A Piece of work can be done by 3 boys and 2 men jointly in 10 days. The same work can be done jointly by 2 boys and 3 men in 8 days. In how many days the same work can be done by 1 boy and 2 men jointly?

Given:

Time by which 3 boys and 2 men finish the work = 10 days

Time by which 2 boys and 3 men finish the work = 8 days

Concept used:

Work = time × efficiency

Calculation:

Let the efficiency of one man be 'm' and one boy be 'b'.

Then, (2m + 3b) × 10  = total work

Also, (3m + 2b) × 8 = total work

Equating the total work in both the cases, 

(2m + 3b) × 10  = (3m + 2b) × 8

⇒ 20m + 30b = 24m + 16b

⇒ 14b = 4m

⇒ 7b = 2m

Number of days in which (2m + b) can complete the work be D.

(2m + b) × D = (2m + 3b) × 10 = total work

Substituting 2m = 7b in the above equation,

⇒ 8b × D = 10b × 10

⇒ D = 100/8

⇒ D = 12.5 days

∴ The time taken to complete is 12.5 days.