Explanation:

Let's assign work rates:

- Let 1 boy from school A = A's 1 day work

- 1 boy from school B = B

- 1 boy from school C = C

Let's set up equations based on the info:

- 4A + 6B = 1/5 (work/day as total work = 1, done in 5 days)

- 5A + 10C = 1/4

- 3B + 4C = 1/10

Let's solve:

1. From equation 2: 5A + 10C = 1/4 ? A + 2C = 1/20

2. From equation 3: 3B + 4C = 1/10

Now use values:

- From above, C = (1/20 - A)/2

- Substitute 3rd into 1st using C's value

- Also, from equation 1: 4A + 6B = 1/5 ? 2A + 3B = 1/10 ? 3B = 1/10 - 2A

But 3B + 4C = 1/10 so,

- (1/10 - 2A) + 4C = 1/10

- 4C = 2A

- C = A/2

Plugging back in A + 2C = 1/20

- A + 2(A/2) = 1/20 ? A + A = 1/20 ? 2A = 1/20 ? A = 1/40

So, 40 boys from school A can do the work in one day.

- Option 1: 80 (too many)

- Option 2: 40 (Correct)

- Option 3: 60 (not matching calculation)

- Option 4: 20 (too few)