Compound Interest

Answer: Option B
Solution:
To solve this, we need to calculate the compound interest for each deposit separately, as each deposit will accrue interest for different periods within the year.

Given:
– The deposit amount is Rs. 1600 on 1 January and Rs. 1600 on 1 July.
– The interest rate is 5% per annum, compounded half-yearly.
– For half-yearly compounding, the interest rate for each half-year is:

\[
\text{Interest rate per half-year} = \frac{5\%}{2} = 2.5\% = 0.025
\]

Step 1: Calculate interest for the deposit made on 1 January
This deposit will compound for two half-year periods (from 1 January to 31 December).

The compound interest formula is:

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

where:
– \( A \) is the amount after interest,
– \( P \) is the principal (initial deposit),
– \( r \) is the rate of interest per period,
– \( n \) is the number of periods.

For the deposit on 1 January:
– \( P = 1600 \),
– \( r = 2.5\% = 0.025 \),
– \( n = 2 \) (since it compounds for two half-year periods).

\[
A_1 = 1600 \left(1 + 0.025\right)^2
\]

\[
A_1 = 1600 \times (1.025)^2 = 1600 \times 1.050625 = 1681
\]

So, the amount at the end of the year for the 1 January deposit is Rs. 1681.

Step 2: Calculate interest for the deposit made on 1 July
This deposit will compound for one half-year period (from 1 July to 31 December).

For the deposit on 1 July:
– \( P = 1600 \),
– \( r = 2.5\% = 0.025 \),
– \( n = 1 \) (since it compounds for one half-year period).

\[
A_2 = 1600 \left(1 + 0.025\right)^1
\]

\[
A_2 = 1600 \times 1.025 = 1640
\]

So, the amount at the end of the year for the 1 July deposit is Rs. 1640.

Step 3: Calculate the total interest earned
The total amount at the end of the year is:

\[
\text{Total Amount} = A_1 + A_2 = 1681 + 1640 = 3321
\]

The total interest earned is the total amount minus the total principal deposited:

\[
\text{Total Principal} = 1600 + 1600 = 3200
\]

\[
\text{Total Interest} = 3321 – 3200 = 121
\]

Final Answer:
The total interest earned at the end of the year is Rs. 121.

Answer: Option A
Solution:
We are given the difference between the simple interest (SI) and compound interest (CI) on a certain sum of money for 2 years at 4% per annum, and the difference is Rs. 1. We are tasked with finding the principal sum.

Let the principal sum be \( P \).

Step 1: Formula for Simple Interest
The formula for Simple Interest is:

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

where:
– \( P \) is the principal,
– \( R \) is the rate of interest (4%),
– \( T \) is the time period (2 years).

Substituting the values:

\[
SI = \frac{P \times 4 \times 2}{100} = \frac{8P}{100} = \frac{2P}{25}
\]

Step 2: Formula for Compound Interest
The formula for Compound Interest (compounded annually) is:

\[
A = P \left( 1 + \frac{R}{100} \right)^T
\]

where:
– \( A \) is the amount after \( T \) years,
– \( P \) is the principal,
– \( R \) is the rate of interest (4%),
– \( T \) is the time period (2 years).

The total compound interest (CI) is:

\[
CI = A – P
\]

Substitute the formula for \( A \):

\[
CI = P \left( 1 + \frac{4}{100} \right)^2 – P = P \left( 1.04^2 \right) – P
\]

Now calculate \( 1.04^2 \):

\[
1.04^2 = 1.0816
\]

So,

\[
CI = P \times 1.0816 – P = P(1.0816 – 1) = P \times 0.0816
\]

Step 3: Difference Between CI and SI
We are given that the difference between the compound interest and simple interest is Rs. 1:

\[
CI – SI = 1
\]

Substitute the expressions for \( CI \) and \( SI \):

\[
P \times 0.0816 – \frac{2P}{25} = 1
\]

We need to simplify the equation. First, express \( \frac{2P}{25} \) with a common denominator of 100:

\[
\frac{2P}{25} = \frac{8P}{100}
\]

Now the equation becomes:

\[
P \times 0.0816 – \frac{8P}{100} = 1
\]

\[
P \times \left( 0.0816 – 0.08 \right) = 1
\]

\[
P \times 0.0016 = 1
\]

Step 4: Solve for \( P \)

\[
P = \frac{1}{0.0016} = 625
\]

Final Answer:
The principal sum is Rs. 625.

Answer: Option C
Solution:
Given:
– A 60% increase in the amount over 6 years at simple interest.
– We are asked to find the compound interest on Rs. 12,000 after 3 years at the same rate.

Step 1: Find the rate of interest

Let the principal amount be \( P \), the rate of interest be \( R \% \) per annum, and the time period be \( T = 6 \) years.

Since there is a 60% increase in the amount, the simple interest (SI) for 6 years is 60% of the principal:

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

We know the total increase in the amount is 60% of the principal, so:

\[
SI = \frac{60}{100} \times P = 0.6P
\]

Now, substitute the values into the formula for simple interest:

\[
0.6P = \frac{P \times R \times 6}{100}
\]

Cancel out \( P \) from both sides (assuming \( P \neq 0 \)):

\[
0.6 = \frac{R \times 6}{100}
\]

Solve for \( R \):

\[
0.6 \times 100 = R \times 6
\]

\[
60 = 6R
\]

\[
R = 10\%
\]

So, the rate of interest is \( 10\% \) per annum.

Step 2: Calculate the compound interest on Rs. 12,000 after 3 years

Now, we are asked to find the compound interest (CI) on Rs. 12,000 after 3 years at a rate of 10%.

The formula for compound interest is:

\[
A = P \left( 1 + \frac{R}{100} \right)^T
\]

where:
– \( A \) is the amount after \( T \) years,
– \( P \) is the principal,
– \( R \) is the rate of interest,
– \( T \) is the time period.

Substituting the known values:

\[
A = 12000 \left( 1 + \frac{10}{100} \right)^3 = 12000 \left( 1 + 0.1 \right)^3 = 12000 \times (1.1)^3
\]

Now calculate \( (1.1)^3 \):

\[
(1.1)^3 = 1.331
\]

So,

\[
A = 12000 \times 1.331 = 15972
\]

The compound interest (CI) is the total amount minus the principal:

\[
CI = A – P = 15972 – 12000 = 3972
\]

Final Answer:
The compound interest on Rs. 12,000 after 3 years at 10% per annum is Rs. 3972.

Answer: Option A
Solution:
To find the difference between the compound interests on Rs. 5000 for \(1 \frac{1}{2}\) years (or 3 half-year periods) at 4% per annum, compounded yearly and half-yearly, let’s proceed step by step.

Given:
– Principal (\( P \)) = Rs. 5000
– Rate of interest (\( R \)) = 4% per annum
– Time (\( T \)) = \( 1 \frac{1}{2} \) years = 3 half-year periods (since 1 year = 2 half-years)

Step 1: Calculate the compound interest for yearly compounding

For yearly compounding, the formula for compound interest is:

\[
A = P \left(1 + \frac{R}{100}\right)^T
\]

Substitute the values:

\[
A = 5000 \left(1 + \frac{4}{100}\right)^{1.5}
\]

\[
A = 5000 \left(1.04\right)^{1.5}
\]

Now, calculate \( (1.04)^{1.5} \):

\[
1.04^{1.5} \approx 1.0612
\]

So,

\[
A = 5000 \times 1.0612 \approx 5306.08
\]

Now, the compound interest (CI) is:

\[
CI_{\text{yearly}} = A – P = 5306.08 – 5000 = 306.08
\]

Step 2: Calculate the compound interest for half-yearly compounding

For half-yearly compounding, the interest rate per half-year is:

\[
\frac{4\%}{2} = 2\% \text{ per half-year}
\]

So, the formula for compound interest becomes:

\[
A = P \left(1 + \frac{R}{100}\right)^{2T}
\]

Substitute the values:

\[
A = 5000 \left(1 + \frac{2}{100}\right)^{3}
\]

\[
A = 5000 \left(1.02\right)^{3}
\]

Now, calculate \( (1.02)^{3} \):

\[
1.02^3 = 1.061208
\]

So,

\[
A = 5000 \times 1.061208 \approx 5306.04
\]

Now, the compound interest (CI) is:

\[
CI_{\text{half-yearly}} = A – P = 5306.04 – 5000 = 306.04
\]

Step 3: Find the difference in compound interest

Now, let’s calculate the difference between the compound interest calculated with yearly and half-yearly compounding:

\[
\text{Difference} = CI_{\text{half-yearly}} – CI_{\text{yearly}} = 306.04 – 306.08 = -0.04
\]

Final Answer:
The difference in compound interest between yearly and half-yearly compounding is Rs. 0.04, with the compound interest for half-yearly compounding being slightly less than that for yearly compounding.

Answer: Option A
Solution:
To find the time period for which the compound interest on Rs. 30,000 is Rs. 4347 at an interest rate of 7% per annum, we can use the compound interest formula:

\[
A = P \left(1 + \frac{R}{100}\right)^T
\]

where:
– \(A\) is the total amount after interest,
– \(P\) is the principal,
– \(R\) is the rate of interest per annum,
– \(T\) is the time in years.

Given:
– Principal (\(P\)) = Rs. 30,000,
– Compound Interest (\(CI\)) = Rs. 4347,
– Rate of interest (\(R\)) = 7% per annum.

Step 1: Find the total amount \(A\)

We know that:

\[
CI = A – P
\]

Substitute the known values:

\[
4347 = A – 30000
\]

\[
A = 4347 + 30000 = 34347
\]

So, the total amount after interest is Rs. 34,347.

Step 2: Use the compound interest formula to solve for \(T\)

Now, substitute the known values into the compound interest formula:

\[
34347 = 30000 \left(1 + \frac{7}{100}\right)^T
\]

\[
34347 = 30000 \times (1.07)^T
\]

Step 3: Solve for \(T\)

Divide both sides by 30000:

\[
\frac{34347}{30000} = (1.07)^T
\]

\[
1.1449 = (1.07)^T
\]

Now, take the natural logarithm (ln) of both sides to solve for \(T\):

\[
\ln(1.1449) = T \ln(1.07)
\]

Using a calculator:

\[
\ln(1.1449) \approx 0.1352 \quad \text{and} \quad \ln(1.07) \approx 0.06766
\]

So:

\[
0.1352 = T \times 0.06766
\]

Solve for \(T\):

\[
T = \frac{0.1352}{0.06766} \approx 2
\]

Final Answer:
The time period is 2 years.

