)^4 \)
\( \Rightarrow \left(\frac{21}{20}\right)^4 = \left(1 + \frac{r}{200}\right)^4 \)
Taking the fourth root of both sides:
\( \Rightarrow \frac{21}{20} = 1 + \frac{r}{200} \)
\( \Rightarrow \frac{r}{200} = \frac{21}{20} - 1 \)
\( \Rightarrow \frac{r}{200} = \frac{1}{20} \)
\( \Rightarrow r = 10 \)
In simple words: Since the bank adds interest twice every year, we divide the yearly rate by 2. The calculations show that the annual interest rate is 10% per annum.

Exam Tip: For half-yearly compounding, always divide the annual rate by 2 in the formula and multiply the years by 2 to get the number of periods - this adjustment is essential for getting the correct answer.

Question 19. Determine the rate of interest for a sum that becomes \( \frac{216}{125} \) times of itself in \( 1\frac{1}{2} \) years, compounded semi-annually.
Answer: For semi-annual compounding, the rate per half-year is \( \frac{r}{2} \). In \( 1\frac{1}{2} \) years, there are 3 half-year periods. If the principal is P, then:
\( \Rightarrow \frac{216}{125}P = P\left(1 + \frac{r}{200}\right)^3 \)
\( \Rightarrow \frac{216}{125} = \left(1 + \frac{r}{200}\right)^3 \)
\( \Rightarrow \left(\frac{6}{5}\right)^3 = \left(1 + \frac{r}{200}\right)^3 \)
Taking the cube root of both sides:
\( \Rightarrow \frac{6}{5} = 1 + \frac{r}{200} \)
\( \Rightarrow \frac{r}{200} = \frac{6}{5} - 1 \)
\( \Rightarrow \frac{r}{200} = \frac{6 - 5}{5} \)
\( \Rightarrow \frac{r}{200} = \frac{1}{5} \)
\( \Rightarrow r = 40 \)
In simple words: When interest compounds twice yearly, we break the time into half-year blocks. The working shows that the yearly rate of interest is 40% per annum.

Exam Tip: Always convert fractional years into the total number of compounding periods - for semi-annual compounding, multiply the years by 2 to get the correct number of periods to use in the formula.

Question 20. At what rate percent p.a. compound interest would Rs.80000 amount to Rs.88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.
Answer: Using the compound interest formula with the given values:
\( \Rightarrow 88200 = 80000\left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \frac{88200}{80000} = \left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \frac{441}{400} = \left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \left(\frac{21}{20}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \)
\( \Rightarrow \frac{21}{20} = 1 + \frac{r}{100} \)
\( \Rightarrow \frac{r}{100} = \frac{21}{20} - 1 \)
\( \Rightarrow \frac{r}{100} = \frac{1}{20} \)
\( \Rightarrow r = 5 \)
Now, to find the amount after 3 years at 5% per annum:
\( A = 80000\left(1 + \frac{5}{100}\right)^3 = 80000\left(\frac{21}{20}\right)^3 = 80000 \times \frac{9261}{8000} = \text{Rs.}92610 \)
In simple words: By comparing what the money grows to in two years, we discover the interest rate is 5% yearly. Using this rate, after three years the original Rs.80000 becomes Rs.92610.

Exam Tip: Break two-part questions into separate steps - first find the rate, then use that rate for the second part. Always show the calculation for the final amount clearly by substituting the rate you just found.

Question 21. A certain sum amounts to Rs.5292 in 2 years and to Rs.5556.60 in 3 years at compound interest. Find the rate and the sum.
Answer: We set up two equations using the given information. For 2 years:
\( \therefore 5292 = P\left(1 + \frac{r}{100}\right)^2 \) ...(Eq. 1)
For 3 years:
\( \therefore 5556.60 = P\left(1 + \frac{r}{100}\right)^3 \) ...(Eq. 2)
Dividing Eq. 2 by Eq. 1:
\( \Rightarrow \frac{5556.60}{5292} = 1 + \frac{r}{100} \)
\( \Rightarrow 1 + \frac{r}{100} = \frac{555660}{529200} \)
\( \Rightarrow \frac{r}{100} = \frac{555660}{529200} - 1 \)
\( \Rightarrow \frac{r}{100} = \frac{26460}{529200} \)
\( \Rightarrow r = 5 \)
Substituting r = 5 back into Eq. 1:
\( \Rightarrow 5292 = P\left(1 + \frac{5}{100}\right)^2 \)
\( \Rightarrow 5292 = P\left(\frac{21}{20}\right)^2 \)
\( \Rightarrow 5292 = P \times \frac{441}{400} \)
\( \Rightarrow P = 5292 \times \frac{400}{441} = \text{Rs.}4800 \)
In simple words: When we know amounts at two different times, we can find the rate by dividing one equation by the other - this cancels out the principal. Once we have the rate, substituting it back reveals the original sum.

Exam Tip: Always divide the later-year equation by the earlier-year equation to eliminate the principal - this is the quickest path to finding the rate when you have two amount values at different times.

Question 22. A certain sum amounts to Rs.798.60 after 3 years and Rs.878.46 after 4 years. Find the interest rate and the sum.
Answer: We create two equations from the given data. After 3 years:
\( \therefore 798.60 = P\left(1 + \frac{r}{100}\right)^3 \) ...(Eq. 1)
After 4 years:
\( \therefore 878.46 = P\left(1 + \frac{r}{100}\right)^4 \) ...(Eq. 2)
Dividing Eq. 2 by Eq. 1:
\( \Rightarrow \frac{878.46}{798.60} = 1 + \frac{r}{100} \)
\( \Rightarrow 1 + \frac{r}{100} = \frac{87846}{79860} \)
\( \Rightarrow \frac{r}{100} = \frac{87846}{79860} - 1 \)
\( \Rightarrow \frac{r}{100} = \frac{7986}{79860} \)
\( \Rightarrow \frac{r}{100} = \frac{1}{10} \)
\( \Rightarrow r = 10 \)
Substituting r = 10 into Eq. 1:
\( \Rightarrow 798.60 = P\left(1 + \frac{10}{100}\right)^3 \)
\( \Rightarrow 798.60 = P\left(\frac{11}{10}\right)^3 \)
\( \Rightarrow 798.60 = P \times \frac{1331}{1000} \)
\( \Rightarrow P = 798.60 \times \frac{1000}{1331} = \text{Rs.}600 \)
In simple words: We find the rate by taking the ratio of amounts from two different years - the principal cancels out automatically. Then we use this rate to calculate the starting amount by working backwards from either amount.

Exam Tip: The division method works because any two compound amounts of the same principal at different times will have a ratio that depends only on the rate and the difference in time periods - use this shortcut to avoid dealing with the principal initially.

Question 23. In what time will Rs.15625 amount to Rs.17576 at 4% per annum compound interest?
Answer: Using the compound interest formula with the known values:
\( A = P\left(1 + \frac{r}{100}\right)^n \)
\( \Rightarrow 17576 = 15625\left(1 + \frac{4}{100}\right)^n \)
\( \Rightarrow 17576 = 15625\left(\frac{104}{100}\right)^n \)
\( \Rightarrow \frac{17576}{15625} = \left(\frac{26}{25}\right)^n \)
\( \Rightarrow \left(\frac{26}{25}\right)^3 = \left(\frac{26}{25}\right)^n \)
Comparing the exponents:
\( \Rightarrow n = 3 \text{ years} \)
In simple words: We divide the final amount by the starting amount and simplify the fraction. If this fraction matches a power we can recognize (like a cube or square), the exponent tells us the number of years.

Exam Tip: Always simplify the fraction between the final and starting amounts before trying to match it to a power - this often makes the problem much easier to spot and solve.

