​3 men and 3 boys can do a piece of work in 10 days. When one man is replaced by a boy, the work is completed in 12 days. In how many days will 2 men and 2 boys complete the work?​

Correct option is D

Given:

3 men and 3 boys can complete the work in 10 days.

When 1 man is replaced by a boy (i.e., 2 men and 4 boys), the work is completed in 12 days.

We need to determine how many days 2 men and 2 boys will take to complete the work.

Concept Used:

The Work Formula based on the $$M_1 D_1 W_1 = M_2 D_2 W_2$$​ principle:

​$$M_1 \times D_1 = M_2 \times D_2$$​​

where:

​$$M_1, D_1$$​​ represent the number of workers and days for the first scenario.

​$$M_2, D_2​$$​ represent the number of workers and days for the second scenario.

Solution:

Let the total work be LCM of 10 and 12, i.e., 60 units.

Work done per day by (3 men + 3 boys)

​$$\frac{60}{10} = 6 \quad \text{(units per day)}$$​​

Work done per day by (2 men + 4 boys)

​$$\frac{60}{12} = 5 \quad \text{(units per day)}$$​​

Let the work done by 1 man per day be M and the work done by 1 boy per day be B.

From the given information:

​$$3M + 3B = 6 \quad \text{(Equation 1)}$$​​

​$$2M + 4B = 5 \quad \text{(Equation 2)}$$​​

From Equation 1 & 2:

M + B = 2; 2M + 4B = 5 => M + 2B = 2.5

Solving above equations:

B=0.5 & M=1.5

Work Done by 2 Men and 2 Boys

​$$\text{Work done per day} = 2M + 2B = 2(1.5) + 2(0.5) =3+1=4(\text{units per day})$$

Since, Total work = 60 units,     Days required = $$\frac{60}{4} = 15 \text{ days}$$

Option (D) is days.

Alternate Method:

Concept Used:

(M₁+B₁)D₁= (M₂+B₂)D₂

where:

M₁, B₁ & D₁ represent the number of men , boys and days for the first scenario.

​M₂,B₂& D₂ represent the number of men , boys and days for the second scenario.

Solution:

(3M + 3B )10 = (2M + 4B)12

5M + 5B=4M + 8B

M = 3B 

$$\frac{M}{B}=\frac{3}{1}$$

Total work = $$(3\times 3 +3\times 1) 10$$ = 120

Then, 

work done per day =2M + 2B = $$2 \times 3+ 2 \times1 = 8$$ 

Days required = $$\frac{120}{8}=15$$

​​Option (D) is days.