Question

A group of labourers promises 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

Total period = 10 days

But work completed in = 12 days

No. of men was absent = 5

Let the number of men in the beginning = x

Now x men can do piece work in = 10 days

1 man will do it in = 10x × days

and (x − 5) will do it in =`(10xx"x")/("x"-5)` days

∴ `(10"x")/("x"-5)=12`

⇒ 10x = 12x − 60

⇒ 12x − 10x = 60 ⇒ 2x = 60

⇒ x =`60/2` = 30

∴ No. of men in the beginning = 30