Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 3 Compound Interest Using Formula

Exercise 3(A)

Question 1. Find the amount and the compound interest on Rs. 12,000 in 3 years at 5% compounded annually.
Answer: We are given that the principal amount \( P \) is Rs. 12,000, the time period \( n \) is 3 years, and the interest rate \( r \) is 5% p.a.
Applying the formula for the total amount:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting our values, we get:
\( A = 12000 \left(1 + \frac{5}{100}\right)^3 \)
\( \implies A = 12000 \left(\frac{21}{20}\right)^3 \)
\( \implies A = \text{Rs. } 13,891.50 \)
To find the compound interest, we subtract the principal from the total amount:
Compound Interest (C.I.) \( = A - P \)
\( \implies \text{C.I.} = \text{Rs. } 13,891.50 - \text{Rs. } 12,000 = \text{Rs. } 1,891.50 \)
In simple words: If you invest Rs. 12,000 for three years with a 5% interest rate compounded each year, you will end up with Rs. 13,891.50. This means you earned Rs. 1,891.50 in interest.

Exam Tip: Make sure to write down the formula clearly before substituting the values, as this helps secure partial marks even if you make a calculation error.

Question 2. Calculate the amount, if Rs. 15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% p.a. and 10% p.a. respectively.
Answer: Here, the initial principal \( P \) is Rs. 15,000, the time frame is 2 years, with different rates of interest for each year: \( r_1 = 8\% \) for the first year and \( r_2 = 10\% \) for the second year.
The formula for calculating the final amount with varying interest rates is:
\( A = P \left(1 + \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \)
Substituting these values into our equation, we obtain:
\( A = 15000 \left(1 + \frac{8}{100}\right) \left(1 + \frac{10}{100}\right) \)
\( \implies A = 15000 \left(\frac{27}{25}\right) \left(\frac{11}{10}\right) \)
\( \implies A = \text{Rs. } 17,820 \)
In simple words: When interest rates change each year, we multiply the principal by the growth factor of each year. Here, a principal of Rs. 15,000 grows to Rs. 17,820 after two years.

Exam Tip: When interest rates are different for consecutive years, do not use the exponent formula; instead, multiply the individual year factors sequentially.

Question 3. Calculate the compound interest accrued on Rs. 6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.
Answer: We are given a principal \( P \) of Rs. 6,000, a time period of 3 years, and different yearly rates of interest: \( r_1 = 5\% \), \( r_2 = 8\% \), and \( r_3 = 10\% \).
The total amount can be found using the formula:
\( A = P \left(1 + \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \left(1 + \frac{r_3}{100}\right) \)
By entering the given numbers, we get:
\( A = 6000 \left(1 + \frac{5}{100}\right) \left(1 + \frac{8}{100}\right) \left(1 + \frac{10}{100}\right) \)
\( \implies A = 6000 \left(\frac{21}{20}\right) \left(\frac{27}{25}\right) \left(\frac{11}{10}\right) \)
\( \implies A = \text{Rs. } 7,484.40 \)
Now we find the compound interest by subtracting the principal from this total amount:
\( \text{C.I.} = A - P \)
\( \implies \text{C.I.} = \text{Rs. } 7,484.40 - \text{Rs. } 6,000 = \text{Rs. } 1,484.40 \)
In simple words: For three years with changing rates of 5%, 8%, and 10%, the final amount becomes Rs. 7,484.40. Subtracting the starting Rs. 6,000 gives us Rs. 1,484.40 as the interest earned.

Exam Tip: Ensure that you reduce fractions to their simplest forms during step-by-step multiplication to avoid long and complicated division at the end.

Question 4. What sum of money will amount to Rs. 5,445 in 2 years at 10% per annum compound interest?
Answer: In this problem, the final amount \( A \) is Rs. 5,445, the duration \( n \) is 2 years, and the rate of interest \( r \) is 10% per annum.
Using the standard formula for compound interest:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting the values we have:
\( 5445 = P \left(1 + \frac{10}{100}\right)^2 \)
\( \implies 5445 = P \left(\frac{11}{10}\right)^2 \)
\( \implies 5445 = P \left(\frac{121}{100}\right) \)
To isolate the principal \( P \), we solve:
\( P = 5445 \times \left(\frac{10}{11}\right)^2 \)
\( \implies P = 5445 \times \frac{100}{121} \)
\( \implies P = 45 \times 100 = \text{Rs. } 4,500 \)
In simple words: To find the original sum that grows to Rs. 5,445, we work backward using the rate and time. The starting sum of money was Rs. 4,500.

Exam Tip: When solving for the principal, writing the fraction as its reciprocal on the other side of the equation is a handy way to prevent multiplication errors.

Question 5. On what sum of money will the compound interest for 2 years at 5% per annum amount to Rs. 768.75?
Answer: We are given that the compound interest is Rs. 768.75, the duration is 2 years, and the yearly interest rate is 5%.
Let the principal amount be \( P \). The final accumulated amount \( A \) is calculated as:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
\( \implies A = P \left(1 + \frac{5}{100}\right)^2 \)
\( \implies A = P \left(\frac{21}{20}\right)^2 = \frac{441}{400}P \)
Since compound interest is the difference between the total amount and the principal:
\( \text{C.I.} = A - P \)
\( \implies \frac{441}{400}P - P = 768.75 \)
\( \implies \frac{41}{400}P = 768.75 \)
Rearranging the terms to find \( P \):
\( P = \frac{768.75 \times 400}{41} \)
\( \implies P = 18.75 \times 400 = \text{Rs. } 7,500 \)
In simple words: We are looking for a starting sum that earns Rs. 768.75 of interest in two years at a rate of 5%. By setting up the equation, we find that the initial sum is Rs. 7,500.

Exam Tip: When dealing with decimals like 768.75, multiplying by 100 or 400 first can turn them into whole numbers, making the division by 41 much cleaner.

Question 6. Find the sum on which the compound interest for 3 years at 10% per annum amounts to Rs. 1,655.
Answer: We are given that the compound interest earned over 3 years at a rate of 10% per annum is Rs. 1,655.
Let the principal be \( P \). The final amount \( A \) is represented by:
\( A = P \left(1 + \frac{10}{100}\right)^3 \)
\( \implies A = P \left(\frac{11}{10}\right)^3 = \frac{1331}{1000}P \)
Since Compound Interest is the total amount minus the principal:
\( \text{C.I.} = A - P \)
\( \implies \frac{1331}{1000}P - P = 1655 \)
\( \implies \frac{331}{1000}P = 1655 \)
Solving for the principal \( P \):
\( P = \frac{1655 \times 1000}{331} \)
\( \implies P = 5 \times 1000 = \text{Rs. } 5,000 \)
In simple words: A certain sum grows over three years at 10% interest. The interest alone comes out to Rs. 1,655. By working through the formula, we find that the starting sum was Rs. 5,000.

Exam Tip: Memorizing basic cubes like \( 11^3 = 1331 \) can speed up calculations significantly in compound interest questions.

Question 7. What principal will amount to Rs. 9,856 in two years, if the rates of interest for successive years are 10% and 12% respectively?
Answer: In this case, the total accumulated amount \( A \) after 2 years is Rs. 9,856. The interest rates for the two consecutive years are \( r_1 = 10\% \) and \( r_2 = 12\% \) respectively.
The final amount formula for varying interest rates is:
\( A = P \left(1 + \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \)
By putting the known values into the equation:
\( 9856 = P \left(1 + \frac{10}{100}\right) \left(1 + \frac{12}{100}\right) \)
\( \implies 9856 = P \left(\frac{11}{10}\right) \left(\frac{28}{25}\right) \)
Now we rearrange the terms to isolate the principal \( P \):
\( P = \frac{9856 \times 10 \times 25}{11 \times 28} \)
\( \implies P = \frac{246400}{308} = \text{Rs. } 8,000 \)
In simple words: If a sum of money grows at a rate of 10% in the first year and 12% in the second year to become Rs. 9,856, the starting amount of money was Rs. 8,000.

Exam Tip: When simplifying terms like \( 11 \times 28 \), try to cancel common factors with the numerator before performing any large multiplications.

Question 8. On a certain sum, the compound interest in 2 years amounts to Rs. 4,240. If the rate of interest for the successive years is 10% and 15% respectively, find the sum.
Answer: Let the principal amount be \( P \). The compound interest accrued after 2 years is Rs. 4,240, meaning the final amount \( A \) is \( P + 4240 \). The interest rates for the consecutive years are \( r_1 = 10\% \) and \( r_2 = 15\% \).
The equation for the total amount is:
\( A = P \left(1 + \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \)
Substituting our values:
\( P + 4240 = P \left(1 + \frac{10}{100}\right) \left(1 + \frac{15}{100}\right) \)
\( \implies P + 4240 = P (1.10) (1.15) \)
\( \implies P + 4240 = 1.265P \)
Subtracting \( P \) from both sides to find \( P \):
\( 1.265P - P = 4240 \)
\( \implies 0.265P = 4240 \)
\( \implies P = \frac{4240}{0.265} = \text{Rs. } 16,000 \)
In simple words: The total amount is the starting sum plus the compound interest of Rs. 4,240. Since the rates are 10% and 15%, the principal grows by a factor of 1.265. Solving this gives the original sum as Rs. 16,000.

Exam Tip: Converting decimals to fractions can sometimes make division easier; for instance, \( 0.265 \) is equivalent to \( \frac{265}{1000} \).

Question 9. At what per cent per annum will Rs. 6,000 amount to Rs. 6,615 in 2 years when interest is compounded annually?
Answer: We are given a principal \( P \) of Rs. 6,000, which grows to a final amount \( A \) of Rs. 6,615 in a time span \( n = 2 \) years. We need to find the rate of interest \( r \).
Using the compound interest amount formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting the given numbers:
\( 6615 = 6000 \left(1 + \frac{r}{100}\right)^2 \)
Now, we isolate the squared term:
\( \left(1 + \frac{r}{100}\right)^2 = \frac{6615}{6000} \)
Simplifying the fraction on the right-hand side by dividing both numerator and denominator by 15:
\( \left(1 + \frac{r}{100}\right)^2 = \frac{441}{400} \)
Taking the square root on both sides:
\( 1 + \frac{r}{100} = \frac{21}{20} \)
Subtracting 1 from both sides:
\( \frac{r}{100} = \frac{21}{20} - 1 \)
\( \implies \frac{r}{100} = \frac{1}{20} \)
\( \implies r = \frac{100}{20} = 5\% \)
Thus, the annual rate of interest is 5%.
In simple words: We want to find the annual interest rate that turns Rs. 6,000 into Rs. 6,615 over two years. By simplifying the fraction and taking square roots, we find the rate is 5% per year.

Exam Tip: When you see a squared term on one side, try simplifying the other side until you get a perfect square ratio, like \( \frac{441}{400} \), which is \( \left(\frac{21}{20}\right)^2 \).

Question 10. The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?
Answer: Let the amounts invested by Pramod and Rohit be Rs. \( x \) and Rs. \( y \) respectively.
Pramod is currently 16 years old and will reach 25 years of age after \( 25 - 16 = 9 \) years.
Rohit is currently 18 years old and will reach 25 years of age after \( 25 - 18 = 7 \) years.
Both of their investments are compounded annually at a rate of 5% p.a., and they receive equal final amounts. Therefore:
\( x \left(1 + \frac{5}{100}\right)^9 = y \left(1 + \frac{5}{100}\right)^7 \)
Dividing both sides of the equation by \( \left(1 + \frac{5}{100}\right)^7 \):
\( x \left(1 + \frac{5}{100}\right)^2 = y \)
\( \implies \frac{x}{y} = \frac{1}{\left(1 + \frac{5}{100}\right)^2} \)
\( \implies \frac{x}{y} = \frac{1}{\left(\frac{21}{20}\right)^2} \)
\( \implies \frac{x}{y} = \frac{400}{441} \)
Hence, the required ratio of investment for Pramod and Rohit is \( 400 : 441 \).
In simple words: Pramod's money has 9 years to grow, while Rohit's money only has 7 years. To end up with the exact same amount at age 25, Pramod needs to invest less today. The ratio of their starting investments must be 400 to 441.

Exam Tip: In age-based investment questions, the person with the longer period must invest less. Use laws of exponents to cancel out matching terms on both sides to keep the math simple.

