OP Malhotra Class 9 Maths Solutions Chapter 2 Compound Interest Chapter Test

Question 1. Nikita invests Rs. 6,000 for two years at a certain rate of interest compounded annually. At end of 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:
Given: Principal for the first year \( (P_1) = \text{Rs. } 6,000 \)
Amount at the end of the first year \( (A_1) = \text{Rs. } 6,720 \)
Period \( (T) = 1 \) year

Interest for the first year \( = A_1 - P_1 = \text{Rs. } (6,720 - 6,000) = \text{Rs. } 720 \)

(i) To find the rate of interest:
We use the formula: \( \text{Rate} = \frac{\text{Interest} \times 100}{\text{Principal} \times \text{Time}} \)
\( \text{Rate} = \frac{720 \times 100}{6000 \times 1} \)
\( \text{Rate} = \frac{72000}{6000} \)
\( \text{Rate} = 12\% \)

So, the rate of interest is \( 12\% \) per annum.

(ii) To find the amount at the end of the second year:
For the second year, the principal \( (P_2) \) is the amount at the end of the first year, which is \( A_1 = \text{Rs. } 6,720 \).
The rate of interest \( (R) \) is \( 12\% \).
The time \( (T) \) for the second year is \( 1 \) year.

Interest for the second year \( = \frac{P_2 \times R \times T}{100} \)
\( = \frac{6720 \times 12 \times 1}{100} \)
\( = \text{Rs. } 806.40 \)

Amount at the end of the second year \( = P_2 + \text{Interest for the second year} \)
\( = \text{Rs. } (6,720 + 806.40) \)
\( = \text{Rs. } 7,526.40 \)
In simple words: First, we found out how much extra money Nikita earned in the first year. Then, we used that to figure out the yearly interest rate. For the second year, that new total amount became the starting money, and we added the interest for that year to get the final total.

🎯 Exam Tip: Remember that in compound interest, the principal changes each year as interest is added. The amount from the previous period becomes the new principal for the next period.

Question 2. 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, calculate the simple interest Rohit pays to Arun:
\( \text{Simple Interest (SI)} = \frac{P \times R \times T}{100} \)
\( = \frac{86000 \times 5 \times 2}{100} \)
\( = \text{Rs. } 8,600 \)

Next, calculate the compound interest Rohit receives from Akshay:
Amount \( (A) = P \left(1 + \frac{R}{100}\right)^n \)
\( A = 86000 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 86000 \left(1 + \frac{1}{20}\right)^2 \)
\( A = 86000 \left(\frac{21}{20}\right)^2 \)
\( A = 86000 \times \frac{21}{20} \times \frac{21}{20} \)
\( A = \text{Rs. } 94,815 \)

Compound Interest \( (CI) = A - P \)
\( CI = 94815 - 86000 \)
\( CI = \text{Rs. } 8,815 \)

Finally, calculate Rohit's profit:
Profit \( = \text{Compound Interest Received} - \text{Simple Interest Paid} \)
\( = \text{Rs. } (8815 - 8600) \)
\( = \text{Rs. } 215 \)
In simple words: Rohit borrowed money with simple interest and lent it out with compound interest. He earned more from compound interest than he paid in simple interest, so he made a small profit. This shows how compound interest helps money grow faster.

🎯 Exam Tip: Always clearly separate the simple interest and compound interest calculations. Make sure to use the correct formulas and apply the principal for compound interest carefully over the years.

Question 3. Mr. Kumar borrowed Rs. 15,000 for two years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays 6,200 at the end of the first year, find the outstanding amount at the end of the second year.

