OP Malhotra Class 9 Maths Solutions Chapter 2 Compound Interest Exercise 2 (C)

Question 1. What amount of money should Mohan invest in a bank in order to get ₹ 1323 in 2 years at 5% compounded annually?
Answer:
Amount after 2 years \( = \) ₹ 1323
Rate \( (R) = 5\% \) p.a.
We know the formula for amount: \( A = P \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( 1323 = P \left(1+\frac{5}{100}\right)^2 \)
\( \implies \) \( 1323 = P \left(\frac{105}{100}\right)^2 \)
\( \implies \) \( 1323 = P \left(\frac{21}{20}\right)^2 \)
\( \implies \) \( P = 1323 \times \left(\frac{20}{21}\right)^2 \)
\( \implies \) \( P = 1323 \times \frac{20}{21} \times \frac{20}{21} \)
\( \implies \) \( P = 3 \times 20 \times 20 \)
\( \implies \) \( P = 1200 \)
Therefore, Principal \( = \) ₹ 1200. This calculation shows that an initial investment of Rs. 1200 will yield the desired amount.
In simple words: You need to find the starting amount (principal) that, when increased by compound interest over two years, reaches a specific final amount of ₹ 1323.

🎯 Exam Tip: Always write down the known values and the formula first. Show each step of calculation clearly to avoid errors, especially when dealing with fractions and powers.

Question 2. Find the sum which amounts to ₹ 1352 in 2 years at 4% compound interest.
Answer:
Amount after 2 years \( = \) ₹ 1352
Rate \( (R) = 4\% \) p.a.
We know that \( A = P \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( 1352 = P \left(1+\frac{4}{100}\right)^2 \)
\( \implies \) \( 1352 = P \left(\frac{104}{100}\right)^2 \)
\( \implies \) \( 1352 = P \left(\frac{26}{25}\right)^2 \)
\( \implies \) \( P = 1352 \times \left(\frac{25}{26}\right)^2 \)
\( \implies \) \( P = 1352 \times \frac{25}{26} \times \frac{25}{26} \)
\( \implies \) \( P = 2 \times 25 \times 25 \)
\( \implies \) \( P = 1250 \)
Therefore, Principal \( = \) ₹ 1250. This is the initial sum that grew to ₹ 1352.
In simple words: We need to find the initial sum of money (principal) that grows to ₹ 1352 when compounded over two years at 4% interest.

🎯 Exam Tip: Remember to simplify the fraction inside the parenthesis \(\left(1+\frac{R}{100}\right)\) before raising it to the power, which can make calculations much easier.

Question 3. What principal will amount to ₹ 9768 in two years. If the rates of interest for the successive years are 10% p.a. and 11% p.a. respectively.
Answer:
Amount after 2 years \( = \) ₹ 9768
Rate of interest for the first year \( (r_1) = 10\% \) p.a.
and for the second year \( (r_2) = 11\% \) p.a.
We know that \( A = P \left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right) \)
\( \implies \) \( 9768 = P \left(1+\frac{10}{100}\right)\left(1+\frac{11}{100}\right) \)
\( \implies \) \( 9768 = P \left(\frac{110}{100}\right)\left(\frac{111}{100}\right) \)
\( \implies \) \( 9768 = P \left(\frac{11}{10}\right)\left(\frac{111}{100}\right) \)
\( \implies \) \( P = 9768 \times \frac{10}{11} \times \frac{100}{111} \)
\( \implies \) \( P = 888 \times \frac{100}{111} \)
\( \implies \) \( P = 8 \times 100 \)
\( \implies \) \( P = 8000 \)
Therefore, Principal \( = \) ₹ 8000. It's important to account for varying interest rates.
In simple words: When the interest rate changes each year, you apply the first year's rate to the principal, and then the second year's rate to the new amount, to find the final amount or work backward to find the principal.

🎯 Exam Tip: For successive interest rates, treat each year's interest calculation as separate, applying the new rate to the amount accumulated at the end of the previous year.