Question 11. At what rate per cent will a sum of Rs. 4,000 yield Rs. 1,324 as compound interest in 3 years?
Answer: We are given a principal \( P \) of Rs. 4,000, a compound interest of Rs. 1,324, and a time period \( n = 3 \) years.
First, we determine the final amount \( A \) by adding the interest to the principal:
\( A = P + \text{C.I.} \)
\( \implies A = \text{Rs. } 4,000 + \text{Rs. } 1,324 = \text{Rs. } 5,324 \)
Now, using the formula for compound interest:
\( A = P \left(1 + \frac{r}{100}\right)^3 \)
Substituting our values:
\( 5324 = 4000 \left(1 + \frac{r}{100}\right)^3 \)
Dividing both sides by 4,000 to isolate the term with \( r \):
\( \left(1 + \frac{r}{100}\right)^3 = \frac{5324}{4000} \)
Simplifying this fraction by dividing both numerator and denominator by 4:
\( \left(1 + \frac{r}{100}\right)^3 = \frac{1331}{1000} \)
Expressing both sides as cubes:
\( \left(1 + \frac{r}{100}\right)^3 = \left(\frac{11}{10}\right)^3 \)
Taking the cube root of both sides:
\( 1 + \frac{r}{100} = \frac{11}{10} \)
Subtracting 1 from both sides:
\( \frac{r}{100} = \frac{11}{10} - 1 \)
\( \implies \frac{r}{100} = \frac{1}{10} \)
\( \implies r = 10\% \)
Hence, the rate of interest is 10% per annum.
In simple words: A starting sum of Rs. 4,000 earns Rs. 1,324 in interest to become Rs. 5,324 after three years. By comparing the ratios and taking cube roots, we find the rate is 10% per year.

Exam Tip: Always calculate the total amount first when given only the compound interest, then set up the ratio to simplify and find the cube root.

Question 12. A person invests Rs. 5,000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to Rs. 6,272. Calculate:
(i) the rate of interest per annum.
(ii) the amount at the end of the third year.
Answer: We are given that the principal amount \( P \) is Rs. 5,000, and it grows to a final amount \( A \) of Rs. 6,272 over a period of \( n = 2 \) years.
(i) To find the rate of interest \( r \), we use the compound interest formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting the given parameters:
\( 6272 = 5000 \left(1 + \frac{r}{100}\right)^2 \)
Isolating the term containing the rate of interest:
\( \left(1 + \frac{r}{100}\right)^2 = \frac{6272}{5000} \)
Dividing both the top and bottom of the fraction by 8 to simplify:
\( \left(1 + \frac{r}{100}\right)^2 = \frac{784}{625} \)
Expressing both sides as perfect squares:
\( \left(1 + \frac{r}{100}\right)^2 = \left(\frac{28}{25}\right)^2 \)
Taking the square root of both sides:
\( 1 + \frac{r}{100} = \frac{28}{25} \)
Solving for the rate \( r \):
\( \frac{r}{100} = \frac{28}{25} - 1 \)
\( \implies \frac{r}{100} = \frac{3}{25} \)
\( \implies r = \frac{3 \times 100}{25} = 12\% \)
Thus, the annual rate of interest is 12%.

(ii) To find the amount at the end of the third year, we use the formula with \( n = 3 \) years:
\( A = P \left(1 + \frac{r}{100}\right)^3 \)
Substituting \( P = 5000 \) and \( r = 12\% \):
\( A = 5000 \left(1 + \frac{12}{100}\right)^3 \)
\( \implies A = 5000 \left(\frac{28}{25}\right)^3 \)
\( \implies A = 5000 \times \frac{21952}{15625} \)
\( \implies A = \frac{21952 \times 8}{25} \)
\( \implies A = \text{Rs. } 7,024.64 \)
So, the total sum at the end of three years is Rs. 7,024.64.
In simple words:
(i) A sum of Rs. 5,000 grows to Rs. 6,272 in two years, which gives an interest rate of 12% per year.
(ii) If we let this money continue to grow for one more year at that same rate of 12%, the final amount becomes Rs. 7,024.64.

Exam Tip: Be careful with calculations in the second part. Alternatively, you can find the third-year amount by simply multiplying the second-year amount of Rs. 6,272 by \( \left(1 + \frac{12}{100}\right) \), which is quicker and less error-prone.

Question 13. In how many years will Rs. 7,000 amount to Rs. 9,317 at 10% per annum compound interest?
Answer: We are given that the principal \( P \) is Rs. 7,000, the final amount \( A \) is Rs. 9,317, and the rate of interest \( r \) is 10% per annum.
Using the formula for compound interest:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting these values:
\( 9317 = 7000 \left(1 + \frac{10}{100}\right)^n \)
Isolating the term with power \( n \):
\( \left(\frac{11}{10}\right)^n = \frac{9317}{7000} \)
Dividing both the numerator and denominator on the right side by 7 to simplify the fraction:
\( \left(\frac{11}{10}\right)^n = \frac{1331}{1000} \)
Writing the right-hand side as a power of \( \frac{11}{10} \):
\( \left(\frac{11}{10}\right)^n = \left(\frac{11}{10}\right)^3 \)
By comparing the exponents on both sides:
\( n = 3 \)
Hence, the required time is 3 years.
In simple words: We need to find how long it takes for Rs. 7,000 to grow to Rs. 9,317 at a 10% interest rate. By simplifying the fraction, we see it matches the cube of our growth rate, meaning it takes exactly 3 years.

Exam Tip: When solving for time \( n \), express the numerical fraction on one side with the same base as the exponential term on the other side. This allows you to solve by direct comparison.

Question 14. Find the time, in years, in which Rs. 4,000 will produce Rs. 630.50 as compound interest at 5% compounded annually.
Answer: We are given a principal \( P \) of Rs. 4,000, compound interest of Rs. 630.50, and an interest rate \( r \) of 5%.
Using the formula for compound interest:
\( \text{C.I.} = P \left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)
By substituting the values we have:
\( 630.50 = 4000 \left[\left(1 + \frac{5}{100}\right)^n - 1\right] \)
Dividing both sides by 4,000:
\( \frac{630.50}{4000} = \left(\frac{21}{20}\right)^n - 1 \)
Simplifying the decimal fraction on the left by multiplying numerator and denominator by 2:
\( \frac{1261}{8000} = \left(\frac{21}{20}\right)^n - 1 \)
Adding 1 to both sides:
\( \left(\frac{21}{20}\right)^n = \frac{1261}{8000} + 1 \)
\( \implies \left(\frac{21}{20}\right)^n = \frac{9261}{8000} \)
Expressing the right side as a power of \( \frac{21}{20} \):
\( \left(\frac{21}{20}\right)^n = \left(\frac{21}{20}\right)^3 \)
Comparing the powers on both sides:
\( n = 3 \)
Therefore, the required time is 3 years.
In simple words: We are looking for the number of years it takes Rs. 4,000 to earn Rs. 630.50 in interest at 5%. By setting up the ratio and simplifying, we find that the time needed is 3 years.

Exam Tip: When a decimal like 630.50 is present, get rid of the decimal by multiplying both numerator and denominator of the fraction by 2 to make calculations easier.

Question 15. Divide Rs. 28,730 between A and B so that when their shares are lent out at 10% compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.
Answer: Let the share of A be Rs. \( y \). Consequently, the share of B will be Rs. \( (28730 - y) \). The rate of interest is given as 10% compounded annually.
Based on the problem's condition, the amount A gets in 3 years is equal to the amount B gets in 5 years:
\( y \left(1 + \frac{10}{100}\right)^3 = (28730 - y) \left(1 + \frac{10}{100}\right)^5 \)
To simplify, we divide both sides of the equation by \( \left(1 + \frac{10}{100}\right)^3 \):
\( y = (28730 - y) \left(1 + \frac{10}{100}\right)^2 \)
\( \implies y = (28730 - y) \left(\frac{11}{10}\right)^2 \)
\( \implies y = (28730 - y) \left(\frac{121}{100}\right) \)
Multiplying both sides by 100 to clear the fraction:
\( 100y = 121(28730 - y) \)
\( \implies 100y = 121 \times 28730 - 121y \)
Adding \( 121y \) to both sides:
\( 221y = 121 \times 28730 \)
Solving for \( y \):
\( y = \frac{121 \times 28730}{221} \)
\( \implies y = 121 \times 130 = \text{Rs. } 15,730 \)
Therefore, the share of A is Rs. 15,730.
The share of B is:
\( \text{Rs. } 28,730 - \text{Rs. } 15,730 = \text{Rs. } 13,000 \)
In simple words: We divide Rs. 28,730 into two parts. Since B's share grows for 5 years and A's share grows for only 3 years, B's starting share must be smaller to reach the same final amount. A gets Rs. 15,730 and B gets Rs. 13,000.

Exam Tip: Divide the higher-power side by the lower-power side right at the beginning of your steps. This removes the need to calculate large numbers like \( 1.1^3 \) and \( 1.1^5 \) separately.

Question 16. A sum of Rs. 44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10% per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of Rs. 44,200?
(ii) What will each receive when 16 years old?
Answer: We are dividing a total of Rs. 44,200 between John (12 years old) and Smith (14 years old) with a 10% annual compound interest rate. Both are to receive equal sums when they reach 16 years of age.
(i) Let John's share be Rs. \( y \). This leaves Smith's share as Rs. \( (44200 - y) \).
John will reach 16 in \( 16 - 12 = 4 \) years, so his investment runs for 4 years.
Smith will reach 16 in \( 16 - 14 = 2 \) years, so his investment runs for 2 years.
Since their final amounts at age 16 are equal:
\( y \left(1 + \frac{10}{100}\right)^4 = (44200 - y) \left(1 + \frac{10}{100}\right)^2 \)
Dividing both sides by \( \left(1 + \frac{10}{100}\right)^2 \):
\( y \left(1 + \frac{10}{100}\right)^2 = 44200 - y \)
\( \implies y \left(\frac{11}{10}\right)^2 = 44200 - y \)
\( \implies \frac{121}{100}y = 44200 - y \)
Multiplying the entire equation by 100 to clear the fraction:
\( 121y = 100(44200 - y) \)
\( \implies 121y = 4420000 - 100y \)
Adding \( 100y \) to both sides:
\( 221y = 4420000 \)
Solving for \( y \):
\( y = \frac{4420000}{221} = \text{Rs. } 20,000 \)
Therefore, John's share is Rs. 20,000.
Smith's share is:
\( \text{Rs. } 44,200 - \text{Rs. } 20,000 = \text{Rs. } 24,200 \).

(ii) To find out what each person receives upon turning 16, we calculate the accumulated amount of John's share over 4 years:
\( A = 20000 \left(1 + \frac{10}{100}\right)^4 \)
\( \implies A = 20000 \left(\frac{11}{10}\right)^4 \)
\( \implies A = 20000 \times \frac{14641}{10000} \)
\( \implies A = 2 \times 14641 = \text{Rs. } 29,282 \)
Thus, each boy will receive Rs. 29,282 when they turn 16.
In simple words:
(i) John's money grows for 4 years, while Smith's grows for only 2 years. To end up with the same amount, John starts with Rs. 20,000 and Smith starts with Rs. 24,200.
(ii) By the time they turn 16, both of their shares will have grown to exactly Rs. 29,282.

Exam Tip: Verify your answer by calculating the final amount using Smith's share as well: \( 24200 \times (1.1)^2 = 24200 \times 1.21 = 29282 \). If both calculations match, your shares are 100% correct.

Question 17. The simple interest on a certain sum of money at 10% per annum is Rs. 6,000 in 2 years. Find:
(i) The sum.
(ii) The amount due at the end of 3 years and at the same rate of interest compounded annually.
(iii) The compound interest earned in 3 years.
Answer: We are given that the simple interest \( I \) on a sum of money is Rs. 6,000 over a time period \( T = 2 \) years at an annual rate of interest \( R = 10\% \).
(i) We find the principal sum \( P \) using the simple interest formula:
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
Rearranging to solve for \( P \):
\( P = \frac{\text{S.I.} \times 100}{R \times T} \)
Substituting the given numbers:
\( P = \frac{6000 \times 100}{10 \times 2} \)
\( \implies P = \frac{600000}{20} = \text{Rs. } 30,000 \)
Thus, the starting sum of money is Rs. 30,000.