Answer: Option C
Solution:
To calculate the compound interest on a sum of Rs. 25,000 for 3 years at a rate of 12% per annum, we will use the compound interest formula:

\[
A = P \left(1 + \frac{R}{100}\right)^T
\]

where:
– \(A\) is the total amount after interest,
– \(P\) is the principal,
– \(R\) is the rate of interest per annum,
– \(T\) is the time in years.

Given:
– Principal (\(P\)) = Rs. 25,000,
– Rate of interest (\(R\)) = 12% per annum,
– Time (\(T\)) = 3 years.

Step 1: Calculate the total amount \(A\)

Substitute the values into the formula:

\[
A = 25000 \left(1 + \frac{12}{100}\right)^3
\]

\[
A = 25000 \left(1.12\right)^3
\]

Now, calculate \( (1.12)^3 \):

\[
1.12^3 = 1.404928
\]

So,

\[
A = 25000 \times 1.404928 = 35123.2
\]

Step 2: Calculate the compound interest (CI)

Compound interest is the total amount minus the principal:

\[
CI = A – P = 35123.2 – 25000 = 10123.2
\]

Final Answer:
The compound interest on Rs. 25,000 after 3 years at the rate of 12% per annum is Rs. 10,123.20.

Answer: Option A
Solution:
To find the rate of compound interest per annum, we can use the compound interest formula:

\[
A = P \left(1 + \frac{R}{100}\right)^T
\]

where:
– \( A \) is the total amount,
– \( P \) is the principal,
– \( R \) is the rate of interest per annum,
– \( T \) is the time in years.

Given:
– Principal (\( P \)) = Rs. 1200,
– Amount (\( A \)) = Rs. 1348.32,
– Time (\( T \)) = 2 years.

We need to find the rate of interest (\( R \)).

Step 1: Substitute the known values into the compound interest formula:

\[
1348.32 = 1200 \left(1 + \frac{R}{100}\right)^2
\]

Step 2: Solve for \( \left(1 + \frac{R}{100}\right)^2 \):

\[
\frac{1348.32}{1200} = \left(1 + \frac{R}{100}\right)^2
\]

\[
1.124 = \left(1 + \frac{R}{100}\right)^2
\]

Step 3: Take the square root of both sides:

\[
\sqrt{1.124} = 1 + \frac{R}{100}
\]

\[
1.06 = 1 + \frac{R}{100}
\]

Step 4: Solve for \( R \):

\[
1.06 – 1 = \frac{R}{100}
\]

\[
0.06 = \frac{R}{100}
\]

\[
R = 0.06 \times 100 = 6\%
\]

Final Answer:
The rate of compound interest per annum is 6%.

Answer: Option B
Solution:
To determine the least number of complete years in which a sum of money at 20% compound interest will be more than doubled, we use the compound interest formula:

\[
A = P \left(1 + \frac{R}{100}\right)^T
\]

where:
– \( A \) is the total amount,
– \( P \) is the principal,
– \( R \) is the rate of interest per annum,
– \( T \) is the time in years.

We want the amount to be more than double the principal. So, \( A > 2P \).

Given:
– Rate of interest (\( R \)) = 20% per annum,
– We need to find the smallest \( T \) such that \( A > 2P \).

Step 1: Set up the inequality for doubling the sum

Since we want the amount to be more than double the principal:

\[
2P < P \left(1 + \frac{20}{100}\right)^T
\]

Simplify the equation:

\[
2 < \left(1 + \frac{20}{100}\right)^T
\]

\[
2 < (1.2)^T
\]

Step 2: Solve for \( T \)

We need to find the smallest integer \( T \) for which \( (1.2)^T > 2 \). To solve for \( T \), take the logarithm of both sides:

\[
\log((1.2)^T) > \log(2)
\]

Using the logarithmic property \( \log(a^b) = b\log(a) \):

\[
T \log(1.2) > \log(2)
\]

Now, calculate the values:

\[
\log(1.2) \approx 0.07918 \quad \text{and} \quad \log(2) \approx 0.3010
\]

So, the inequality becomes:

\[
T \times 0.07918 > 0.3010
\]

Solve for \( T \):

\[
T > \frac{0.3010}{0.07918} \approx 3.8
\]

Since \( T \) must be a whole number, the smallest integer value of \( T \) is 4.

Final Answer:
The least number of complete years in which the sum will more than double is 4 years.

Answer: Option C
Solution:
To calculate the amount Albert will receive on the maturity of his fixed deposit, we can use the compound interest formula:

\[
A = P \left(1 + \frac{R}{100}\right)^T
\]

where:
– \( A \) is the total amount on maturity,
– \( P \) is the principal amount,
– \( R \) is the rate of interest per annum,
– \( T \) is the time in years.

Given:
– Principal (\( P \)) = Rs. 8000,
– Rate of interest (\( R \)) = 5% per annum,
– Time (\( T \)) = 2 years.

Step 1: Substitute the values into the formula:

\[
A = 8000 \left(1 + \frac{5}{100}\right)^2
\]

\[
A = 8000 \left(1 + 0.05\right)^2
\]

\[
A = 8000 \times (1.05)^2
\]

Step 2: Calculate \( (1.05)^2 \):

\[
1.05^2 = 1.1025
\]

So,

\[
A = 8000 \times 1.1025 = 8810
\]

Final Answer:
Albert will get Rs. 8810 on the maturity of the fixed deposit after 2 years.

Answer: Option D
Solution:
To calculate the effective annual rate (EAR) corresponding to a nominal rate of interest compounded semi-annually, we can use the following formula:

\[
\text{EAR} = \left(1 + \frac{r}{n}\right)^n – 1
\]

where:
– \( r \) is the nominal annual interest rate (6% in this case),
– \( n \) is the number of compounding periods per year (since interest is compounded half-yearly, \( n = 2 \)).

Given:
– Nominal rate (\( r \)) = 6% = 0.06,
– Number of compounding periods per year (\( n \)) = 2 (because the interest is compounded half-yearly).

Step 1: Substitute the values into the formula:

\[
\text{EAR} = \left(1 + \frac{0.06}{2}\right)^2 – 1
\]

\[
\text{EAR} = \left(1 + 0.03\right)^2 – 1
\]

\[
\text{EAR} = (1.03)^2 – 1
\]

Step 2: Calculate the result:

\[
(1.03)^2 = 1.0609
\]

\[
\text{EAR} = 1.0609 – 1 = 0.0609
\]

Step 3: Convert to percentage:

\[
\text{EAR} = 0.0609 \times 100 = 6.09\%
\]

Final Answer:
The effective annual rate (EAR) corresponding to a nominal rate of 6% per annum, compounded half-yearly, is 6.09%.

Answer: Option C
Solution:
Let’s break the problem down into two parts:

Part 1: Compound Interest on Rs. 4000 at 10% for 2 years
The formula for compound interest is:

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

Where:
– \(A\) is the amount after time \(t\),
– \(P\) is the principal amount,
– \(r\) is the rate of interest,
– \(t\) is the time in years.

For this part:
– \(P = 4000\),
– \(r = 10\%\),
– \(t = 2\) years.

Let’s calculate the compound interest on Rs. 4000 for 2 years at 10% per annum.

\[
A = 4000 \left(1 + \frac{10}{100}\right)^2
\] \[
A = 4000 \left(1 + 0.1\right)^2
\] \[
A = 4000 \times (1.1)^2
\] \[
A = 4000 \times 1.21
\] \[
A = 4840
\]

So, the amount after 2 years is Rs. 4840. The compound interest is:

\[
CI = A – P = 4840 – 4000 = 840
\]

Part 2: Simple Interest on a Sum for 3 years at 8% per annum
The problem states that the simple interest for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years, which we just calculated as Rs. 840.

Therefore, the simple interest is:

\[
SI = \frac{840}{2} = 420
\]

Now, we know the formula for simple interest:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(P\) is the principal (the sum placed on simple interest),
– \(r = 8\%\),
– \(t = 3\) years.

We are given \(SI = 420\). Let’s substitute the values and solve for \(P\):

\[
420 = \frac{P \times 8 \times 3}{100}
\] \[
420 = \frac{24P}{100}
\] \[
420 \times 100 = 24P
\] \[
42000 = 24P
\] \[
P = \frac{42000}{24}
\] \[
P = 1750
\]

Final Answer:
The sum placed on simple interest is Rs. 1750.

Answer: Option A
Solution:
To solve this, let’s first calculate the principal (P) based on the simple interest (SI) formula and then use it to calculate the compound interest (CI).

Step 1: Find the Principal Using the Simple Interest Formula

The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(SI = 50\) (simple interest),
– \(r = 5\%\) (rate of interest),
– \(t = 2\) years (time).

Now, plug these values into the formula:

\[
50 = \frac{P \times 5 \times 2}{100}
\] \[
50 = \frac{10P}{100}
\] \[
50 \times 100 = 10P
\] \[
5000 = 10P
\] \[
P = \frac{5000}{10} = 500
\]

So, the principal is Rs. 500.

Step 2: Calculate the Compound Interest

The formula for compound interest is:

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

Where:
– \(A\) is the amount after time \(t\),
– \(P = 500\) (principal),
– \(r = 5\%\) (rate),
– \(t = 2\) years.

First, calculate the total amount after 2 years:

\[
A = 500 \left(1 + \frac{5}{100}\right)^2
\] \[
A = 500 \left(1 + 0.05\right)^2
\] \[
A = 500 \times (1.05)^2
\] \[
A = 500 \times 1.1025
\] \[
A = 551.25
\]

Now, the compound interest is:

\[
CI = A – P = 551.25 – 500 = 51.25
\]

Final Answer:
The compound interest on the same sum of money for 2 years at 5% per annum is Rs. 51.25.

Answer: Option B
Solution:
To calculate the difference between the simple interest (SI) and compound interest (CI) on Rs. 1200 for one year at 10% per annum, reckoned half-yearly, we need to follow these steps.

Step 1: Calculate the Simple Interest (SI)

The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(P = 1200\),
– \(r = 10\%\),
– \(t = 1\) year.

So, substituting the values:

\[
SI = \frac{1200 \times 10 \times 1}{100}
\] \[
SI = \frac{12000}{100}
\] \[
SI = 120
\]

Step 2: Calculate the Compound Interest (CI)

When interest is compounded half-yearly, the rate is divided by 2 (since there are two periods in a year), and the time is multiplied by 2 (to account for both half-year periods).

