4 boys from school A and 6 boysfrom 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 boysfrom school C together can do in 10 days. Then how manyboys from schoolA can set up the exhibition in one day?

Let "Ba" = number of boys from school A,   "Bb" = number of boys from school B  and "Bc" number of boys from school C

it is given that (5 Ba +10 Bc) can set the exhibition in 4 days on (3 Bb + 4 Bc) can set in the 10 days

then $$ (5 Ba +10 Bc) \times 4 = (3 Bb +40 Bc) \times 10 $$

$$\Rightarrow 20 Ba + 40 Bc = 30 Bb +40 Bc $$

$$\Rightarrow 20 Ba = 30 Bb $$

$$\Rightarrow Bb = \dfrac{2}{3} Ba $$

also given that (4 Ba + 6 Bb) can setup exhibition in 5day 

put the value of Bb  in the  above equestoin 

 $$ (4 Ba + 6 Bb) = 4 Ba + 6 (\dfrac{2}{3}\times Ba) $$

$$\Rightarrow 4Ba +4Ba = 8 Ba $$

then 8 boys a from school A can setup the exhibition in 5 days 

so, $$8\times 5 =40 $$

Required number of boys from school A = 40  days