(ii) Now we calculate the final amount \( A \) on this principal of Rs. 30,000 after \( n = 3 \) years with a 10% interest rate compounded yearly:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting our values:
\( A = 30000 \left(1 + \frac{10}{100}\right)^3 \)
\( \implies A = 30000 \left(\frac{11}{10}\right)^3 \)
\( \implies A = 30000 \times \frac{1331}{1000} \)
\( \implies A = 30 \times 1331 = \text{Rs. } 39,930 \)
So, the amount due at the end of 3 years is Rs. 39,930.

(iii) The compound interest earned during these 3 years is the final amount minus our starting principal:
\( \text{C.I.} = A - P \)
\( \implies \text{C.I.} = \text{Rs. } 39,930 - \text{Rs. } 30,000 = \text{Rs. } 9,930 \)
The compound interest earned is Rs. 9,930.
In simple words:
(i) A sum that earns Rs. 6,000 in simple interest at 10% over two years must be Rs. 30,000.
(ii) If this Rs. 30,000 is compounded annually at 10% for three years, it grows to Rs. 39,930.
(iii) Subtracting the starting Rs. 30,000 from Rs. 39,930 shows that the compound interest earned is Rs. 9,930.

Exam Tip: When a question has multiple linked sub-parts, the answer to the first part is used in the subsequent parts. Always double-check your initial calculation to avoid carrying over errors.

Question 18. Find the difference between compound interest and simple interest on Rs. 8,000 in 2 years and at 5% per annum.
Answer: We are given a principal \( P \) of Rs. 8,000, an interest rate \( R = 5\% \) per annum, and a time period of 2 years.
First, we find the Simple Interest (S.I.):
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
Substituting our values:
\( \text{S.I.} = \frac{8000 \times 5 \times 2}{100} \)
\( \implies \text{S.I.} = \text{Rs. } 800 \)
Next, we find the Compound Interest (C.I.):
First, we determine the final amount \( A \) using the compound interest formula:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
Substituting our values:
\( A = 8000 \left(1 + \frac{5}{100}\right)^2 \)
\( \implies A = 8000 \left(\frac{21}{20}\right)^2 \)
\( \implies A = 8000 \times \frac{441}{400} \)
\( \implies A = 20 \times 441 = \text{Rs. } 8,820 \)
Subtracting the principal to find the compound interest:
\( \text{C.I.} = A - P \)
\( \implies \text{C.I.} = \text{Rs. } 8,820 - \text{Rs. } 8,000 = \text{Rs. } 820 \)
Finally, we find the difference between compound interest and simple interest:
\( \text{Difference} = \text{C.I.} - \text{S.I.} \)
\( \implies \text{Difference} = \text{Rs. } 820 - \text{Rs. } 800 = \text{Rs. } 20 \)
The difference is Rs. 20.
In simple words: Simple interest on Rs. 8,000 for two years at 5% is Rs. 800, while compound interest is Rs. 820. The difference between the two types of interest is Rs. 20.

Exam Tip: Remember that for 2 years, the difference between C.I. and S.I. can also be calculated using the direct shortcut formula \( \text{C.I.} - \text{S.I.} = P \left(\frac{R}{100}\right)^2 \). This is a great way to verify your answer quickly.

Exercise 3(B)

Question 1. The difference between simple interest and compound interest on a certain sum is Rs. 54.40 for 2 years at 8 per cent per annum. Find the sum.
Answer: Let the principal sum be Rs. \( x \). The interest rate \( R \) is 8% per annum, and the time period \( T \) is 2 years.
We calculate the Simple Interest (S.I.):
\( \text{S.I.} = \frac{x \times 8 \times 2}{100} = \frac{16x}{100} = \frac{4x}{25} \)
Next, we find the Compound Interest (C.I.):
\( \text{C.I.} = A - P = x \left(1 + \frac{8}{100}\right)^2 - x \)
\( \implies \text{C.I.} = x \left[\left(1 + \frac{2}{25}\right)^2 - 1\right] \)
\( \implies \text{C.I.} = x \left[\left(\frac{27}{25}\right)^2 - 1\right] \)
\( \implies \text{C.I.} = x \left[\frac{729}{625} - 1\right] \)
\( \implies \text{C.I.} = \frac{104x}{625} \)
We are given that the difference between compound interest and simple interest is Rs. 54.40:
\( \text{C.I.} - \text{S.I.} = 54.40 \)
\( \implies \frac{104x}{625} - \frac{4x}{25} = 54.40 \)
To solve, we make the denominators equal by multiplying the second term's numerator and denominator by 25:
\( \frac{104x}{625} - \frac{4x \times 25}{25 \times 25} = 54.40 \)
\( \implies \frac{104x - 100x}{625} = 54.40 \)
\( \implies \frac{4x}{625} = 54.40 \)
Now we solve for \( x \):
\( x = \frac{54.40 \times 625}{4} \)
\( \implies x = 13.6 \times 625 = \text{Rs. } 8,500 \)
Therefore, the principal sum is Rs. 8,500.
In simple words: The difference between compound interest and simple interest on our unknown sum is Rs. 54.40. By expressing both types of interest in terms of the variable \( x \) and solving, we find the principal sum is Rs. 8,500.

Exam Tip: Keep your fractions un-evaluated until the final steps of subtraction to avoid decimal rounding errors during intermediate steps.

Question 2. A sum of money, invested at compound interest, amounts to Rs. 19,360 in 2 years and to Rs. 23,425.60 in 4 years. Find the rate per cent and the original sum of money.
Answer: Let the principal amount be Rs. \( X \) and the rate of interest be \( R\% \).
From the given information, after 2 years, the amount accumulated is Rs. 19,360:
\( X \left(1 + \frac{R}{100}\right)^2 = 19360 \quad \text{(Equation 1)} \)
After 4 years, the total amount is Rs. 23,425.60:
\( X \left(1 + \frac{R}{100}\right)^4 = 23425.60 \quad \text{(Equation 2)} \)
Dividing Equation (2) by Equation (1) to eliminate \( X \):
\( \frac{X \left(1 + \frac{R}{100}\right)^4}{X \left(1 + \frac{R}{100}\right)^2} = \frac{23425.60}{19360} \)
\( \implies \left(1 + \frac{R}{100}\right)^2 = \frac{23425.60}{19360} \)
Multiplying both numerator and denominator by 100 to remove the decimals:
\( \left(1 + \frac{R}{100}\right)^2 = \frac{2342560}{1936000} \)
Simplifying this fraction by dividing both numerator and denominator by 160:
\( \left(1 + \frac{R}{100}\right)^2 = \frac{14641}{12100} \)
Expressing both sides as squares:
\( \left(1 + \frac{R}{100}\right)^2 = \left(\frac{121}{110}\right)^2 \)
Taking the square root on both sides:
\( 1 + \frac{R}{100} = \frac{121}{110} \)
Simplifying the fraction on the right:
\( 1 + \frac{R}{100} = \frac{11}{10} \)
Subtracting 1 from both sides:
\( \frac{R}{100} = \frac{11}{10} - 1 \)
\( \implies \frac{R}{100} = \frac{1}{10} \)
\( \implies R = 10\% \)
So, the annual interest rate is 10%.
Now, substitute \( R = 10\% \) into Equation (1) to find \( X \):
\( X \left(1 + \frac{10}{100}\right)^2 = 19360 \pmb{} \)
\( \implies X \left(\frac{11}{10}\right)^2 = 19360 \)
\( \implies X \times \frac{121}{100} = 19360 \)
\( \implies X = \frac{19360 \times 100}{121} \)
\( \implies X = 160 \times 100 = \text{Rs. } 16,000 \)
Therefore, the original sum of money is Rs. 16,000.
In simple words: By writing equations for 2 years and 4 years and dividing them, we cancel the starting sum to find the rate of interest, which is 10%. Substituting this back gives us the original sum of Rs. 16,000.

Exam Tip: When dividing two exponential equations with different powers (like 4 and 2), subtract the powers to simplify. Here, \( 4 - 2 = 2 \), which leaves a squared term.

Question 3. A sum of money lent out at compound interest amounts to three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.
Answer: Let the initial principal be \( x \), and let \( R \) be the rate of interest per annum.
Case I: The sum becomes 3 times itself in 8 years. Here, Amount \( A = 3x \), and Time \( T = 8 \) years. Using the formula:
\( A = P \left(1 + \frac{R}{100}\right)^T \)
\( \implies 3x = x \left(1 + \frac{R}{100}\right)^8 \)
Dividing both sides by \( x \):
\( 3 = \left(1 + \frac{R}{100}\right)^8 \)
Taking the 8th root of both sides:
\( 3^{\frac{1}{8}} = 1 + \frac{R}{100} \quad \text{(Equation 1)} \)
Case II: The sum becomes 27 times itself in an unknown time \( T \). Here, Amount \( A = 27x \). Using the formula:
\( 27x = x \left(1 + \frac{R}{100}\right)^T \)
Dividing both sides by \( x \):
\( 27 = \left(1 + \frac{R}{100}\right)^T \)
Taking the \( T \)-th root of both sides:
\( 27^{\frac{1}{T}} = 1 + \frac{R}{100} \quad \text{(Equation 2)} \)
Equating the expressions for \( 1 + \frac{R}{100} \) from Equation (1) and Equation (2):
\( 3^{\frac{1}{8}} = 27^{\frac{1}{T}} \)
Since \( 27 = 3^3 \), we can write:
\( 3^{\frac{1}{8}} = \left(3^3\right)^{\frac{1}{T}} \)
\( \implies 3^{\frac{1}{8}} = 3^{\frac{3}{T}} \)
Since the bases are identical, their exponents must be equal:
\( \frac{1}{8} = \frac{3}{T} \)
\( \implies T = 8 \times 3 = 24 \)
Therefore, the money will become 27 times of itself in 24 years.
In simple words: The money triples every 8 years. To become 27 times the original sum, which is \( 3 \times 3 \times 3 \) (three triplings), it will take three intervals of 8 years. This equals 24 years.

Exam Tip: For questions involving multiples, express the multiplier (like 27) as a power of the initial multiplier (like 3). This makes equating the exponents extremely direct and quick.

Question 4. On what sum of money will the compound interest for 2 years be the same as simple interest on Rs. 9,430 for 10 years, both at the rate of 5 percent per annum?
Answer: We are given that simple interest is earned on a principal of Rs. 9,430 for a period of 10 years at a rate of 5% per annum.
First, let us calculate the Simple Interest (S.I.):
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
Substituting the given values:
\( \text{S.I.} = \frac{9430 \times 5 \times 10}{100} \)
\( \implies \text{S.I.} = 94.3 \times 50 = \text{Rs. } 4,715 \)
Next, let the sum of money for which compound interest is calculated be Rs. \( x \). The compound interest (C.I.) earned in 2 years at 5% p.a. is equal to this simple interest:
\( \text{C.I.} = \text{Rs. } 4,715 \)
The compound interest formula is:
\( \text{C.I.} = x \left[\left(1 + \frac{R}{100}\right)^T - 1\right] \)
Substituting \( R = 5\% \) and \( T = 2 \) years:
\( 4715 = x \left[\left(1 + \frac{5}{100}\right)^2 - 1\right] \)
\( \implies 4715 = x \left[\left(\frac{21}{20}\right)^2 - 1\right] \)
\( \implies 4715 = x \left[\frac{441}{400} - 1\right] \)
\( \implies 4715 = x \left[\frac{41}{400}\right] \)
Solving for \( x \):
\( x = \frac{4715 \times 400}{41} \)
\( \implies x = 115 \times 400 = \text{Rs. } 46,000 \)
Hence, the required principal sum of money is Rs. 46,000.
In simple words: First, we find that the simple interest earned on Rs. 9,430 over 10 years at 5% is Rs. 4,715. Then, we find what starting sum would earn that same Rs. 4,715 as compound interest in 2 years at 5%. That sum is Rs. 46,000.

Exam Tip: Make sure to read the question carefully to distinguish between which variables apply to the simple interest calculation and which apply to the compound interest part.

