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A contractor had to finish a work in $40$ days and he employed some men to do the work. They finished one-fifth of the work in $20$ days. When $80$ more men were added, the work was finished on the specified time. How many men were employed in the beginning?

My Attempt, enter image description here

Where did I made the error. The answer in my book is $80$ men..

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    I get the same answer ($\frac {80} 3$); your book is wrong and you are right. Unless you made a mistake in transcribing the problem. – 2017-01-21
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    @MyGlasses, I guess the first 80 were lazy. ;) – 2017-01-21
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    A minor observation but $80/3=26,666..$ workers would be quite a gruesome site.. – 2017-01-21

1 Answers 1

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We have formula -

$\frac{M1 \times D1}{F1} = \frac{M2 \times D2}{F2}$

M1, M2 are number of men before and after. D1, D2 are number of days. F1 and F2 are fractions of work.

Let x number of men initially.

$\frac{x \times 20}{\frac 15} = \frac{(x + 80) \times 20}{\frac 45}$

$ 4x = x + 80$

$ 3x = 80$

$ x = \frac{80}{3}$

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    I think some mistake in question. – 2017-01-21
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    I think original question is something like this. http://www.readbd.com/question/24318/a-contractor-employs-60-person-for-doing-a-job-in-80-days-dot-after-20-days-it-was-found-that-only-one-fifth-of-work-was-finished-dot-how-many-more-pe – 2017-01-21
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    You can try my formula with any other similar question or question in that link to see its work perfect. – 2017-01-21