Problematic Job

We let the amount of days to complete the work be $x$ . We have

$\frac{x}{6} = \frac{x}{5} - 8$

$\implies x=240$

Then, we let $k$ to be the amount of total workers if the work is needed to be done 28 days earlier,

$\frac{240}{k} = \frac{240}{5} - 28$

$\frac{240}{k} = 20$

$\implies k=12$

Therefore, the amount of worker need to be hired additionally is $\boxed7$ , which is our desired answer.

Assume that the men work at the same rate. When I say part of the work done $= 1$ , I mean that the work has been completed.

$5$ men complete the job in $x$ days.

$6$ men complete the job in $x-8$ days.

$y$ men complete the job in $x-28$ days

Part of the job done by $5$ men in $x$ days $= 1$

Part of the job done by $5$ men in $1$ day $= \frac{1}{x}$

Part of the job done by $1$ man in $1$ day $= \frac{1}{5x}$ . . . (i)

Similarly,

Part of the job done by $6$ men in $x-8$ days $= 1$

Part of the job done by $1$ man in $1$ day $= \frac{1}{6(x-8)}$ . . . (ii)

And,

Part of the job done by $y$ men in $x-28$ days $= 1$

Part of the job done by $1$ man in $1$ day $= \frac{1}{y(x-28)}$ . . . (iii)

From (i) and (ii),

$\frac{1}{5x} = \frac{1}{6(x-8)}$

$\iff 5x = 6x-48$

$\iff x = 48$

That is, $5$ men take $48$ days to complete the job.

Again from (i) and (iii),

$\frac{1}{5x}=\frac{1}{y(x-28)}$

$\iff 5 \times 48 = 20y$

$\iff y=12$

That is, $12$ men complete the job 28 days earlier. Therefore number of additional men who have to be hired is $\boxed{7}$

Let $t$ denotes the initial time needed to complete the job when there are still only 5 men hired and $m$ denotes the number of additional men needed to complete the job 28 days earlier. Then $(t-8)$ must be the time when an additional men has been hired and analogous, $(t-28)$ must be the time when $m$ men has been hired. Now, since this problem is a problem about proportion, we can illustrate the proportion to make it easier to solve as below:

$5 \rightarrow t$

$6 \rightarrow (t-8)$

$(5+m) \rightarrow (t-28)$

The relation between the number of men or workers is inversely proportional to the days needed to complete the job. Consequently, the product of number of men and the days needed is always the same in every condition. For instance,

$5t=6(t-8)=m(t-28)$

Solving for $t$ at the right and middle side gives us $t=48$ . The last step is to substitute this value to above equation.

$5t=(5+m)(t-28) \Leftrightarrow 5+m= \frac {5.48}{48-28} = 12 \Leftrightarrow m= \boxed{7}$

$5$ men can do the work in $x$ days (say)

Therefore, $6$ men can do the work in $\frac{5x}{6}$ days.

So, $x$ - $\frac{5x}{6}$ = $8$ . Solving we find $x$ = $48$ .

The work is done in $48$ days $5$ men

" " " " " $1$ day by $48 \times 5$ men

" " " " " $20$ days by \frac{48 \times 5}{20} = \(12 .

Therefore, $12$ - $5$ = $\boxed{7}$ aditional men are required

Begin with the equation $t=\frac{w}{r}$ , where $t=$ time, $w=$ amount of work, and $r=$ rate. Now let $r=xr_0$ , where $x$ is the number of people and thus $r_0$ is the rate for each person.

From the initial information, we have $\frac{w}{5r_0}-8=\frac{w}{6r_0}$ . Rearranging yields $8=\frac{w}{5r_0}-\frac{w}{6r_0}=\frac{1}{30r_0}$ . Thus we have $r_0=\frac{1}{240}$ .