Answer:
Given: Principal for the first year \( (P_1) = \text{Rs. } 15,000 \)
Rate for the first year \( (R_1) = 8\% \)
Rate for the second year \( (R_2) = 10\% \)
Repayment after 1st year \( = \text{Rs. } 6,200 \)

For the 1st year:
Simple Interest \( (SI_1) = \frac{P_1 \times R_1 \times 1}{100} \)
\( = \frac{15000 \times 8 \times 1}{100} \)
\( = \text{Rs. } 1,200 \)

Amount at the end of the 1st year \( (A_1) = P_1 + SI_1 \)
\( = 15000 + 1200 \)
\( = \text{Rs. } 16,200 \)

Remaining amount after repayment at the end of the 1st year:
\( = A_1 - \text{Repayment} \)
\( = 16200 - 6200 \)
\( = \text{Rs. } 10,000 \)

For the 2nd year:
Principal for the 2nd year \( (P_2) = \text{Rs. } 10,000 \)
Rate for the 2nd year \( (R_2) = 10\% \)

Simple Interest \( (SI_2) = \frac{P_2 \times R_2 \times 1}{100} \)
\( = \frac{10000 \times 10 \times 1}{100} \)
\( = \text{Rs. } 1,000 \)

Outstanding amount at the end of the 2nd year \( = P_2 + SI_2 \)
\( = 10000 + 1000 \)
\( = \text{Rs. } 11,000 \)
In simple words: First, we calculated the money owed after the first year, adding the interest. Then, Mr. Kumar paid some money back, reducing his debt. We then calculated the interest for the second year on this smaller debt and added it to find the total outstanding amount.

🎯 Exam Tip: When there are varying interest rates or partial repayments, it's crucial to calculate the amount year by year. The amount at the end of one year becomes the principal for the next year, adjusted for any payments made.

Question 4. In what period of time will Rs. 12,000 yield Rs. 3,972 as compound interest at 10% per annum, if compounded on an yearly basis?

Answer:
Given: Principal \( (P) = \text{Rs. } 12,000 \)
Compound Interest \( (CI) = \text{Rs. } 3,972 \)
Rate \( (R) = 10\% \) p.a.

First, calculate the Amount \( (A) \):
\( A = P + CI \)
\( A = 12000 + 3972 \)
\( A = \text{Rs. } 15,972 \)

Now, use the compound interest formula to find the time \( (n) \):
\( A = P \left(1 + \frac{R}{100}\right)^n \)
\( 15972 = 12000 \left(1 + \frac{10}{100}\right)^n \)
\( \frac{15972}{12000} = \left(1 + \frac{1}{10}\right)^n \)
\( \frac{15972}{12000} = \left(\frac{11}{10}\right)^n \)

Simplify the fraction on the left:
Dividing both by 12, we get:
\( \frac{1331}{1000} = \left(\frac{11}{10}\right)^n \)

We know that \( 1331 = 11^3 \) and \( 1000 = 10^3 \).
So, \( \left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^n \)

By comparing the powers, we find:
\( n = 3 \)

Therefore, the time period is 3 years.
In simple words: We first added the principal and compound interest to get the total amount. Then, we used the compound interest formula and simplified it to find how many times the money grew by the given rate. This helped us find the number of years.

🎯 Exam Tip: When solving for 'n' (time period) in compound interest problems, simplifying the fraction \( \frac{A}{P} \) to a power of \( \left(1 + \frac{R}{100}\right) \) is key. Knowing common powers of small numbers (like \( 11^3=1331 \) and \( 10^3=1000 \)) helps speed up the calculation.

Question 5. On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to Rs. 25 if the rate of interest charged for both is 5%?

Answer:
Given: Difference between CI and SI \( = \text{Rs. } 25 \)
Rate \( (R) = 5\% \) p.a.
Time \( (n) = 2 \) years

Let's assume the Principal \( (P) = \text{Rs. } 100 \).