Question 24(i). In what time will Rs.1500 yield Rs.496.50 as compound interest at 10% per annum compounded annually?
Answer: The compound interest formula relating interest earned to principal and rate is:
\( C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)
Substituting the given values:
\( \Rightarrow 496.50 = 1500\left[\left(1 + \frac{10}{100}\right)^n - 1\right] \)
\( \Rightarrow 496.50 = 1500\left(\frac{110}{100}\right)^n - 1500 \)
\( \Rightarrow 496.50 + 1500 = 1500\left(\frac{11}{10}\right)^n \)
\( \Rightarrow \frac{1996.50}{1500} = \left(\frac{11}{10}\right)^n \)
\( \Rightarrow \frac{1331}{1000} = \left(\frac{11}{10}\right)^n \)
\( \Rightarrow \left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^n \)
\( \Rightarrow n = 3 \text{ years} \)
In simple words: The compound interest earned is the total amount that accumulates minus the original principal. By working with this interest amount directly, we can find how long the money sat in the bank earning interest.

Exam Tip: When given the compound interest amount (not the total amount), add it back to the principal first to get the total, then apply the standard formula - this step is often missed by students.

Question 24(ii). Find the time (in years) in which Rs.12500 will produce Rs.3246.40 as compound interest at 8% per annum, interest compounded annually.
Answer: Using the compound interest formula:
\( C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)
Substituting the values:
\( \Rightarrow 3246.40 = 12500\left[\left(1 + \frac{8}{100}\right)^n - 1\right] \)
\( \Rightarrow 3246.40 = 12500\left(\frac{108}{100}\right)^n - 12500 \)
\( \Rightarrow 3246.40 + 12500 = 12500\left(\frac{27}{25}\right)^n \)
\( \Rightarrow \frac{15746.40}{12500} = \left(\frac{27}{25}\right)^n \)
\( \Rightarrow \frac{19683}{15625} = \left(\frac{27}{25}\right)^n \)
\( \Rightarrow \left(\frac{27}{25}\right)^3 = \left(\frac{27}{25}\right)^n \)
\( \Rightarrow n = 3 \text{ years} \)
In simple words: We take the interest that was earned and the principal, then use them to set up the compound interest equation. Simplifying leads us to see that the money grew for exactly 3 years.

Exam Tip: Always reduce fractions to their simplest form before comparing exponents - a fraction like \( \frac{19683}{15625} \) may look confusing until you recognize it as \( \left(\frac{27}{25}\right)^3 \) in disguise.

Question 25. Rs16000 invested at 10% p.a., compounded semi-annually, amounts to Rs18522. Find the time period of investment.
Answer: The rate is 10% per annum, which means 5% for each half-year period. Let n represent the number of half-years.

Using the compound interest formula \( A = P(1 + \frac{r}{100})^n \):

\( 18522 = 16000(1 + \frac{5}{100})^n \)

\( 18522 = 16000(\frac{105}{100})^n \)

\( \frac{18522}{16000} = (\frac{21}{20})^n \)

\( \frac{9261}{8000} = (\frac{21}{20})^n \)

\( (\frac{21}{20})^3 = (\frac{21}{20})^n \)

\( n = 3 \text{ half-years} \)

Therefore, the time period is 1½ years.
In simple words: The money grows to the target amount in 3 half-year periods, which equals one and a half years total.

Exam Tip: Remember to convert the annual rate to a half-yearly rate by dividing by 2, and ensure n represents the number of half-year periods, not full years.

Question 26. What sum will amount to Rs2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years?
Answer: When the interest rates change in successive years, the formula becomes \( A = P(1 + \frac{r_1}{100})(1 + \frac{r_2}{100}) \).

Substituting the given values:

\( 2782.50 = P(1 + \frac{5}{100})(1 + \frac{6}{100}) \)

\( 2782.50 = P(\frac{21}{20})(\frac{53}{50}) \)

\( P = \frac{2782.50 \times 20 \times 50}{21 \times 53} \)

\( P = \text{Rs}2500 \)
In simple words: When rates differ each year, multiply the growth factors for each year together, then solve for the starting amount.

Exam Tip: Always check if rates vary by year - if they do, use the product method rather than the power formula. Verify your answer by multiplying back through both years.

Question 27. A sum of money is invested at compound interest payable annually. The interest in two successive years is Rs225 and Rs240. Find : (i) the rate of interest. (ii) the original sum. (iii) the interest earned in the third year.
Answer:
Given: Interest for year 1 = Rs225; Interest for year 2 = Rs240; Difference = Rs15.

Since the difference is Rs15, this represents interest earned on the first year's interest (Rs225) for one year.

(i) Finding the rate:

Using \( R = \frac{S.I. \times 100}{P \times T} \):

\( R = \frac{15 \times 100}{225 \times 1} = \frac{20}{3} = 6\frac{2}{3}\% \)

(ii) Finding the principal:

Using \( P = \frac{S.I. \times 100}{R \times T} \):

\( P = \frac{225 \times 100}{\frac{20}{3} \times 1} = \frac{22500}{\frac{20}{3}} = \frac{67500}{20} = \text{Rs}3375 \)

(iii) Finding interest in third year:

Amount after 2 years = Rs3375 + Rs225 + Rs240 = Rs3840

Interest in year 3 = \( \frac{3840 \times \frac{20}{3} \times 1}{100} = \frac{3840 \times 20}{300} = \text{Rs}256 \)
In simple words: The rising interest each year shows the rate - find it by calculating what percentage turned Rs225 into an extra Rs15. Once you know the rate, work backwards to find the principal and forwards to find year 3's interest.

Exam Tip: The key insight is recognizing that the difference between successive years' interest equals interest on the prior year's interest - this unlocks the rate immediately without needing the principal first.

Question 28. On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to Rs25 if the rate of interest charged for both is 5% p.a.?
Answer: Let the sum be Rs x. Rate = 5% p.a.; Period = 2 years.

Simple Interest:

\( S.I. = \frac{P \times R \times T}{100} = \frac{x \times 5 \times 2}{100} = \frac{x}{10} \)

Compound Interest:

\( A = x(1 + \frac{5}{100})^2 = x(1 + \frac{1}{20})^2 = x \times (\frac{21}{20})^2 = \frac{441x}{400} \)

\( C.I. = A - P = \frac{441x}{400} - x = \frac{41x}{400} \)

Finding the difference:

\( C.I. - S.I. = \frac{41x}{400} - \frac{x}{10} = \frac{41x - 40x}{400} = \frac{x}{400} \)

Given that this difference equals Rs25:

\( \frac{x}{400} = 25 \)

\( x = 25 \times 400 = \text{Rs}10000 \)
In simple words: Compound interest pays interest on the interest too, while simple interest does not. Over 2 years, this extra earning grows proportionally with the principal.

Exam Tip: Always calculate C.I. and S.I. separately, then find their difference - never try to subtract algebraically before simplifying. Check your answer by verifying the difference at your computed principal equals Rs25.

Question 29. The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is Rs15. Find the sum of money lent out.
Answer: Let the sum be Rs x. Rate = 10% p.a. or 5% half-yearly; Period = 1 year or 2 half-years.

Compound Interest (half-yearly):

\( A = x(1 + \frac{5}{100})^2 = x(1 + \frac{1}{20})^2 = x \times (\frac{21}{20})^2 = \frac{441x}{400} \)

\( C.I. = A - P = \frac{441x}{400} - x = \frac{41x}{400} \)

Simple Interest:

\( S.I. = \frac{x \times 10 \times 1}{100} = \frac{x}{10} \)

Finding the difference:

\( C.I. - S.I. = \frac{41x}{400} - \frac{x}{10} = \frac{41x - 40x}{400} = \frac{x}{400} \)

Given that this difference equals Rs15:

\( \frac{x}{400} = 15 \)

\( x = 15 \times 400 = \text{Rs}6000 \)
In simple words: Half-yearly compounding causes the money to grow faster than simple annual interest because interest itself starts earning interest twice in one year.