Now we let $x_{+}$ be the number of people that are needed (total) to complete the job $28$ days earlier. Then we have $\frac{240w}{5}-28=\frac{240w}{x_{+}}\Rightarrow x_{+}=\frac{240w}{\frac{240w}{5}-28}$ . So we must solve for $w$ before we get our answer.

Consider our earlier equation, $\frac{w}{5r_0}-8=\frac{w}{6r_0}$ . Substituting our value of $r_0$ yields $48w-8=40w\Rightarrow w=1$ .

So now we have $x_{+}=\frac{240(1)}{\frac{240(1)}{5}-28}=\frac{240}{20}=12$ , so that the number of people we need is $12-5=\boxed{7}$ .

Let 5 men complete the work in x days. Therefore, 1 man completes the work in 5x days. If an additional man is hired, number of men is 6. 6 men complete the work in x-8 days. Therefore, 1 man completes the work in 6(x-8) days. now, 5x=6(x-8)......so, x=48 48 days------5 men 1 day----------240 men 20 days--------240/20=12 men Therefore, number of additional men required=12-5=7

Establish the common product first which is Man * Days (Common Product for inverse relationship) Let x number of days for a complete work 5x = 6(x-8) ; x = 48; Let n be number of Men 5x = n(x-28); Substitute x in the equation 5(48) = n (48-28) 20n = 240 n = 12; 12-5 = 7 additional men

Say "x" is the number of hours in which 5 men completes the works then the equations as simple as this 5/6=x-8/x ; here x comes out to be 48.

Now how many men can complete the work in x-28 = 48-28 = 20 hrs

5/5+w = 20/48 where w stands for No of workers required and here W comes out to be 7

Simple insn't it ???

Let D be number of days in which job has to be done. So, Amount of work = (5 D) From problem, (5 D)=[6 (D-8)] This gives, D=48 If you want to complete the job 28 days earlier, amount of days you need is 48-28=20 days Let x number of men are required to complete job in 20 days. Therefore, 5 48=x*20 Which gives x=12 So additional men required are 12-5=7

This is a relatively simple problem. If $n$ is the number of days it took for the 5 men, these equations followed her:

$5n=6(n-8)$

$5n=6n-48$

$n=48$

By subtracting 28 from 48, you get 20. Then, you multiply 48 by 5 (the number of men) and divide the product by 20. You should get an answer of 12. Finally, you subtract 5 from 12 and get an answer of 7.

let 5 men does the job in X days. So 1 man will do it in 5X days Now 6 men does the job in (X-8) days So 1 man will do it in 6(X-8) days equating these two 5X=6(X-8) X=48 now 28 days earlier means the work is to be done in 48-28=20 days 5 48 days is required by 1 man So 20 days is required by 5 48/20=12 men So additional 7 man is to be hired

Let the no. of days taken to finish the work by 5 men be x.

For x days,5 men are required and for (x-8) days,6 men are required.

Using method of inverse proportion, we get,

\(6(x-8)=5x \implies 6x-48=5x \implies x=48

Again,using unitary method and inverse proportion we get,

48 days ....... 5 men \(\implies\) 1 day .... $5\times 48 = 240$ men $\implies$ 28 days earlier(48-28=20 days).... $\frac{240}{20} = 12$ men.

Now, additional men hired $=12-5 = \boxed{7}$

Assume that every man does 1 unit of work every day.Then,let the number of days that 5 men need to finish the job be $x$ .So,the total work amount is $5x$ .If we hire one additional man,then we have 6 men and we need $x-8$ days to finish.So the total work amount is $6 \times (x-8) = 6x-48$ .Because the total work amount is the same, $6x-48 = 5x$ .So $(6x-48)-5x = 0 = x-48$ .Hence $x = 48$ .So the total work amount is $5 \times 48 = 240$ .If the job has to be completed 28 days earlier, then we need $\frac{240}{48-28} = \frac{240}{20} = 12$ men.So the number of additional men who have to be hired is $12 - 5 = 7$ .