Question 5. Kamal and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand received Rs. 15 more than Kamal. Find the amount of money lent by each.
Answer:
Let us assume that the principal amount is Rs. 100, the rate of interest is 5%, and the time period is 2 years.
For Kamal, we calculate the Simple Interest (S.I.) as follows:
S.I. = \( \frac{100 \times 5 \times 2}{100} \) = Rs. 10
For Anand, we find the total accumulated amount (A) using the compound interest formula:
A = \( P \left( 1 + \frac{R}{100} \right)^T \)

\( \implies \) A = \( 100 \left( 1 + \frac{5}{100} \right)^2 \)

\( \implies \) A = \( 100 \times \frac{21}{20} \times \frac{21}{20} \)

\( \implies \) A = \( \frac{441}{4} \)
Now, the Compound Interest (C.I.) earned by Anand is:
C.I. = \( \frac{441}{4} - 100 = \frac{41}{4} \)
Next, we determine the difference between Anand's compound interest and Kamal's simple interest:
Difference = \( \frac{41}{4} - 10 \)

\( \implies \) Difference = \( \frac{41 - 40}{4} \)

\( \implies \) Difference = Rs. \( \frac{1}{4} \)
Using the unitary method, we can determine the actual principal:
When the interest difference is Rs. \( \frac{1}{4} \), the assumed principal is Rs. 100.
When the interest difference is Rs. 1, the principal is \( 100 \times 4 \).
When the interest difference is Rs. 15, the actual principal is:
Principal = \( 100 \times 4 \times 15 \) = Rs. 6,000
Consequently, the sum of money lent by both Kamal and Anand is Rs. 6,000.
In simple words: Assume a starting principal of Rs. 100 to find the relative difference between compound and simple interest. Once we find that difference, we use a simple ratio to scale it up to the real difference of Rs. 15, giving us the actual principal of Rs. 6,000.

Exam Tip: Using an assumed principal of Rs. 100 is an effective technique to simplify calculations before scaling up with the unitary method.

Question 6. Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find the compound interest on the same sum and at the same rate for 2 years.
Answer:
First, we are given the following values for the simple interest:
Simple Interest (S.I.) = Rs. 450
Rate (R) = 4%
Time (T) = 2 years
We use the simple interest formula to determine the unknown principal (P):
P = \( \frac{\text{S.I.} \times 100}{R \times T} \)

\( \implies \) P = \( \frac{450 \times 100}{4 \times 2} \)

\( \implies \) P = Rs. 5,625
Next, we calculate the accumulated amount (A) for compound interest using the same principal, rate, and time:
P = Rs. 5,625, R = 4%, T = 2 years
A = \( P \left( 1 + \frac{R}{100} \right)^T \)

\( \implies \) A = \( 5,625 \left( 1 + \frac{4}{100} \right)^2 \)

\( \implies \) A = \( 5,625 \left( \frac{26}{25} \right)^2 \)

\( \implies \) A = \( \frac{3,802,500}{625} \)

\( \implies \) A = Rs. 6,084
Finally, we compute the Compound Interest (C.I.) by subtracting the principal from the total amount:
C.I. = A - P

\( \implies \) C.I. = Rs. 6,084 - Rs. 5,625

\( \implies \) C.I. = Rs. 459
Thus, the compound interest is Rs. 459.
In simple words: First, find the initial principal amount using the given simple interest details. Then, plug this principal into the compound interest formula to find the final amount, and subtract the principal to get the compound interest of Rs. 459.

Exam Tip: Be sure to compute the final interest instead of stopping at the total amount, as marks are awarded for the final compound interest value.

Question 7. Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by Rs. 228. Find the sum.
Answer:
Let the principal sum be represented by P.
For simple interest, we are given the rate R = 4% and time T = 4 years:
S.I. = \( \frac{P \times 4 \times 4}{100} \)

\( \implies \) S.I. = \( \frac{4P}{25} \)
For compound interest, we are given the rate R = 5% and time T = 3 years:
C.I. = \( P \left[ \left( 1 + \frac{5}{100} \right)^3 - 1 \right] \)

\( \implies \) C.I. = \( P \left[ \left( \frac{21}{20} \right)^3 - 1 \right] \)

\( \implies \) C.I. = \( P \left[ \frac{9,261}{8,000} - 1 \right] \)

\( \implies \) C.I. = \( \frac{1,261}{8,000}P \)
We are given that the simple interest exceeds the compound interest by Rs. 228:
S.I. - C.I. = 228

\( \implies \frac{4P}{25} - \frac{1,261P}{8,000} = 228 \)
To find a common denominator, we multiply the first fraction's numerator and denominator by 320:

\( \implies \frac{4 \times 320P - 1,261P}{8,000} = 228 \)

\( \implies \frac{1,280P - 1,261P}{8,000} = 228 \)

\( \implies \frac{19P}{8,000} = 228 \)

\( \implies \) 19P = \( 228 \times 8,000 \)

\( \implies \) P = \( \frac{228 \times 8,000}{19} \)

\( \implies \) P = Rs. 96,000
Therefore, the principal sum is Rs. 96,000.
In simple words: Write expressions for both simple interest and compound interest in terms of the principal. Find their difference, set it equal to Rs. 228, and solve the equation to find the original sum of Rs. 96,000.

Exam Tip: Take extra care when working out the subtraction with different denominators - finding the correct common multiple avoids algebraic errors.

Question 8. Compound interest on a certain sum of money at 5% per annum for two years is Rs. 246. Calculate simple interest on the same sum for 3 years at 6% per annum.
Answer:
We are given that the compound interest (C.I.) is Rs. 246, the rate (R) is 5%, and the time (T) is 2 years.
We can relate compound interest to the principal (P) using:
C.I. = A - P

\( \implies \) 246 = \( P \left[ \left( 1 + \frac{5}{100} \right)^2 - 1 \right] \)

\( \implies \) 246 = \( P \left[ \left( \frac{21}{20} \right)^2 - 1 \right] \)

\( \implies \) 246 = \( P \left[ \frac{441}{400} - 1 \right] \)

\( \implies \) 246 = \( P \left( \frac{61}{400} \right) \)
Now, we solve for P:
P = \( \frac{246 \times 400}{41} \)

\( \implies \) P = Rs. 2,400
Now that we have the principal, we can determine the simple interest for a period of 3 years at a rate of 6% per annum:
P = Rs. 2,400, R = 6%, T = 3 years
S.I. = \( \frac{2,400 \times 6 \times 3}{100} \)

\( \implies \) S.I. = Rs. 432
Thus, the simple interest is Rs. 432.
In simple words: First, find the principal sum of Rs. 2,400 from the compound interest equation. Then, calculate the simple interest on this principal at 6% for 3 years to get Rs. 432.

Exam Tip: Be sure to write down the general formula for compound interest as a function of principal first, which ensures step marks even if an arithmetic slip-up occurs later.

Question 9. A certain sum of money amounts to Rs. 23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest.
Answer:
Let us represent the initial principal sum as x.
Given the total simple interest amount = Rs. 23,400, rate R = 10%, and time T = 3 years:
The simple interest (I) is given by:
I = \( \frac{x \times 10 \times 3}{100} \)

\( \implies \) I = \( \frac{3x}{10} \)
The total accumulated amount is the sum of the principal and interest:
Amount = Principal + Interest

\( \implies \) 23,400 = \( x + \frac{3x}{10} \)

\( \implies \) 23,400 = \( \frac{13x}{10} \)

\( \implies \) 13x = \( 23,400 \times 10 \)

\( \implies \) x = \( \frac{234,000}{13} \)

\( \implies \) x = 18,000
Thus, the principal sum is Rs. 18,000.
Now, we must find the compound interest amount on this same sum after 2 years at 10% per annum compounding annually:
P = Rs. 18,000, r = 10%, n = 2 years
A = \( P \left( 1 + \frac{r}{100} \right)^n \)

\( \implies \) A = \( 18,000 \left( 1 + \frac{10}{100} \right)^2 \)

\( \implies \) A = \( 18,000 \left( \frac{11}{10} \right)^2 \)

\( \implies \) A = \( 18,000 \left( \frac{121}{100} \right) \)

\( \implies \) A = 21,780
Therefore, the amount after 2 years under compound interest is Rs. 21,780.
In simple words: Use the simple interest relationship to work backward and identify the starting principal as Rs. 18,000. Once known, apply the standard compounding formula to find the new final amount of Rs. 21,780.

Exam Tip: Remember that simple interest is calculated on the original principal, whereas compound interest grows based on the changing amount each year.

Question 10. Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying Rs. 12,600 at the end of the first year and Rs. 17,640 at the end of the second year. Find the sum borrowed.
Answer:
Let us analyze the two separate payments made to settle the loan:
First, we look at the payment of Rs. 12,600 made at the end of the first year:
Amount (A) = Rs. 12,600, n = 1 year, and rate (r) = 5%
Using the compound interest formula:
A = \( P \left( 1 + \frac{r}{100} \right)^n \)

\( \implies 12,600 = P \left( 1 + \frac{5}{100} \right)^1 \)

\( \implies 12,600 = P \left( \frac{21}{20} \right) \)

\( \implies P = \frac{20}{21} \times 12,600 \)

\( \implies P = \text{Rs. } 12,000 \)
Second, we look at the payment of Rs. 17,640 made at the end of the second year:
Amount (A) = Rs. 17,640, n = 2 years, and rate (r) = 5%
Using the compound interest formula:
A = \( P \left( 1 + \frac{r}{100} \right)^n \)

\( \implies 17,640 = P \left( 1 + \frac{5}{100} \right)^2 \)

\( \implies 17,640 = P \left( \frac{21}{20} \right)^2 \)

\( \implies P = \frac{20}{21} \times \frac{20}{21} \times 17,640 \)

\( \implies P = \text{Rs. } 16,000 \)
To find the total sum borrowed, we sum these two individual present values:
Total Borrowed Sum = Rs. 12,000 + Rs. 16,000 = Rs. 28,000
Thus, the total sum borrowed is Rs. 28,000.
In simple words: Find the starting values of each payment separately by moving back in time using the interest rate. Adding these present values together gives the total borrowed amount of Rs. 28,000.

Exam Tip: For loan installments, calculate the present value of each payment separately and add them up to find the initial sum borrowed.

Exercise 3(C)

Question 1. If the interest is compounded half-yearly, calculate the amount when principal is Rs. 7,400; the rate of interest is 5% per annum and the duration is one year.
Answer:
We are given:
Principal (P) = Rs. 7,400
Rate (r) = 5% p.a.
Time (n) = 1 year
Since the compounding is done half-yearly, we adjust the rate and time periods:
Amount (A) = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( 7,400 \left( 1 + \frac{5}{2 \times 100} \right)^{1 \times 2} \)

\( \implies \) A = \( 7,400 \left( 1 + \frac{5}{200} \right)^2 \)

\( \implies \) A = \( 7,400 \left( \frac{41}{40} \right)^2 \)

\( \implies \) A = \( 7,400 \times \frac{1,681}{1,600} \)

\( \implies \) A = Rs. 7,774.63
Thus, the final amount is Rs. 7,774.63.
In simple words: When compounding interest twice a year, divide the yearly rate by 2 and multiply the number of years by 2. This gives a final amount of Rs. 7,774.63.

Exam Tip: Remember to use the formula \( P \left(1 + \frac{r}{200}\right)^{2n} \) for half-yearly compounding so that both rate and time periods are correctly adjusted.