Exam Tip: When compounding is half-yearly, divide the annual rate by 2 and double the number of periods. The extra earnings from half-yearly compounding grow larger as the principal grows.

Question 30. The amount at compound interest which is calculated yearly on a certain sum of money is Rs1250 in one year and Rs1375 after two years. Calculate the rate of interest.
Answer: Amount after 1 year = Rs1250; Amount after 2 years = Rs1375.

The increase in the second year = Rs1375 - Rs1250 = Rs125.

This Rs125 represents the interest earned on Rs1250 (the amount at the end of year 1) during the second year.

Using the rate formula:

\( R = \frac{S.I. \times 100}{P \times T} = \frac{125 \times 100}{1250 \times 1} = \frac{12500}{1250} = 10\% \)
In simple words: Each year the interest is calculated on the total built up so far. When you know the amounts at the end of years 1 and 2, the increase tells you what interest rate was applied to the year 1 amount.

Exam Tip: Do not assume the principal is Rs1250 - that is the amount after year 1. The interest in year 2 applies to Rs1250, making it easy to find the rate directly without first calculating the principal.

Question 31. The simple interest on a certain sum for 3 years is Rs225 and the compound interest on the same sum at the same rate for 2 years is Rs153. Find the rate of interest and principal.
Answer: Let principal = Rs P and rate = R% p.a.

From the first condition (Simple Interest for 3 years):

\( \frac{P \times R \times 3}{100} = 225 \)

\( P \times R = \frac{225 \times 100}{3} = 7500 \) ... (i)

From the second condition (Compound Interest for 2 years):

\( P[(1 + \frac{R}{100})^2 - 1] = 153 \)

\( P[(\frac{100 + R}{100})^2 - 1] = 153 \)

\( P[\frac{(100 + R)^2 - 100^2}{100^2}] = 153 \)

\( P[\frac{R^2 + 200R}{100^2}] = 153 \)

Substituting \( P = \frac{7500}{R} \) from equation (i):

\( \frac{7500}{R} \times \frac{R(R + 200)}{100^2} = 153 \)

\( \frac{7500(R + 200)}{100^2} = 153 \)

\( R + 200 = \frac{153 \times 10000}{7500} = 204 \)

\( R = 4\% \)

Substituting back into equation (i):

\( P = \frac{7500}{4} = \text{Rs}1875 \)
In simple words: Use the S.I. condition to link the principal and rate, then use the C.I. condition to solve for the actual values by substitution.

Exam Tip: Always form two equations from the two separate conditions (S.I. and C.I.) and use substitution. Verify your answer by checking both conditions are satisfied with your computed P and R.

Question 32. Find the difference between compound interest on Rs8000 for 1½ years at 10% p.a. when compounded annually and semi-annually.
Answer:
Case 1: When compounded annually

For 1½ years, the rate for the first year is 10%, and for the next ½ year the rate is 5% (since ½ year earns half the annual rate).

\( A = 8000(1 + \frac{10}{100})(1 + \frac{5}{100}) \)

\( A = 8000(\frac{110}{100})(\frac{105}{100}) \)

\( A = 8000 \times \frac{11}{10} \times \frac{21}{20} = Rs(40 \times 11 \times 21) = Rs9240 \)

\( C.I. = 9240 - 8000 = Rs1240 \)

Case 2: When compounded semi-annually

Rate = 5% per half-year; n = 3 (three half-year periods in 1½ years).

\( A = 8000(1 + \frac{5}{100})^3 = 8000(\frac{105}{100})^3 \)

\( A = 8000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} = Rs(21 \times 21 \times 21) = Rs9261 \)

\( C.I. = 9261 - 8000 = Rs1261 \)

Difference: Rs1261 - Rs1240 = Rs21
In simple words: Semi-annual compounding produces slightly more interest because interest starts growing more frequently. The difference may seem small but compounds more over longer periods.

Exam Tip: For fractional years with annual compounding, treat the fraction separately - the first full year uses the annual rate, and the fraction uses a proportional rate. Count all periods carefully when switching between annual and semi-annual compounding.

Question 33. A sum of money is lent out at compound interest for two years at 20% p.a., C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, C.I. being reckoned half-yearly, it would have fetched Rs482 more by way of interest. Calculate the sum of money lent out.
Answer: Let the sum lent out = Rs x.

Case 1: When C.I. is reckoned yearly

Rate = 20% p.a.; n = 2 years.

\( A = x(1 + \frac{20}{100})^2 = x(\frac{6}{5})^2 = \frac{36x}{25} \)

\( C.I. = \frac{36x}{25} - x = \frac{11x}{25} \)

Case 2: When C.I. is reckoned half-yearly

Rate = 10% per half-year (20% ÷ 2); n = 4 half-year periods (2 years × 2).

\( A = x(1 + \frac{10}{100})^4 = x(\frac{11}{10})^4 = \frac{14641x}{10000} \)

\( C.I. = \frac{14641x}{10000} - x = \frac{4641x}{10000} \)

Finding the difference:

\( \frac{4641x}{10000} - \frac{11x}{25} = 482 \)

Converting to a common denominator (10000):

\( \frac{4641x}{10000} - \frac{4400x}{10000} = 482 \)

\( \frac{241x}{10000} = 482 \)

\( x = \frac{482 \times 10000}{241} = \text{Rs}20000 \)
In simple words: Half-yearly compounding produces more interest over the same period because the interest is calculated and added back more frequently. The extra earnings grow with the size of the principal.

Exam Tip: When rates are compounded half-yearly, divide the annual rate by 2 and multiply the time period by 2 (convert years to half-years). Always set up the difference equation carefully - subtract the smaller (yearly) C.I. from the larger (half-yearly) C.I.

Question 34. A sum of money amounts to Rs.13230 in one year and to Rs.13891.50 in 1½ years at compound interest, compounded semi-annually. Find the sum and rate of interest per annum.
Answer: Let the principal be Rs. x and the rate be r% per annum. Since compound interest is calculated half-yearly, the half-yearly rate becomes r/2 percent. For one year (2 half-yearly periods), using the compound interest formula \( A = P(1 + \frac{r}{100})^n \):

\( 13230 = x(1 + \frac{r}{200})^2 \) ...(i)

For 1½ years (3 half-yearly periods):

\( 13891.50 = x(1 + \frac{r}{200})^3 \) ...(ii)

Dividing equation (ii) by equation (i):

\( \frac{13891.50}{13230} = \frac{x(1 + \frac{r}{200})^3}{x(1 + \frac{r}{200})^2} \)

\( \frac{138915}{132300} = 1 + \frac{r}{200} \)

\( \frac{r}{200} = \frac{138915}{132300} - 1 = \frac{6615}{132300} \)

\( r = \frac{6615 \times 200}{132300} = 10 \)

Substituting r = 10 in equation (i):

\( 13230 = x(1 + \frac{10}{200})^2 = x(1 + \frac{1}{20})^2 = x(\frac{21}{20})^2 = x \times \frac{441}{400} \)

\( x = \frac{13230 \times 400}{441} = 12000 \)

Therefore, the sum is Rs. 12000 and the rate of interest per annum is 10%.
In simple words: When you divide the second amount by the first amount, you get one plus half the yearly rate. From this, we find the rate is 10%. Then putting this back into the first equation gives us the starting amount of Rs. 12000.

Exam Tip: Always divide the two equations to eliminate the principal; this is the key shortcut to finding the rate quickly in compound interest problems.

Exercise 2.3

Question 1. The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years.
Answer: Using the population growth formula for varying rates:

\( V = V_0(1 + \frac{r_1}{100})(1 + \frac{r_2}{100}) \)

Substituting the given values:

\( V = 200000(1 + \frac{10}{100})(1 + \frac{15}{100}) \)

\( V = 200000 \times \frac{110}{100} \times \frac{115}{100} \)

\( V = 20 \times 110 \times 115 = 253000 \)

The population at the end of two years will be 253000.
In simple words: Multiply the first year's growth factor (110/100) by the second year's growth factor (115/100) and the starting population. This gives you the final population.

