A sum of money at a certain rate per annum of simple interest doubles in the 5 years and at a difference rate becomes three times in 12 years. The lower rate of interest per annum is

Options

A. 15%
B. 20%
C. `15(3)/(4) %`
D. `16(2)/(3) %`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understanding the Problem
We have two scenarios:
1. A sum of money doubles in 5 years at a certain rate of simple interest.
2. The same sum of money becomes three times in 12 years at a different rate of simple interest.

We need to find the lower rate of interest per annum.

### Step 2: Setting Up the First Scenario
Let the principal amount be \( P \).

According to the first scenario, the amount doubles in 5 years. Therefore:
- Amount after 5 years = \( 2P \)
- Interest earned in 5 years = \( 2P - P = P \)

Using the formula for simple interest:
\[ \text{Simple Interest (SI)} = \frac{P \times R \times T}{100} \]
Where:
- \( SI = P \)
- \( R \) = rate of interest per annum
- \( T = 5 \) years

Substituting the values, we get:
\[ P = \frac{P \times R \times 5}{100} \]

### Step 3: Simplifying the First Equation
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ 1 = \frac{R \times 5}{100} \]
Multiplying both sides by 100:
\[ 100 = 5R \]
Dividing by 5:
\[ R = 20\% \]

### Step 4: Setting Up the Second Scenario
Now, for the second scenario, the amount becomes three times in 12 years. Therefore:
- Amount after 12 years = \( 3P \)
- Interest earned in 12 years = \( 3P - P = 2P \)

Using the same formula for simple interest:
\[ 2P = \frac{P \times r \times 12}{100} \]
Where \( r \) is the new rate of interest.

### Step 5: Simplifying the Second Equation
Again, we can cancel \( P \) from both sides:
\[ 2 = \frac{r \times 12}{100} \]
Multiplying both sides by 100:
\[ 200 = 12r \]
Dividing by 12:
\[ r = \frac{200}{12} = \frac{50}{3} \approx 16.67\% \]

### Conclusion
The lower rate of interest per annum is:
\[ \text{Lower Rate} = \frac{50}{3}\% \text{ or } 16\frac{2}{3}\% \]