Question 2. Find the difference between the compound interest compounded yearly and half-yearly on Rs. 10,000 for 18 months at 10% per annum.
Answer:
(i) When the interest is compounded yearly:
We are given:
Principal (P) = Rs. 10,000
Time (n) = 18 months = \( 1\frac{1}{2} \) years
Rate (r) = 10% p.a.
For the first full year:
A = \( P \left( 1 + \frac{r}{100} \right)^n \)

\( \implies \) A = \( 10,000 \left( 1 + \frac{10}{100} \right)^1 \)

\( \implies \) A = \( 10,000 \left( \frac{11}{10} \right) \)

\( \implies \) A = Rs. 11,000
For the remaining \( \frac{1}{2} \) year:
The new principal becomes P = Rs. 11,000, rate (r) = 10%, and time (n) = \( \frac{1}{2} \) year:
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( 11,000 \left( 1 + \frac{10}{2 \times 100} \right)^{\frac{1}{2} \times 2} \)

\( \implies \) A = \( 11,000 \left( \frac{21}{20} \right)^1 \)

\( \implies \) A = Rs. 11,550
The Compound Interest (C.I.) is:
C.I. = Rs. 11,550 - Rs. 10,000 = Rs. 1,550
(ii) When the interest is compounded half-yearly:
We are given:
Principal (P) = Rs. 10,000, Time (n) = \( 1\frac{1}{2} \) years = \( \frac{3}{2} \) years, and Rate (r) = 10% p.a.:
Using the half-yearly compounding formula:
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( 10,000 \left( 1 + \frac{10}{2 \times 100} \right)^{\frac{3}{2} \times 2} \)

\( \implies \) A = \( 10,000 \left( \frac{21}{20} \right)^3 \)

\( \implies \) A = Rs. 11,576.25
The Compound Interest (C.I.) is:
C.I. = Rs. 11,576.25 - Rs. 10,000 = Rs. 1,576.25
To find the difference between the two compounding methods:
Difference = Rs. 1,576.25 - Rs. 1,550 = Rs. 26.25
Thus, the difference between the two interests is Rs. 26.25.
In simple words: Calculate the interest using yearly compounding first, breaking down fractional years. Then, calculate it using half-yearly compounding. Subtract the two to find a difference of Rs. 26.25.

Exam Tip: For fractional years under annual compounding, calculate the amount for the whole years first, and then use that amount as the principal for the remaining fraction of the year.

Question 3. A man borrowed Rs. 16,000 for 3 years under the following terms:
- 20% simple interest for the first 2 years.
- 20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.
Find the total amount to be paid at the end of the three years.
Answer:
For the first 2 years, interest is calculated as simple interest:
S.I. = \( \frac{P \times N \times R}{100} \)

\( \implies \) S.I. = \( \frac{16,000 \times 2 \times 20}{100} \)

\( \implies \) S.I. = Rs. 6,400
The total amount due at the end of the first 2 years is:
Amount = S.I. + P

\( \implies \) Amount = Rs. 6,400 + Rs. 16,000

\( \implies \) Amount = Rs. 22,400
Thus, the amount due at the end of 2 years is Rs. 22,400.
For the final 1 year, the interest is compounded half-yearly on this amount:
P = Rs. 22,400, r = 20%, n = 1 year
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( 22,400 \left( 1 + \frac{20}{200} \right)^2 \)

\( \implies \) A = \( 22,400 \left( \frac{11}{10} \right)^2 \)

\( \implies \) A = Rs. 27,104
Therefore, the total amount to be paid at the end of 3 years is Rs. 27,104.
In simple words: First, calculate the simple interest for 2 years and add it to the principal to get Rs. 22,400. Then, treat this as the new principal to calculate compound interest half-yearly for the 3rd year, giving a final amount of Rs. 27,104.

Exam Tip: Be careful to transition the final amount from the simple interest phase as the starting principal for the compound interest phase.

Question 4. What sum of money will amount to Rs. 27,783 in one and a half years at 10% per annum compounded half-yearly?
Answer:
We are given:
Amount (A) = Rs. 27,783
Time (n) = \( 1\frac{1}{2} \) years = \( \frac{3}{2} \) years
Rate (r) = 10% p.a. compounded half-yearly
Using the half-yearly compounding formula:
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies 27,783 = P \left( 1 + \frac{10}{200} \right)^{\frac{3}{2} \times 2} \)

\( \implies 27,783 = P \left( \frac{21}{20} \right)^3 \)
Now, we solve for the principal P:
P = \( 27,783 \left( \frac{20}{21} \right)^3 \)

\( \implies \) P = Rs. 24,000
Thus, the principal sum is Rs. 24,000.
In simple words: Put the given values into the half-yearly compounding formula. Solve for the principal by multiplying the final amount by the reciprocal cube of the fraction, resulting in Rs. 24,000.

Exam Tip: When solving for the principal with fraction powers like \( \left(\frac{21}{20}\right)^3 \), write out the steps clearly to avoid errors during cancellation.

Question 5. Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets Rs. 33 more than Ashok in 18 months, calculate the money invested.
Answer:
(i) For Ashok (where interest is compounded yearly):
Let the principal invested be represented by Rs. y.
Time (n) = 18 months = \( 1\frac{1}{2} \) years, Rate (r) = 20% p.a.
For the first year:
A = \( P \left( 1 + \frac{r}{100} \right)^n = y \left( 1 + \frac{20}{100} \right)^1 = \left( \frac{6}{5} \right)y \)
For the next \( \frac{1}{2} \) year:
Principal = Rs. \( \frac{6}{5}y \), n = \( \frac{1}{2} \) year, and r = 20%
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( \left( \frac{6}{5} \right)y \left( 1 + \frac{20}{2 \times 100} \right)^{\frac{1}{2} \times 2} \)

\( \implies \) A = \( \left( \frac{6}{5} \right)y \left( \frac{11}{10} \right)^1 \)

\( \implies \) A = Rs. \( \left( \frac{66}{50} \right)y \)
(ii) For Geeta (where interest is compounded half-yearly):
Principal = Rs. y, Time (n) = \( 1\frac{1}{2} \) years, and Rate (r) = 20% p.a.:
Using the half-yearly compounding formula:
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( y \left( 1 + \frac{20}{2 \times 100} \right)^{\frac{3}{2} \times 2} \)

\( \implies \) A = \( y \left( \frac{11}{10} \right)^3 \)

\( \implies \) A = Rs. \( \left( \frac{1,331}{1,000} \right)y \)
According to the problem, Geeta's accumulated amount exceeds Ashok's by Rs. 33:
\( \frac{1,331}{1,000}y - \frac{66}{50}y = 33 \)

\( \implies \left( \frac{1,331 - 1,320}{1,000} \right)y = 33 \)

\( \implies \left( \frac{11}{1,000} \right)y = 33 \)

\( \implies \) y = \( \frac{33 \times 1,000}{11} \)

\( \implies \) y = Rs. 3,000
Thus, the sum of money invested by each person is Rs. 3,000.
In simple words: Represent the investment as a variable. Compute the final amount for Ashok (compounded annually with a fractional year) and Geeta (compounded half-yearly). Set their difference to Rs. 33 to solve for the investment of Rs. 3,000.

Exam Tip: For complex comparison problems, define a single variable for the principal and carry out the calculations for both options in parallel before setting up the final equation.

Question 6. At what rate of interest per annum will a sum of Rs. 62,500 earn a compound interest of Rs. 5,100 in one year? The interest is to be compounded half-yearly.
Answer:
We are given:
Principal (P) = Rs. 62,500
Compound Interest (C.I.) = Rs. 5,100
Time (n) = 1 year
The total accumulated amount (A) is:
A = P + C.I. = 62,500 + 5,100 = Rs. 67,600
Since compounding is done half-yearly, we use the formula:
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies 67,600 = 62,500 \left( 1 + \frac{r}{200} \right)^{1 \times 2} \)

\( \implies \left( 1 + \frac{r}{200} \right)^2 = \frac{67,600}{62,500} \)

\( \implies \left( 1 + \frac{r}{200} \right)^2 = \frac{676}{625} \)
Taking the square root on both sides:

\( \implies 1 + \frac{r}{200} = \frac{26}{25} \)

\( \implies \frac{r}{200} = \frac{26}{25} - 1 \)

\( \implies \frac{r}{200} = \frac{1}{25} \)

\( \implies r = \frac{200}{25} \)

\( \implies r = 8 \)
Therefore, the rate of interest is 8% per annum.
In simple words: Calculate the total amount first, then set up the half-yearly compounding formula. Take the square root on both sides to isolate and solve for the rate of interest, which is 8% per annum.

Exam Tip: Simplify the fraction on the right-hand side first so that identifying perfect squares becomes much easier when taking the square root.

Question 7. In what time will Rs. 1,500 yield Rs. 496.50 as compound interest at 20% per year compounded half-yearly?
Answer:
We are given:
Principal (P) = Rs. 1,500
Compound Interest (C.I.) = Rs. 496.50
Rate (r) = 20% p.a.
Since the interest is compounded half-yearly:
C.I. = \( P \left[ \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} - 1 \right] \)

\( \implies 496.50 = 1,500 \left[ \left( 1 + \frac{20}{2 \times 100} \right)^{n \times 2} - 1 \right] \)

\( \implies \frac{496.50}{1,500} = \left( 1 + \frac{20}{200} \right)^{2n} - 1 \)

\( \implies \frac{331}{1,000} = \left( \frac{11}{10} \right)^{2n} - 1 \)

\( \implies \left( \frac{11}{10} \right)^{2n} = \frac{331}{1,000} + 1 \)

\( \implies \left( \frac{11}{10} \right)^{2n} = \frac{1,331}{1,000} \)
We express the right side as a power of \( \frac{11}{10} \):

\( \implies \left( \frac{11}{10} \right)^{2n} = \left( \frac{11}{10} \right)^3 \)
Comparing the exponents on both sides:
2n = 3

\( \implies n = 1\frac{1}{2} \) years
Thus, the required time is \( 1\frac{1}{2} \) years.
In simple words: Set up the compounding formula with the given values. Solve the equation to match bases on both sides, which shows that 2n equals 3, meaning the time required is 1.5 years.

Exam Tip: Expressing numerical fractions as perfect powers of the base fraction (like writing \( \frac{1331}{1000} \) as \( \left(\frac{11}{10}\right)^3 \)) is crucial to equating exponents in time-finding problems.

Question 8. Calculate the C.I. on Rs. 3,500 at 6% per annum for 3 years, the interest being compounded half-yearly. Do not use mathematical tables. Use the necessary information from the following:
\( (1.06)^3 = 1.191016 \); \( (1.03)^3 = 1.092727 \)
\( (1.06)^6 = 1.418519 \); \( (1.03)^6 = 1.194052 \)
Answer:
We are given:
Principal (P) = Rs. 3,500
Rate (r) = 6% p.a.
Time (n) = 3 years
Since interest is compounded half-yearly:
C.I. = \( P \left[ \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} - 1 \right] \)

\( \implies \) C.I. = \( 3,500 \left[ \left( 1 + \frac{6}{2 \times 100} \right)^{3 \times 2} - 1 \right] \)

\( \implies \) C.I. = \( 3,500 \left[ \left( 1 + \frac{3}{100} \right)^6 - 1 \right] \)

\( \implies \) C.I. = \( 3,500 \left[ (1.03)^6 - 1 \right] \)
We use the given value \( (1.03)^6 = 1.194052 \) from the problem details:

\( \implies \) C.I. = \( 3,500 [1.194052 - 1] \)

\( \implies \) C.I. = \( 3,500 \times 0.194052 \)

\( \implies \) C.I. = Rs. 679.18
Hence, the compound interest is Rs. 679.18.
In simple words: Apply the compound interest formula with compounding periods doubled and rate halved. Use the provided decimal value for \( (1.03)^6 \) to perform the final multiplication, resulting in Rs. 679.18.

Exam Tip: Be sure to choose the correct exponential value among those provided in the question; since the rate is halved to 3%, you must select the value with base 1.03.

Question 9. Find the difference between compound interest and simple interest on Rs. 12,000 in \( 1\frac{1}{2} \) years at 10% per annum, compounded yearly.
Answer:
We are given:
Principal (P) = Rs. 12,000
Time (n) = \( 1\frac{1}{2} \) years
Rate (r) = 10% p.a.
First, we find the simple interest:
S.I. = \( \frac{P \times R \times T}{100} \)

\( \implies \) S.I. = \( \frac{12,000 \times 10 \times \frac{3}{2}}{100} \)

\( \implies \) S.I. = Rs. 1,800
Second, we calculate the compound interest compounded yearly:
For the first 1 year:
A = \( P \left( 1 + \frac{r}{100} \right)^n \)

\( \implies \) A = \( 12,000 \left( 1 + \frac{10}{100} \right)^1 \)

\( \implies \) A = Rs. 13,200
For the remaining \( \frac{1}{2} \) year:
The new principal is P = Rs. 13,200, r = 10%, and n = \( \frac{1}{2} \) year:
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( 13,200 \left( 1 + \frac{10}{2 \times 100} \right)^{\frac{1}{2} \times 2} \)

\( \implies \) A = \( 13,200 \left( \frac{21}{20} \right)^1 \)

\( \implies \) A = Rs. 13,860
The Compound Interest (C.I.) is:
C.I. = Rs. 13,860 - Rs. 12,000 = Rs. 1,860
Finally, we compute the difference between both interests:
Difference = C.I. - S.I. = Rs. 1,860 - Rs. 1,800 = Rs. 60
Thus, the difference is Rs. 60.
In simple words: First find the simple interest of Rs. 1,800. Next, calculate the compound interest of Rs. 1,860 by working out the fractional year's growth. Subtracting the two gives a final difference of Rs. 60.