Exam Tip: When growth rates differ year to year, always multiply the individual growth multipliers rather than adding the percentages together.

Question 2. The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years?
Answer: Using the compound growth formula:

\( V = V_0(1 + \frac{r}{100})^n \)

Substituting the values:

\( V = 15625(1 + \frac{4}{100})^3 = 15625 \times (\frac{104}{100})^3 = 15625 \times (\frac{26}{25})^3 \)

\( V = 15625 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25} = \frac{15625 \times 17576}{15625} = 17576 \)

The increase in population over 3 years is 17576 - 15625 = 1951.
In simple words: The population grows by a factor of (26/25) each year for 3 years. The new population is 17576, so the increase is 1951 people.

Exam Tip: Always subtract the original population from the final population to get the actual increase; do not confuse the final value with the increase itself.

Question 3. The population of a city increases each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find:
(i) its population 2 years hence
(ii) its population 2 years ago.
Answer:
(i) Using the compound growth formula:

\( V = V_0(1 + \frac{r}{100})^n \)

\( V = 6760000(1 + \frac{4}{100})^2 = 6760000 \times (\frac{104}{100})^2 = 6760000 \times (\frac{26}{25})^2 \)

\( V = 6760000 \times \frac{26}{25} \times \frac{26}{25} = \frac{6760000 \times 676}{625} = 10816 \times 676 = 7311616 \)

The population after 2 years will be 7311616.

(ii) Let the population 2 years ago be P. The present population becomes the final value. Using the same formula:

\( 6760000 = P(1 + \frac{4}{100})^2 = P(\frac{26}{25})^2 = P \times \frac{676}{625} \)

\( P = \frac{6760000 \times 625}{676} = 625 \times 10000 = 6250000 \)

The population 2 years ago was 6250000.
In simple words: To find future population, multiply by the growth factor twice. To find past population, divide the current population by the growth factor twice.

Exam Tip: Remember that finding past value requires reversing the formula - divide instead of multiply. The fraction should be inverted from growth to decay.

Question 4. The cost of a refrigerator is Rs.9000. Its value depreciates at the rate of 5% every year. Find the total depreciation in its value at the end of 2 years.
Answer: Using the depreciation formula:

\( V = V_0(1 - \frac{r}{100})^n \)

Substituting the values:

\( V = 9000(1 - \frac{5}{100})^2 = 9000 \times (\frac{95}{100})^2 = 9000 \times (\frac{19}{20})^2 \)

\( V = 9000 \times \frac{19}{20} \times \frac{19}{20} = \frac{9000 \times 361}{400} = 8122.5 \)

The depreciation is Rs. 9000 - Rs. 8122.5 = Rs. 877.5.
In simple words: The machine's value drops to 95% of its worth each year. After 2 years, it is worth Rs. 8122.5, so the loss is Rs. 877.5.

Exam Tip: For depreciation questions, use (1 - r/100) in the formula. The difference between starting and ending value gives total depreciation.

Question 5. Dinesh purchased a scooter for Rs.24000. The value of scooter is depreciating at the rate of 5% per annum. Calculate its value after 3 years.
Answer: Using the depreciation formula:

\( V = V_0(1 - \frac{r}{100})^n \)

\( V = 24000(1 - \frac{5}{100})^3 = 24000 \times (\frac{95}{100})^3 = 24000 \times (\frac{19}{20})^3 \)

\( V = 24000 \times \frac{19}{20} \times \frac{19}{20} \times \frac{19}{20} = \frac{24000 \times 6859}{8000} = 3 \times 6859 = 20577 \)

The value of the scooter after 3 years will be Rs. 20577.
In simple words: The scooter loses 5% of its value each year. After three years, it decays by the factor (19/20) three times, leaving it worth Rs. 20577.

Exam Tip: For longer depreciation periods, check your fraction reduction - simplifying 95/100 to 19/20 makes calculation easier and reduces errors.

Question 6. A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?
Answer: Let the production 2 years ago be P. The current production is the final value. Using the growth formula:

\( V = V_0(1 + \frac{r}{100})^n \)

\( 2187 = P(1 + \frac{8}{100})^2 = P(\frac{108}{100})^2 = P(\frac{27}{25})^2 = P \times \frac{729}{625} \)

\( P = \frac{2187 \times 625}{729} = 3 \times 625 = 1875 \)

The yearly produce of wheat 2 years ago was 1875 quintals.
In simple words: Since production grows by 8% each year, we reverse the process by dividing. The factor (27/25) applied twice takes 1875 to 2187.

Exam Tip: When finding past values, divide by the growth factor raised to the same power. Always verify by multiplying back: 1875 × (27/25)² should equal 2187.

Question 7. The value of a property decreases every year at the rate of 5%. If its present value is Rs.411540, what was its value three years ago?
Answer: Let the value 3 years ago be V₀. The present value is the final value. Using the depreciation formula:

\( V = V_0(1 - \frac{r}{100})^n \)

\( 411540 = V_0(1 - \frac{5}{100})^3 = V_0(\frac{95}{100})^3 = V_0(\frac{19}{20})^3 \)

\( V_0 = \frac{411540 \times 20 \times 20 \times 20}{19 \times 19 \times 19} = \frac{411540 \times 8000}{6859} = 60 \times 8000 = 480000 \)

The value of the property 3 years ago was Rs. 480000.
In simple words: The property loses 5% each year, so each year it becomes 19/20 of what it was. To go back 3 years, we divide the current value by (19/20)³.

Exam Tip: Set up the depreciation equation with the known present value on the left and the unknown past value multiplied by the decay factor on the right, then solve for the unknown.

Question 8. Ahmed purchased an old scooter for Rs.16000. If the cost of the scooter after 2 years depreciates to Rs.14440, find the rate of depreciation.
Answer: Let the rate of depreciation be r% per annum. Using the depreciation formula:

\( V = V_0(1 - \frac{r}{100})^n \)

\( 14440 = 16000(1 - \frac{r}{100})^2 \)

\( \frac{14440}{16000} = (1 - \frac{r}{100})^2 \)

\( \frac{1444}{1600} = (1 - \frac{r}{100})^2 \)

\( (\frac{38}{40})^2 = (1 - \frac{r}{100})^2 \)

Taking the square root of both sides:

\( 1 - \frac{r}{100} = \frac{38}{40} \)

\( \frac{r}{100} = 1 - \frac{38}{40} = \frac{2}{40} \)

\( r = 5 \)

The rate of depreciation is 5% per annum.
In simple words: Divide the final value by the starting value to get the square of the decay factor. Take the square root to find the yearly factor, then work out the rate.

Exam Tip: Always take square roots carefully - both positive and negative roots exist mathematically, but only the positive one makes sense for depreciation rates (must be less than 100%).

Question 9. A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-2015. Find the annual rate of growth of production of cars.
Answer: Let the annual rate of growth be r% per annum. The time span from 2011-2012 to 2014-2015 is 3 years. Using the growth formula:

\( V = V_0(1 + \frac{r}{100})^n \)

\( 92610 = 80000(1 + \frac{r}{100})^3 \)

\( \frac{92610}{80000} = (1 + \frac{r}{100})^3 \)

\( \frac{9261}{8000} = (1 + \frac{r}{100})^3 \)

\( (\frac{21}{20})^3 = (1 + \frac{r}{100})^3 \)

Taking the cube root of both sides:

\( 1 + \frac{r}{100} = \frac{21}{20} \)

\( \frac{r}{100} = \frac{21}{20} - 1 = \frac{1}{20} \)

\( r = 5 \)

The annual rate of growth of production is 5%.
In simple words: Divide the final production by the starting production and take the cube root to find the yearly growth factor. Subtract 1 to get the growth rate.

Exam Tip: When extracting roots from fractions, factor the numerator and denominator separately to recognize perfect powers. For example, 9261 = 21³ and 8000 = 20³.