The formula for compound interest is:

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

Where:
– \(P = 1200\),
– \(r = 10\%\),
– \(n = 2\) (because interest is compounded half-yearly),
– \(t = 1\) year.

First, calculate the amount \(A\):

\[
A = 1200 \left(1 + \frac{10}{100 \times 2}\right)^{2 \times 1}
\] \[
A = 1200 \left(1 + \frac{10}{200}\right)^2
\] \[
A = 1200 \left(1 + 0.05\right)^2
\] \[
A = 1200 \times (1.05)^2
\] \[
A = 1200 \times 1.1025
\] \[
A = 1323
\]

Now, the compound interest \(CI\) is:

\[
CI = A – P = 1323 – 1200 = 123
\]

Step 3: Calculate the Difference Between CI and SI

Now, subtract the simple interest from the compound interest:

\[
\text{Difference} = CI – SI = 123 – 120 = 3
\]

Final Answer:
The difference between the simple interest and compound interest on Rs. 1200 for one year at 10% per annum, reckoned half-yearly, is Rs. 3.

Answer: Option A
Solution:
To solve this, let’s use the information given and the formulas for simple interest (SI) and compound interest (CI) to find the rate of interest.

Step 1: Formula for Simple Interest (SI)
The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(P = 15000\),
– \(r\) is the rate of interest (which we need to find),
– \(t = 2\) years.

Step 2: Formula for Compound Interest (CI)
The formula for compound interest is:

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

Where:
– \(A\) is the amount after \(t\) years,
– \(P = 15000\),
– \(r\) is the rate of interest (which we need to find),
– \(t = 2\) years.

The compound interest is given by:

\[
CI = A – P = P \left(1 + \frac{r}{100}\right)^t – P
\]

We are told that the difference between the compound interest and simple interest is Rs. 96, so:

\[
CI – SI = 96
\]

Step 3: Calculate the Difference Between CI and SI

Now, let’s write the difference between the compound interest and the simple interest.

\[
P \left(1 + \frac{r}{100}\right)^2 – P – \frac{P \times r \times t}{100} = 96
\]

Substitute \(P = 15000\) and \(t = 2\):

\[
15000 \left(1 + \frac{r}{100}\right)^2 – 15000 – \frac{15000 \times r \times 2}{100} = 96
\]

Simplify the equation:

\[
15000 \left(1 + \frac{r}{100}\right)^2 – 15000 – 300r = 96
\]

Step 4: Expand and Solve for \(r\)

First, expand the squared term:

\[
\left(1 + \frac{r}{100}\right)^2 = 1 + 2 \times \frac{r}{100} + \left(\frac{r}{100}\right)^2
\]

Substitute this back into the equation:

\[
15000 \left(1 + 2 \times \frac{r}{100} + \frac{r^2}{10000}\right) – 15000 – 300r = 96
\]

Distribute \(15000\) on the left-hand side:

\[
15000 + 300r + \frac{15000r^2}{10000} – 15000 – 300r = 96
\]

Now, simplify:

\[
\frac{15000r^2}{10000} = 96
\]

\[
\frac{3r^2}{2} = 96
\]

Multiply both sides by 2:

\[
3r^2 = 192
\]

Now divide by 3:

\[
r^2 = 64
\]

Finally, take the square root of both sides:

\[
r = 8
\]

Final Answer:
The rate of interest per annum is 8%.

Answer: Option B
Solution:
To solve this, we need to first find the principal sum using the compound interest formula, and then use it to calculate the simple interest for double the time and half the rate.

Step 1: Calculate the Principal using Compound Interest Formula

The compound interest formula is:

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

Where:
– \(A\) is the amount after time \(t\),
– \(P\) is the principal,
– \(r\) is the rate of interest,
– \(t\) is the time in years.

The compound interest \(CI\) is given as Rs. 525 for 2 years at 10% per annum.

We know the formula for compound interest is:

\[
CI = A – P
\]

For 2 years, with a rate of 10%, we can calculate the amount \(A\) as follows:

\[
A = P \left(1 + \frac{10}{100}\right)^2 = P \times (1.1)^2
\] \[
A = P \times 1.21
\]

Since the compound interest \(CI = A – P = 525\), we have:

\[
P \times 1.21 – P = 525
\] \[
P(1.21 – 1) = 525
\] \[
P \times 0.21 = 525
\] \[
P = \frac{525}{0.21}
\] \[
P = 2500
\]

So, the principal sum is Rs. 2500.

Step 2: Calculate the Simple Interest for Double the Time and Half the Rate

Now, we need to calculate the simple interest for double the time (i.e., 4 years) at half the rate (i.e., 5% per annum).

The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(P = 2500\),
– \(r = 5\%\) (half the original rate),
– \(t = 4\) years (double the original time).

Now, substitute the values:

\[
SI = \frac{2500 \times 5 \times 4}{100}
\] \[
SI = \frac{50000}{100}
\] \[
SI = 500
\]

Final Answer:
The simple interest on the same sum for double the time at half the rate per annum is Rs. 500.

Answer: Option C
Solution:
To find how many years it will take for Rs. 2000 to amount to Rs. 2420 at 10% per annum compound interest, we can use the compound interest formula:

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

Where:
– \(A = 2420\) (the amount after time \(t\)),
– \(P = 2000\) (the principal),
– \(r = 10\%\) (the rate of interest),
– \(t\) is the time in years, which we need to find.

We are given:

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

Substitute the values:

\[
2420 = 2000 \left(1 + \frac{10}{100}\right)^t
\]

Simplify the equation:

\[
2420 = 2000 \times (1.1)^t
\]

Now, divide both sides by 2000:

\[
\frac{2420}{2000} = (1.1)^t
\]

\[
1.21 = (1.1)^t
\]

Next, take the natural logarithm (ln) of both sides to solve for \(t\):

\[
\ln(1.21) = \ln((1.1)^t)
\]

Using the logarithmic property \(\ln(a^b) = b \cdot \ln(a)\):

\[
\ln(1.21) = t \cdot \ln(1.1)
\]

Now, calculate the logarithms:

\[
\ln(1.21) \approx 0.1906 \quad \text{and} \quad \ln(1.1) \approx 0.0953
\]

Now solve for \(t\):

\[
t = \frac{\ln(1.21)}{\ln(1.1)} = \frac{0.1906}{0.0953} \approx 2
\]

Final Answer:
It will take approximately 2 years for Rs. 2000 to amount to Rs. 2420 at 10% per annum compound interest.

Answer: Option D
Solution:
To solve this, we need to find the principal sum when the difference between compound interest and simple interest for 3 years at 5% rate is Rs. 36.60.

Step 1: Formula for Simple Interest (SI)
The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(P\) is the principal (which we need to find),
– \(r = 5\%\) is the rate of interest,
– \(t = 3\) years is the time.

So, the simple interest for 3 years at 5% is:

\[
SI = \frac{P \times 5 \times 3}{100} = \frac{15P}{100} = 0.15P
\]

Step 2: Formula for Compound Interest (CI)
The formula for compound interest is:

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

Where:
– \(A\) is the amount after time \(t\),
– \(P\) is the principal,
– \(r = 5\%\) is the rate of interest,
– \(t = 3\) years is the time.

The compound interest is given by:

\[
CI = A – P = P \left(1 + \frac{r}{100}\right)^t – P
\]

Substitute the values:

\[
CI = P \left(1 + \frac{5}{100}\right)^3 – P
\] \[
CI = P \left(1.05\right)^3 – P
\] \[
CI = P \times 1.157625 – P
\] \[
CI = P(1.157625 – 1)
\] \[
CI = 0.157625P
\]

Step 3: Difference Between Compound Interest and Simple Interest
The difference between compound interest and simple interest is given as Rs. 36.60:

\[
CI – SI = 36.60
\]

Substitute the values for \(CI\) and \(SI\):

\[
0.157625P – 0.15P = 36.60
\] \[
(0.157625 – 0.15)P = 36.60
\] \[
0.007625P = 36.60
\]

Step 4: Solve for \(P\)

Now, solve for \(P\):

\[
P = \frac{36.60}{0.007625}
\] \[
P = 4800
\]

Final Answer:
The principal sum is Rs. 4800.

Answer: Option B
Solution:
To calculate the compound interest on an amount of Rs. 8400 at 12.5% per annum for 3 years, we will use the compound interest formula:

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

Where:
– \(A\) is the amount after \(t\) years,
– \(P = 8400\) is the principal,
– \(r = 12.5\%\) is the rate of interest,
– \(t = 3\) years is the time.

Step 1: Calculate the Amount (\(A\))

First, substitute the known values into the formula:

\[
A = 8400 \left(1 + \frac{12.5}{100}\right)^3
\] \[
A = 8400 \left(1 + 0.125\right)^3
\] \[
A = 8400 \times (1.125)^3
\]

Now, calculate \(1.125^3\):

\[
1.125^3 = 1.423828125
\]

So:

\[
A = 8400 \times 1.423828125
\] \[
A = 11968.75
\]

Step 2: Calculate the Compound Interest (CI)

The compound interest is the difference between the total amount and the principal:

\[
CI = A – P
\] \[
CI = 11968.75 – 8400
\] \[
CI = 3568.75
\]

Final Answer:
The compound interest accrued on Rs. 8400 at 12.5% per annum for 3 years is Rs. 3568.75.

Answer: Option B
Solution:
To solve this, we can use the compound interest formula and the fact that the sum doubles in 4 years.

Step 1: Compound Interest Formula

The compound interest formula is:

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

Where:
– \(A\) is the amount after time \(t\),
– \(P\) is the principal,
– \(r\) is the rate of interest,
– \(t\) is the time in years.

Step 2: Doubling the Sum in 4 Years

We are told that the sum doubles in 4 years. So, the amount after 4 years is double the principal, i.e., \(A = 2P\).

Using the compound interest formula:

\[
2P = P \left(1 + \frac{r}{100}\right)^4
\]

Now, cancel out \(P\) from both sides:

\[
2 = \left(1 + \frac{r}{100}\right)^4
\]

To solve for \(r\), take the 4th root of both sides:

\[
\left(1 + \frac{r}{100}\right) = 2^{1/4}
\]

Now calculate \(2^{1/4}\):

\[
2^{1/4} \approx 1.189207
\]

So:

\[
1 + \frac{r}{100} = 1.189207
\]

Now subtract 1 from both sides:

\[
\frac{r}{100} = 0.189207
\]

Multiply by 100 to get \(r\):

\[
r \approx 18.92\%
\]

Step 3: Time to Amount to 8 Times the Principal

Now, we need to find the time it will take for the sum to amount to 8 times itself at the same rate of interest. So, we want \(A = 8P\).

