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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.
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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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