Number of students participating in Chess and Carrom fom Class E is what percent more or less than the number of students participating in Chess from C and D together?

Given:

n(C) = 240

n(D) = 125

n(E) = 390

Concept used: 

n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

Calculation:

Let, A represent students participating in chess

B represents students participating in carrom.

⇒ Number of students participating in Chess and Carrom from Class E,

⇒ 390 - 130 = 260

In class C,

⇒ 240 = 8m + 5m + 110

⇒ m = \(\dfrac{130}{13}\) = 10

⇒ Chess participants in class C = 8m = 80 students

In Class D,

⇒ 125 = 2m + 3m + 45

⇒ 5m = 80

⇒ m = 16

⇒ Chess participants in class D = 2m = 32 students

Chess participants from class C and D together = 80 + 32 = 112 students

Number of students participating in Chess and Carrom from Class E = 260

⇒ Difference = 260 - 112 = 148

⇒ Required ratio = \(\dfrac{148}{112} \times 100\) = 132.143% = 132\(\frac{1}{7}\)%

∴ The required number of participants from Class E to C & D is 132\(\frac{1}{7}\)% more.