Question

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

Answer 1

40 boys from school A can set up the exhibition in one day.

Step-by-step explanation:

Let’s assume,

“Ba” = no. of boys from school A

“Bb” = no. of boys from school B  

“Bc” = no. of boys from school C

It is given that (5Ba+10Bc) can set the exhibition in 4 days or (3Bb+4Bc) can set the exhibition in 10 days, i.e., we can write the eq. as,

(5Ba + 10Bc) * 4 = (3Bb + 4 Bc) * 10

⇒ 20Ba + 40Bc = 30Bb + 40Bc

⇒ 20Ba = 30Bb

⇒ Bb = $\frac{2}{3}$ * Ba …… (i)  

Also given that (4Ba+6Bb) can set up an exhibition in 5 days

So, substituting from (i), we get

4Ba + 6Bb = 4Ba + [6 * {$\frac{2}{3}$ * Ba}] = 4Ba + 4Ba = 8Ba

⇒ 8 boys from school A can set up the exhibition in 5 days

Therefore, in order to find the required no. of boys to set up an exhibition from school A in 1 day, we will be applying the Man-Days formula, i.e.,

8 * 5 = [Required no. of boys from school A] * 1 day

⇒ Required no. of boys from school A = 40

Thus, 40 boys from school A can set up the exhibition in 1 day.

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