Question : 4 boys from school A and 6 boys from 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?
Option 1: 60
Option 2: 40
Option 3: 20
Option 4: 80
Correct Answer: 40
Solution :
Total work of first group = total work of second group = total work of third group
(4A + 6B) × 5 = (5A + 10C) × 4 = (3B + 4C) × 10
⇒ (4A + 6B) × 5 = (3B + 4C) × 10
⇒ 4A + 6B = 6B + 8C
⇒ 4A = 8C
⇒ A : C = 8 : 4
⇒ A : C = 2 : 1....................................(1)
Again,
⇒ (5A + 10C) × 4 = (3B + 4C) × 10
⇒ 20A + 40C = 30B + 40C
⇒ 20A = 30B
⇒ A : B = 30 : 20
⇒ A : B = 3 : 2.......................................(2)
From equation (1) and equation (2), we get,
A : B : C : = 6 : 4 : 3
Total work = (4A + 6B) × 5 = (4 × 6 + 6 × 4) × 5 = 48 × 5 = 240
Let the number of students from school A complete the work be $x$.
$6x = 240$
⇒ $x = \frac{240}{6}$
$\therefore x = 40$
Hence, the correct answer is 40.
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