4 boys from school A and 6 boysfrom school B together can set up an exhibition in 5 days, which 5 boys from school A and 10 boys from school C together can do in 4 days or 3 boys from school B and 4 boys from school C together can do in 10 days. Then how many boys from school A can set up the exhibition in one day?

40

4A + 6B = 5 days,     

5A + 10C = 4 days,  

3B + 4C =10 days,    

⇒ as we know, (4A + 6B) x 5 = (5A + 10C ) x 4 = (3B + 4C ) x 10

⇒ (4A + 6B) x 5 = (3B + 4C ) x 10

⇒  4A + 6B = 6B + 8C

⇒ 4A = 8C

⇒ A : C = 2: 1       ..(1)

Again,

⇒ (5A + 10C ) x 4 = (3B + 4C ) x 10

⇒ 10A + 20C = 15B + 20C

⇒ 10A = 15B

⇒ A : B = 3 : 2     ..(2)

From equation (1) & (2),

A : B : C = 6 : 4 : 3

⇒ Total work = (4A + 6B) x 5 = (4(6) + 6(4)) x 5 = (24 + 24) x 5= 240 units.

⇒ No. of students from School A to complete the work in 1 day = \(\frac{240}{6x}\) = 1,

⇒ x = 40 

Therefore 40 students from school A are needed to complete work in one day.