Exam Tip: Be careful with the phrasing of the compounding interval (yearly vs. half-yearly) to ensure you calculate the compound interest correctly.

Question 10. Find the difference between compound interest and simple interest on Rs. 12,000 in \( 1\frac{1}{2} \) years at 10% per annum, compounded half-yearly.
Answer:
We are given:
Principal (P) = Rs. 12,000
Time (n) = \( 1\frac{1}{2} \) years
Rate (r) = 10% p.a.
First, we find the simple interest:
S.I. = \( \frac{P \times R \times T}{100} \)

\( \implies \) S.I. = \( \frac{12,000 \times 10 \times \frac{3}{2}}{100} \)

\( \implies \) S.I. = Rs. 1,800
Second, we calculate the compound interest when compounded half-yearly:
Time (n) = \( \frac{3}{2} \) years
A = \( P \left( 1 + \frac{r}{2 \times 100} \right)^{n \times 2} \)

\( \implies \) A = \( 12,000 \left( 1 + \frac{10}{2 \times 100} \right)^{\frac{3}{2} \times 2} \)

\( \implies \) A = \( 12,000 \left( \frac{21}{20} \right)^3 \)

\( \implies \) A = Rs. 13,891.50
The Compound Interest (C.I.) is:
C.I. = Rs. 13,891.50 - Rs. 12,000 = Rs. 1,891.50
Finally, we compute the difference between both interests:
Difference = C.I. - S.I. = Rs. 1,891.50 - Rs. 1,800 = Rs. 91.50
Thus, the difference is Rs. 91.50.
In simple words: First find the simple interest of Rs. 1,800. Next, find the compound interest of Rs. 1,891.50 with compounding done every six months. The subtraction yields a difference of Rs. 91.50.

Exam Tip: Note that half-yearly compounding results in a slightly higher compound interest than yearly compounding due to the more frequent compounding periods.

Exercise 3(D)

Question 1. The cost of a machine is supposed to depreciate each year at 12% of its value at the beginning of the year. If the machine is valued at Rs. 44,000 at the beginning of 2008, find its value:
(i) at the end of 2009.
(ii) at the beginning of 2007.
Answer:
We are given:
Value of machine in 2008 = Rs. 44,000
Depreciation rate (r) = 12%
(i) To find the value of the machine at the end of 2009:
The time period from the beginning of 2008 to the end of 2009 is n = 2 years.
Value = \( P \left( 1 - \frac{r}{100} \right)^n \)

\( \implies \) Value = \( 44,000 \left( 1 - \frac{12}{100} \right)^2 \)

\( \implies \) Value = \( 44,000 \times \left( \frac{88}{100} \right)^2 \)

\( \implies \) Value = Rs. 34,073.60
Thus, the value of the machine at the end of 2009 is Rs. 34,073.60.
(ii) To find the value of the machine at the beginning of 2007 (P):
The time period from the beginning of 2007 to the beginning of 2008 is n = 1 year.
A = \( P \left( 1 - \frac{r}{100} \right)^n \)

\( \implies 44,000 = P \left( 1 - \frac{12}{100} \right)^1 \)

\( \implies 44,000 = P \left( \frac{88}{100} \right) \)
Now, we solve for P:
P = \( \frac{44,000 \times 100}{88} \)

\( \implies \) P = Rs. 50,000
Thus, the value of the machine at the beginning of 2007 is Rs. 50,000.
In simple words: Since value decreases over time, we use a minus sign in the formula. Moving forward 2 years gives a lower value of Rs. 34,073.60, whereas moving backward 1 year gives a higher starting value of Rs. 50,000.

Exam Tip: Remember to use a minus sign \( \left(1 - \frac{r}{100}\right) \) inside the formula whenever dealing with problems involving depreciation or decay.

Question 2. The value of an article decreases for two years at the rate of 10% per year and then in the third year it increases by 10%. Find the original value of the article, if its value at the end of 3 years is Rs. 40,095.
Answer:
Let the original value of the article be denoted by x.
The value of the article decreases for the first two years at a rate of 10% per year:
The value of the article at the end of the 1st year is:
x - 10% of x = 0.90x
The value of the article at the end of the 2nd year is:
0.90x - 10% of (0.90x) = 0.81x
During the 3rd year, the value of the article increases by 10%:
The value of the article at the end of the 3rd year is:
0.81x + 10% of (0.81x) = 0.891x
We are given that the final value of the article at the end of 3 years is Rs. 40,095:
0.891x = 40,095

\( \implies \) x = 45,000
Therefore, the original value of the article is Rs. 45,000.
In simple words: Set the initial value as x. Apply the decreases and subsequent increase year-by-year to write a final expression of 0.891x. Equating this to Rs. 40,095 gives the original value of Rs. 45,000.

Exam Tip: Be apply the percentage changes on the updated value at the end of each year rather than on the original value x.

Question 3. According to a census taken towards the end of the year 2005, the population of a rural town was found to be 64,000. If the population of this town had a growth rate of 5% per annum, in how many years did the population reach 74,088?
Answer:
Given:
Initial population \( (P) = 64,000 \)
Growth rate \( (r) = 5\% \) per year
Assume that the population reaches 74,088 after \( n \) years, so the final population is \( A = 74,088 \).

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

Substituting the given values:
\( 74,088 = 64,000 \left(1 + \frac{5}{100}\right)^n \)

Dividing both sides by 64,000:
\( \frac{74,088}{64,000} = \left(1 + \frac{1}{20}\right)^n \)

Simplifying the fraction on the left side by dividing the numerator and denominator by 8:
\( \frac{9,261}{8,000} = \left(\frac{21}{20}\right)^n \)

Expressing the fraction on the left side as a cubic power:
\( \left(\frac{21}{20}\right)^3 = \left(\frac{21}{20}\right)^n \)

By comparing the exponents on both sides, we find:
\( n = 3 \) years
In simple words: To find the time, we use the formula for population growth just like compound interest. By simplifying the fraction, we match the bases on both sides to find that the number of years is 3.

Exam Tip: Always reduce fractions to their lowest terms. Recognizing that 9,261 and 8,000 are the cubes of 21 and 20 respectively makes matching the exponential bases straightforward.

Question 4. The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 2,85,120.
Answer:
Let the population at the beginning of 1998 be \( P \).
The final population at the end of 1999 is given as \( A = 2,85,120 \).
The rate of decrease in the first year is \( r_1 = -12\% \) and the rate of increase in the second year is \( r_2 = +8\% \).

Using the formula for successive population change:
\( A = P \left(1 - \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \)

Substituting the given values:
\( 2,85,120 = P \left(1 - \frac{12}{100}\right) \left(1 + \frac{8}{100}\right) \)

Simplifying inside the brackets:
\( 2,85,120 = P \left(\frac{88}{100}\right) \left(\frac{108}{100}\right) \)
\( 2,85,120 = P \left(\frac{22}{25}\right) \left(\frac{27}{25}\right) \)

Solving for \( P \):
\( P = \frac{2,85,120 \times 25 \times 25}{22 \times 27} \)
\( P = 3,00,000 \)

Thus, the population of the town at the beginning of 1998 was 3,00,000.
In simple words: When the population first drops and then grows, we multiply the original number by both the drop factor and the growth factor to find the final number. By working backward, we can find that the starting population was 3,00,000.

Exam Tip: Be careful with the signs: use a minus sign for depreciation/decrease and a plus sign for appreciation/increase in successive rate problems.

Question 5. A sum of money, invested at compound interest, amounts to Rs. 16,500 in 1 year and to Rs. 19,965 in 3 years. Find the rate per cent and the original sum of money invested.
Answer:
Let the principal sum be Rs. \( P \) and the rate of interest be \( r\% \) per annum.
The amount after 1 year is Rs. 16,500 and after 3 years is Rs. 19,965.

For the first year:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Substituting the 1-year values:
\( 16,500 = P \left(1 + \frac{r}{100}\right)^1 \) --- (1)

For 3 years:
\( A = P \left(1 + \frac{r}{100}\right)^n \)

Substituting the 3-year values:
\( 19,965 = P \left(1 + \frac{r}{100}\right)^3 \) --- (2)

By dividing equation (2) by equation (1), we get:
\( \frac{19,965}{16,500} = \frac{P \left(1 + \frac{r}{100}\right)^3}{P \left(1 + \frac{r}{100}\right)^1} \)
\( \frac{121}{100} = \left(1 + \frac{r}{100}\right)^2 \)
\( \left(\frac{11}{10}\right)^2 = \left(1 + \frac{r}{100}\right)^2 \)

By comparing both sides of the equation:
\( \frac{11}{10} = 1 + \frac{r}{100} \)

Subtracting 1 from both sides:
\( \frac{11}{10} - 1 = \frac{r}{100} \)
\( \frac{1}{10} = \frac{r}{100} \)
\( r = 10\% \)

Now, substitute the value of \( r \) back into equation (1) to solve for \( P \):
\( 16,500 = P \left(1 + \frac{10}{100}\right) \)
\( 16,500 = P \left(\frac{11}{10}\right) \)
\( P = \frac{16,500 \times 10}{11} \)
\( P = \text{Rs. } 15,000 \)

So, the rate of interest is 10% and the original sum is Rs. 15,000.
In simple words: We set up two equations for the amounts at 1 year and 3 years. Dividing them cancels out the starting money, allowing us to find the rate of 10% first, and then we find the original sum of Rs. 15,000.

Exam Tip: Dividing equations is a very efficient way to solve compound interest problems when the principal and rate are both unknown, as it eliminates the principal variable immediately.

Question 6. The difference between the compound interest and the simple interest on Rs. 7,500 for two years is Rs. 12 at the same rate of interest per annum. Find the rate of interest.
Answer:
Here, the principal \( P = \text{Rs. } 7,500 \) and the duration \( n = 2 \) years.
Assume the yearly interest rate is \( y\% \).

First, find the Simple Interest (S.I.):
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{7,500 \times y \times 2}{100} \)
\( \text{S.I.} = 150y \)

Next, find the Compound Interest (C.I.):
\( \text{C.I.} = P \left(1 + \frac{R}{100}\right)^n - P \)
\( \text{C.I.} = 7,500 \left(1 + \frac{y}{100}\right)^2 - 7,500 \)

According to the problem, the difference between compound interest and simple interest is Rs. 12:
\( \text{C.I.} - \text{S.I.} = 12 \)

Substituting our S.I. and C.I. expressions:
\( 7,500 \left(1 + \frac{y}{100}\right)^2 - 7,500 - 150y = 12 \)

Expanding the squared term inside the parentheses:
\( 7,500 \left(1 + \frac{y^2}{10,000} + \frac{2y}{100}\right) - 7,500 - 150y = 12 \)

Multiplying through by 7,500:
\( 7,500 + \frac{7,500y^2}{10,000} + 150y - 7,500 - 150y = 12 \)

Cancelling out the equal terms of opposite signs:
\( \frac{3y^2}{4} = 12 \)

Solving for \( y^2 \):
\( 3y^2 = 48 \)
\( y^2 = 16 \)

Since the rate must be positive:
\( y = 4\% \)

Therefore, the rate of interest is 4% per annum.
In simple words: We calculate both the simple interest and compound interest for 2 years using the rate 'y'. By subtracting them, we get a simple equation that easily tells us the rate is 4%.

Exam Tip: For a 2-year period, you can also use the direct formula: \( \text{Difference} = P \left(\frac{R}{100}\right)^2 \) to quickly verify your steps in the exam.

Question 7. A sum of money lent at compound interest amounts to three times itself in 10 years. Find in how many years will the same sum of money become twenty-seven times of itself at the same rate of interest p.a.
Answer:
Let the initial principal amount be Rs. \( y \) and the rate of interest be \( r\% \).
From the first condition, the money becomes three times its value in 10 years, which means the amount is Rs. \( 3y \).