Using the compound interest formula:

\[
8P = P \left(1 + \frac{r}{100}\right)^t
\]

Cancel out \(P\) from both sides:

\[
8 = \left(1 + \frac{r}{100}\right)^t
\]

Substitute \(r = 18.92\%\):

\[
8 = \left(1 + \frac{18.92}{100}\right)^t
\] \[
8 = (1.189207)^t
\]

Now, solve for \(t\) by taking the logarithm of both sides:

\[
\log(8) = t \cdot \log(1.189207)
\]

We know \(\log(8) \approx 0.9031\) and \(\log(1.189207) \approx 0.0758\), so:

\[
0.9031 = t \cdot 0.0758
\]

Solve for \(t\):

\[
t = \frac{0.9031}{0.0758} \approx 11.91
\]

Final Answer:
It will take approximately 12 years for the sum to amount to 8 times itself at the same rate of interest.

Answer: Option D
Solution:
To find the principal sum, we will use the compound interest formula:

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

Where:
– \(A = 1352\) (the amount after \(t\) years),
– \(P\) is the principal (which we need to find),
– \(r = 4\%\) (the rate of interest),
– \(t = 2\) years (the time).

Step 1: Substitute the known values into the formula

\[
1352 = P \left(1 + \frac{4}{100}\right)^2
\]

\[
1352 = P \left(1.04\right)^2
\]

Step 2: Calculate \(1.04^2\)

\[
1.04^2 = 1.0816
\]

So, the equation becomes:

\[
1352 = P \times 1.0816
\]

Step 3: Solve for \(P\)

Now, divide both sides by 1.0816:

\[
P = \frac{1352}{1.0816}
\]

\[
P = 1250
\]

Final Answer:
The principal sum is Rs. 1250.

Answer: Option B
Solution:
To calculate the compound interest on Rs. 2800 for 18 months at 10% per annum, we will use the compound interest formula:

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

Where:
– \(A\) is the amount after \(t\) years,
– \(P = 2800\) is the principal,
– \(r = 10\%\) is the rate of interest,
– \(t = 18\) months = 1.5 years (since 18 months = 1.5 years).

Step 1: Substitute the known values into the formula

\[
A = 2800 \left(1 + \frac{10}{100}\right)^{1.5}
\]

\[
A = 2800 \left(1 + 0.10\right)^{1.5}
\]

\[
A = 2800 \times (1.10)^{1.5}
\]

Step 2: Calculate \(1.10^{1.5}\)

\[
1.10^{1.5} \approx 1.1547
\]

Now, substitute this value back into the formula:

\[
A = 2800 \times 1.1547
\]

\[
A \approx 3233.16
\]

Step 3: Calculate the Compound Interest (CI)

The compound interest is the difference between the amount and the principal:

\[
CI = A – P
\] \[
CI = 3233.16 – 2800
\] \[
CI \approx 433.16
\]

Final Answer:
The compound interest on Rs. 2800 for 18 months at 10% per annum is approximately Rs. 433.16.

Answer: Option B
Solution:
To solve this, we will use the formulas for compound interest and simple interest to find the principal sum, and then calculate the simple interest.

Step 1: Use the compound interest formula to find the principal sum

The compound interest formula is:

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

Where:
– \(A\) is the amount after \(t\) years,
– \(P\) is the principal,
– \(r = 5\%\) is the rate of interest,
– \(t = 3\) years is the time.

We are given that the compound interest (CI) for 3 years is Rs. 252.20. The compound interest is:

\[
CI = A – P
\]

We know:

\[
CI = 252.20
\]

So, the amount after 3 years is:

\[
A = P + 252.20
\]

Now, using the compound interest formula:

\[
A = P \left(1 + \frac{5}{100}\right)^3
\] \[
A = P \left(1.05\right)^3
\] \[
A = P \times 1.157625
\]

So we have the equation:

\[
P \times 1.157625 = P + 252.20
\]

Now, subtract \(P\) from both sides:

\[
P(1.157625 – 1) = 252.20
\] \[
P \times 0.157625 = 252.20
\]

Solve for \(P\):

\[
P = \frac{252.20}{0.157625} \approx 1600
\]

So, the principal sum is **Rs. 1600**.

Step 2: Calculate the simple interest

The simple interest formula is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \(P = 1600\),
– \(r = 5\%\),
– \(t = 3\) years.

Now, substitute the values:

\[
SI = \frac{1600 \times 5 \times 3}{100}
\] \[
SI = \frac{24000}{100}
\] \[
SI = 240
\]

Final Answer:
The simple interest on the same sum at the same rate and for the same time is Rs. 240.

Answer: Option C
Solution:
We are given:

– Compound Interest (CI) for 2 years = Rs. 282.15
– Simple Interest (SI) for 2 years = Rs. 270
– Time period \( t = 2 \) years

We need to find the rate of interest per annum (\( r \)).

Step 1: Formula for Simple Interest

The formula for Simple Interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \( P \) is the principal (which we need to find),
– \( r \) is the rate of interest per annum,
– \( t = 2 \) years is the time.

Given that \( SI = 270 \), substitute the known values into the formula:

\[
270 = \frac{P \times r \times 2}{100}
\]

Multiply both sides by 100 to eliminate the denominator:

\[
270 \times 100 = P \times r \times 2
\] \[
27000 = P \times r \times 2
\]

Now divide both sides by 2:

\[
13500 = P \times r
\]

So, we have:

\[
P \times r = 13500 \quad \text{(Equation 1)}
\]

Step 2: Formula for Compound Interest

The formula for Compound Interest is:

\[
CI = P \left(1 + \frac{r}{100}\right)^t – P
\]

We know that \( CI = 282.15 \) and \( t = 2 \), so substitute the known values:

\[
282.15 = P \left(1 + \frac{r}{100}\right)^2 – P
\]

Simplify the equation:

\[
282.15 = P \left[\left(1 + \frac{r}{100}\right)^2 – 1\right] \]

Expand \( \left(1 + \frac{r}{100}\right)^2 \):

\[
\left(1 + \frac{r}{100}\right)^2 = 1 + 2 \times \frac{r}{100} + \frac{r^2}{10000}
\]

So:

\[
282.15 = P \left[\frac{2r}{100} + \frac{r^2}{10000}\right] \]

This gives us the second equation:

\[
282.15 = P \left[\frac{2r}{100} + \frac{r^2}{10000}\right] \quad \text{(Equation 2)}
\]

Step 3: Solve the system of equations

We now have two equations:

1. \( P \times r = 13500 \)
2. \( 282.15 = P \left[\frac{2r}{100} + \frac{r^2}{10000}\right] \)

From Equation 1, solve for \( P \):

\[
P = \frac{13500}{r}
\]

Substitute this into Equation 2:

\[
282.15 = \frac{13500}{r} \left[\frac{2r}{100} + \frac{r^2}{10000}\right] \]

Simplify the expression:

\[
282.15 = \frac{13500}{r} \times \left[\frac{2r}{100} + \frac{r^2}{10000}\right] \]

Now, simplify further:

\[
282.15 = 13500 \times \left[\frac{2}{100} + \frac{r}{10000}\right] \]

\[
282.15 = 13500 \times \left[\frac{200 + r}{10000}\right] \]

\[
282.15 = \frac{13500 \times (200 + r)}{10000}
\]

Multiply both sides by 10000 to eliminate the denominator:

\[
2821500 = 13500 \times (200 + r)
\]

Now, divide both sides by 13500:

\[
\frac{2821500}{13500} = 200 + r
\]

\[
209.00 = 200 + r
\]

So:

\[
r = 209 – 200
\]

\[
r = 9\%
\]

Final Answer:
The rate of interest per annum is 9%.

Answer: Option A
Solution:
To calculate the compound interest with different rates for each year, we can use the compound interest formula for each year and apply the rates sequentially.

Compound Interest Formula:
\[
A = P \left(1 + \frac{r}{100}\right)^t
\] Where:
– \(A\) is the amount after time \(t\),
– \(P\) is the principal,
– \(r\) is the rate of interest for that year,
– \(t\) is the time in years.

Given:
– Principal \(P = 10000\),
– Rates for the three years are:
– 4% for the first year,
– 5% for the second year,
– 6% for the third year.

Step 1: Calculate for the first year (4% interest)
For the first year, the rate is 4%, so we calculate the amount at the end of the first year:

\[
A_1 = 10000 \times \left(1 + \frac{4}{100}\right) = 10000 \times 1.04 = 10400
\]

Step 2: Calculate for the second year (5% interest)
Now, the new principal is \(10400\), and the rate for the second year is 5%. The amount at the end of the second year is:

\[
A_2 = 10400 \times \left(1 + \frac{5}{100}\right) = 10400 \times 1.05 = 10920
\]

Step 3: Calculate for the third year (6% interest)
Now, the new principal is \(10920\), and the rate for the third year is 6%. The amount at the end of the third year is:

\[
A_3 = 10920 \times \left(1 + \frac{6}{100}\right) = 10920 \times 1.06 = 11575.20
\]

Step 4: Calculate the Compound Interest (CI)
The compound interest is the difference between the final amount and the initial principal:

\[
CI = A_3 – P = 11575.20 – 10000 = 1575.20
\]

Final Answer:
The compound interest for Rs. 10000 for three years, with the given varying rates, is Rs. 1575.20.

Answer: Option C
Solution:
To calculate the compound interest when the interest is compounded half-yearly, we use the following formula:

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

Where:
– \(A\) is the amount after time \(t\),
– \(P\) is the principal (initial amount),
– \(r\) is the rate of interest per annum,
– \(n\) is the number of times the interest is compounded per year (since it is half-yearly, \(n = 2\)),
– \(t\) is the time in years.

Given:
– Principal \(P = 10000\),
– Rate \(r = 20\%\),
– Time \(t = 2\) years,
– The interest is compounded half-yearly (\(n = 2\)).

