If certain amount becomes double of itself in 10 years at the certain rate of compound interest then in what time period that amount will become 4times of itself?

Options

A. 20 year
B. 10 year
C. 22 year
D. 25 year

Answer

To solve the problem step by step, we will follow the logic used in the video transcript.

### Step 1: Understand the Problem
We have a principal amount \( P \) that doubles in 10 years at a certain rate of compound interest. We need to find out how long it will take for the same amount to become four times itself.

### Step 2: Set Up the Initial Equation
Since the amount doubles in 10 years, we can write:
\[ A = P(1 + \frac{r}{100})^n \]
Where:
- \( A \) is the amount after time \( n \),
- \( P \) is the principal,
- \( r \) is the rate of interest,
- \( n \) is the time in years.

For our case:
\[ 2P = P(1 + \frac{r}{100})^{10} \]

### Step 3: Simplify the Equation
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ 2 = (1 + \frac{r}{100})^{10} \]

### Step 4: Take the 10th Root
To isolate \( 1 + \frac{r}{100} \), we take the 10th root of both sides:
\[ 1 + \frac{r}{100} = 2^{\frac{1}{10}} \]

### Step 5: Set Up the New Equation for 4 Times the Principal
Now we want to find out when the amount becomes \( 4P \):
\[ 4P = P(1 + \frac{r}{100})^t \]
Again, we can cancel \( P \):
\[ 4 = (1 + \frac{r}{100})^t \]

### Step 6: Substitute the Value of \( 1 + \frac{r}{100} \)
From our previous step, we know:
\[ 1 + \frac{r}{100} = 2^{\frac{1}{10}} \]
Substituting this into the equation gives:
\[ 4 = (2^{\frac{1}{10}})^t \]

### Step 7: Rewrite 4 as a Power of 2
We can express 4 as \( 2^2 \):
\[ 2^2 = (2^{\frac{1}{10}})^t \]

### Step 8: Equate the Exponents
Since the bases are the same, we can equate the exponents:
\[ 2 = \frac{t}{10} \]

### Step 9: Solve for \( t \)
Multiplying both sides by 10 gives:
\[ t = 20 \]

### Conclusion
Thus, the time period for the amount to become four times itself is **20 years**.