Question 10. The value of a machine worth Rs.500000 is depreciating at the rate of 10% every year. In how many years will its value be reduced to Rs.364500?
Answer: Let the number of years be n. Using the depreciation formula:

\( V = V_0(1 - \frac{r}{100})^n \)

\( 364500 = 500000(1 - \frac{10}{100})^n = 500000(\frac{90}{100})^n = 500000(\frac{9}{10})^n \)

\( \frac{364500}{500000} = (\frac{9}{10})^n \)

\( \frac{729}{1000} = (\frac{9}{10})^n \)

\( (\frac{9}{10})^3 = (\frac{9}{10})^n \)

Therefore, n = 3.

The value of the machine will be reduced from Rs. 500000 to Rs. 364500 in 3 years.
In simple words: Divide the final value by the starting value. Recognize that 729/1000 = (9/10)³. This tells you the answer is 3 years.

Exam Tip: When both sides of the equation have the same base raised to different powers, the exponents must be equal. Practice recognizing perfect cubes and higher powers to solve these quickly.

Question 11. Mahindra set up a factory by investing Rs2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, calculate his total profit.
Answer: When different rates of interest apply, the total amount follows the formula \( A = P\left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right) \). Substituting the given values: \( A = \text{Rs}2500000\left(1 + \frac{5}{100}\right)\left(1 + \frac{10}{100}\right) = \text{Rs}2500000\left(\frac{21}{20}\right)\left(\frac{11}{10}\right) = \text{Rs}2500000 \times \frac{231}{200} = \text{Rs}2887500 \). The profit earned is the difference between the final amount and the initial principal: Profit = Rs2887500 - Rs2500000 = Rs387500.
In simple words: When money grows at different rates each year, you multiply together all the growth factors. The profit is what you have left over after subtracting what you started with.

Exam Tip: Use the compound growth formula with multiple rates rather than calculating year-by-year interest separately. Always subtract the principal from the final amount to find profit.

Question 12. The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?
Answer: Let the initial value be \( V_0 \). Using the growth formula, after 3 years: \( V = V_0\left(1 + \frac{25}{100}\right)^3 = V_0\left(\frac{5}{4}\right)^3 = V_0 \times \frac{125}{64} = \frac{125V_0}{64} \). The change in value is: \( C = \frac{125V_0}{64} - V_0 = \frac{125V_0 - 64V_0}{64} = \frac{61V_0}{64} \). To find the percentage increase: \( \text{Percentage increase} = \frac{\text{Change in value}}{\text{Original value}} \times 100 = \frac{61V_0}{64} \div V_0 \times 100 = \frac{61}{64} \times 100 = \frac{6100}{64} = \frac{1525}{16} = 95\frac{5}{16}\% \).
In simple words: To find how much something grows as a percentage, work out what it becomes, subtract the starting amount, then divide that gain by the starting amount and multiply by 100.

Exam Tip: Always express the final percentage as a mixed number if required by the question format. Double-check by verifying that your fraction simplifies correctly before converting to a percentage.

Question 13. Mr. Durani bought a plot of land for Rs180000 and a car for Rs320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?
Answer: For the land, which appreciates at 30% p.a., we apply the growth formula: \( V = V_0\left(1 + \frac{r}{100}\right)^n \). After 3 years: \( V = \text{Rs}180000\left(1 + \frac{30}{100}\right)^3 = \text{Rs}180000\left(\frac{130}{100}\right)^3 = \text{Rs}180000\left(\frac{13}{10}\right)^3 = \text{Rs}180000 \times \frac{2197}{1000} = \text{Rs}395460 \). For the car, which loses 20% in year one and 15% p.a. in years two and three, the value after 3 years is: \( V = V_0\left(1 - \frac{r_1}{100}\right)\left(1 - \frac{r_2}{100}\right)\left(1 - \frac{r_3}{100}\right) = \text{Rs}320000\left(1 - \frac{20}{100}\right)\left(1 - \frac{15}{100}\right)\left(1 - \frac{15}{100}\right) = \text{Rs}320000 \times \frac{80}{100} \times \frac{85}{100} \times \frac{85}{100} = \text{Rs}320000 \times \frac{80 \times 85 \times 85}{1000000} = \text{Rs}184960 \). The total starting value was Rs180000 + Rs320000 = Rs500000. After 3 years, the total value is Rs395460 + Rs184960 = Rs580420. The overall profit is Rs580420 - Rs500000 = Rs80420.
In simple words: Calculate what the land is worth after growing for three years. Calculate what the car is worth after it shrinks in value. Add them together and compare the total to what was spent at the start.

Exam Tip: Watch carefully for different depreciation rates in different years - apply each year's rate to the value at the start of that year, not the original value. Remember to add the final values of both assets before subtracting the total initial investment.

Question 1. The compound interest on Rs1000 at 10% p.a. compounded annually for 2 years is
(a) Rs190
(b) Rs200
(c) Rs210
(d) Rs1210
Answer: (c) Rs210
In simple words: When money earns interest that then earns its own interest (compound interest), you use the formula C.I. = P[(1 + r/100)^n - 1]. Here, Rs1000 earning 10% for 2 years gives Rs1000 × [(1.1)^2 - 1] = Rs1000 × [1.21 - 1] = Rs1000 × 0.21 = Rs210.

Exam Tip: Compound interest differs from simple interest because in the second year, interest is earned on both the original amount AND the first year's interest. Always use the formula rather than calculating year-by-year if the question asks for the total interest earned.

Question 2. If Sukriti borrows Rs8000 for two years at the rate of 10% per annum compound interest, then the amount to be paid by her at the end of two years to clear the debt is
(a) Rs8800
(b) Rs9600
(c) Rs9680
(d) Rs102400
Answer: (c) Rs9680
In simple words: Use the amount formula A = P(1 + r/100)^n where P = Rs8000, r = 10%, and n = 2. You get A = Rs8000 × (1.1)^2 = Rs8000 × 1.21 = Rs9680, which is the total amount (principal plus interest) owed.

Exam Tip: Be careful to distinguish between compound interest earned (the gain only) and the total amount due (principal plus interest). The question asks for the amount to be paid, which is the total, not just the interest portion.

Question 3. If a man invests Rs12000 for two years at the rate of 10% per annum compound interest, then the compound interest earned by him at the end of two years is
(a) Rs2400
(b) Rs2520
(c) Rs2000
(d) Rs1800
Answer: (b) Rs2520
In simple words: The compound interest is found using C.I. = P[(1 + r/100)^n - 1] = Rs12000 × [1.21 - 1] = Rs12000 × 0.21 = Rs2520. This is just the interest earned, not including the original investment.

Exam Tip: When the question asks for "compound interest earned," it wants only the interest portion, not the full amount. This is different from questions asking for the "amount to be paid," which would include the principal.

Question 4. Mr. Rao bought 1-year, Rs10000 certificate of deposit that paid interest at an annual rate of 8% compounded semi-annually. The interest received by him on maturity is
(a) Rs816
(b) Rs864
(c) Rs800
(d) Rs10816
Answer: (a) Rs816
In simple words: When interest compounds semi-annually (twice a year), the annual rate is divided by 2, giving 4% per half-year. In 1 year there are 2 conversion periods. Using C.I. = P[(1 + r/100)^n - 1] with r = 4 and n = 2: C.I. = Rs10000 × [(1.04)^2 - 1] = Rs10000 × [1.0816 - 1] = Rs10000 × 0.0816 = Rs816.

Exam Tip: For semi-annual compounding, always divide the annual rate by 2 and double the number of periods. Keep careful track of which periods you are counting - a 1-year certificate with semi-annual compounding means 2 conversion periods, not 1.

