To solve the equation \( x:2 + x:4 + x:6 + \ldots + x:9900 = 99 \), we first need to understand the notation \( x:n \). This notation typically means \( \frac{x}{n} \). So, the given equation can be rewritten as:

\[ \frac{x}{2} + \frac{x}{4} + \frac{x}{6} + \ldots + \frac{x}{9900} = 99 \]

Next, we observe that the terms form a series where the denominators are the even numbers from 2 to 9900. We can factor out \( x \) from each term:

\[ x \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{9900} \right) = 99 \]

Now, we need to find the sum of the series inside the parentheses:

\[ S = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{9900} \]

This is a harmonic series of the even numbers. The sum of the reciprocals of the first \( n \) even numbers can be approximated by:

\[ S \approx \frac{1}{2} \left( \ln(n) + \gamma \right) \]

where \( \gamma \) is the Euler-Mascheroni constant, approximately 0.577, and \( n \) is the number of terms. In this case, \( n = 4950 \) because there are 4950 even numbers from 2 to 9900.

However, for simplicity, we can approximate the sum of the series by considering the sum of the harmonic series up to \( 2n \) and then dividing by 2:

\[ S \approx \frac{1}{2} \left( \ln(9900) + \gamma \right) \]

Using \( \ln(9900) \approx 9.2 \):

\[ S \approx \frac{1}{2} (9.2 + 0.577) \approx \frac{1}{2} \times 9.777 \approx 4.8885 \]

Now, we have:

\[ x \cdot 4.8885 = 99 \]

Solving for \( x \):

\[ x = \frac{99}{4.8885} \approx 20.25 \]

Therefore, the value of \( x \) that satisfies the equation is approximately \( 20.25 \).