The population of town B is 300%more than that of town A.For the next two years. the population ofA increases by x% per year and that of B decreases by the same percentage per year. After 2 years, if the population of A and B become equal, then the value of x is .........

Options

A. `30(2/3)`
B. `33(1/3)`
C. `40`
D. `25`

Answer

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

### Step 1: Define the populations of towns A and B
Let the population of town A be \( P_A = 100 \). According to the problem, the population of town B is 300% more than that of town A.

\[ P_B = P_A + 300\% \text{ of } P_A = 100 + 300 = 400 \]

### Step 2: Set up the population growth and decline equations
For the next two years, the population of A increases by \( x\% \) per year, and the population of B decreases by \( x\% \) per year.

The population of A after 2 years can be calculated using the formula for compound interest:

\[ P_A' = P_A \times \left(1 + \frac{x}{100}\right)^2 \]

The population of B after 2 years will be:

\[ P_B' = P_B \times \left(1 - \frac{x}{100}\right)^2 \]

### Step 3: Set the populations equal after 2 years
According to the problem, after 2 years, the populations of A and B become equal:

\[ P_A' = P_B' \]

Substituting the expressions we derived:

\[ 100 \times \left(1 + \frac{x}{100}\right)^2 = 400 \times \left(1 - \frac{x}{100}\right)^2 \]

### Step 4: Simplify the equation
Dividing both sides by 100:

\[ \left(1 + \frac{x}{100}\right)^2 = 4 \times \left(1 - \frac{x}{100}\right)^2 \]

### Step 5: Expand both sides
Expanding both sides gives:

\[ \left(1 + \frac{x}{100}\right)^2 = 1 + 2\left(\frac{x}{100}\right) + \left(\frac{x}{100}\right)^2 \]
\[ 4 \times \left(1 - \frac{x}{100}\right)^2 = 4 \left(1 - 2\left(\frac{x}{100}\right) + \left(\frac{x}{100}\right)^2\right) = 4 - 8\left(\frac{x}{100}\right) + 4\left(\frac{x}{100}\right)^2 \]

### Step 6: Set the expanded equations equal
Setting the two expansions equal to each other:

\[ 1 + 2\left(\frac{x}{100}\right) + \left(\frac{x}{100}\right)^2 = 4 - 8\left(\frac{x}{100}\right) + 4\left(\frac{x}{100}\right)^2 \]

### Step 7: Rearrange the equation
Rearranging gives:

\[ 0 = 3 - 10\left(\frac{x}{100}\right) + 3\left(\frac{x}{100}\right)^2 \]

### Step 8: Multiply through by 100² to eliminate fractions
\[ 0 = 30000 - 1000x + 3x^2 \]

### Step 9: Rearrange into standard quadratic form
\[ 3x^2 - 1000x + 30000 = 0 \]

### Step 10: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 3, b = -1000, c = 30000 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-1000)^2 - 4 \times 3 \times 30000 = 1000000 - 360000 = 640000 \]

Now applying the quadratic formula:

\[ x = \frac{1000 \pm \sqrt{640000}}{6} \]
\[ x = \frac{1000 \pm 800}{6} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{1800}{6} = 300 \)
2. \( x = \frac{200}{6} = \frac{100}{3} \approx 33.33 \)

### Final Answer
The value of \( x \) that satisfies the condition is:

\[ x = \frac{100}{3} \text{ or } 33 \frac{1}{3} \% \]