Question 4. On what sum of money does the difference between the simple interest and compound interest in 2 years at 5% per annum is Rs. 15?
Answer:
Let principal \( (P) = \) ₹ 100
Rate \( (R) = 5\% \) p.a.
Period \( (n) = 2 \) years
Difference between S.I. and C.I. \( = \) ₹ 15
First, calculate Simple Interest (S.I.):
\( S.I = \frac{PRT}{100} \)
\( \implies \) \( S.I = \frac{100 \times 5 \times 2}{100} \)
\( \implies \) \( S.I = \) ₹ 10
Next, calculate Compound Interest (C.I.):
\( A = P \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( A = 100 \left(1+\frac{5}{100}\right)^2 \)
\( \implies \) \( A = 100 \left(\frac{105}{100}\right)^2 \)
\( \implies \) \( A = 100 \left(\frac{21}{20}\right)^2 \)
\( \implies \) \( A = 100 \times \frac{21}{20} \times \frac{21}{20} \)
\( \implies \) \( A = \frac{441}{4} \)
So, \( C.I. = A - P \)
\( \implies \) \( C.I. = \frac{441}{4} - 100 \)
\( \implies \) \( C.I. = \frac{441 - 400}{4} \)
\( \implies \) \( C.I. = \frac{41}{4} \)
Now, find the difference between C.I. and S.I. for the assumed principal of ₹ 100:
Difference \( = C.I. - S.I. \)
\( \implies \) Difference \( = \frac{41}{4} - 10 \)
\( \implies \) Difference \( = \frac{41 - 40}{4} \)
\( \implies \) Difference \( = \frac{1}{4} \)
If the difference is \( \frac{1}{4} \), then the principal is ₹ 100.
If the difference is ₹ 1, then principal \( = \frac{100}{\frac{1}{4}} = 100 \times 4 = \) ₹ 400.
If the actual difference is ₹ 15, then principal \( = \frac{100 \times 4 \times 15}{1} \)
\( \implies \) Principal \( = \) ₹ 6000. It's a common practice to use an assumed principal for such problems.
In simple words: Compound interest grows faster than simple interest, and we're looking for the initial principal where the difference in earnings over two years at 5% is exactly ₹ 15.

🎯 Exam Tip: When the problem involves the difference between C.I. and S.I., assuming a principal (like ₹100) simplifies calculations, allowing you to find a scaling factor for the actual principal.

Question 5. The difference between simple and compound interest on the same sum of money at \( 6\frac{2}{3}\% \) for 3 years is 184. Determine the sum.
Answer:
Difference between C.I. and S.I. \( = \) ₹ 184
Let principal \( (Sum) = \) Rs. 100
Rate \( (R) = 6\frac{2}{3}\% = \frac{20}{3}\% \)
Period \( (n) = 3 \) years
Simple interest \( (S.I.) = \frac{PRT}{100} \)
\( \implies \) \( S.I. = \frac{100 \times \frac{20}{3} \times 3}{100} \)
\( \implies \) \( S.I. = \) ₹ 20
By compound interest:
\( A = P \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( A = 100 \left(1+\frac{20/3}{100}\right)^3 \)
\( \implies \) \( A = 100 \left(1+\frac{20}{300}\right)^3 \)
\( \implies \) \( A = 100 \left(1+\frac{1}{15}\right)^3 \)
\( \implies \) \( A = 100 \left(\frac{16}{15}\right)^3 \)
\( \implies \) \( A = 100 \times \frac{16 \times 16 \times 16}{15 \times 15 \times 15} \)
\( \implies \) \( A = 100 \times \frac{4096}{3375} \)
\( \implies \) \( A = \frac{409600}{3375} = \frac{16384}{135} \)
Now, \( C.I. = A - P \)
\( \implies \) \( C.I. = \frac{16384}{135} - 100 \)
\( \implies \) \( C.I. = \frac{16384 - 13500}{135} \)
\( \implies \) \( C.I. = \frac{2884}{135} \)
Difference between C.I. and S.I. for the assumed principal of ₹ 100:
Difference \( = C.I. - S.I. \)
\( \implies \) Difference \( = \frac{2884}{135} - 20 \)
\( \implies \) Difference \( = \frac{2884 - (20 \times 135)}{135} \)
\( \implies \) Difference \( = \frac{2884 - 2700}{135} \)
\( \implies \) Difference \( = \frac{184}{135} \)
If the difference is \( \frac{184}{135} \), then principal \( = \) ₹ 100.
If the difference is ₹ 1, then principal \( = \frac{100}{\frac{184}{135}} = \frac{100 \times 135}{184} \)
If the actual difference is ₹ 184, then principal \( = \frac{100 \times 135 \times 184}{184} \)
\( \implies \) Principal \( = \) ₹ 13500. This calculation requires careful fractional arithmetic.
In simple words: Similar to Question 4, but for three years, calculate both simple and compound interest using an assumed principal, then use their difference to find the initial sum for the given difference of ₹ 184.