Calculate Simple Interest \( (SI) \) for \( P = \text{Rs. } 100 \):
\( SI = \frac{P \times R \times n}{100} \)
\( SI = \frac{100 \times 5 \times 2}{100} \)
\( SI = \text{Rs. } 10 \)

Calculate Compound Interest \( (CI) \) for \( P = \text{Rs. } 100 \):
Amount \( (A) = P \left(1 + \frac{R}{100}\right)^n \)
\( A = 100 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 100 \left(1 + \frac{1}{20}\right)^2 \)
\( A = 100 \left(\frac{21}{20}\right)^2 \)
\( A = 100 \times \frac{21}{20} \times \frac{21}{20} \)
\( A = \frac{441}{4} \)
\( A = \text{Rs. } 110.25 \)

\( CI = A - P \)
\( CI = 110.25 - 100 \)
\( CI = \text{Rs. } 10.25 \)

Now, find the difference between CI and SI for the assumed principal of Rs. 100:
Difference \( = CI - SI \)
\( = 10.25 - 10 \)
\( = \text{Rs. } 0.25 \) (or \( \frac{1}{4} \))

We know that if the difference is Rs. \( 0.25 \), the principal is Rs. \( 100 \).
So, if the difference is Rs. \( 1 \), the principal will be \( \frac{100}{0.25} = 100 \times 4 = \text{Rs. } 400 \).

If the difference is Rs. \( 25 \), the principal will be:
\( = 400 \times 25 \)
\( = \text{Rs. } 10,000 \)
In simple words: We guessed a starting amount of Rs. 100 and calculated how much more compound interest would be than simple interest for that amount. Then, using simple math, we figured out what the original amount must have been for the actual difference to be Rs. 25.

🎯 Exam Tip: For problems involving the difference between CI and SI, assuming a principal (like Rs. 100) often simplifies calculations. After finding the difference for the assumed principal, use direct proportionality to find the actual principal for the given difference.

Question 6. A sum of Rs. 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become
(a) Rs. 48,000
(b) Rs. 96,000
(c) Rs. 1,90,000
(d) Rs. 1,92,000
Answer: (d) Rs. 1,92,000
Let the initial Principal \( (P) = \text{Rs. } 12,000 \).

We are given that the sum doubles in 5 years.
So, after 5 years, Amount \( A = 2 \times P = 2 \times 12000 = \text{Rs. } 24,000 \).

Using the compound interest formula: \( A = P \left(1 + \frac{R}{100}\right)^n \)
\( 2P = P \left(1 + \frac{R}{100}\right)^5 \)
\( 2 = \left(1 + \frac{R}{100}\right)^5 \)

We need to find the amount after 20 years.
We can write 20 years as \( 4 \times 5 \) years.

Amount after 20 years \( = P \left(1 + \frac{R}{100}\right)^{20} \)
\( = P \left[\left(1 + \frac{R}{100}\right)^5\right]^4 \)

Substitute the value \( \left(1 + \frac{R}{100}\right)^5 = 2 \):
Amount after 20 years \( = P \times (2)^4 \)
\( = 12000 \times 16 \)
\( = \text{Rs. } 1,92,000 \)
In simple words: The money doubles every 5 years. Since 20 years is four times 5 years, the money will double four times. So, the original amount gets multiplied by 2, four times, which is \( 2 \times 2 \times 2 \times 2 = 16 \) times the starting amount.

🎯 Exam Tip: For problems where the amount multiplies by a fixed factor over a specific period, you can find the amount after a longer period by raising that factor to the power of how many such periods fit into the total time.

Question 7. The difference between compound interest (C.I.) and simple interest (S.I.) on a sum of Rs. 4000 at 5% p.a. for 2 years is:
(a) Rs. 20
(b) Rs. 10
(c) Rs. 50
(d) Rs. 60
Answer: (b) Rs. 10
Given: Principal \( (P) = \text{Rs. } 4,000 \)
Rate \( (R) = 5\% \) p.a.
Time \( (n) = 2 \) years

First, calculate Simple Interest \( (SI) \):
\( SI = \frac{P \times R \times n}{100} \)
\( SI = \frac{4000 \times 5 \times 2}{100} \)
\( SI = \text{Rs. } 400 \)

