Simple Interest All Competence Aptitude

Let's denote the principal amount as \(P\).

According to the problem, the ratio of the principal amount and the simple interest obtained from it is \(10 : 3\). This means that the interest earned is \(\dfrac{3}{10}\) of the principal amount.

We can set up the equation based on the simple interest formula:

\[ \text{Simple Interest} = \dfrac{P \times R \times T}{100} \]

Given that the rate of interest is \(10\% = 0.10\), and the ratio of the principal to the interest is \(10 : 3\), we can express this relationship as:

\[ \dfrac{3}{10}P = \dfrac{P \times 10 \times T}{100} \]

Now, let's solve for \(T\), the time in years:

\[ \dfrac{3}{10}P = \dfrac{10PT}{100} \]\[ 3 = \dfrac{10T}{100} \]\[ 3 = \dfrac{T}{10} \]\[ T = 30 \]

So, the time for which the money was invested is \(30\) years.

However, it's unlikely that an investment would be held for 30 years at a 10% interest rate. This suggests a misunderstanding or a typo in the problem.

Let's revisit the problem:

The principal and the interest are in the ratio \(10 : 3\). This means that for every \(10\) units of the principal, \(3\) units are the interest.

Let's consider an example where the principal is \(10\) units. According to the given ratio, the interest would be \(3\) units.

Now, let's use the simple interest formula \(I = \dfrac{P \times R \times T}{100}\) to find the time for this particular case:

\[ 3 = \dfrac{10 \times 10 \times T}{100} \]\[ 3 = \dfrac{100T}{100} \]\[ 3 = T \]

So, for a principal of \(10\) units, the time is \(3\) years.

Therefore, the correct answer is \textbf{3 years}.

Let's denote the principal amount as \(P\) and the rate of interest per annum as \(R\).

According to the problem, the simple interest on a sum for 5 years is two-fifth of the sum. This can be expressed as the simple interest \(I\) being \(\dfrac{2}{5}\) of the principal amount \(P\):

\[I = \dfrac{2}{5}P\]

We know that the simple interest (\(I\)) is given by the formula:

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

Given that the time (\(T\)) is 5 years, we can substitute \(T = 5\) into the formula:

\[ \dfrac{2}{5}P = P \times R \times 5\]

Now, let's solve for \(R\), the rate of interest per annum:

\[ \dfrac{2}{5}P = 5PR\]

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

\[ \dfrac{2}{5} = 5R\]

Now, divide both sides by \(5\):

\[ \dfrac{2}{5 \times 5} = R\]

\[ \dfrac{2}{25} = R\]

So, the rate of interest per annum is \(0.08\), which is \(8\%\) in percentage.

Therefore, the correct answer is \textbf{0.08}.

Let's denote the principal amount as \(P\) and the rate of interest per annum as \(R\).

According to the problem, a certain sum of money becomes three times of itself in 20 years at simple interest. This means that the amount after 20 years (\(A\)) is three times the principal amount (\(P\)):

\[A = 3P\]

We know that the amount (\(A\)) after \(T\) years with simple interest is given by the formula:

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

Given that \(A = 3P\) and \(T = 20\) years, we can substitute these values into the formula:

\[3P = P(1 + 20R)\]

Now, let's solve for \(R\), the rate of interest per annum:

\[3 = 1 + 20R\]

\[20R = 3 - 1\]

\[20R = 2\]

\[R = \dfrac{2}{20} = 0.1\]

So, the rate of interest per annum is \(0.1\), which is \(10\%\) in percentage.

Now, we need to find how many years it takes for the sum to become double of itself at the same rate of simple interest.

Let's denote this time as \(T'\).

We know that \(A = 2P\) for this scenario.

So, we can set up the equation:

\[2P = P(1 + 0.1T')\]

\[2 = 1 + 0.1T'\]

\[0.1T' = 2 - 1\]

\[0.1T' = 1\]

\[T' = \dfrac{1}{0.1}\]

\[T' = 10\]

So, it takes \(\textbf{10 years}\) for the sum to become double of itself at the same rate of simple interest.

Therefore, the correct answer is \textbf{10 years}.

Let's denote the principal amount as \(P\) and the rate of interest per annum as \(R\).

