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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.
Solution
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
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