Next, calculate Compound Interest \( (CI) \):
Amount \( (A) = P \left(1 + \frac{R}{100}\right)^n \)
\( A = 4000 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 4000 \left(1 + \frac{1}{20}\right)^2 \)
\( A = 4000 \left(\frac{21}{20}\right)^2 \)
\( A = 4000 \times \frac{21}{20} \times \frac{21}{20} \)
\( A = 10 \times 21 \times 21 \)
\( A = 10 \times 441 \)
\( A = \text{Rs. } 4,410 \)

\( CI = A - P \)
\( CI = 4410 - 4000 \)
\( CI = \text{Rs. } 410 \)

Finally, find the difference between CI and SI:
Difference \( = CI - SI \)
\( = 410 - 400 \)
\( = \text{Rs. } 10 \)
In simple words: We calculated how much money would be earned with simple interest and how much with compound interest for the given amount and time. The difference between these two earnings is the answer. Compound interest always grows a little faster.

🎯 Exam Tip: Remember the shortcut for the difference between CI and SI for 2 years: \( D = P \left(\frac{R}{100}\right)^2 \). For this question, \( D = 4000 \left(\frac{5}{100}\right)^2 = 4000 \left(\frac{1}{20}\right)^2 = 4000 \times \frac{1}{400} = \text{Rs. } 10 \). This can save time.

Question 8. If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is Rs. 31, the sum is
(a) Rs. 500
(b) Rs. 50
(c) Rs. 1000
Answer: (c) Rs. 1000
Given: Difference between CI and SI \( = \text{Rs. } 31 \)
Rate \( (R) = 10\% \) p.a.
Time \( (n) = 3 \) years

Let's assume the Principal \( (P) = \text{Rs. } 100 \).

Calculate Simple Interest \( (SI) \) for \( P = \text{Rs. } 100 \):
\( SI = \frac{P \times R \times n}{100} \)
\( SI = \frac{100 \times 10 \times 3}{100} \)
\( SI = \text{Rs. } 30 \)

Calculate Compound Interest \( (CI) \) for \( P = \text{Rs. } 100 \):
Amount \( (A) = P \left(1 + \frac{R}{100}\right)^n \)
\( A = 100 \left(1 + \frac{10}{100}\right)^3 \)
\( A = 100 \left(\frac{11}{10}\right)^3 \)
\( A = 100 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \)
\( A = \frac{1331}{10} \)
\( A = \text{Rs. } 133.10 \)

\( CI = A - P \)
\( CI = 133.10 - 100 \)
\( CI = \text{Rs. } 33.10 \)

Now, find the difference between CI and SI for the assumed principal of Rs. 100:
Difference \( = CI - SI \)
\( = 33.10 - 30 \)
\( = \text{Rs. } 3.10 \) (or \( \frac{31}{10} \))

We know that if the difference is Rs. \( 3.10 \), the principal is Rs. \( 100 \).
So, if the difference is Rs. \( 1 \), the principal will be \( \frac{100}{3.10} = \frac{100}{\frac{31}{10}} = \frac{100 \times 10}{31} = \frac{1000}{31} \).

If the given difference is Rs. \( 31 \), the actual principal will be:
\( = \frac{1000}{31} \times 31 \)
\( = \text{Rs. } 1,000 \)
In simple words: We pretended the starting money was Rs. 100 and calculated how much the compound interest was more than the simple interest. Since the actual difference was Rs. 31, and our calculation showed a difference of Rs. 3.10 for Rs. 100, we simply multiplied our starting amount (Rs. 100) by 10 to get the correct starting money (Rs. 1000).

🎯 Exam Tip: When dealing with the difference between CI and SI for 3 years, the assumed principal method is very effective. Ensure careful calculation of both SI and CI for the assumed principal to avoid errors in proportionality.

Question 9. On what sum of money will the compound interest for 2 years at 5% per annum be Rs. 164?
(a) 1600
(b) Rs. 1500
(d) Rs. 1700
Answer: (a) 1600
Given: Compound Interest \( (CI) = \text{Rs. } 164 \)
Rate \( (R) = 5\% \) p.a.
Time \( (n) = 2 \) years

Let's assume the Principal \( (P) = \text{Rs. } 100 \).

