A sum of money lent out at C.I at a certain r% (rate of growth) doubles itself in 5 years. Find in how many years will the money become eight times of itself at the same rate of growth.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We know that a sum of money doubles itself in 5 years at a certain rate of interest (r%). We need to find out how many years it will take for the same sum of money to become eight times itself at the same rate.

### Step 2: Set up the initial condition
Let the principal amount be \( P = X \). According to the problem, the amount after 5 years is double the principal:
\[ A = 2X \]
Using the compound interest formula:
\[ A = P \left(1 + \frac{r}{100}\right)^T \]
Substituting the values we have:
\[ 2X = X \left(1 + \frac{r}{100}\right)^5 \]

### Step 3: Simplify the equation
We can cancel \( X \) from both sides (assuming \( X \neq 0 \)):
\[ 2 = \left(1 + \frac{r}{100}\right)^5 \]

### Step 4: Find the relationship for eight times
Now, we need to find out when the amount becomes eight times the principal:
\[ A = 8X \]
Using the same compound interest formula:
\[ 8X = X \left(1 + \frac{r}{100}\right)^T \]
Again, cancel \( X \):
\[ 8 = \left(1 + \frac{r}{100}\right)^T \]

### Step 5: Relate the two equations
From the first equation, we have:
\[ \left(1 + \frac{r}{100}\right)^5 = 2 \]
From the second equation, we can express 8 as:
\[ 8 = 2^3 \]
Thus, we can write:
\[ \left(1 + \frac{r}{100}\right)^T = 2^3 \]

### Step 6: Substitute the first equation into the second
We can substitute \( \left(1 + \frac{r}{100}\right)^5 = 2 \) into the expression for \( 8 \):
\[ \left(1 + \frac{r}{100}\right)^T = \left(1 + \frac{r}{100}\right)^{5 \cdot \frac{T}{5}} = 2^3 \]

### Step 7: Set the exponents equal
Since both sides have the same base, we can equate the exponents:
\[ T = 5 \cdot 3 = 15 \]

### Conclusion
Thus, the time taken for the money to become eight times itself is:
\[ \boxed{15 \text{ years}} \]

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