Question 5. The compound interest on Rs5000 at 20% per annum for 1\(\frac{1}{2}\) years compounded half-yearly is
(a) Rs6655
(b) Rs1655
(c) Rs1500
(d) Rs1565
Answer: (b) Rs1655
In simple words: For semi-annual compounding, divide the 20% rate by 2 to get 10% per half-year. In 1.5 years there are 3 half-year periods. Using C.I. = P[(1 + r/100)^n - 1]: C.I. = Rs5000 × [(1.10)^3 - 1] = Rs5000 × [1.331 - 1] = Rs5000 × 0.331 = Rs1655.

Exam Tip: When time is given as a mixed number (like 1½ years) with semi-annual compounding, convert it to the number of half-years: 1½ years = 3 half-years. This is crucial for getting the correct value of n in the formula.

Question 6. If the number of conversion periods ≥ 2, then the compound interest is
(a) less than simple interest
(b) equal to simple interest
(c) greater than or equal to simple interest
(d) greater than simple interest
Answer: (d) greater than simple interest
In simple words: Compound interest always exceeds simple interest whenever there are 2 or more conversion periods. In compound interest, each period's interest earns interest itself in later periods. In simple interest, interest is always calculated on the starting amount only, so it stays smaller.

Exam Tip: This is a fundamental concept to remember - compound interest grows faster than simple interest. The more conversion periods, the bigger the advantage of compound interest. For exactly 1 period, C.I. and S.I. are equal, but for 2 or more periods, C.I. is strictly greater.

Question 7. The present population of a city is 12,00,000. If it increases at the rate of 8% every year then the population of the city after 2 years is
(a) 199680
(b) 1399680
(c) 1500000
(d) 1299680
Answer: (b) 1399680
In simple words: Population growth follows the same formula as compound interest: V = V₀(1 + r/100)^n. Here, V = 1200000 × (1.08)^2 = 1200000 × 1.1664 = 1399680. After 2 years at 8% annual growth, the population reaches this number.

Exam Tip: Population growth, asset depreciation, and interest calculations all use the same exponential growth/decay formulas. Once you master the formula for one context, you can apply it to others.

Question 8. Consider the following two statements.
Statement 1: A sum of Rs 1,000 invested at 10% p.a. rate of interest will earn Rs 100 interest in first year.
Statement 2: Simple and compound interest is the same for first conversion period at the same rate of interest.
Which of the following is valid?
(a) Both the statements are true
(b) Both the statements are false
(c) Statement 1 is true, and Statement 2 is false
(d) Statement 1 is false, and Statement 2 is true
Answer: (a) Both the statements are true
In simple words: For Statement 1, using C.I. = P[(1 + r/100)^n - 1] with P = Rs1000, r = 10%, n = 1: C.I. = Rs1000 × [1.1 - 1] = Rs100. This is true. For Statement 2, S.I. = (P × R × T)/100 = (1000 × 10 × 1)/100 = Rs100. When there is only one period, compound and simple interest are identical because compound interest has no chance to earn interest on interest yet. Both statements are correct.

Exam Tip: Remember that C.I. and S.I. are equal only for the first conversion period - after that, compound interest pulls ahead. This equality in the first year is a key checkpoint for verifying your formulas are working correctly.

Question. Assertion (A): The population of a town in 2015 was 10,000. It grew by 10% every year. So, in the year 2020, the population was 15,000.
Reason (R): Formula used in such problems is V = V₀(1 + r/100)ⁿ
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (2) Assertion (A) is false, Reason (R) is true.
In simple words: The formula works correctly for calculating population growth. However, when you use it with the numbers given, the actual population in 2020 turns out to be about 16,105, not 15,000. So the formula is right, but the claim about the population amount is wrong.

Exam Tip: When checking Assertion-Reason questions, evaluate each part on its own merits first - is the formula correct? Is the calculation correct? Then check if one explains the other.

Question. Assertion (A): Two friends invest the same amount of money for the same time (> 2 years) at the same rate of interest. One earns simple interest, but the other earns compound interest. Then both will get the same amount of money back at the end of investment.
Reason (R): The principal for each conversion period increases for the compound interest calculation.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (2) Assertion (A) is false, Reason (R) is true.
In simple words: Simple interest stays the same each year. Compound interest adds the earlier interest to the amount, so you earn interest on your interest. This means one person gets more money than the other. The reason given is correct - the principal does grow with compound interest - but the assertion is wrong.

Exam Tip: Remember that compound interest always beats simple interest when the time period exceeds one year. The difference increases as time passes.

Chapter Test

Question 1. Rs10,000 was lent for one year at 10% per annum. By how much more will the interest be, if the sum was lent at 10% per annum, interest being compounded half-yearly?
Answer: When interest is compounded yearly, P = Rs10,000, r = 10%, T = 1.
\[ C.I. = \frac{10000 \times 10 \times 1}{100} = Rs1000. \]

When interest is compounded half-yearly, P = Rs10,000, r = 10/2% = 5%, T = 2.

For the first half-year:
\[ C.I. = \frac{10000 \times 5 \times 1}{100} = Rs500. \]

Amount after first half-year = Rs1000 + Rs500 = Rs1500.

Principal for second half-year = Rs1500.
\[ C.I. = \frac{10500 \times 5 \times 1}{100} = Rs525. \]

Amount after second half-year = Rs1500 + Rs525 = Rs2025.
\[ C.I. = \text{Final amount} - \text{Principal} = Rs2025 - Rs1000 = Rs1025. \]

Difference in C.I. in both cases = Rs1025 - Rs1000 = Rs25.

The interest will be Rs25 more if the sum was lent at 10% per annum, with interest being compounded half-yearly.
In simple words: When you break the year into two half-year periods, you earn interest on the interest from the first half. This extra earning adds up to Rs25 more than if you earned interest just once a year.

Exam Tip: When compounding happens more frequently (half-yearly instead of yearly), the total interest earned increases. Always calculate step-by-step for each compounding period to avoid mistakes.

Question 2. A man invests Rs3,072 for two years at compound interest. After one year the money amounts to Rs3,264. Find the rate of interest and the amount due at the end of 2nd year.
Answer: Let the rate of interest be r% per annum.

Given, Rs3,072 amounts to Rs3,264 in one year.

\[ \therefore \text{Compound interest} = \text{Final amount} - \text{Principal} = Rs3264 - Rs3072 = Rs192. \]

\[ \therefore 192 = \frac{3072 \times r \times 1}{100} \]

\[ \Rightarrow r = \frac{19200}{3072} \]

\[ \Rightarrow r = 6.25 \]

Principal for second year = Rs3,264.

\[ C.I. = \frac{3264 \times 6.25 \times 1}{100} = \frac{20400}{100} = Rs204. \]

Amount after second year = Rs3,264 + Rs204 = Rs3,468.

The rate of interest is 6.25% and the amount after the second year is Rs3,468.
In simple words: The interest earned in the first year tells you the rate. Once you know the rate, use it to find the interest for the second year. Each year's principal grows because you add the previous year's interest to it.

Exam Tip: When the amount after a specific year is given, use that to find the rate first - it's usually much simpler than trying to solve both unknowns at once.

Question 3. What sum will amount to Rs28,090 in two years at 6% per annum compound interest? Also find the compound interest.
Answer: Let principal be Rs P.

We know, \( A = P\left(1 + \frac{r}{100}\right)^n \)

Putting values in formula we get:

\[ \Rightarrow 28090 = P\left(1 + \frac{6}{100}\right)^2 \]

\[ \Rightarrow 28090 = P\left(\frac{106}{100}\right)^2 \]

\[ \Rightarrow 28090 = P\left(\frac{53}{50}\right)^2 \]

\[ \Rightarrow 28090 = P \times \frac{53}{50} \times \frac{53}{50} \]

\[ \Rightarrow P = \frac{28090 \times 50 \times 50}{53 \times 53} \]

\[ \Rightarrow P = \frac{28090 \times 2500}{2809} \]

\[ \Rightarrow P = Rs25000. \]

\[ C.I. = \text{Final amount} - \text{Principal} = Rs28090 - Rs25000 = Rs3090. \]

The principal is Rs25,000 and compound interest is Rs3,090.
In simple words: Work backwards from the final amount using the compound interest formula. Rearrange to find what the starting amount must have been. Then subtract the original amount from the final amount to get the interest earned.

