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?
Answer & explanation
Correct answer: option 1
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.