Step-by-step Solution:

Let the number of students in the three classes be \( 2x, 3x, \) and \( 5x \) respectively.
After increasing 20 students in each class, the new numbers become:
First class: \( 2x + 20 \)
Second class: \( 3x + 20 \)
Third class: \( 5x + 20 \)
Given that the new ratio is 4:5:7, we can set up the equations: \[ \frac{2x + 20}{4} = \frac{3x + 20}{5} = \frac{5x + 20}{7} = k \] From the first equation: \[ 2x + 20 = 4k \quad \Rightarrow \quad 2x = 4k - 20 \quad \Rightarrow \quad x = 2k - 10 \] From the second equation: \[ 3x + 20 = 5k \] Substituting \( x = 2k - 10 \): \[ 3(2k - 10) + 20 = 5k \] \[ 6k - 30 + 20 = 5k \] \[ 6k - 10 = 5k \] \[ k = 10 \] Now, substituting \( k = 10 \) into \( x = 2k - 10 \): \[ x = 2(10) - 10 = 10 \] Thus, the original number of students in each class:
First class: \( 2x = 2(10) = 20 \)
Second class: \( 3x = 3(10) = 30 \)
Third class: \( 5x = 5(10) = 50 \)
Total number of students before increase: \( 20 + 30 + 50 = 100 \)