A sum of money becomes 3 times of itself in 5 years at certain simple interest. In what time it will becomes 9 times of itself.

Answer

To solve the problem step by step, we will first determine the rate of interest and then use it to find out the time required for the sum of money to become 9 times of itself.

### Step 1: Understand the problem
We know that a sum of money becomes 3 times itself in 5 years. We denote the principal amount as \( P \). Therefore, the amount after 5 years is \( 3P \).

### Step 2: Write the formula for Simple Interest
The formula for the amount \( A \) in terms of principal \( P \), rate of interest \( R \), and time \( T \) is:
\[ A = P + \text{SI} \]
where SI (Simple Interest) is given by:
\[ \text{SI} = \frac{P \cdot R \cdot T}{100} \]
Thus, we can rewrite the amount as:
\[ A = P + \frac{P \cdot R \cdot T}{100} \]

### Step 3: Substitute the known values for the first scenario
For the first scenario, we have:
\[ 3P = P + \frac{P \cdot R \cdot 5}{100} \]
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ 3 = 1 + \frac{R \cdot 5}{100} \]

### Step 4: Solve for \( R \)
Rearranging the equation gives:
\[ 3 - 1 = \frac{R \cdot 5}{100} \]
\[ 2 = \frac{R \cdot 5}{100} \]
Multiplying both sides by 100:
\[ 200 = R \cdot 5 \]
Dividing both sides by 5:
\[ R = 40\% \]

### Step 5: Set up the equation for the second scenario
Now, we need to find the time \( T \) when the amount becomes 9 times the principal:
\[ 9P = P + \frac{P \cdot R \cdot T}{100} \]
Again, cancel \( P \) from both sides:
\[ 9 = 1 + \frac{R \cdot T}{100} \]

### Step 6: Substitute \( R \) and solve for \( T \)
Substituting \( R = 40\% \):
\[ 9 = 1 + \frac{40 \cdot T}{100} \]
Rearranging gives:
\[ 9 - 1 = \frac{40 \cdot T}{100} \]
\[ 8 = \frac{40 \cdot T}{100} \]
Multiplying both sides by 100:
\[ 800 = 40T \]
Dividing both sides by 40:
\[ T = 20 \text{ years} \]

### Final Answer
The time it will take for the sum of money to become 9 times itself is **20 years**.

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