Let B's initial salary be represented by $S_B$.

A's initial salary, $S_A$, is 40% more than B's. This can be written as:

$S_A = S_B + 0.40 \times S_B = 1.40 \times S_B$

Salary Percentage Changes

B's salary increases by 20%. The new salary for B, $S_{B\_new}$, is:

$S_{B\_new} = S_B + 0.20 \times S_B = 1.20 \times S_B$

A's salary increases by x%. The new salary for A, $S_{A\_new}$, is:

$S_{A\_new} = S_A + \frac{x}{100} \times S_A = S_A \times (1 + \frac{x}{100})$

Substitute the expression for $S_A$:

$S_{A\_new} = (1.40 \times S_B) \times (1 + \frac{x}{100})$

New Salary Comparison

The problem states that A's new salary is 25% more than B's new salary. This means:

$S_{A\_new} = S_{B\_new} + 0.25 \times S_{B\_new} = 1.25 \times S_{B\_new}$

Solving for x

Now, we equate the two expressions for $S_{A\_new}$:

$(1.40 \times S_B) \times (1 + \frac{x}{100}) = 1.25 \times S_{B\_new}$

Substitute the expression for $S_{B\_new}$:

$(1.40 \times S_B) \times (1 + \frac{x}{100}) = 1.25 \times (1.20 \times S_B)$

Divide both sides by $S_B$ (assuming $S_B \neq 0$):

$1.40 \times (1 + \frac{x}{100}) = 1.25 \times 1.20$

Calculate the right side:

$1.40 \times (1 + \frac{x}{100}) = 1.50$

Isolate the term with x:

$1 + \frac{x}{100} = \frac{1.50}{1.40}$

$1 + \frac{x}{100} = \frac{15}{14}$

Solve for $\frac{x}{100}$:

$\frac{x}{100} = \frac{15}{14} - 1 = \frac{15 - 14}{14} = \frac{1}{14}$

Solve for x:

$x = \frac{100}{14} = \frac{50}{7} \approx 7.1428...$

Therefore, the value of x is approximately 7.14%.