Two substances A and B are present such that `[A_(0)]=4[B_(0]` and half-life of A is 5 minutes and that of B is 15 minutes. If they start decaying at the same time following first order kinetics after how much time the concentration of both of them would be same ?

Explanation

To solve the problem, we need to determine the time at which the concentrations of substances A and B become equal, given their initial concentrations and half-lives.

### Step-by-Step Solution:

1. **Define Initial Concentrations**:
Given that the initial concentration of A is four times that of B:
\[ [A_0] = 4[B_0] \]

2. **Half-Lives**:
The half-life of substance A is 5 minutes, and the half-life of substance B is 15 minutes.

3. **Concentration After n Half-Lives**:
The concentration of A after \( n_1 \) half-lives is given by:
\[ [A] = [A_0] \left(\frac{1}{2}\right)^{n_1} = 4[B_0] \left(\frac{1}{2}\right)^{n_1} \]
The concentration of B after \( n_2 \) half-lives is given by:
\[ [B] = [B_0] \left(\frac{1}{2}\right)^{n_2} \]

4. **Setting Concentrations Equal**:
We want to find the time when the concentrations of A and B are equal:
\[ 4[B_0] \left(\frac{1}{2}\right)^{n_1} = [B_0] \left(\frac{1}{2}\right)^{n_2} \]
Dividing both sides by \([B_0]\) (assuming \([B_0] \neq 0\)):
\[ 4 \left(\frac{1}{2}\right)^{n_1} = \left(\frac{1}{2}\right)^{n_2} \]

5. **Rearranging the Equation**:
This can be rewritten as:
\[ \frac{4}{1} = \left(\frac{1}{2}\right)^{n_2 - n_1} \]
Since \(4 = 2^2\), we can express this as:
\[ 2^2 = \left(\frac{1}{2}\right)^{n_2 - n_1} \]
This implies:
\[ 2 = n_1 - n_2 \]

6. **Relating n1 and n2**:
We know that the time taken for \( n_1 \) half-lives of A and \( n_2 \) half-lives of B can be expressed as:
\[ t = n_1 \cdot t_{1/2,A} = n_2 \cdot t_{1/2,B} \]
Substituting the values of half-lives:
\[ n_1 \cdot 5 = n_2 \cdot 15 \]
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{15}{5} = 3 \quad \Rightarrow \quad n_1 = 3n_2 \]

7. **Substituting Back**:
From \( n_1 - n_2 = 2 \) and \( n_1 = 3n_2 \):
\[ 3n_2 - n_2 = 2 \quad \Rightarrow \quad 2n_2 = 2 \quad \Rightarrow \quad n_2 = 1 \]
Therefore, substituting back gives:
\[ n_1 = 3 \cdot 1 = 3 \]

8. **Calculating Time**:
Now substituting \( n_1 \) back into the time equation:
\[ t = n_1 \cdot t_{1/2,A} = 3 \cdot 5 = 15 \text{ minutes} \]

### Final Answer:
The concentration of both substances A and B will be the same after **15 minutes**.