- 4.2 days
- 3 days
- 3.6 days
- 4 days

Option 3 : 3.6 days

**Given:**

4 Men(M) and 6 Boys(B) can do a piece of work = 9 days

16 Men(M) and 35 Boys(B) can do a piece of work = 2 days

**Formula used:**

If M_{1} can do W_{1} work in D_{1} days and M_{2} man can do W_{2} work in D_{2} days, then

(M_{1}D_{1})/W_{1} = (M_{2}D_{2})/W_{2}

**Calculation:**

We have,

4M + 6B = 9 days ----(1)

16M + 35B = 2 days ----(2)

According to the question

(4M + 6B) × 9 = (16M + 35B) × 2

⇒ 36M + 54B = 32M + 70B

⇒ 4M = 16B

⇒ M = 4B

⇒ M ∶ B = 4 ∶ 1

So, Total work = (4 × 4 + 6 × 1) × 9

⇒ 22 × 9 = 198 units

Work done by 10 Men and 15 Boys = (10M + 15B)

⇒ (10 × 4 + 15 × 1) = 55 units

Time taken by 10 Men and 15 Boys = 198/55

⇒ 3.6 days

**∴ The time taken by 10 Men and 15 Boys is 3.6 days.**

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