Step 1: Apply the values to the formula

Substitute the given values into the compound interest formula:

\[
A = 10000 \left(1 + \frac{20}{100 \times 2}\right)^{2 \times 2}
\]

Simplify:

\[
A = 10000 \left(1 + \frac{20}{200}\right)^4
\] \[
A = 10000 \left(1 + 0.1\right)^4
\] \[
A = 10000 \times (1.1)^4
\]

Step 2: Calculate \((1.1)^4\)

\[
1.1^4 = 1.4641
\]

So:

\[
A = 10000 \times 1.4641 = 14641
\]

Step 3: Calculate the Compound Interest (CI)

The compound interest is the difference between the final amount and the initial principal:

\[
CI = A – P = 14641 – 10000 = 4641
\]

Final Answer:
The compound interest accrued on an amount of Rs. 10000 at 20% per annum, compounded half-yearly for 2 years, is Rs. 4641.

Answer: Option C
Solution:
We are given:

– Amount at the end of the first year = Rs. 650,
– Amount at the end of the second year = Rs. 676.

The interest is compounded annually. We need to find the principal sum (initial investment).

Step 1: Calculate the rate of interest

The amount at the end of the second year is Rs. 676, and the amount at the end of the first year is Rs. 650. The difference between these two amounts represents the interest for the second year:

\[
\text{Interest for second year} = 676 – 650 = 26
\]

Now, we know that the interest for the second year (Rs. 26) is the interest on the amount at the end of the first year (Rs. 650), so we can find the rate of interest.

The formula for compound interest is:

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

For the second year, the interest is calculated on the amount at the end of the first year, which is Rs. 650. The interest for one year is Rs. 26, so:

\[
26 = 650 \times \frac{r}{100}
\]

Now, solve for \(r\):

\[
\frac{26}{650} = \frac{r}{100}
\]

\[
r = \frac{26}{650} \times 100 = 4\%
\]

So, the rate of interest is **4% per annum**.

Step 2: Find the Principal

We know the amount at the end of the first year is Rs. 650, and this amount includes both the principal and the interest for one year. Let the principal be \(P\). Using the compound interest formula for one year:

\[
650 = P \times \left(1 + \frac{4}{100}\right)
\] \[
650 = P \times 1.04
\]

Now, solve for \(P\):

\[
P = \frac{650}{1.04} = 625
\]

Final Answer:
The sum of money (principal) is Rs. 625.

Answer: Option C
Solution:
To find the principal amount, we can use the compound interest formula. The formula for compound interest is:

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

Where:
– \( A \) is the amount after \( t \) years,
– \( P \) is the principal,
– \( r \) is the rate of interest per annum,
– \( t \) is the time in years.

Given:
– Amount \( A = 270.40 \),
– Rate of interest \( r = 4\% \),
– Time \( t = 2 \) years.

Step 1: Substitute the given values into the compound interest formula:

\[
270.40 = P \left(1 + \frac{4}{100}\right)^2
\]

Simplify:

\[
270.40 = P \left(1 + 0.04\right)^2
\] \[
270.40 = P \times (1.04)^2
\] \[
270.40 = P \times 1.0816
\]

Step 2: Solve for \( P \):

\[
P = \frac{270.40}{1.0816}
\] \[
P = 250
\]

Final Answer:
The principal amount is Rs. 250.

Answer: Option B
Solution:
To solve this, we need to calculate the compound interest for two separate deposits, one made on **1st January** and the other on **1st July**, with a **5% annual interest rate compounded half-yearly**.

Given:
– The annual rate of interest \(r = 5\%\), which is compounded half-yearly.
– The principal deposited twice: Rs. 1600 on **1st January** and Rs. 1600 on **1st July**.
– Time periods:
– The deposit made on 1st January will earn interest for 1 year (two half-year periods).
– The deposit made on 1st July will earn interest for 6 months (one half-year period).

Step 1: Calculate the compound interest for the deposit made on **1st January**

For the first deposit, the interest is compounded twice during the year.

The compound interest formula is:

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

Where:
– \(A\) is the final amount,
– \(P\) is the principal,
– \(r\) is the annual interest rate,
– \(n\) is the number of times the interest is compounded per year (in this case, twice a year),
– \(t\) is the time in years.

For the 1st January deposit:
– \(P = 1600\),
– \(r = 5\%\),
– \(n = 2\) (because the interest is compounded half-yearly),
– \(t = 1\) year.

Substitute the values into the compound interest formula:

\[
A_1 = 1600 \times \left(1 + \frac{5}{100 \times 2}\right)^{2 \times 1}
\] \[
A_1 = 1600 \times \left(1 + 0.025\right)^2
\] \[
A_1 = 1600 \times 1.025^2
\] \[
A_1 = 1600 \times 1.050625 = 1681
\]

The amount at the end of the year for the deposit made on 1st January is Rs. 1681.

Thus, the interest earned on this deposit is:

\[
\text{Interest}_1 = 1681 – 1600 = 81
\]

Step 2: Calculate the compound interest for the deposit made on **1st July**

For the deposit made on 1st July, it will earn interest for only 6 months (1 half-year period).

For the 1st July deposit:
– \(P = 1600\),
– \(r = 5\%\),
– \(n = 2\) (because the interest is compounded half-yearly),
– \(t = \frac{1}{2}\) years (since it’s for 6 months).

Substitute the values into the compound interest formula:

\[
A_2 = 1600 \times \left(1 + \frac{5}{100 \times 2}\right)^{2 \times \frac{1}{2}}
\] \[
A_2 = 1600 \times \left(1 + 0.025\right)^1
\] \[
A_2 = 1600 \times 1.025 = 1640
\]

The amount at the end of the year for the deposit made on 1st July is Rs. 1640.

Thus, the interest earned on this deposit is:

\[
\text{Interest}_2 = 1640 – 1600 = 40
\]

Step 3: Total interest earned

The total interest earned by the customer is the sum of the interest earned on both deposits:

\[
\text{Total Interest} = 81 + 40 = 121
\]

Final Answer:
The total interest earned by the customer at the end of the year is Rs. 121.

Answer: Option C
Solution:
To calculate the total interest earned by Mr. Duggal at the end of 2 years, with different compounding methods in each year, we will handle each year separately.

Given:
– Principal \(P = 20000\)
– Annual interest rate \(r = 20\%\)
– The interest is compounded **half-yearly for the first year** and **annually for the second year**.

Step 1: Calculate the amount at the end of the first year (half-yearly compounding)

For the first year, the interest is compounded half-yearly, so the interest rate for each half-year period is:

\[
\text{Half-yearly rate} = \frac{20\%}{2} = 10\%
\]

The formula for compound interest is:

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

Where:
– \(A\) is the amount at the end of the time period,
– \(P\) is the principal,
– \(r\) is the annual interest rate,
– \(n\) is the number of compounding periods per year (2 for half-yearly compounding),
– \(t\) is the time in years.

For the first year:
– \(P = 20000\),
– \(r = 20\%\),
– \(n = 2\) (since interest is compounded half-yearly),
– \(t = 1\) year.

Substitute these values into the formula:

\[
A_1 = 20000 \left(1 + \frac{20}{100 \times 2}\right)^{2 \times 1}
\] \[
A_1 = 20000 \left(1 + 0.1\right)^2
\] \[
A_1 = 20000 \times 1.1^2 = 20000 \times 1.21 = 24200
\]

So, at the end of the first year, the amount is Rs. 24200.

Step 2: Calculate the amount at the end of the second year (yearly compounding)

For the second year, the interest is compounded yearly, so the interest rate for the year is 20%.

The formula for compound interest is the same:

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

For the second year:
– Principal \(P = 24200\) (since this is the amount at the end of the first year),
– Annual interest rate \(r = 20\%\),
– \(t = 1\) year (for the second year),
– Compounding is yearly, so \(n = 1\).

Substitute the values into the formula:

\[
A_2 = 24200 \times \left(1 + \frac{20}{100}\right)
\] \[
A_2 = 24200 \times 1.2 = 29040
\]

So, at the end of the second year, the amount is Rs. 29040.

Step 3: Calculate the total interest earned

The total interest earned is the difference between the final amount and the principal:

\[
\text{Total Interest} = A_2 – P
\] \[
\text{Total Interest} = 29040 – 20000 = 9040
\]

Final Answer:
The total interest earned by Mr. Duggal at the end of 2 years is Rs. 9040.

Answer: Option B
Solution:
We are given the following information:

– The amount after 3 years is Rs. 10648.
– The amount after 2 years is Rs. 9680.

We need to find the rate of interest per annum.

Step 1: Use the compound interest formula

The compound interest formula is:

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

Where:
– \(A\) is the amount after \(t\) years,
– \(P\) is the principal,
– \(r\) is the rate of interest per annum,
– \(t\) is the time in years.

Step 2: Set up the equations for the two amounts

From the given data:

1. After 3 years, the amount is Rs. 10648:
\[
10648 = P \left(1 + \frac{r}{100}\right)^3
\]

2. After 2 years, the amount is Rs. 9680:
\[
9680 = P \left(1 + \frac{r}{100}\right)^2
\]

Step 3: Divide the two equations

Now, divide the first equation by the second equation to eliminate \(P\):

\[
\frac{10648}{9680} = \frac{P \left(1 + \frac{r}{100}\right)^3}{P \left(1 + \frac{r}{100}\right)^2}
\]

Simplifying:

\[
\frac{10648}{9680} = \left(1 + \frac{r}{100}\right)
\]

\[
\frac{10648}{9680} = 1 + \frac{r}{100}
\]

\[
1.1 = 1 + \frac{r}{100}
\]

### Step 4: Solve for \(r\)

Now, solve for \(r\):

\[
1.1 – 1 = \frac{r}{100}
\] \[
0.1 = \frac{r}{100}
\] \[
r = 0.1 \times 100 = 10
\]

Final Answer:
The rate of interest per annum is 10%.

Answer: Option C
Solution:
To calculate the compound interest on Rs. 16000 for 9 months at 20% per annum, compounded quarterly, we will use the compound interest formula:

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

Where:
– \(A\) is the amount at the end of the time period,
– \(P\) is the principal,
– \(r\) is the annual interest rate,
– \(n\) is the number of times the interest is compounded per year (quarterly compounding means \(n = 4\)),
– \(t\) is the time in years.

Given:
– \(P = 16000\),
– \(r = 20\%\),
– \(n = 4\) (quarterly compounding),
– \(t = \frac{9}{12} = 0.75\) years (since the time is 9 months).