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

For the first scenario:
\( 3y = y \left(1 + \frac{r}{100}\right)^{10} \)
\( 3 = \left(1 + \frac{r}{100}\right)^{10} \) --- (1)

For the second condition, let the sum grow to 27 times its initial value, or Rs. \( 27y \), after \( n \) years:
\( 27y = y \left(1 + \frac{r}{100}\right)^n \)
\( 27 = \left(1 + \frac{r}{100}\right)^n \)
\( (3)^3 = \left(1 + \frac{r}{100}\right)^n \)

Now, substitute the value of 3 from equation (1) into this equation:
\( \left[\left(1 + \frac{r}{100}\right)^{10}\right]^3 = \left(1 + \frac{r}{100}\right)^n \)
\( \left(1 + \frac{r}{100}\right)^{30} = \left(1 + \frac{r}{100}\right)^n \)

By comparing the exponents on both sides of the equation:
\( n = 30 \) years

Thus, the money will become 27 times itself in 30 years.
In simple words: Since the money triples every 10 years, growing to 27 times means tripling 3 times over (\( 3 \times 3 \times 3 = 27 \)). This takes three intervals of 10 years, giving 30 years in total.

Exam Tip: For base comparison problems under compound interest, write the larger multiplier as a power of the smaller multiplier to equate their bases directly.

Question 8. Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying Rs. 19,360 at the end of the second year and Rs. 31,944 at the end of the third year he clears the debt; find the sum borrowed by him.
Answer:
Let the original sum borrowed by Mr. Sharma be \( P \).
First, let us find the total accumulated amount after the first two years:
\( A_1 = P \left(1 + \frac{R}{100}\right)^2 \)
\( A_1 = P \left(1 + \frac{10}{100}\right)^2 \)

Since a payment of Rs. 19,360 is made at the end of the second year, the remaining balance that acts as the principal for the third year is:
\( P_{\text{third year}} = A_1 - 19,360 \)

To completely clear the debt, a final payment of Rs. 31,944 is made at the end of the third year. This means the amount due after the third year is equal to Rs. 31,944:
\( A_2 = P_{\text{third year}} \left(1 + \frac{R}{100}\right)^1 \)
\( 31,944 = \left[P \left(1 + \frac{10}{100}\right)^2 - 19,360\right] \left(1 + \frac{10}{100}\right)^1 \)

Simplifying the term \( \left(1 + \frac{10}{100}\right) = \frac{11}{10} \):
\( 31,944 = \left[P \left(\frac{11}{10}\right)^2 - 19,360\right] \left(\frac{11}{10}\right) \)

Multiplying both sides by \( \frac{10}{11} \):
\( 29,040 = P \left(\frac{11}{10}\right)^2 - 19,360 \)

Adding 19,360 to both sides:
\( P \left(\frac{11}{10}\right)^2 = 29,040 + 19,360 \)
\( P \left(\frac{121}{100}\right) = 48,400 \)

Solving for \( P \):
\( P = \frac{48,400 \times 100}{121} \)
\( P = \text{Rs. } 40,000 \)

Hence, Mr. Sharma originally borrowed Rs. 40,000.
In simple words: The debt grows with interest over 2 years, then a payment reduces it. The remaining balance earns interest for 1 more year, ending at Rs. 31,944. Working backward reveals the original loan was Rs. 40,000.

Exam Tip: Break down installment and repayment questions year-by-year. Calculate the interest up to the payment point, subtract the payment, and then use that new balance as the next year's principal.

Question 9. The difference between the compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is Rs. 15. Find the sum of money lent out.
Answer:
Let the sum of money invested be Rs. \( y \).

First, calculate the Simple Interest (S.I.) for one year:
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{y \times 10 \times 1}{100} \)
\( \text{S.I.} = \text{Rs. } \frac{y}{10} \)

Next, calculate the Compound Interest (C.I.) compounded half-yearly for one year:
\( \text{C.I.} = P \left[\left(1 + \frac{R}{2 \times 100}\right)^{N \times 2} - 1\right] \)
\( \text{C.I.} = y \left[\left(1 + \frac{10}{200}\right)^{1 \times 2} - 1\right] \)
\( \text{C.I.} = y \left[\left(1 + \frac{1}{20}\right)^2 - 1\right] \)
\( \text{C.I.} = y \left[\left(\frac{21}{20}\right)^2 - 1\right] \)
\( \text{C.I.} = y \left[\frac{441}{400} - 1\right] \)
\( \text{C.I.} = \text{Rs. } \frac{41}{400}y \)

We are given that the difference between C.I. and S.I. is Rs. 15:
\( \text{C.I.} - \text{S.I.} = 15 \)
\( \frac{41}{400}y - \frac{y}{10} = 15 \)

Taking the common denominator:
\( \frac{41y - 40y}{400} = 15 \)
\( \frac{y}{400} = 15 \)
\( y = 15 \times 400 \)
\( y = \text{Rs. } 6,000 \)

Therefore, the sum of money lent out is Rs. 6,000.
In simple words: Simple interest for a year is 10%. With half-yearly compounding, the rate is split to 5% twice, giving slightly more interest. The tiny difference between these two rates equals Rs. 15, which helps us find the main sum of Rs. 6,000.

Exam Tip: For interest compounded half-yearly, remember to divide the annual rate by 2 and multiply the number of years by 2 before putting them into the formula.

Question 10. The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?
Answer:
Let the sums invested by Pramod and Rohit be Rs. \( x \) and Rs. \( y \) respectively.
Pramod is 16 years old, so his investment will grow for:
\( 25 - 16 = 9 \) years

Rohit is 18 years old, so his investment will grow for:
\( 25 - 18 = 7 \) years

The maturity amount for both must be identical at the age of 25:
\( x \left(1 + \frac{5}{100}\right)^9 = y \left(1 + \frac{5}{100}\right)^7 \)

Dividing both sides to find the ratio \( \frac{x}{y} \):
\( \frac{x}{y} = \frac{\left(1 + \frac{5}{100}\right)^7}{\left(1 + \frac{5}{100}\right)^9} \)
\( \frac{x}{y} = \frac{1}{\left(1 + \frac{5}{100}\right)^2} \)
\( \frac{x}{y} = \frac{1}{\left(1 + \frac{1}{20}\right)^2} \)
\( \frac{x}{y} = \frac{1}{\left(\frac{21}{20}\right)^2} \)
\( \frac{x}{y} = \frac{400}{441} \)

Hence, Pramod and Rohit should invest their money in the ratio of 400 : 441.
In simple words: Since Pramod is younger, his money stays in the bank longer (9 years) than Rohit's (7 years). To end up with the same final amount, Pramod needs to invest less money initially, in the ratio of 400 to 441.

Exam Tip: Set up the equality of the two amounts and divide the terms using laws of exponents (\( a^m / a^n = a^{m-n} \)) to quickly find the simplified ratio.

Exercise 3(E)

Question 1. Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half-yearly.
Answer:
First Part: For the simple interest scenario, we have S.I. = Rs. 450, a time of 2 years, and a rate of 4% per annum.
Using the simple interest formula to find the principal \( P \):
\( P = \frac{\text{S.I.} \times 100}{R \times T} \)
\( P = \frac{450 \times 100}{4 \times 2} \)
\( P = \text{Rs. } 5,625 \)

Second Part: For the compound interest scenario compounded semi-annually over 1 year on the same principal:
Principal \( P = \text{Rs. } 5,625 \)
Annual Rate \( r = 4\% \)
Time \( n = 1 \) year

Since the interest is compounded half-yearly, the rate is halved and the number of periods is doubled:
\( A = P \left(1 + \frac{r}{2 \times 100}\right)^{n \times 2} \)
\( A = 5,625 \left(1 + \frac{4}{200}\right)^{1 \times 2} \)
\( A = 5,625 \left(1 + \frac{1}{50}\right)^2 \)
\( A = 5,625 \left(\frac{51}{50}\right)^2 \)
\( A = 5,625 \times 1.0404 \)
\( A = \text{Rs. } 5,852.25 \)

Now, calculate the Compound Interest (C.I.):
\( \text{C.I.} = A - P \)
\( \text{C.I.} = 5,852.25 - 5,625 \)
\( \text{C.I.} = \text{Rs. } 227.25 \)
In simple words: We first find the starting money (Rs. 5,625) using the simple interest information. Then, we use this sum to find the compound interest for 1 year with half-yearly compounding, which gives Rs. 227.25.

Exam Tip: In multi-step questions, double-check that you transfer the principal calculated in the first step accurately into the second step's compound interest formula.

Question 2. Find the compound interest to the nearest rupee on Rs. 10,800 for 2 1/2 years at 10% per annum.
Answer:
We are given that the principal \( P = \text{Rs. } 10,800 \), the total time is \( 2 \frac{1}{2} \) years, and the annual rate is \( 10\% \).
First, calculate the compound interest amount for the first 2 full years:
\( A = P \left(1 + \frac{r}{100}\right)^n \)
\( A = 10,800 \left(1 + \frac{10}{100}\right)^2 \)
\( A = 10,800 \left(\frac{11}{10}\right)^2 \)
\( A = 10,800 \left(\frac{121}{100}\right) \)
\( A = \text{Rs. } 13,068 \)

Now, treat Rs. 13,068 as the principal for the remaining half year. Calculate the interest compounded for this \( \frac{1}{2} \) year:
\( A_{\text{final}} = P_{\text{new}} \left(1 + \frac{r}{2 \times 100}\right)^{n \times 2} \)
\( A_{\text{final}} = 13,068 \left(1 + \frac{10}{200}\right)^{\frac{1}{2} \times 2} \)
\( A_{\text{final}} = 13,068 \left(1 + \frac{1}{20}\right)^1 \)
\( A_{\text{final}} = 13,068 \left(\frac{21}{20}\right) \)
\( A_{\text{final}} = \text{Rs. } 13,721.40 \)

Rounding to the nearest rupee, the final amount is Rs. 13,721.
Now, calculate the Compound Interest (C.I.):
\( \text{C.I.} = A_{\text{final}} - P \)
\( \text{C.I.} = 13,721 - 10,800 \)
\( \text{C.I.} = \text{Rs. } 2,921 \)
In simple words: We calculate the money accumulated over the first 2 full years, which is Rs. 13,068. We then use this new amount to calculate the interest for the final half-year, finding a total compound interest of Rs. 2,921.

Exam Tip: For fractional year problems, calculate the amount for the whole year portion first, and then apply simple interest (or half-year compound interest) to the remaining fraction using that amount as the new principal.

Question 3. The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is Rs. 97,200, find:
(i) its value after 2 years.
(ii) its value when it was purchased.
Answer:
(i) Let the current value of the machine be \( P = \text{Rs. } 97,200 \) and the rate of depreciation be \( 10\% \).
The machine's worth after a period of 2 years is:
\( \text{Worth} = P \left(1 - \frac{r}{100}\right)^n \)
\( \text{Worth} = 97,200 \left(1 - \frac{10}{100}\right)^2 \)
\( \text{Worth} = 97,200 \left(\frac{9}{10}\right)^2 \)
\( \text{Worth} = 97,200 \left(\frac{81}{100}\right) \)
\( \text{Worth} = \text{Rs. } 78,732 \)

(ii) Let the value when purchased \( 2 \) years ago be \( P \) and the current value be \( A = \text{Rs. } 97,200 \).
To find the purchase cost from 2 years back:
\( A = P \left(1 - \frac{r}{100}\right)^n \)
\( 97,200 = P \left(1 - \frac{10}{100}\right)^2 \)
\( 97,200 = P \left(\frac{9}{10}\right)^2 \)
\( 97,200 = P \left(\frac{81}{100}\right) \)

Solving for \( P \):
\( P = \frac{97,200 \times 100}{81} \)
\( P = \text{Rs. } 1,20,000 \)
In simple words: (i) A machine costing Rs. 97,200 today will lose 10% value each year, leaving it worth Rs. 78,732 in 2 years. (ii) Working backward, since it depreciated to Rs. 97,200 over the last 2 years, its original price was Rs. 1,20,000.

