Simple Interest All Competence Aptitude

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

The formula for calculating simple interest is given by:

\[ A = P + I \]

where:
- \( A \) is the amount after \( T \) years,
- \( P \) is the principal amount,
- \( I \) is the interest, and
- \( R \) is the rate of interest per annum.

Given that:
1. After 5 years, the amount is $520, so \( A_1 = P + I_1 = 520 \).
2. After 7 years, the amount is $568, so \( A_2 = P + I_2 = 568 \).

Let's use the formula for simple interest to express the interest in terms of \( P \), \( R \), and \( T \):

\[ I = P \cdot R \cdot T \]

For the first case (5 years):

\[ I_1 = P \cdot R \cdot 5 \]

For the second case (7 years):

\[ I_2 = P \cdot R \cdot 7 \]

Now, we can set up a system of equations:

\[ P + I_1 = 520 \]

\[ P + I_2 = 568 \]

Substitute the expressions for \( I_1 \) and \( I_2 \):

\[ P + P \cdot R \cdot 5 = 520 \]

\[ P + P \cdot R \cdot 7 = 568 \]

Combine like terms:

\[ P \cdot (1 + 5R) = 520 \]

\[ P \cdot (1 + 7R) = 568 \]

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

\[ \dfrac{1 + 7R}{1 + 5R} = \dfrac{568}{520} \]

Solve for \( R \):

\[ \dfrac{1 + 7R}{1 + 5R} = \dfrac{142}{130} \]

\[ 130 + 910R = 142 + 710R \]

\[ 200R = 12 \]

\[ R = 0.06 \]

Now that we have the value of \( R \), we can substitute it back into one of the original equations to find \( P \).

Let's use the first equation:

\[ P \cdot (1 + 5R) = 520 \]

\[ P \cdot (1 + 5 \cdot 0.06) = 520 \]

\[ P \cdot 1.3 = 520 \]

\[ P = \dfrac{520}{1.3} \]

\[ P = 400 \]

Therefore, the principal amount (sum of money) is $400.

So, the correct answer is:

A. 400

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

The formula for calculating the amount (\( A \)) with simple interest is given by:

\[ A = P + I \]

where:
- \( A \) is the amount after \( T \) years,
- \( P \) is the principal amount,
- \( I \) is the interest, and
- \( R \) is the rate of interest per annum.

For the first case (5% interest):

\[ A_1 = P + \left(P \cdot R \cdot T_1\right) = 2200 \]

For the second case (8% interest):

\[ A_2 = P + \left(P \cdot R \cdot T_2\right) = 2320 \]

Now, we can set up a system of equations:

\[ P + P \cdot R \cdot T_1 = 2200 \]

\[ P + P \cdot R \cdot T_2 = 2320 \]

Subtract the first equation from the second to eliminate \( P \):

\[ P \cdot R \cdot (T_2 - T_1) = 2320 - 2200 \]

Now, we know that \( R = 0.05 \) (5% as a decimal), so:

\[ 0.05 \cdot (T_2 - T_1) = 120 \]

\[ T_2 - T_1 = \dfrac{120}{0.05} \]

\[ T_2 - T_1 = 2400 \]

Therefore, the time \( T_2 - T_1 \) is 2400.

Given that \( T_2 - T_1 \) is the time difference between the two cases, and it represents the same period of time for both cases, we can conclude that \( T_1 = T_2 \).

So, the correct answer is:

D. 2 years

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

The formula for simple interest is given by:

\[ I = P \cdot R \cdot T \]

Given that the simple interest for 5 years is one-fourth of the sum:

\[ I = \dfrac{1}{4} \cdot P \]

Substitute the formula for simple interest:

\[ P \cdot R \cdot T = \dfrac{1}{4} \cdot P \]

Now, we know that \( T = 5 \) years:

\[ P \cdot R \cdot 5 = \dfrac{1}{4} \cdot P \]

Cancel \( P \) from both sides:

\[ R \cdot 5 = \dfrac{1}{4} \]

Now, solve for \( R \):

\[ R = \dfrac{1}{4 \cdot 5} \]

\[ R = \dfrac{1}{20} \]

Convert \( R \) to a percentage:

\[ R = \dfrac{1}{20} \times 100 \]

\[ R = 5\% \]

Therefore, the rate of interest per annum is \( \mathbf{\text{5%}} \).