Exam Tip: Always simplify fractions early - reducing 106/100 to 53/50 makes the arithmetic much cleaner and reduces chance of calculation errors.

Question 4. Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was Rs422, find:
(i) the equal sums
(ii) compound interest for each sum.
Answer: (i) Let the sum be Rs P.

C.I. when sum is lent at 5% for 2 years:

\[ C.I. = P\left[\left(1 + \frac{5}{100}\right)^2 - 1\right] \]

\[ = P\left[\left(1 + \frac{1}{20}\right)^2 - 1\right] \]

\[ = P\left[\left(\frac{21}{20}\right)^2 - 1\right] \]

\[ = P\left[\frac{441}{400} - 1\right] \]

\[ = P\left[\frac{441 - 400}{400}\right] \]

\[ = P \times \frac{41}{400} \]

\[ = \frac{41P}{400} \]

C.I. when sum is lent at 6% for 2 years:

\[ C.I. = P\left[\left(1 + \frac{6}{100}\right)^2 - 1\right] \]

\[ = P\left[\left(1 + \frac{3}{50}\right)^2 - 1\right] \]

\[ = P\left[\left(\frac{53}{50}\right)^2 - 1\right] \]

\[ = P\left[\frac{2809}{2500} - 1\right] \]

\[ = P \times \frac{2809 - 2500}{2500} \]

\[ = \frac{309P}{2500} \]

Given, difference in C.I. = Rs422.

\[ \therefore \frac{309P}{2500} - \frac{41P}{400} = 422 \]

\[ \Rightarrow \frac{309P \times 4 - 41P \times 25}{10000} = 422 \]

\[ \Rightarrow \frac{1236P - 1025P}{10000} = 422 \]

\[ \Rightarrow \frac{211P}{10000} = 422 \]

\[ \Rightarrow P = \frac{422 \times 10000}{211} \]

\[ \Rightarrow P = Rs20000. \]

The equal sum is Rs20,000.

(ii) C.I. when sum is lent at 5% for 2 years = \( \frac{41 \times 20000}{400} \)

\[ C.I. = 41 \times 50 \]

\[ = Rs2050. \]

C.I. when sum is lent at 6% for 2 years = \( \frac{309 \times 20000}{2500} \)

\[ C.I. = 309 \times 8 \]

\[ = Rs2472. \]

C.I. is Rs2,050 when sum is lent at 5% for 2 years and C.I. is Rs2,472 when sum is lent at 6% for 2 years.
In simple words: Set up the compound interest formula for both rates. Write them as a difference equal to 422. This gives you one equation with one unknown (P). Solve for P, then plug it back to find each interest amount.

Exam Tip: When dealing with "difference in compound interest," make sure you subtract the smaller from the larger to get a positive result. Always simplify the compound interest expressions before writing the difference equation.

Question 5. The compound interest on a sum of money for 2 years is Rs1,331.20 and the simple interest on the same sum for the same period at the same rate is Rs1,280. Find the sum and the rate of interest per annum.
Answer: Let the sum be Rs x and rate be r%.

Given, S.I. for 2 years = Rs1,280.

\[ \therefore \frac{P \times R \times T}{100} = 1280 \]

\[ \Rightarrow \frac{x \times r \times 2}{100} = 1280 \]

\[ \Rightarrow x \times r = \frac{1280 \times 100}{2} \]

\[ \Rightarrow x \times r = 64000 \ldots (i) \]

Given, C.I. for 2 years = Rs1,331.20

\[ C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] \]

\[ \Rightarrow 1331.20 = x\left[\left(1 + \frac{r}{100}\right)^2 - 1\right] \]

\[ \Rightarrow 1331.20 = x\left[\left(\frac{100 + r}{100}\right)^2 - 1\right] \]

\[ \Rightarrow 1331.20 = x\left[\frac{(100 + r)^2}{100^2} - 1\right] \]

\[ \Rightarrow 1331.20 = x\left[\frac{(100 + r)^2 - 100^2}{100^2}\right] \]

\[ \Rightarrow 1331.20 = x\left[\frac{100^2 + r^2 + 200r - 100^2}{100^2}\right] \]

\[ \Rightarrow 1331.20 = x\left[\frac{r^2 + 200r}{100^2}\right] \]

\[ \Rightarrow 1331.20 \times 100^2 = x \times r \times (r + 200) \]

\[ \Rightarrow 13312000 = 64000(r + 200) \ldots \text{(Using (i))} \]

\[ \Rightarrow r + 200 = \frac{13312000}{64000} \]

\[ \Rightarrow r + 200 = 208 \]

\[ \Rightarrow r = 8 \]

Putting value of r in Eq. (i) we get:

\[ \Rightarrow x \times 8 = 64000 \]

\[ \Rightarrow x = \frac{64000}{8} \]

\[ \Rightarrow x = Rs8000. \]

The sum is Rs8,000 and rate is 8%.
In simple words: The simple interest formula gives you one equation connecting the sum and rate. The compound interest formula gives you another. Use these two equations to find both unknowns. The algebra is lengthy but straightforward.

Exam Tip: Write out the relationship x × r = 64,000 from the simple interest part first. This substitution in the compound interest equation drastically reduces the complexity of the algebra.

Question 6. On what sum will the difference between the simple and compound interest for 3 years at 10% p.a. is Rs232.50?
Answer: Let the sum be Rs P.

\[ S.I. = \frac{P \times 10 \times 3}{100} = \frac{3P}{10} \]

By formula,

\[ C.I. = P\left[\left(1 + \frac{r}{100}\right)^n - 1\right] \]

Putting values in formula we get:

\[ C.I. = P\left[\left(1 + \frac{10}{100}\right)^3 - 1\right] \]

\[ = P\left[\left(\frac{110}{100}\right)^3 - 1\right] \]

\[ = P\left[\left(\frac{11}{10}\right)^3 - 1\right] \]

\[ = P\left[\frac{1331}{1000} - 1\right] \]

\[ = P\left[\frac{1331 - 1000}{1000}\right] \]

\[ = P \times \frac{331}{1000} \]

\[ = \frac{331P}{1000} \]

Given, difference between C.I. and S.I. = Rs232.50

\[ \therefore \frac{331P}{1000} - \frac{3P}{10} = 232.50 \]

\[ \Rightarrow \frac{331P - 300P}{1000} = 232.50 \]

\[ \Rightarrow \frac{31P}{1000} = 232.50 \]

\[ \Rightarrow P = \frac{232.50 \times 1000}{31} \]

\[ \Rightarrow P = \frac{232500}{31} \]

\[ \Rightarrow P = Rs7500. \]

The sum is Rs7,500.
In simple words: Calculate simple interest and compound interest using the same principal, rate, and time. The difference between them gives you an equation. Solve for P.

Exam Tip: The difference between C.I. and S.I. only appears when the time period is more than one year. It grows rapidly as time increases, making it useful for solving problems where the difference is given but the principal is unknown.

Question 7. The simple interest on a certain sum for 3 years is Rs1,080 and the compound interest on the same sum at the same rate for 2 years is Rs741.60. Find:
Answer: [This question appears incomplete in the source document. The sub-parts to be found and the complete working are not provided. Based on typical structure of such problems, the expected sub-parts would be: (i) the sum (ii) the rate of interest. However, without the complete answer section in the source, a full generated answer cannot be reliably provided here. The question heading has been preserved verbatim as per rules.]

Exam Tip: When given both S.I. and C.I. for different time periods, use the S.I. formula and C.I. formula separately to establish two equations in two unknowns (P and r), then solve simultaneously.

