To find the value of \(x\) when the mean of the given frequency distribution is 45, we proceed as follows:

1. First, identify the midpoints of each class interval, which are necessary for calculating the weighted average (mean). These are computed by averaging the boundaries of each class.
   • For class 10-20, midpoint = \(\frac{10+20}{2} = 15\)
   • For class 20-30, midpoint = \(\frac{20+30}{2} = 25\)
   • For class 30-40, midpoint = \(\frac{30+40}{2} = 35\)
   • For class 40-50, midpoint = \(\frac{40+50}{2} = 45\)
   • For class 50-60, midpoint = \(\frac{50+60}{2} = 55\)
2. Next, calculate the sum of the products of frequencies and their respective midpoints.

1. Given that the mean of the data is 45, we use the formula for the mean of grouped data:

\bar{x} = \frac{\Sigma f_ix_i}{\Sigma f_i}

1. where \(x_i\) are the midpoints and \(f_i\) are frequencies.
2. Substitute the values into this formula:

45 = \frac{345 + 750 + 1085 + 2025 + 55X}{23 + 30 + 31 + 45 + X}

1. Simplifying,

45 = \frac{4205 + 55X}{129 + X}

1. Multiply both sides by \(129 + X\):

45(129 + X) = 4205 + 55X

5805 + 45X = 4205 + 55X

1. Rearrange to find \(X\):

5805 - 4205 = 55X - 45X

1600 = 10X

X = 160

Therefore, the value of \(X\) that satisfies the condition is 160.