Question

A certain number of men can finish a job in 90 days. If there were 16 more men, the work could have been completed 18 days earlier. How many men were there initially
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Answer 1

Answer:

80

Step-by-step explanation:

Statement of the given problem,

A certain number of men could complete a piece of work in 90 days. If there were 8 men less, it would take 10 days more. How many men were there originally?

Basic assumptions,

Let W denotes the whole given work.

Let N denotes the original number of men, who can complete the work W in 90 days.

Hence from above data we get following relation,

W/(N*90) = W/[(N - 8)*(90 + 10)] [= constant = amount of work done by 1 man in 1 day]

or (N - 8)*(90 + 10) = N*90 or 10*N = 800 or N = 80 [Ans]

Answer 2

Given:

No. of days the men can finish the job = 90

No. of extra men = 16

No. of days the work could be completed if 16 men were present = 18

To find:

The no. of men that were initially present.

Solution:

Let $x$ be the no. of men that were present initially. These men could complete the work in 90 days. So, in one day, they were able to complete $\frac{1}{90x}$ part of the work.

Amount of work completed by $x$ no. of men in 1 day = $\frac{1}{90x}...(1)$

If $16$ extra men were there, they could complete the work 18 days earlier.

No. of days taken by the $(x+16)$ men to complete the work = $90-18=72$

So, amount of work that could be completed in one day by $(x+16)$ men = $\frac{1}{72(x+16)}...(2)$

To determine the no. of men that were present initially, the amount of work done by both groups are equal. Hence, equating equation (1) and equation (2),

$\frac{1}{90x}=\frac{1}{72(x+16)}$

$90x=72(x+16)$

$90x=72x+1152$

$90x-72x=1152$

$18x=1152$

$x=\frac{1152}{18} =64$

Hence, there were 64 men initially.

64 men were there initially to finish the job in 90 days.