Question 7. The simple interest on a sum for 3 years is Rs.1080 and the compound interest for 2 years is Rs.741.60. Find (i) the rate of interest (ii) the principal.
Answer: Let the amount be Rs.x and the rate be r%.

From the simple interest formula, we have:
\( \frac{x \times r \times 3}{100} = 1080 \)
\( x \times r = 36000 \)...(i)

Using the compound interest formula:
\( 741.60 = x\left[\left(1 + \frac{r}{100}\right)^2 - 1\right] \)
\( 741.60 = x\left[\frac{(100+r)^2 - 100^2}{100^2}\right] \)
\( 741.60 = x\left[\frac{r^2 + 200r}{100^2}\right] \)
\( 741.60 \times 100^2 = x \times r \times (r + 200) \)
\( 7416000 = 36000(r + 200) \) (using (i))
\( r + 200 = 206 \)
\( r = 6 \)

(i) The rate of interest is 6%.

(ii) Substituting r = 6 in equation (i):
\( x \times 6 = 36000 \)
\( x = Rs.6000 \)

The principal is Rs.6000.
In simple words: When you know both simple and compound interest amounts, you can set up two equations. Solve them together to find the rate first, then use that rate to work out the principal.

Exam Tip: Always use the SI formula to get one equation and the CI formula to get another - solving them simultaneously helps find both unknowns efficiently.

Question 8. In what time will Rs.2400 amount to Rs.2646 at 10% p.a. compounded semi-annually?
Answer: Since interest is compounded semi-annually, the rate becomes \( \frac{10\%}{2} = 5\% \).

Let the time be n half-years.

Using the compound amount formula:
\( 2646 = 2400\left(1 + \frac{5}{100}\right)^n \)
\( \frac{2646}{2400} = \left(\frac{105}{100}\right)^n \)
\( \frac{441}{400} = \left(\frac{21}{20}\right)^n \)
\( \left(\frac{21}{20}\right)^2 = \left(\frac{21}{20}\right)^n \)
\( n = 2 \)

Time = 2 half-years = 1 year.

Therefore, Rs.2400 becomes Rs.2646 in 1 year at 10% p.a. compounded semi-annually.
In simple words: When interest compounds twice a year, divide the yearly rate in half. Then use the compound formula to find how many half-year periods you need.

Exam Tip: Remember to convert the time period to match the compounding frequency - if it compounds semi-annually, express your answer in half-years and then convert back to years.

Question 9. Sudarshan invested Rs.60000 in a finance company and received Rs.79860 after 1\(\frac{1}{2}\) years. Find the rate of interest per annum compounded half-yearly.
Answer: Let the rate of interest be r% p.a., which means \( \frac{r}{2}\% \) when compounded half-yearly.

Since 1\(\frac{1}{2}\) years = 3 half-years, using the compound amount formula:
\( 79860 = 60000\left(1 + \frac{r}{200}\right)^3 \)
\( \frac{79860}{60000} = \left(1 + \frac{r}{200}\right)^3 \)
\( \frac{1331}{1000} = \left(1 + \frac{r}{200}\right)^3 \)
\( \left(\frac{11}{10}\right)^3 = \left(1 + \frac{r}{200}\right)^3 \)
\( \frac{11}{10} = 1 + \frac{r}{200} \)
\( \frac{11}{10} - 1 = \frac{r}{200} \)
\( \frac{1}{10} = \frac{r}{200} \)
\( r = 20 \)

Therefore, the rate of interest is 20% per annum.
In simple words: When compounding happens twice yearly, use half the yearly rate in your formula. Work backwards from the final amount to find what rate produces that result.

Exam Tip: Take cube roots carefully when you see a power of 3 on both sides - recognizing perfect cubes like 1331 = 11³ and 1000 = 10³ saves calculation time.

Question 10. The population of a city is 320000. If the annual birth rate is 9.2% and the annual death rate is 1.7%, calculate the population of the town after 3 years.
Answer: Net growth rate = 9.2% - 1.7% = 7.5%.

Using the growth formula:
\( V = V_0\left(1 + \frac{r}{100}\right)^n \)

Substituting the values:
\( V = 320000\left(1 + \frac{7.5}{100}\right)^3 \)
\( = 320000 \times \left(\frac{107.5}{100}\right)^3 \)
\( = 320000 \times \left(\frac{1075}{1000}\right)^3 \)
\( = 320000 \times \left(\frac{43}{40}\right)^3 \)
\( = 320000 \times \frac{43}{40} \times \frac{43}{40} \times \frac{43}{40} \)
\( = 320000 \times \frac{79507}{64000} \)
\( = 5 \times 79507 \)
\( = 397535 \)

The population of the town after 3 years is 397535.
In simple words: Find the net growth rate by subtracting deaths from births. Then apply the growth formula year after year to see how the population increases.

Exam Tip: Always reduce fractions to lowest terms before cubing them - this makes the arithmetic much simpler and less error-prone.

Question 11. The cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If its present worth is Rs.315600, find: (i) its purchase price (ii) its value after 3 years.
Answer:
(i) Using the depreciation formula:
\( V = V_0\left(1 - \frac{r}{100}\right)^n \)

Let the initial value be \( V_0 \). Substituting the given values:
\( 315600 = V_0\left(1 - \frac{20}{100}\right)^2 \)
\( 315600 = V_0\left(\frac{80}{100}\right)^2 \)
\( 315600 = V_0\left(\frac{4}{5}\right)^2 \)
\( 315600 = V_0 \times \frac{4}{5} \times \frac{4}{5} \)
\( 315600 = V_0 \times \frac{16}{25} \)
\( V_0 = 315600 \times \frac{25}{16} \)
\( V_0 = Rs.493125 \)

The purchase price of the car was Rs.493125.

(ii) Using the depreciation formula for 3 years from the present value:
\( V = 315600\left(1 - \frac{20}{100}\right)^3 \)
\( = 315600 \times \left(\frac{80}{100}\right)^3 \)
\( = 315600 \times \left(\frac{4}{5}\right)^3 \)
\( = 315600 \times \frac{64}{125} \)
\( = \frac{20198400}{125} \)
\( = Rs.161587.20 \)

The value of the car after 3 years will be Rs.161587.20.
In simple words: Depreciation means value reduces by a fixed percentage each year. Work backwards using the formula to find the original price, and work forwards to find the future value.

Exam Tip: Be clear about the starting point - when finding future value from present worth, use the present value as your principal, not the original purchase price.

Question 12. Amar Singh started a business with an initial investment of Rs.400000. In the first year, he incurred a loss of 4%. However, during the second year, he earned a profit of 5% which in third year rose to 10%. Calculate his net profit for the entire period of 3 years.
Answer: Given:
Initial investment = Rs.400000
Loss in first year = 4%
Profit in second year = 5%
Profit in third year = 10%

The amount after 3 years is calculated as:
\( \text{Amount after 3 years} = Rs.400000 \times \left(1 - \frac{4}{100}\right) \times \left(1 + \frac{5}{100}\right) \times \left(1 + \frac{10}{100}\right) \)
\( = Rs.400000 \times \frac{96}{100} \times \frac{105}{100} \times \frac{110}{100} \)
\( = Rs.400000 \times \frac{24}{25} \times \frac{21}{20} \times \frac{11}{10} \)
\( = Rs.400000 \times \frac{5544}{5000} \)
\( = Rs.(80 \times 5544) \)
\( = Rs.443520 \)

Net profit = Final amount - Initial investment
\( = Rs.443520 - Rs.400000 \)
\( = Rs.43520 \)

Therefore, his net profit after 3 years is Rs.43520.
In simple words: Apply loss in year 1, then apply profits in years 2 and 3 one after another to the growing amount. The difference between the final amount and the starting investment is the net profit.

Exam Tip: Multiply the factors in sequence - losses use (1 - r/100) and profits use (1 + r/100). Always subtract the original investment from the final amount to get net profit, not the final percentage change.