🎯 Exam Tip: When the rate is a mixed fraction (e.g., \(6\frac{2}{3}\%\)), convert it to an improper fraction before using it in the formulas to simplify calculations and avoid errors.

Question 6. On what sum of money will the difference between the simple interest and the compound interest for 2 years at 5% per annum be equal to ₹50.
Answer:
Let sum \( (P) = \) ₹ 100
Rate \( (R) = 5\% \) p.a.
Period \( (n) = 2 \) years
Simple Interest \( (S.I.) = \frac{PRT}{100} \)
\( \implies \) \( S.I. = \frac{100 \times 5 \times 2}{100} \)
\( \implies \) \( S.I. = \) ₹ 10
By compound interest:
\( A = P \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( A = 100 \left(1+\frac{5}{100}\right)^2 \)
\( \implies \) \( A = 100 \left(\frac{105}{100}\right)^2 \)
\( \implies \) \( A = 100 \left(\frac{21}{20}\right)^2 \)
\( \implies \) \( A = 100 \times \frac{21}{20} \times \frac{21}{20} \)
\( \implies \) \( A = \frac{441}{4} \)
So, \( C.I. = A - P \)
\( \implies \) \( C.I. = \frac{441}{4} - 100 \)
\( \implies \) \( C.I. = \frac{441 - 400}{4} \)
\( \implies \) \( C.I. = \frac{41}{4} \)
Difference between C.I. and S.I. for assumed principal of ₹ 100:
Difference \( = C.I. - S.I. \)
\( \implies \) Difference \( = \frac{41}{4} - 10 \)
\( \implies \) Difference \( = \frac{41 - 40}{4} \)
\( \implies \) Difference \( = \frac{1}{4} \)
If the difference is \( \frac{1}{4} \), then principal \( = \) ₹ 100.
If the difference is ₹ 1, then principal \( = \frac{100}{\frac{1}{4}} = 100 \times 4 = \) ₹ 400.
If the actual difference is ₹ 50, then principal \( = \frac{100 \times 4 \times 50}{1} \)
\( \implies \) Principal \( = \) ₹ 20,000. It's crucial to set up the proportionality correctly.
In simple words: We are looking for the original amount of money where the compound interest earned over two years is ₹50 more than the simple interest earned over the same period.

🎯 Exam Tip: When using the unitary method to find the actual principal, ensure the ratio of difference to principal is consistent. For example, if a difference of 'x' corresponds to principal 'P', then a difference of 'y' corresponds to \(P \times \frac{y}{x}\).

Question 7. Find the rate of interest per annum, if compounded yearly.
(i) Principal \( = \) ₹ 196, Amount \( = \) ₹ 225, time \( = 2 \) years
(ii) Principal \( = \) ₹ 3136, Compound interest \( = \) ₹ 345, Time \( = 2 \) years
Answer:
(i) Principal \( (P) = \) ₹ 196
Amount \( (A) = \) Rs. 225
Time \( (n) = 2 \) years
We know that, \( \frac{A}{P} = \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( \frac{225}{196} = \left(1+\frac{R}{100}\right)^2 \)
\( \implies \) \( \left(\frac{15}{14}\right)^2 = \left(1+\frac{R}{100}\right)^2 \)
Comparing both sides,
\( \implies \) \( 1+\frac{R}{100} = \frac{15}{14} \)
\( \implies \) \( \frac{R}{100} = \frac{15}{14} - 1 \)
\( \implies \) \( \frac{R}{100} = \frac{15 - 14}{14} \)
\( \implies \) \( \frac{R}{100} = \frac{1}{14} \)
\( \implies \) \( R = \frac{100}{14} = \frac{50}{7} \)
\( \implies \) Rate \( = 7\frac{1}{7}\% \) p.a.

