The table given below shows the number of players participating in two games in six different schools?

Number of players participating in game L from school D and E is how much percent less than the number of players participating in game M from school H and I ?

22.68 percent

Number of Players in game L from D & E = 82 + 68 = 150

Number of players in game M from H & I = 86 + 108 = 194

Required percentage = \(\frac{194 - 150 }{194}\) × 100

= \(\frac{44 }{194}\) × 100 

= 22.68%