According to the problem, a certain sum of money becomes double of itself in 10 years at simple interest. This means that the amount after 10 years (\(A\)) is double the principal amount (\(P\)):

\[A = 2P\]

We know that the amount (\(A\)) after \(T\) years with simple interest is given by the formula:

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

Given that \(A = 2P\) and \(T = 10\) years, we can substitute these values into the formula:

\[2P = P(1 + 10R)\]

Now, let's solve for \(R\), the rate of interest per annum:

\[2 = 1 + 10R\]

\[10R = 2 - 1\]

\[10R = 1\]

\[R = \dfrac{1}{10} = 0.1\]

So, the rate of interest per annum is \(0.1\), which is \(10\%\) in percentage.

Now, we need to find how many years it takes for the sum to become treble of itself at the same rate of simple interest.

Let's denote this time as \(T'\).

We know that \(A = 3P\) for this scenario.

So, we can set up the equation:

\[3P = P(1 + 0.1T')\]

\[3 = 1 + 0.1T'\]

\[0.1T' = 3 - 1\]

\[0.1T' = 2\]

\[T' = \dfrac{2}{0.1}\]

\[T' = 20\]

So, it takes \(\textbf{20 years}\) for the sum to become treble of itself at the same rate of simple interest.

Therefore, the correct answer is \textbf{20 years}.

Let's denote the principal amount as \(P\).

We're given two scenarios:

1. The simple interest earned on a certain sum for 8 months at 4% per annum is 129 less than the simple interest earned on the same sum for 15 months at 5% per annum.

2. We need to find the sum.

For the first scenario, we'll calculate the simple interest for both scenarios using the formula:

\[I = \dfrac{P \times R \times T}{100}\]

For 8 months at 4% per annum:\[I_1 = \dfrac{P \times 4 \times \dfrac{8}{12}}{100}\]

For 15 months at 5% per annum:\[I_2 = \dfrac{P \times 5 \times \dfrac{15}{12}}{100}\]

According to the problem, \(I_2\) is 129 more than \(I_1\). So, we can write:

\[I_2 - I_1 = 129\]

Substituting the expressions for \(I_1\) and \(I_2\), we get:

\[\dfrac{P \times 5 \times \dfrac{15}{12}}{100} - \dfrac{P \times 4 \times \dfrac{8}{12}}{100} = 129\]

Let's solve this equation to find the value of \(P\):

\[ \dfrac{5 \times 15 \times P}{12 \times 100} - \dfrac{4 \times 8 \times P}{12 \times 100} = 129 \]

\[ \dfrac{75P}{1200} - \dfrac{32P}{1200} = 129 \]

\[ \dfrac{75P - 32P}{1200} = 129 \]

\[ \dfrac{43P}{1200} = 129 \]

Now, let's isolate \(P\):

\[ 43P = 129 \times 1200 \]

\[ 43P = 154800 \]

\[ P = \dfrac{154800}{43} \]

\[ P ≈ 3600 \]

So, the sum is approximately \(\text{Rs.} \ 3600\).

Therefore, the correct answer is \(\textbf{3600}\).

To find out when their debts will be equal, we can set up equations for the amount of money each person owes over time.

Let's denote:
- A's initial principal amount as \( P_A = 800 \)
- B's initial principal amount as \( P_B = 910 \)
- A's annual interest rate as \( r_A = 12\% \) or \( 0.12 \)
- B's annual interest rate as \( r_B = 10\% \) or \( 0.10 \)
- The time taken for their debts to be equal as \( t \) years

The amount A owes after \( t \) years is given by:\[ A(t) = P_A + P_A \times r_A \times t \]

Similarly, the amount B owes after \( t \) years is given by:\[ B(t) = P_B + P_B \times r_B \times t \]

We want to find out when \( A(t) = B(t) \).

So, we have:\[ P_A + P_A \times r_A \times t = P_B + P_B \times r_B \times t \]

Plugging in the given values:\[ 800 + 800 \times 0.12 \times t = 910 + 910 \times 0.10 \times t \]

Solving for \( t \):\[ 800 + 96t = 910 + 91t \]\[ 96t - 91t = 910 - 800 \]\[ 5t = 110 \]\[ t = \dfrac{110}{5} \]\[ t = 22 \]

So, it will take 22 years for their debts to be equal. Therefore, the answer is:

\[ \text{In 22 years} \]

Let's denote:
- Loan amount as \( L \)
- Total interest amount as \( I \)
- The rate of interest as \( r \)
- The time period as \( t = 5 \) years

Given that the ratio of the loan amount to the total interest amount is \( 5:2 \), we can express this as:

\[ \dfrac{L}{I} = \dfrac{5}{2} \]

Now, we know that the total interest is calculated using the formula \( I = L \times r \times t \).