(ii) Principal \( (P) = \) ₹ 3136
Compound Interest \( (C.I.) = \) ₹ 345
Time \( (n) = 2 \) years
Amount \( (A) = P + C.I. \)
\( \implies \) \( A = 3136 + 345 \)
\( \implies \) \( A = \) ₹ 3481
We know that, \( \frac{A}{P} = \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( \frac{3481}{3136} = \left(1+\frac{R}{100}\right)^2 \)
\( \implies \) \( \left(\frac{59}{56}\right)^2 = \left(1+\frac{R}{100}\right)^2 \)
Comparing both sides,
\( \implies \) \( 1+\frac{R}{100} = \frac{59}{56} \)
\( \implies \) \( \frac{R}{100} = \frac{59}{56} - 1 \)
\( \implies \) \( \frac{R}{100} = \frac{59 - 56}{56} \)
\( \implies \) \( \frac{R}{100} = \frac{3}{56} \)
\( \implies \) \( R = \frac{100 \times 3}{56} = \frac{300}{56} = \frac{75}{14} \)
\( \implies \) Rate \( = 5\frac{5}{14}\% \) p.a. Finding the square root of fractions like 3481/3136 requires recognizing perfect squares.
In simple words: When you know the starting amount, the ending amount, and the time, you can work backwards using the compound interest formula to find the annual interest rate.

🎯 Exam Tip: For problems involving square roots, try to identify perfect squares in the numerator and denominator (e.g., 225 = 15\(^2\), 196 = 14\(^2\)) to simplify calculations quickly.

Question 8. Hari purchased Relief Bonds for ₹ 1000, a sum which will fetch him ₹ 2000 after 5 years. Find the rate of interest if the interest is compounded half-yearly. (Given that \( \sqrt[10]{2} = 1.072 \))
Answer:
Principal \( (P) = \) ₹ 1000
Amount \( (A) = \) ₹ 2000
Period \( (n) = 5 \) years \( = 10 \) half years (since compounded half-yearly)
We know that, \( \frac{A}{P} = \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( \frac{2000}{1000} = \left(1+\frac{R}{100}\right)^{10} \)
\( \implies \) \( \frac{2}{1} = \left(1+\frac{R}{100}\right)^{10} \)
\( \implies \) \( 2 = \left(1+\frac{R}{100}\right)^{10} \)
Take the 10th root on both sides:
\( \implies \) \( \sqrt[10]{2} = 1+\frac{R}{100} \)
Given \( \sqrt[10]{2} = 1.072 \)
\( \implies \) \( 1.072 = 1+\frac{R}{100} \)
\( \implies \) \( \frac{R}{100} = 1.072 - 1.000 \)
\( \implies \) \( \frac{R}{100} = 0.072 \)
\( \implies \) \( R = 0.072 \times 100 \)
\( \implies \) \( R = 7.2 \)
Rate half-yearly \( = 7.2\% \)
And rate annually \( = 7.2 \times 2 = 14.4\% \) p.a. This demonstrates the conversion from half-yearly to annual rate.
In simple words: When interest is compounded half-yearly, the period doubles and the annual rate is halved before applying the compound interest formula to find the interest rate.

🎯 Exam Tip: For half-yearly compounding, remember to double the number of years for 'n' and divide the annual rate by two for 'R' in the formula. Use the provided roots/log values to simplify calculations.

Question 9. ₹ 8000 became ₹ 9261 in a certain interval of time at the rate of 5% per annum C.I. Find the time.
Answer:
Principal \( (P) = \) ₹ 8000
Amount \( (A) = \) ₹ 9261
Rate \( (R) = 5\% \) p.a.
We know that, \( \frac{A}{P} = \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( \frac{9261}{8000} = \left(1+\frac{5}{100}\right)^n \)
\( \implies \) \( \frac{9261}{8000} = \left(\frac{105}{100}\right)^n \)
\( \implies \) \( \frac{9261}{8000} = \left(\frac{21}{20}\right)^n \)
Recognize that \( 9261 = 21^3 \) and \( 8000 = 20^3 \):
\( \implies \) \( \left(\frac{21}{20}\right)^3 = \left(\frac{21}{20}\right)^n \)
Comparing powers on both sides, we get \( n = 3 \). This step is key for finding the time.
Therefore, Period \( = 3 \) years.
In simple words: You're trying to figure out how many years it takes for an initial amount of ₹ 8000 to grow to ₹ 9261 at a 5% compound interest rate.

🎯 Exam Tip: When solving for time 'n', try to express both sides of the equation with the same base. This often involves recognizing perfect squares or cubes, which are common in these types of problems.