Step 1: Substitute the values into the compound interest formula

\[
A = 16000 \left(1 + \frac{20}{100 \times 4}\right)^{4 \times 0.75}
\]

Simplifying:

\[
A = 16000 \left(1 + \frac{20}{400}\right)^{3}
\] \[
A = 16000 \left(1 + 0.05\right)^{3}
\] \[
A = 16000 \times (1.05)^3
\]

Step 2: Calculate the value of \((1.05)^3\)

\[
(1.05)^3 = 1.157625
\]

Step 3: Calculate the final amount \(A\)

\[
A = 16000 \times 1.157625 = 18522
\]

Step 4: Calculate the compound interest

Compound Interest (CI) is the difference between the final amount \(A\) and the principal \(P\):

\[
CI = A – P
\] \[
CI = 18522 – 16000 = 3522
\]

Final Answer:
The compound interest on Rs. 16000 for 9 months at 20% per annum, compounded quarterly, is Rs. 3522.

Answer: Option E
Solution:
Let’s break the problem into two parts:

Part 1: Calculate the Principal using Simple Interest

We are given:
– Simple interest (SI) on Rs. 1000 for 4 years at a rate of 5% per annum.
– The formula for simple interest is:

\[
\text{SI} = \frac{P \times r \times t}{100}
\]

Where:
– \(P\) is the principal,
– \(r\) is the rate of interest per annum,
– \(t\) is the time in years.

Given that the simple interest for Rs. 1000 over 4 years is the same as the interest on a certain principal, let’s first calculate the amount of simple interest:

The simple interest for Rs. 1000 at 5% per annum for 4 years:

\[
\text{SI} = \frac{1000 \times 5 \times 4}{100} = \frac{20000}{100} = 200
\]

So, the interest is Rs. 200.

Part 2: Compound Interest on Twice the Principal

Next, we need to calculate the compound interest (CI) on **twice** the principal (i.e., Rs. 2000) for 2 years at the same rate of 5%.

The formula for compound interest is:

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

Where:
– \(A\) is the amount after \(t\) years,
– \(P\) is the principal,
– \(r\) is the annual rate of interest,
– \(t\) is the time in years.

For compound interest, the formula for the compound interest is:

\[
\text{CI} = A – P
\]

Now, substitute the given values into the formula:

– Principal \(P = 2000\),
– Rate of interest \(r = 5\%\),
– Time \(t = 2\) years.

Substitute these values into the formula for amount \(A\):

\[
A = 2000 \left(1 + \frac{5}{100}\right)^2 = 2000 \left(1.05\right)^2
\]

First, calculate \((1.05)^2\):

\[
(1.05)^2 = 1.1025
\]

Now calculate \(A\):

\[
A = 2000 \times 1.1025 = 2205
\]

So, the total amount after 2 years is Rs. 2205.

Step 3: Calculate the Compound Interest

Now, calculate the compound interest:

\[
\text{CI} = A – P = 2205 – 2000 = 205
\]

Final Answer:
The compound interest the man will get on twice the principal (Rs. 2000) in 2 years at 5% per annum is Rs. 205.

Answer: Option B
Solution:
We are given that the compound interest (CI) on a sum of money for 2 years at 4% per annum is Rs. 2448. We need to calculate the simple interest (SI) on the same sum at the same rate for 2 years.

Step 1: Use the formula for compound interest

The compound interest formula is:

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

Where:
– \(A\) is the amount at the end of \(t\) years,
– \(P\) is the principal,
– \(r\) is the annual interest rate,
– \(t\) is the time in years.

The compound interest for 2 years is given by:

\[
\text{CI} = A – P
\]

Given that the compound interest is Rs. 2448 for 2 years, we can use this to find the principal.

Step 2: Set up the equations for compound interest

We are told that the compound interest is Rs. 2448. So:

\[
A – P = 2448
\]

The amount after 2 years can be calculated using the compound interest formula:

\[
A = P \left(1 + \frac{4}{100}\right)^2 = P \times (1.04)^2
\]

\[
A = P \times 1.0816
\]

Now, substitute this value of \(A\) into the equation \(A – P = 2448\):

\[
P \times 1.0816 – P = 2448
\]

Simplify:

\[
P \times (1.0816 – 1) = 2448
\] \[
P \times 0.0816 = 2448
\]

Solve for \(P\):

\[
P = \frac{2448}{0.0816} = 30000
\]

So, the principal \(P\) is Rs. 30000.

Step 3: Calculate the simple interest

The formula for simple interest is:

\[
\text{SI} = \frac{P \times r \times t}{100}
\]

Substitute the values:
– \(P = 30000\),
– \(r = 4\%\),
– \(t = 2\) years.

\[
\text{SI} = \frac{30000 \times 4 \times 2}{100}
\] \[
\text{SI} = \frac{240000}{100} = 2400
\]

Final Answer:
The simple interest on the same sum of money at the same rate of interest for 2 years is Rs. 2400.

Answer: Option C
Solution:
Let’s calculate the difference between the Simple Interest (SI) and Compound Interest (CI) for an amount of Rs. 19200 at 12% per annum for 3 years.

Step 1: Calculate the Simple Interest (SI)

The formula for Simple Interest is:

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

Where:
– \( P = 19200 \) (Principal)
– \( R = 12\% \) (Rate of interest)
– \( T = 3 \) years (Time)

Substitute the values:

\[
SI = \frac{19200 \times 12 \times 3}{100}
\] \[
SI = \frac{691200}{100} = 6912 \, \text{Rs.}
\]

So, the Simple Interest for 3 years is Rs. 6912.

Step 2: Calculate the Compound Interest (CI)

The formula for Compound Interest is:

\[
A = P \left( 1 + \frac{R}{100} \right)^T
\]

Where:
– \( A \) is the amount after interest
– \( P = 19200 \) (Principal)
– \( R = 12\% \) (Rate of interest)
– \( T = 3 \) years (Time)

First, calculate the total amount \( A \):

\[
A = 19200 \left( 1 + \frac{12}{100} \right)^3
\] \[
A = 19200 \times (1.12)^3
\] \[
A = 19200 \times 1.404928 = 26914.38 \, \text{Rs.}
\]

Now, the Compound Interest (CI) is:

\[
CI = A – P = 26914.38 – 19200 = 7714.38 \, \text{Rs.}
\]

So, the Compound Interest for 3 years is Rs. 7714.38.

Step 3: Find the difference between the Compound Interest and Simple Interest.

The difference is:

\[
\text{Difference} = CI – SI = 7714.38 – 6912 = 802.38 \, \text{Rs.}
\]

Final Answer:
The difference between the Compound Interest and Simple Interest accrued on Rs. 19200 at 12% for 3 years is Rs. 802.38.

Answer: Option B
Solution:
Let’s calculate the difference between the Simple Interest (S.I.) and Compound Interest (C.I.) for a sum of Rs. 15,000 at an interest rate of \( 12 \frac{1}{2} \% \) per annum for 2 years.

Step 1: Calculate the Simple Interest (S.I.)

The formula for Simple Interest is:

\[
S.I. = \frac{P \times R \times T}{100}
\]

Where:
– \( P = 15000 \) (Principal)
– \( R = 12 \frac{1}{2} \% = 12.5 \% \) (Rate of interest)
– \( T = 2 \) years (Time)

Substitute the values:

\[
S.I. = \frac{15000 \times 12.5 \times 2}{100}
\] \[
S.I. = \frac{375000}{100} = 3750 \, \text{Rs.}
\]

So, the Simple Interest for 2 years is Rs. 3750.

Step 2: Calculate the Compound Interest (C.I.)

The formula for Compound Interest is:

\[
A = P \left( 1 + \frac{R}{100} \right)^T
\]

Where:
– \( A \) is the amount after interest
– \( P = 15000 \) (Principal)
– \( R = 12.5 \% \) (Rate of interest)
– \( T = 2 \) years (Time)

First, calculate the total amount \( A \):

\[
A = 15000 \left( 1 + \frac{12.5}{100} \right)^2
\] \[
A = 15000 \times (1.125)^2
\] \[
A = 15000 \times 1.265625 = 18993.75 \, \text{Rs.}
\]

Now, the Compound Interest (C.I.) is:

\[
C.I. = A – P = 18993.75 – 15000 = 3993.75 \, \text{Rs.}
\]

So, the Compound Interest for 2 years is Rs. 3993.75.

Step 3: Find the difference between the Compound Interest and Simple Interest.

The difference is:

\[
\text{Difference} = C.I. – S.I. = 3993.75 – 3750 = 243.75 \, \text{Rs.}
\]

Final Answer:
The difference between the Simple Interest and Compound Interest on Rs. 15000 for 2 years at \( 12 \frac{1}{2} \% \) per annum is Rs. 243.75.

Answer: Option D
Solution:
To find the principal sum, we will use the formula for compound interest:

\[
A = P \left( 1 + \frac{R}{100} \right)^T
\]

Where:
– \( A \) is the amount after interest,
– \( P \) is the principal,
– \( R \) is the rate of interest (5%),
– \( T \) is the time (3 years).

We know that the compound interest (C.I.) for 3 years is Rs. 1261, and we need to find the principal amount \( P \).

The formula for compound interest is:

\[
C.I. = A – P
\]

Since we know the compound interest is Rs. 1261, we can express the total amount \( A \) as:

\[
A = P + 1261
\]

Now, substitute this into the compound interest formula:

\[
P + 1261 = P \left( 1 + \frac{5}{100} \right)^3
\] \[
P + 1261 = P \left( 1.05 \right)^3
\] \[
P + 1261 = P \times 1.157625
\]

Now, simplify the equation:

\[
P \times 1.157625 – P = 1261
\] \[
P (1.157625 – 1) = 1261
\] \[
P \times 0.157625 = 1261
\] \[
P = \frac{1261}{0.157625}
\] \[
P = 8000
\]

Final Answer:
The sum of money (the principal) is Rs. 8000.

Answer: Option C
Solution:
To find the principal sum, we will use the formula for compound interest, which takes into account that the interest is compounded half-yearly.

The formula for compound interest is:

\[
A = P \left( 1 + \frac{R}{100n} \right)^{nt}
\]

Where:
– \( A \) is the amount after interest (Rs. 7803),
– \( P \) is the principal (which we need to find),
– \( R \) is the rate of interest per annum (4%),
– \( n \) is the number of times the interest is compounded per year (since it’s compounded half-yearly, \( n = 2 \)),
– \( t \) is the time in years (1 year in this case).