Substituting this into the ratio \( \dfrac{L}{I} \), we get:

\[ \dfrac{L}{L \times r \times t} = \dfrac{5}{2} \]

\[ \dfrac{1}{r \times t} = \dfrac{5}{2} \]

\[ 2 = 5 \times r \times t \]

\[ r \times t = \dfrac{2}{5} \]

Given that \( t = 5 \) years:

\[ r = \dfrac{2}{5 \times 5} \]

\[ r = \dfrac{2}{25} \]

So, the ratio of the loan amount to the interest rate is \( L : r \), which is:

\[ L : r = 1 : \dfrac{2}{25} \]

Multiplying both sides by 25 to eliminate the fraction, we get:

\[ L : r = 25 : 2 \]

Therefore, the correct answer is:

\[ \text{The ratio of loan amount : interest rate is } 25 : 2 \]

Given that the ratio of principal to simple interest for 1 year is \( 25:1 \), we can express this as:

\[ \dfrac{P}{SI} = \dfrac{25}{1} \]

Here, \( P \) represents the principal and \( SI \) represents the simple interest.

We know that simple interest (\( SI \)) is calculated using the formula:

\[ SI = P \times r \times t \]

Given that the ratio of \( P \) to \( SI \) is \( 25:1 \) and the time period (\( t \)) is \( 1 \) year, we can deduce that \( SI = \dfrac{1}{25}P \).

So, substituting this value of \( SI \) into the formula for simple interest:

\[ \dfrac{1}{25}P = P \times r \times 1 \]

\[ \dfrac{1}{25} = r \]

\[ r = \dfrac{1}{25} \]

\[ r = 0.04 \]

To convert this into percentage:

\[ r = 0.04 \times 100\% = 4\% \]

So, the rate of interest is \( 4\% \).

Therefore, the correct answer is:

\[ \text{The rate of interest is } 4\% \]

Let's denote:
- The original rate of interest as \( r \)
- The principal amount (sum) as \( P \)

According to the given condition, if the sum had been invested at a rate \( 3\% \) higher, the interest would have been \( 72 \) more. This can be expressed as an equation:

\[ P \times (r + 0.03) \times 2 - P \times r \times 2 = 72 \]

Simplifying this equation, we get:

\[ 2P(r + 0.03) - 2Pr = 72 \]

\[ 2Pr + 0.06P - 2Pr = 72 \]

\[ 0.06P = 72 \]

Now, we can solve for \( P \):

\[ P = \dfrac{72}{0.06} \]

\[ P = 1200 \]

So, the sum (principal amount) is \( \$1200 \).

Therefore, the correct answer is:

\[ \text{The sum is } \$1200 \]

Let's denote:
- The principal amount as \( P = 2100 \)
- The interest rate as \( r \)
- The interest earned in 2 years at the original rate as \( I_1 \)
- The interest earned in 2 years at the new rate as \( I_2 \)

We are given that the amount \( 2100 \) becomes \( 2352 \) in 2 years at simple interest. Therefore, the interest earned in 2 years at the original rate is:

\[ I_1 = 2352 - 2100 = 252 \]

We need to find the new interest earned when the interest rate is decreased by \( 1\% \). The new interest rate would be \( r - 0.01 \).

The formula for simple interest is:

\[ I = P \times r \times t \]

So, the new interest earned in 2 years at the decreased rate is:

\[ I_2 = P \times (r - 0.01) \times 2 \]

We're given that \( I_1 = 252 \), so:

\[ 252 = 2100 \times (r - 0.01) \times 2 \]

Solving for \( r - 0.01 \):

\[ 252 = 4200(r - 0.01) \]

\[ 252 = 4200r - 42 \]

\[ 4200r = 252 + 42 \]

\[ 4200r = 294 \]

\[ r = \dfrac{294}{4200} \]

\[ r = 0.07 \]

So, the new interest rate (\( r \)) is \( 7\% \).

Now, we can calculate the new interest (\( I_2 \)):

\[ I_2 = 2100 \times (0.07) \times 2 \]

\[ I_2 = 210 \]

So, the new interest earned when the interest rate is decreased by \( 1\% \) is \( \$210 \).

Therefore, the correct answer is:

\[ \text{The new interest is } \$210 \]