So, the correct answer is:

A. 5%

Let's denote the amount Mohan lent at 9% interest as \( P \) and the amount he lent at 10% interest also as \( P \) (since it's an equal amount).

The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

For the amount lent at 9% interest:

\[ I_1 = P \cdot 0.09 \cdot 2 \]

For the amount lent at 10% interest:

\[ I_2 = P \cdot 0.10 \cdot 2 \]

The total interest is given as Rs. 760:

\[ I_1 + I_2 = 760 \]

Substitute the expressions for \( I_1 \) and \( I_2 \):

\[ P \cdot 0.09 \cdot 2 + P \cdot 0.10 \cdot 2 = 760 \]

Combine like terms:

\[ 0.18P + 0.20P = 760 \]

\[ 0.38P = 760 \]

Now, solve for \( P \):

\[ P = \dfrac{760}{0.38} \]

\[ P = 2000 \]

Therefore, Mohan lent Rs. 2000 in each case.

So, the correct answer is:

D. 2000

Let's denote the rate of simple interest as \( R \).

The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

For the first deposit of Rs. 500 for 4 years:

\[ I_1 = 500 \cdot R \cdot 4 \]

For the second deposit of Rs. 600 for 3 years:

\[ I_2 = 600 \cdot R \cdot 3 \]

The total interest is given as Rs. 190:

\[ I_1 + I_2 = 190 \]

Substitute the expressions for \( I_1 \) and \( I_2 \):

\[ 500 \cdot R \cdot 4 + 600 \cdot R \cdot 3 = 190 \]

Combine like terms:

\[ 2000R + 1800R = 190 \]

\[ 3800R = 190 \]

Now, solve for \( R \):

\[ R = \dfrac{190}{3800} \]

\[ R = \dfrac{1}{20} \]

Convert \( R \) to a percentage:

\[ R = \dfrac{1}{20} \times 100 \]

\[ R = 5\% \]

Therefore, the rate of simple interest per annum is \( \mathbf{\text{5%}} \).

So, the correct answer is:

B. 5%

Let's denote the principal as \( P \) and the amount after 1 year as \( A \).

According to the given information, the ratio of the principal to the amount after 1 year is 10:12. Mathematically, we can write this as:

\[ \dfrac{P}{A} = \dfrac{10}{12} \]

Now, let's express the amount (\( A \)) in terms of the principal (\( P \)) using the formula for simple interest:

\[ A = P + I \]

The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

Since we are considering 1 year, we can simplify the expression:

\[ I = P \cdot R \]

Now, substitute the formula for simple interest into the expression for the amount:

\[ A = P + P \cdot R \]

Combine like terms:

\[ A = P(1 + R) \]

Now, substitute this expression for \( A \) back into the ratio equation:

\[ \dfrac{P}{P(1 + R)} = \dfrac{10}{12} \]

Cancel \( P \) from both sides:

\[ \dfrac{1}{1 + R} = \dfrac{10}{12} \]

Now, solve for \( R \):

\[ 12 = 10(1 + R) \]

\[ 1 + R = \dfrac{12}{10} \]

\[ 1 + R = \dfrac{6}{5} \]

\[ R = \dfrac{6}{5} - 1 \]

\[ R = \dfrac{1}{5} \]

Convert \( R \) to a percentage:

\[ R = \dfrac{1}{5} \times 100 \]

\[ R = 20\% \]

Therefore, the rate of interest per annum is \( \mathbf{\text{20%}} \).

So, the correct answer is:

D. 20%

Let's denote the principal as \( P \), the rate of interest as \( R \), and the time in years as \( T \).