Exam Tip: Pay close attention to the time frame. Use standard depreciation for future value, but set the current value as the 'Amount (A)' when calculating the original past value.

Question 4. Anuj and Rajesh each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. Rajesh received Rs. 64 more than Anuj. Find the money lent by each and interest received.
Answer:
Let the sum of money lent by both individuals be Rs. \( y \).
For Anuj, the investment earns simple interest:
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{y \times 8 \times 2}{100} \)
\( \text{S.I.} = \frac{4y}{25} \)

For Rajesh, the investment earns compound interest:
\( \text{C.I.} = P \left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)
\( \text{C.I.} = y \left[\left(1 + \frac{8}{100}\right)^2 - 1\right] \)
\( \text{C.I.} = y \left[\left(\frac{27}{25}\right)^2 - 1\right] \)
\( \text{C.I.} = y \left[\frac{729}{625} - 1\right] \)
\( \text{C.I.} = \frac{104y}{625} \)

We are given that Rajesh earned Rs. 64 more than Anuj, which is the difference between C.I. and S.I.:
\( \text{C.I.} - \text{S.I.} = 64 \)
\( \frac{104y}{625} - \frac{4y}{25} = 64 \)

Making the denominators equal:
\( \frac{104y - 100y}{625} = 64 \)
\( \frac{4y}{625} = 64 \)
\( y = \frac{64 \times 625}{4} \)
\( y = \text{Rs. } 10,000 \)

The interest earned by Anuj:
\( \text{Interest (Anuj)} = \frac{4 \times 10,000}{25} = \text{Rs. } 1,600 \)

The interest earned by Rajesh:
\( \text{Interest (Rajesh)} = \frac{104 \times 10,000}{625} = \text{Rs. } 1,664 \)
In simple words: Anuj and Rajesh lent Rs. 10,000 each. Anuj gets Rs. 1,600 from simple interest, while Rajesh gets Rs. 1,664 because compound interest earns interest on interest.

Exam Tip: Be sure to calculate the actual interest received by *each* person at the end of the steps if the question specifically asks for "interest received by each".

Question 5. Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest on Rs. 4,715 for 5 years, both at the rate of 5% per annum.
Answer:
First Part: For the simple interest calculation, we are given principal \( P = \text{Rs. } 4,715 \), time \( T = 5 \) years, and rate \( R = 5\% \) p.a.
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{4,715 \times 5 \times 5}{100} \)
\( \text{S.I.} = \text{Rs. } 1,178.75 \)

Second Part: The compound interest (C.I.) is specified to be 4 times this simple interest amount:
\( \text{C.I.} = 4 \times \text{S.I.} \)
\( \text{C.I.} = 4 \times 1,178.75 = \text{Rs. } 4,715 \)

For the compound interest calculation, the time is 2 years and the rate is 5% p.a.
Let the principal for the compound interest be \( P_{\text{CI}} \):
\( \text{C.I.} = P_{\text{CI}} \left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)
\( 4,715 = P_{\text{CI}} \left[\left(1 + \frac{5}{100}\right)^2 - 1\right] \)
\( 4,715 = P_{\text{CI}} \left[\left(\frac{21}{20}\right)^2 - 1\right] \)
\( 4,715 = P_{\text{CI}} \left[\frac{441}{400} - 1\right] \)
\( 4,715 = P_{\text{CI}} \left[\frac{41}{400}\right] \)

Solving for \( P_{\text{CI}} \):
\( P_{\text{CI}} = \frac{4,715 \times 400}{41} \)
\( P_{\text{CI}} = \text{Rs. } 46,000 \)
In simple words: First we find the simple interest on Rs. 4,715 which is Rs. 1,178.75. Multiplying this by 4 gives us the required compound interest (Rs. 4,715). We then find that the sum needed to earn this interest over 2 years at 5% is Rs. 46,000.

Exam Tip: Be careful not to confuse the two different principals in this question. Use the first given principal only to find the simple interest, then set up the equation to find the new principal.

Question 6. A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to Rs. 4,950, find the sum invested.
Answer:
We are given that the compound interest earned solely during the second year is Rs. 4,950, and the interest rate for the second year is \( 15\% \).
The interest for the second year is calculated on the principal at the start of the second year (which is the amount at the end of the first year, \( A_1 \)):
\( \text{C.I.}_{\text{2nd year}} = A_1 \times \frac{r_{\text{2nd year}}}{100} \)
\( 4,950 = A_1 \left[\left(1 + \frac{15}{100}\right)^1 - 1\right] \)
\( 4,950 = A_1 \left(\frac{3}{20}\right) \)

Solving for \( A_1 \):
\( A_1 = \frac{4,950 \times 20}{3} \)
\( A_1 = \text{Rs. } 33,000 \)

Since the amount at the end of the first year (\( A_1 \)) is Rs. 33,000, we can now find the original principal \( P \) using the first year's rate of \( 10\% \):
\( A_1 = P \left(1 + \frac{r_1}{100}\right) \)
\( 33,000 = P \left(1 + \frac{10}{100}\right) \)
\( 33,000 = P \left(\frac{11}{10}\right) \)

Solving for \( P \):
\( P = \frac{33,000 \times 10}{11} \)
\( P = \text{Rs. } 30,000 \)
In simple words: The interest in the second year is Rs. 4,950 at a 15% rate, which means the money at the start of the second year was Rs. 33,000. Since this money grew by 10% in the first year, the starting investment was Rs. 30,000.

Exam Tip: Remember that interest for any specific year is calculated on the principal of *that* year, which is the total amount accumulated by the end of the preceding year.

Question 7. A sum of money is invested at 10% per annum compounded half-yearly. If the difference of amounts at the end of 6 months and 12 months is Rs. 189, find the sum of money invested.
Answer:
Let the invested sum of money be Rs. \( y \), and the interest rate is \( 10\% \) per annum compounded semi-annually.
Amount after the first 6 months (one half-year period, so \( n = \frac{1}{2} \) year):
\( A_1 = P \left(1 + \frac{r}{2 \times 100}\right)^{n \times 2} \)
\( A_1 = y \left(1 + \frac{10}{200}\right)^{\frac{1}{2} \times 2} \)
\( A_1 = y \left(1 + \frac{1}{20}\right)^1 \)
\( A_1 = \frac{21}{20}y \)

Amount after the first 12 months (one full year, meaning two half-year periods, so \( n = 1 \) year):
\( A_2 = y \left(1 + \frac{10}{200}\right)^{1 \times 2} \)
\( A_2 = y \left(1 + \frac{1}{20}\right)^2 \)
\( A_2 = \frac{441}{400}y \)

We are given that the difference between these two amounts is Rs. 189:
\( A_2 - A_1 = 189 \)
\( \frac{441}{400}y - \frac{21}{20}y = 189 \)

Subtracting the fractions by finding a common denominator:
\( \frac{441y - 420y}{400} = 189 \)
\( \frac{21y}{400} = 189 \)

Solving for \( y \):
\( y = \frac{189 \times 400}{21} \)
\( y = 9 \times 400 \)
\( y = \text{Rs. } 3,600 \)
In simple words: The money is compounded half-yearly, meaning it grows at 5% every six months. The difference in the total amount between 6 months and 12 months is Rs. 189, which corresponds to the interest earned in the second half-year, showing the starting sum was Rs. 3,600.

Exam Tip: When compounding half-yearly, make sure the power in the formula matches the number of half-year conversion periods (6 months is 1 period, 12 months is 2 periods).

Question 8. Rohit borrows Rs. 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit's profit in the transaction at the end of two years.
Answer:
Given:
Principal \( P = \text{Rs. } 86,000 \)
Time \( T = 2 \) years
Rate \( R = 5\% \) p.a.

First, let's find the Simple Interest (S.I.) that Rohit owes to Arun:
\( \text{S.I.} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{86,000 \times 5 \times 2}{100} \)
\( \text{S.I.} = \text{Rs. } 8,600 \)

Next, calculate the Compound Interest (C.I.) that Rohit receives from Akshay:
\( \text{C.I.} = P \left[\left(1 + \frac{r}{100}\right)^n - 1\right] \)
\( \text{C.I.} = 86,000 \left[\left(1 + \frac{5}{100}\right)^2 - 1\right] \)
\( \text{C.I.} = 86,000 \left[\left(\frac{21}{20}\right)^2 - 1\right] \)
\( \text{C.I.} = 86,000 \left[\frac{441}{400} - 1\right] \)
\( \text{C.I.} = 86,000 \left[\frac{41}{400}\right] \)
\( \text{C.I.} = 215 \times 41 \)
\( \text{C.I.} = \text{Rs. } 8,815 \)

The profit Rohit earns is the difference between the compound interest received and the simple interest paid:
\( \text{Profit} = \text{C.I.} - \text{S.I.} \)
\( \text{Profit} = 8,815 - 8,600 \)
\( \text{Profit} = \text{Rs. } 215 \)
In simple words: Rohit pays Rs. 8,600 in simple interest to Arun, but receives Rs. 8,815 in compound interest from Akshay. Since compound interest earns more, Rohit makes a net profit of Rs. 215.

Exam Tip: In 'borrowing-lending' questions, profit is always the difference between the compound interest earned and the simple interest paid.

Question 9. The simple interest on a certain sum of money for 3 years at 5% per annum is Rs. 1,200. Find the amount due and the compound interest on this sum of money at the same rate and after 2 years. Interest is reckoned annually.
Answer:
Let the principal sum of money be Rs. \( x \).
The interest rate is \( 5\% \) per year, simple interest is Rs. 1,200, and the duration is 3 years.
Using the simple interest formula to find the principal \( x \):
\( \text{S.I.} = \frac{x \times R \times T}{100} \)
\( 1,200 = \frac{x \times 5 \times 3}{100} \)
\( 1,200 = \frac{15x}{100} \)
\( x = \frac{1,200 \times 100}{15} \)
\( x = \text{Rs. } 8,000 \)

Now, calculate the final amount and compound interest on this same sum (Rs. 8,000) for a 2-year period:
Principal \( P = \text{Rs. } 8,000 \)
Rate \( r = 5\% \) p.a.
Time \( n = 2 \) years

\( A = P \left(1 + \frac{r}{100}\right)^n \)
\( A = 8,000 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 8,000 (1.05)^2 \)
\( A = 8,000 \times 1.1025 \)
\( A = \text{Rs. } 8,820 \)

Calculate the Compound Interest (C.I.):
\( \text{C.I.} = A - P \)
\( \text{C.I.} = 8,820 - 8,000 \)
\( \text{C.I.} = \text{Rs. } 820 \)
In simple words: We first find that the original sum of money is Rs. 8,000 using the simple interest. Then, we find that at 5% compound interest for 2 years, it grows to Rs. 8,820, yielding Rs. 820 in interest.

Exam Tip: Be careful with the time values in the question: it is 3 years for Simple Interest, but only 2 years for the Compound Interest part.

Question 10. Nikita invests Rs. 6,000 for two years at a certain rate of interest compounded annually. At the end of the first year, it amounts to Rs. 6,720. Calculate:
(i) the rate of interest,
(ii) the amount at the end of the second year.
Answer:
(i) For the first year, using the compound interest formula with \( n = 1 \):
\( A = P \left(1 + \frac{r}{100}\right)^n \)
\( 6,720 = 6,000 \left(1 + \frac{x}{100}\right)^1 \)
\( 6,720 = 6,000 + 60x \)

Subtracting 6,000 from both sides:
\( 720 = 60x \)
\( x = \frac{720}{60} \)
\( x = 12 \)

Thus, the annual rate of interest is 12%.

(ii) To find the total amount at the end of the second year:
\( A = P \left(1 + \frac{r}{100}\right)^2 \)
\( A = 6,000 \left(1 + \frac{12}{100}\right)^2 \)
\( A = 6,000 \left(\frac{112}{100}\right)^2 \)
\( A = 6,000 \times 1.2544 \)
\( A = \text{Rs. } 7,526.40 \)
In simple words: (i) The investment grows from Rs. 6,000 to Rs. 6,720 in the first year, which means the interest rate is 12%. (ii) Using this 12% rate for the second year, the total amount grows to Rs. 7,526.40.

Exam Tip: The amount at the end of the first year can also serve as the principal for the second year, so you can alternatively calculate: \( 6,720 \times 1.12 \) to find the second-year amount quickly.