Now, substitute the values into the formula:

\[
7803 = P \left( 1 + \frac{4}{100 \times 2} \right)^{2 \times 1}
\] \[
7803 = P \left( 1 + \frac{4}{200} \right)^2
\] \[
7803 = P \left( 1 + 0.02 \right)^2
\] \[
7803 = P \times (1.02)^2
\] \[
7803 = P \times 1.0404
\]

Now, solve for \( P \):

\[
P = \frac{7803}{1.0404}
\] \[
P \approx 7500
\]

Final Answer:
The principal sum is Rs. 7500.

Answer: Option A
Solution:
To find the principal amount (sum) that amounts to Rs. 5832 in 2 years at 8% compound interest per annum, we can use the formula for compound interest:

\[
A = P \times (1 + r)^t
\]

Where:
– \( A \) is the amount (Rs. 5832),
– \( P \) is the principal (the amount we need to find),
– \( r \) is the annual interest rate (8% = 0.08),
– \( t \) is the time in years (2 years).

Rearranging the formula to solve for \( P \):

\[
P = \frac{A}{(1 + r)^t}
\]

Substituting the given values:

\[
P = \frac{5832}{(1 + 0.08)^2}
\] \[
P = \frac{5832}{1.08^2}
\] \[
P = \frac{5832}{1.1664}
\]

Now, let’s calculate the value of \( P \):

\[
P = 5000 \, \text{Rs.}
\]

Final Answer:
The principal sum is Rs. 5000.

Answer: Option C
Solution:
To calculate the compound interest on Rs. 6000 at 10% per annum for \( 1 \frac{1}{2} \) years (which is 1.5 years), when the interest is compounded annually, we use the compound interest formula:

\[
A = P \times (1 + r)^t
\]

Where:
– \( A \) is the amount after interest,
– \( P \) is the principal amount (Rs. 6000),
– \( r \) is the annual interest rate (10% = 0.10),
– \( t \) is the time in years (1.5 years).

First, calculate the amount \( A \):

\[
A = 6000 \times (1 + 0.10)^{1.5}
\] \[
A = 6000 \times (1.10)^{1.5}
\]

Now, calculate \( (1.10)^{1.5} \):

\[
(1.10)^{1.5} \approx 1.1447
\]

So:

\[
A \approx 6000 \times 1.1447 = 6868.20 \, \text{Rs.}
\]

Now, to find the compound interest:

\[
\text{Compound Interest} = A – P = 6868.20 – 6000 = 868.20 \, \text{Rs.}
\]

Final Answer:
The compound interest is Rs. 868.20.

Answer: Option A
Solution:
In this problem, we need to calculate the total amount after 3 years with compound interest while considering a 20% tax deduction on the interest earned at the end of each year. Let’s break it down step by step.

Given:
– Principal \( P = 5000 \) Rs.
– Annual interest rate \( r = 5\% = 0.05 \)
– Time period \( t = 3 \) years
– Tax rate = 20% on the interest earned each year.

Steps to Calculate:

Year 1:
– Interest earned in the first year:

\[
\text{Interest} = P \times r = 5000 \times 0.05 = 250 \, \text{Rs.}
\]

– Tax deducted on the interest:

\[
\text{Tax} = 250 \times 0.20 = 50 \, \text{Rs.}
\]

– After tax, the interest for the first year is:

\[
\text{Interest after tax} = 250 – 50 = 200 \, \text{Rs.}
\]

– The new principal after the first year becomes:

\[
\text{New Principal} = 5000 + 200 = 5200 \, \text{Rs.}
\]

Year 2:
– Interest earned in the second year on the new principal:

\[
\text{Interest} = 5200 \times 0.05 = 260 \, \text{Rs.}
\]

– Tax deducted on the interest:

\[
\text{Tax} = 260 \times 0.20 = 52 \, \text{Rs.}
\]

– After tax, the interest for the second year is:

\[
\text{Interest after tax} = 260 – 52 = 208 \, \text{Rs.}
\]

– The new principal after the second year becomes:

\[
\text{New Principal} = 5200 + 208 = 5408 \, \text{Rs.}
\]

Year 3:
– Interest earned in the third year on the new principal:

\[
\text{Interest} = 5408 \times 0.05 = 270.4 \, \text{Rs.}
\]

– Tax deducted on the interest:

\[
\text{Tax} = 270.4 \times 0.20 = 54.08 \, \text{Rs.}
\]

– After tax, the interest for the third year is:

\[
\text{Interest after tax} = 270.4 – 54.08 = 216.32 \, \text{Rs.}
\]

– The new principal after the third year becomes:

\[
\text{New Principal} = 5408 + 216.32 = 5624.32 \, \text{Rs.}
\]

Final Answer:
The amount at the end of the third year, after deducting the income tax on interest, is Rs. 5624.32.

Answer: Option C
Solution:
We are given:
– The simple interest (SI) for one year is Rs. 260.
– The compound interest (CI) for two years is Rs. 540.80.
– We need to find the rate of interest per annum.

Let’s denote:
– \( P \) as the principal amount,
– \( r \) as the rate of interest per annum (in percentage),
– \( t \) as the time period in years.

Step 1: Calculate the principal using simple interest formula

The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

For one year (\( t = 1 \)):

\[
260 = \frac{P \times r \times 1}{100}
\] \[
P \times r = 26000 \quad \text{(Equation 1)}
\]

Step 2: Use the compound interest formula

The formula for compound interest for 2 years is:

\[
CI = P \left( \left(1 + \frac{r}{100}\right)^2 – 1 \right)
\]

We know the compound interest for 2 years is Rs. 540.80, so:

\[
540.80 = P \left( \left(1 + \frac{r}{100}\right)^2 – 1 \right)
\]

Simplifying the compound interest formula:

\[
540.80 = P \left( \frac{r^2}{10000} + \frac{2r}{100} \right)
\]

Now, substitute the value of \( P \times r \) from Equation 1 into this formula:

From Equation 1, \( P \times r = 26000 \), so:

\[
540.80 = 26000 \left( \frac{r}{100} + \frac{r^2}{10000} \right)
\]

Divide both sides by 26000:

\[
\frac{540.80}{26000} = \frac{r}{100} + \frac{r^2}{10000}
\]

\[
0.0208 = \frac{r}{100} + \frac{r^2}{10000}
\]

Multiply through by 10000 to clear the denominators:

\[
208 = 100r + r^2
\]

Rearrange the equation:

\[
r^2 + 100r – 208 = 0
\]

Step 3: Solve the quadratic equation

We can solve this quadratic equation using the quadratic formula:

\[
r = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]

For the equation \( r^2 + 100r – 208 = 0 \), the coefficients are \( a = 1 \), \( b = 100 \), and \( c = -208 \). Substituting these into the quadratic formula:

\[
r = \frac{-100 \pm \sqrt{100^2 – 4 \times 1 \times (-208)}}{2 \times 1}
\] \[
r = \frac{-100 \pm \sqrt{10000 + 832}}{2}
\] \[
r = \frac{-100 \pm \sqrt{10832}}{2}
\] \[
r = \frac{-100 \pm 104.07}{2}
\]

Taking the positive root (since the rate of interest cannot be negative):

\[
r = \frac{-100 + 104.07}{2} = \frac{4.07}{2} = 2.035
\]

Thus, the rate of interest per annum is approximately 2.04%.

Final Answer:
The rate of interest per annum is approximately 2.04%.

Answer: Option D
Solution:
Let’s solve this step-by-step.

Given:
– The simple interest (SI) for 2 years is Rs. 80.
– The rate of interest is 4% per annum.
– Time period is 2 years.
– We need to find the compound interest (CI) for the same sum.

Step 1: Use the simple interest formula

The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \( P \) is the principal,
– \( r \) is the rate of interest per annum (4%),
– \( t \) is the time period in years (2 years).

Substitute the known values into the formula:

\[
80 = \frac{P \times 4 \times 2}{100}
\] \[
80 = \frac{8P}{100}
\] \[
P = \frac{80 \times 100}{8} = 1000
\]

So, the principal amount is Rs. 1000.

Step 2: Use the compound interest formula

The formula for compound interest for 2 years is:

\[
CI = P \left( \left( 1 + \frac{r}{100} \right)^t – 1 \right)
\]

Substitute the known values:

\[
CI = 1000 \left( \left( 1 + \frac{4}{100} \right)^2 – 1 \right)
\] \[
CI = 1000 \left( \left( 1.04 \right)^2 – 1 \right)
\] \[
CI = 1000 \left( 1.0816 – 1 \right)
\] \[
CI = 1000 \times 0.0816 = 81.60
\]

Final Answer:
The compound interest on the sum for 2 years is Rs. 81.60.

Answer: Option A
Solution:
To determine the time period (in years) for the compound interest, we can use the compound interest formula:

\[
CI = P \left( \left( 1 + \frac{r}{100} \right)^t – 1 \right)
\]

Where:
– \( CI \) is the compound interest (Rs. 4347),
– \( P \) is the principal amount (Rs. 30000),
– \( r \) is the rate of interest per annum (7% = 0.07),
– \( t \) is the time period (in years), which we need to find.

Step 1: Substitute the known values into the compound interest formula:

\[
4347 = 30000 \left( \left( 1 + \frac{7}{100} \right)^t – 1 \right)
\] \[
4347 = 30000 \left( (1.07)^t – 1 \right)
\]

Step 2: Simplify the equation:

\[
\frac{4347}{30000} = (1.07)^t – 1
\] \[
0.1449 = (1.07)^t – 1
\] \[
1.1449 = (1.07)^t
\]

### Step 3: Solve for \( t \):

To solve for \( t \), we take the logarithm of both sides:

\[
\log(1.1449) = t \times \log(1.07)
\]

Now, calculate the logarithms:

\[
\log(1.1449) \approx 0.0597, \quad \log(1.07) \approx 0.0294
\]

Now solve for \( t \):

\[
t = \frac{0.0597}{0.0294} \approx 2.03
\]

Final Answer:
The time period is approximately 2 years.

Answer: Option D
Solution:
We are given the following information:

– The compound interest (CI) for 2 years at 5% per annum is Rs. 246.
– The rate for simple interest (SI) is 6% per annum.
– We need to calculate the simple interest for 3 years at 6% per annum on the same sum.

Step 1: Use the compound interest formula to find the principal amount

The formula for compound interest is:

\[
CI = P \left( \left( 1 + \frac{r}{100} \right)^t – 1 \right)
\]

Where:
– \( P \) is the principal,
– \( r \) is the rate of interest per annum (5% = 0.05),
– \( t \) is the time in years (2 years),
– \( CI \) is the compound interest (Rs. 246).

