Question

A group of labourers promised to do a piece of work in 10 days, but five of them become absent. If the remaining labourers complete the work in 12 days, find their original number in the group ​

Answer 1

Answer:

Total period = 10 days

But work completed in = 12 days

No. of men were absent = 5

Let the number of men in the beginning = x

Now x men can do a piece work in = 10 days

1 man will do it in = 10x x days

And (x – 5) will do it in = (10 × x)/(x – 5) days

∴ 10x/(x – 5) = 12 ⇒ 10x = 12x – 60

⇒ 12x – 10x = 60 ⇒ 2x = 60

⇒ x = 60/2 = 30

∴ No. of men in the beginning = 30

Answer 2

Step-by-step explanation:

Let the number of labour =x

Initially work should be  completed in 10 days

so for one labour it takes 10×x days  

But when the labour is 5 less it will complete in 12 days

so for one labour it takes 12×(x−5) Days

hence 10x=12×(x−5)

⟹10x=12x−60

⟹10x−12x=−60

⟹−2x=−60

⟹x=30

Hence in starting number of labour =30.