Question 10. In how many years will a sum of ₹ 3000 at 20% per annum compounded semi-annually become 3993.
Answer:
Principal \( (P) = \) ₹ 3000
Amount \( (A) = \) ₹ 3993
Rate \( (R) = 20\% \) p.a. or \( 10\% \) half-yearly (since compounded semi-annually)
We know that, \( \frac{A}{P} = \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( \frac{3993}{3000} = \left(1+\frac{10}{100}\right)^n \)
\( \implies \) \( \frac{1331}{1000} = \left(\frac{110}{100}\right)^n \)
\( \implies \) \( \frac{1331}{1000} = \left(\frac{11}{10}\right)^n \)
Recognize that \( 1331 = 11^3 \) and \( 1000 = 10^3 \):
\( \implies \) \( \left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^n \)
Comparing powers on both sides, we get \( n = 3 \). This 'n' represents half-year periods.
Therefore, Period \( = 3 \) half years or \( 1\frac{1}{2} \) years.
In simple words: For semi-annual compounding, the interest period is half a year, and the rate is halved, so you're finding the number of these half-year periods required for the investment to grow.

🎯 Exam Tip: Be careful to convert the annual rate and time period correctly when interest is compounded semi-annually (rate is halved, time period 'n' is doubled in terms of compounding intervals).

Question 11. A sum of money put out at compound interest amounts in 2 years to ₹ 578.40 and in 3 years to 614.55. Find the rate of interest.
Answer:
Amount for 3 years \( = \) ₹ 614.55
Amount for 2 years \( = \) ₹ 578.40
Subtracting the amount for 2 years from the amount for 3 years gives the interest earned in the 3rd year.
Interest for 1 year \( = 614.55 - 578.40 \)
\( \implies \) Interest for 1 year \( = \) ₹ 36.15
This ₹ 36.15 is the interest earned on ₹ 578.40 for 1 year (from the end of year 2 to the end of year 3).
We can use the simple interest formula to find the rate for this single year:
\( Rate = \frac{\text{Simple interest} \times 100}{P \times \text{Time}} \)
Here, the principal \( (P) \) for this 1 year period is the amount at the end of 2 years, which is ₹ 578.40.
\( \implies \) \( Rate = \frac{36.15 \times 100}{578.40 \times 1} \)
\( \implies \) \( Rate = \frac{3615}{578.40} \)
\( \implies \) \( Rate = \frac{36150}{5784} \)
\( \implies \) \( Rate = \frac{25}{4}\% \)
\( \implies \) Rate \( = 6\frac{1}{4}\% \) p.a. Understanding this relationship between successive amounts is crucial.
In simple words: When you know the amounts after different periods, the interest earned in one year can be found by subtracting the previous year's amount from the current year's amount. This interest is then used with the previous year's amount as the principal to calculate the rate.

🎯 Exam Tip: In compound interest, the amount at the end of a previous year acts as the principal for the next year's interest calculation. Use this fact to find the interest for a single year between given amounts.

Question 12. A sum compounded annually becomes \( \frac{25}{16} \) times of itself in 2 years. Determine the rate of interest per annum?
Answer:
Let the principal be \( P \).
Then Amount \( (A) = P \times \frac{25}{16} \)
Period \( (n) = 2 \) years
Let \( R \) be the rate of interest per annum.
We know that, \( \frac{A}{P} = \left(1+\frac{R}{100}\right)^n \)
\( \implies \) \( \frac{P \times \frac{25}{16}}{P} = \left(1+\frac{R}{100}\right)^2 \)
\( \implies \) \( \frac{25}{16} = \left(1+\frac{R}{100}\right)^2 \)
\( \implies \) \( \left(\frac{5}{4}\right)^2 = \left(1+\frac{R}{100}\right)^2 \)
Comparing both sides,
\( \implies \) \( 1+\frac{R}{100} = \frac{5}{4} \)
\( \implies \) \( \frac{R}{100} = \frac{5}{4} - 1 \)
\( \implies \) \( \frac{R}{100} = \frac{5-4}{4} \)
\( \implies \) \( \frac{R}{100} = \frac{1}{4} \)
\( \implies \) \( R = \frac{1}{4} \times 100 \)
\( \implies \) \( R = 25 \)
Therefore, Rate \( = 25\% \) p.a. This demonstrates a direct way to find the rate.
In simple words: If you know how many times the principal grows in a certain number of years, you can use the compound interest formula to set up an equation and solve for the annual interest rate.

🎯 Exam Tip: When the amount is given as a multiple of the principal (e.g., \(\frac{25}{16}\) times itself), you can directly substitute \( \frac{A}{P} \) in the formula and solve for R, which simplifies the process.