Calculate the Amount \( (A) \) for \( P = \text{Rs. } 100 \):
\( A = P \left(1 + \frac{R}{100}\right)^n \)
\( A = 100 \left(1 + \frac{5}{100}\right)^2 \)
\( A = 100 \left(1 + \frac{1}{20}\right)^2 \)
\( A = 100 \left(\frac{21}{20}\right)^2 \)
\( A = 100 \times \frac{21}{20} \times \frac{21}{20} \)
\( A = \frac{441}{4} \)
\( A = \text{Rs. } 110.25 \)

Calculate Compound Interest \( (CI) \) for \( P = \text{Rs. } 100 \):
\( CI = A - P \)
\( CI = 110.25 - 100 \)
\( CI = \text{Rs. } 10.25 \) (or \( \frac{41}{4} \))

We know that if the CI is Rs. \( 10.25 \), the principal is Rs. \( 100 \).
So, if the CI is Rs. \( 1 \), the principal will be \( \frac{100}{10.25} = \frac{100}{\frac{41}{4}} = \frac{100 \times 4}{41} = \frac{400}{41} \).

If the given CI is Rs. \( 164 \), the actual principal will be:
\( = \frac{400}{41} \times 164 \)
\( = 400 \times 4 \) (since \( 164 \div 41 = 4 \))
\( = \text{Rs. } 1,600 \)
In simple words: We started by assuming a principal amount of Rs. 100 and calculated the compound interest for it. Then, we used a simple ratio: if a certain interest comes from Rs. 100, what principal would give us the actual interest of Rs. 164?

🎯 Exam Tip: When given the compound interest and asked to find the principal, using an assumed principal (like Rs. 100) and then applying proportionality is a reliable and structured method. Ensure all calculations for the assumed principal are accurate.

Question 10. A sum of money becomes eight times in 3 years. If the rate is compounded annually, in how much time will the same amount at the same compound rate becomes sixteen times?
(a) 6 years
(b) 4 years
(c) 8 years
(d) 5 years
Answer: (b) 4 years
Let the principal be \( P \).
The amount becomes 8 times the principal in 3 years.
So, \( A = 8P \) when \( n = 3 \) years.

Using the compound interest formula: \( A = P \left(1 + \frac{R}{100}\right)^n \)
\( 8P = P \left(1 + \frac{R}{100}\right)^3 \)
\( 8 = \left(1 + \frac{R}{100}\right)^3 \)

We know that \( 8 = 2^3 \).
So, \( 2^3 = \left(1 + \frac{R}{100}\right)^3 \)
This means \( 1 + \frac{R}{100} = 2 \) ... (Equation 1)

Now, we want to find the time \( n \) when the amount becomes sixteen times the principal:
\( A = 16P \)
Using the formula again:
\( 16P = P \left(1 + \frac{R}{100}\right)^n \)
\( 16 = \left(1 + \frac{R}{100}\right)^n \)

Substitute the value of \( \left(1 + \frac{R}{100}\right) \) from Equation 1:
\( 16 = (2)^n \)

We know that \( 16 = 2^4 \).
So, \( 2^4 = 2^n \)

By comparing the powers, we find:
\( n = 4 \)

Therefore, the sum will become sixteen times in 4 years.
In simple words: If money grows 8 times in 3 years, it means the growth factor per year is such that multiplying it three times gives 8. We found that this factor is 2. So, the money doubles each year. To become 16 times, it needs to double four times \( (2 \times 2 \times 2 \times 2 = 16) \), which will take 4 years.

🎯 Exam Tip: For problems involving money multiplying by powers (doubling, tripling, etc.), it's often easiest to find the base multiplication factor over the given time. Then, express the target multiplication factor as a power of this base factor to quickly determine the new time period.