To find the value of \( x \) given that the mean of the data is 40, we will use the formula for the mean of a grouped frequency distribution:

\(\text{Mean} = \frac{\sum{f_i \cdot x_i}}{\sum{f_i}}\)

Here, \( f_i \) denotes the frequency of the class, and \( x_i \) is the mid-point of each class interval.

Let's calculate \( x_i \) for each class interval:

• Mid-point of 10-20 = \(\frac{10 + 20}{2} = 15\)
• Mid-point of 20-30 = \(\frac{20 + 30}{2} = 25\)
• Mid-point of 30-40 = \(\frac{30 + 40}{2} = 35\)
• Mid-point of 40-50 = \(\frac{40 + 50}{2} = 45\)
• Mid-point of 50-60 = \(\frac{50 + 60}{2} = 55\)

We can now calculate the sum of the products of frequencies and their respective mid-points:

\(\sum{f_i \cdot x_i} = 30 \cdot 15 + 27 \cdot 25 + 44 \cdot 35 + 29 \cdot 45 + x \cdot 55\)

\(= 450 + 675 + 1540 + 1305 + 55x\)

\(= 3970 + 55x\)

The total frequency is:

\(\sum{f_i} = 30 + 27 + 44 + 29 + x\)

\(= 130 + x\)

Since the mean is 40, we equate it to the expression we derived for the mean:

\(\frac{3970 + 55x}{130 + x} = 40\)

Cross-multiplying gives:

\(3970 + 55x = 40(130 + x)\)

\(3970 + 55x = 5200 + 40x\)

Simplify and solve for \( x \):

\(55x - 40x = 5200 - 3970\)

\(15x = 1230\)

\(x = \frac{1230}{15} = 82\)

Therefore, the value of \( x \) is 82.