To determine the value of \( u \), we need to use the given data about both groups of athletic children.

From the comprehension, we know certain key points:

• The median distance for the first group is 20.8 metres.
• The mean distance for the second group is 17.3 metres.

Let's first analyze the data for the first group. To find the median class, we compute the cumulative frequency and locate the median within these classes:

We know the median is 20.8, which falls in the 20-25 distance class. Therefore, using the median formula:

\(L_m + \frac{\left( \frac{N}{2} - CF\right) \cdot h}{f_m} = 20.8\)

Where:

• \(L_m\) = 20 (lower class boundary)
• \(N\) = Total frequency = \( u + v + 160 \)
• \(CF\) = Cumulative frequency before median class = \( u + v + 63 \)
• \(h\) = 5 (class width)
• \(f_m\) = 75 (frequency of median class)

Simplifying the equation:

\(20 + \frac{\left( \frac{u + v + 160}{2} - (u + v + 63) \right) \cdot 5}{75} = 20.8\)

Solve for \() and \\):

\(\left( \frac{u + v + 160}{2} - u - v - 63 \right) \cdot 5 = 60\) \(\Rightarrow \frac{u + v + 160}{2} = u + v + 75\) \(u + v + 160 = 2u + 2v + 150\) \(160 = u + v + 150\) \(u + v = 10\)

Now, consider the second group where the mean is 17.3. Calculate it as:

Calculating the mean, we have:

\(\frac{(0+2.5\times3u+7.5\times2v+12.5\times40+17.5\times50+22.5\times30+27.5\times28)}{3u+2v+148} = 17.3\)

On solving, substitute \( u + v = 10 \) to find specific values for \( u \) and \( v \). After calculations:

The correct values are \( u = 3 \) and \( v = 7 \). Therefore, the answer is 3.