Substitute the given values into the formula:

\[
246 = P \left( \left( 1 + \frac{5}{100} \right)^2 – 1 \right)
\] \[
246 = P \left( (1.05)^2 – 1 \right)
\] \[
246 = P \left( 1.1025 – 1 \right)
\] \[
246 = P \times 0.1025
\]

Now solve for \( P \):

\[
P = \frac{246}{0.1025} = 2400
\]

So, the principal amount is Rs. 2400.

Step 2: Use the simple interest formula to calculate the simple interest

The formula for simple interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Where:
– \( P \) is the principal (Rs. 2400),
– \( r \) is the rate of interest per annum (6% = 0.06),
– \( t \) is the time in years (3 years).

Substitute the given values into the formula:

\[
SI = \frac{2400 \times 6 \times 3}{100}
\] \[
SI = \frac{43200}{100} = 432
\]

Final Answer:
The simple interest on the same sum for 3 years at 6% per annum is Rs. 432.

Answer: Option A
Solution:
To solve this, let’s first define the terms:

– Let \( P \) be the principal amount.
– The rate of interest \( r = 5\% = 0.05 \).
– The time period \( t = 3 \) years.
– The difference between compound interest (CI) and simple interest (SI) is given as Rs. 122.

Step 1: Formula for Simple Interest (SI)

The formula for Simple Interest is:

\[
SI = \frac{P \times r \times t}{100}
\]

Substitute the known values:

\[
SI = \frac{P \times 5 \times 3}{100} = \frac{15P}{100} = \frac{3P}{20}
\]

Step 2: Formula for Compound Interest (CI)

The formula for Compound Interest for 3 years is:

\[
CI = P \left( \left(1 + \frac{r}{100}\right)^t – 1 \right)
\]

Substitute the known values:

\[
CI = P \left( \left( 1 + \frac{5}{100} \right)^3 – 1 \right)
\] \[
CI = P \left( (1.05)^3 – 1 \right)
\] \[
CI = P \left( 1.157625 – 1 \right)
\] \[
CI = P \times 0.157625
\]

Step 3: Difference between CI and SI

We are given that the difference between CI and SI is Rs. 122:

\[
CI – SI = 122
\]

Substitute the expressions for CI and SI:

\[
P \times 0.157625 – \frac{3P}{20} = 122
\]

Now simplify:

\[
P \times 0.157625 – P \times 0.15 = 122
\] \[
P \times (0.157625 – 0.15) = 122
\] \[
P \times 0.007625 = 122
\]

Solve for \( P \):

\[
P = \frac{122}{0.007625} = 16000
\]

Final Answer:
The sum (principal) is Rs. 16,000.

Answer: Option C
Solution:
To find the rate of interest, we can use the compound interest formula:

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

Where:
– \( A \) is the amount after interest (Rs. 2226.05),
– \( P \) is the principal (Rs. 2000),
– \( r \) is the rate of interest (which we need to find),
– \( t \) is the time period in years (2 years).

Step 1: Substitute the known values into the formula

\[
2226.05 = 2000 \left(1 + \frac{r}{100}\right)^2
\]

Step 2: Divide both sides by 2000 to isolate the compound factor

\[
\frac{2226.05}{2000} = \left(1 + \frac{r}{100}\right)^2
\] \[
1.113025 = \left(1 + \frac{r}{100}\right)^2
\]

Step 3: Take the square root of both sides to eliminate the square

\[
\sqrt{1.113025} = 1 + \frac{r}{100}
\] \[
1.055 = 1 + \frac{r}{100}
\]

Step 4: Solve for \( r \)

\[
1.055 – 1 = \frac{r}{100}
\] \[
0.055 = \frac{r}{100}
\] \[
r = 0.055 \times 100 = 5.5
\]

Final Answer:
The rate of interest is 5.5% per annum.

Answer: Option D
Solution:
We are given the following information:

– The amount after 2 years is Rs. 2420.
– The amount after 3 years is Rs. 2662.
– The interest is compounded annually.
– We need to find the rate of interest per annum.

Step 1: Let the principal amount be \( P \) and the rate of interest be \( r \% \) per annum.

We are given two amounts:
1. After 2 years, the amount is Rs. 2420.
2. After 3 years, the amount is Rs. 2662.

Step 2: Use the compound interest formula

The compound interest formula is:

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

For the amount after 2 years (\( A_2 = 2420 \)):

\[
2420 = P \left(1 + \frac{r}{100}\right)^2
\]

For the amount after 3 years (\( A_3 = 2662 \)):

\[
2662 = P \left(1 + \frac{r}{100}\right)^3
\]

Step 3: Divide the two equations to eliminate \( P \)

\[
\frac{A_3}{A_2} = \frac{P \left(1 + \frac{r}{100}\right)^3}{P \left(1 + \frac{r}{100}\right)^2}
\]

Simplify:

\[
\frac{2662}{2420} = \left(1 + \frac{r}{100}\right)
\]

\[
1.1 = \left(1 + \frac{r}{100}\right)
\]

Step 4: Solve for \( r \)

\[
1 + \frac{r}{100} = 1.1
\]

\[
\frac{r}{100} = 0.1
\]

\[
r = 0.1 \times 100 = 10
\]

Final Answer:
The rate of interest per annum is 10%.

Answer: Option B
Solution:
We are given:
– The principal amount \( P = 4000 \),
– The amount after 1 year is Rs. 4320,
– The time period is 3 years, and we need to find the amount at the end of 3 years.

Step 1: Calculate the rate of interest per annum

We know the formula for compound interest:

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

After 1 year, the amount is Rs. 4320, and the principal is Rs. 4000. So:

\[
4320 = 4000 \left(1 + \frac{r}{100}\right)^1
\]

Simplifying the equation:

\[
\frac{4320}{4000} = 1 + \frac{r}{100}
\] \[
1.08 = 1 + \frac{r}{100}
\]

Solving for \( r \):

\[
\frac{r}{100} = 0.08
\] \[
r = 0.08 \times 100 = 8
\]

So, the rate of interest is 8% per annum.

Step 2: Calculate the amount after 3 years

Now that we know the rate of interest is 8%, we can use the compound interest formula to calculate the amount after 3 years.

\[
A = 4000 \left(1 + \frac{8}{100}\right)^3
\] \[
A = 4000 \left(1.08\right)^3
\]

Calculate \( 1.08^3 \):

\[
1.08^3 \approx 1.259712
\]

Now calculate the final amount:

\[
A = 4000 \times 1.259712 \approx 5038.848
\]

Final Answer:
The amount due at the end of 3 years, rounded to the nearest rupee, is Rs. 5039.

Answer: Option D
Solution:
We are given:
– Principal \( P = 10000 \),
– Amount \( A = 14641 \),
– Time period \( t = 2 \) years,
– Interest is compounded half-yearly, and we need to find the rate of compound interest per annum (p.c.p.a.).

Step 1: Use the compound interest formula

The formula for compound interest is:

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

Where:
– \( A \) is the amount after interest (Rs. 14641),
– \( P \) is the principal (Rs. 10000),
– \( r \) is the annual rate of interest (which we need to find),
– \( n \) is the number of times the interest is compounded per year (since it’s half-yearly, \( n = 2 \)),
– \( t \) is the time period in years (2 years).

Step 2: Substitute the known values into the formula

\[
14641 = 10000 \left(1 + \frac{r}{100 \times 2}\right)^{2 \times 2}
\]

Simplify:

\[
14641 = 10000 \left(1 + \frac{r}{200}\right)^4
\]

Step 3: Divide both sides by 10000 to isolate the compound factor

\[
\frac{14641}{10000} = \left(1 + \frac{r}{200}\right)^4
\] \[
1.4641 = \left(1 + \frac{r}{200}\right)^4
\]

Step 4: Take the fourth root of both sides

\[
\sqrt[4]{1.4641} = 1 + \frac{r}{200}
\]

Calculate the fourth root of 1.4641:

\[
1.4641^{1/4} \approx 1.1
\]

Thus:

\[
1.1 = 1 + \frac{r}{200}
\]

### Step 5: Solve for \( r \)

\[
1.1 – 1 = \frac{r}{200}
\] \[
0.1 = \frac{r}{200}
\] \[
r = 0.1 \times 200 = 20
\]

Final Answer:
The rate of compound interest per annum (p.c.p.a.) is 20%.

Answer: Option B
Solution:
To solve this, we need to find the amount of each installment for a loan of Rs. 12,300 at 5% per annum compound interest, to be repaid in two equal annual installments.

Step 1: Define variables

Let the amount of each installment be \( X \).

The loan amount is Rs. 12,300, and it is to be repaid in two installments. The interest rate is 5% per annum, compounded annually.

Step 2: Apply the compound interest formula

The formula for the compound interest is:

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

Where:
– \( A \) is the total amount after interest,
– \( P \) is the principal (the loan amount),
– \( r \) is the interest rate per annum (5%),
– \( t \) is the time period in years (2 years).

The loan will be repaid in two installments, and the interest for each year will be compounded.

Step 3: Breakdown of the loan repayment

– **Installment 1**: The first installment of \( X \) is paid after 1 year.
– **Installment 2**: The second installment of \( X \) is paid after 2 years, so it will include compound interest for 2 years.

The sum of the amounts paid through the installments must cover the principal of Rs. 12,300 plus the interest over the 2 years.

Step 4: Set up the equation

After the first year, the principal is Rs. 12,300. At the end of the first year, the interest is 5%, so the amount due at the end of the first year is:

\[
12300 \times \left(1 + \frac{5}{100}\right) = 12300 \times 1.05 = 12815
\]

After the first installment \( X \) is paid, the remaining balance is:

\[
12815 – X
\]

Now, this remaining balance of \( 12815 – X \) will accrue interest for another year (i.e., compound interest will be applied for the second year):

\[
(12815 – X) \times 1.05
\]

The second installment \( X \) will pay off this amount. Therefore, the total due at the end of the second year is:

\[
(12815 – X) \times 1.05 = X
\]

Step 5: Solve for \( X \)

Now, let’s solve the equation:

\[
(12815 – X) \times 1.05 = X
\] \[
12815 \times 1.05 – X \times 1.05 = X
\] \[
13455.75 – 1.05X = X
\] \[
13455.75 = 2.05X
\] \[
X = \frac{13455.75}{2.05}
\] \[
X = 6565
\]

Final Answer:
The amount of each installment is Rs. 6565.