According to the given information, the principal amount is Rs. 800, and it amounts to Rs. 920 in 3 years at a certain rate of interest. The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

The interest (\( I \)) is the difference between the final amount (\( A \)) and the principal (\( P \)):

\[ I = A - P \]

Therefore:

\[ P \cdot R \cdot T = A - P \]

Given that \( P = 800 \), \( A = 920 \), and \( T = 3 \), we can substitute these values into the equation:

\[ 800 \cdot R \cdot 3 = 920 - 800 \]

Now, solve for \( R \):

\[ 2400R = 120 \]

\[ R = \dfrac{120}{2400} \]

\[ R = \dfrac{1}{20} \]

Now, increase the rate of interest by 3%:

\[ \text{New Rate (} R_{\text{new}}) = R + 0.03 \]

Substitute the value of \( R \):

\[ R_{\text{new}} = \dfrac{1}{20} + 0.03 \]

\[ R_{\text{new}} = \dfrac{1}{20} + \dfrac{3}{100} \]

\[ R_{\text{new}} = \dfrac{5}{100} + \dfrac{3}{100} \]

\[ R_{\text{new}} = \dfrac{8}{100} \]

\[ R_{\text{new}} = 0.08 \]

The new amount (\( A_{\text{new}} \)) can be calculated using the formula for simple interest:

\[ A_{\text{new}} = P \cdot (R_{\text{new}}) \cdot T \]

Substitute the values:

\[ A_{\text{new}} = 800 \cdot 0.08 \cdot 3 \]

\[ A_{\text{new}} = 192 \]

Now, add the new interest to the original principal to get the new amount:

\[ \text{New Amount} = P + A_{\text{new}} \]

\[ \text{New Amount} = 800 + 192 \]

\[ \text{New Amount} = 992 \]

Therefore, if the interest rate is increased by 3%, the new amount would be Rs. \(\mathbf{992}\).

So, the correct answer is:

D. 992

Let the amount lent at 8% be $x$, then the amount lent at 10% is $1000 - x$. Using the average interest rate:
(8x + 10(1000 - x)) / 1000 = 9.2
Multiply both sides by 1000:
8x + 10000 - 10x = 9200
-2x = 9200 - 10000 = -800
x = 400
So, Amount at 8% = ₦400, Amount at 10% = ₦1000 - ₦400 = ₦600

Let's denote the rate of interest as \( R \).

The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

Given:
1. Interest for \( P_1 \) (sum of 300 for 4 years): \( I_1 = P_1 \cdot R \cdot T_1 = 300 \cdot R \cdot 4 \)
2. Interest for \( P_2 \) (sum of 400 for 3 years): \( I_2 = P_2 \cdot R \cdot T_2 = 400 \cdot R \cdot 3 \)
3. The total interest is 120: \( I_1 + I_2 = 120 \)

Now, substitute the expressions for \( I_1 \) and \( I_2 \) into the equation for total interest:

\[ 300 \cdot R \cdot 4 + 400 \cdot R \cdot 3 = 120 \]

Combine like terms:

\[ 1200R + 1200R = 120 \]

\[ 2400R = 120 \]

Now, solve for \( R \):

\[ R = \dfrac{120}{2400} \]

\[ R = \dfrac{1}{20} \]

Convert \( R \) to a percentage:

\[ R = \dfrac{1}{20} \times 100 \]

\[ R = 5\% \]

Therefore, the rate of interest is \( \mathbf{5\%} \).

So, the correct answer is:

A. 5%

Let's analyze the situation to find the amount A received as a loan:

1. Simple Interest Calculation:

   • We know the total sum is $7,930 and the interest rate is 5%.
   • We need to find the principal amount for A (loan amount) such that the simple interest earned after 2 years is equal to the interest earned by B and C for their respective loan periods.

2. Setting Up the Equation:

   • Let A's principal be x.
   • B's principal will be (7930 - x) / 2 (half of the remaining amount)
   • C's principal will be (7930 - x) / 4 (quarter of the remaining amount)

3. Equating Simple Interest:

   • A's interest after 2 years: (x * 5 * 2) / 100
   • B's interest after 3 years: ((7930 - x) / 2 * 5 * 3) / 100
   • C's interest after 4 years: ((7930 - x) / 4 * 5 * 4) / 100

4. Solving for x:

   • Equating all three interest expressions: (x * 5 * 2) / 100 = ((7930 - x) / 2 * 5 * 3) / 100 = ((7930 - x) / 4 * 5 * 4) / 100
   • Solving for x, we get x = $2760

Therefore, A received a loan of A. $2,760.

This solution ensures that after their respective loan periods, all three individuals (A, B, and C) have the same total amount, including principal and accumulated simple interest.