Question

15 men can finish a piece of work in 8 days. 18 women can finish the same piece of work in 12 days and 16 children can finish it in 20 days 12 women and 8 children started working together and worked for 10 days. if the remaining work need to be finished in 2 days, how many men should be added to the existing number of workers

Answer 1

11.66 is ans 
15m in 8days , 18w in 12days , 16c in 20days 
1m complete wrk in 120days , 1w in 216days , 1c in 320days 
12w and 8c complete the work in 360/29 days 
So in 10 days they complete 29/36 wrk
Remaining wrk is 7/36 
1m complete 1 work in 120days 
1m complete 7/36 work in 70/3 days 
So x men complete 7/36 work in 2 days 
X= 70/6 = 11.66 = 12(approx)

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Answer 2

Given:

15 Men complete a piece of work in 8 days.

18 Women complete the same work in 12 days.

16 Children complete the same work in 20 days.

12 Women and 8 children started working together and worked for 10 days.

To Find:

The number of men that should be added to the existing number of workers so that the remaining work is finished in 2 days.

Solution:

Let M denotes men, W denotes women and C denotes children.

According to the question, we have

$\[\begin{array}{l}15M \times 8 = 18W \times 12 = 16C \times 20\\ \\\Rightarrow 15M = 27W = 40C\\ \\\Rightarrow 1W = \frac{{40}}{{27}}C{\rm{ \,and\, 1}}M{\rm{ = }}\frac{{40}}{{15}}C\end{array}\]$

Therefore,

$\[\begin{array}{l}12W + 8C = 12 \times \left( {\frac{{40}}{{27}}C} \right) + 8C\\ \\\Rightarrow 12W + 8C = \left( {\frac{{160}}{9} + 8} \right)C\\\\ \Rightarrow 12W + 8C = \frac{{232}}{9}C\end{array}\]$

Since 15 men can complete a piece of work in 8 days, therefore,

Total work $\[ = 15 M\times 8 = 120M\]$

Let the number of men that should be added to finish the remaining work in 2 days be $x$.

Now, according to the information given, we have

$\[\begin{array}{l}\left( {12W + 8C} \right) \times 10 + \left[ {\left( {12W + 8C} \right) + xM} \right] \times 2 = 120M\\\\ \Rightarrow \left( {\frac{{232}}{9}C} \right) \times 10 + \left[ {\frac{{232}}{9}C + x\left( {\frac{{40}}{{15}}C} \right)} \right] \times 2 = 120 \times \frac{{40}}{{15}}C\end{array}\]$

Dividing both sides by $2C$, we get

$\[\begin{array}{l} \Rightarrow \left( {\frac{{232}}{9}} \right) \times 5 + \left[ {\frac{{232}}{9} + \frac{8}{3}x} \right] = 60 \times \frac{8}{3}\\\\ \Rightarrow \frac{{1160}}{9} + \frac{{232 + 24x}}{9} = 160\\\\ \Rightarrow 1392 + 24x = 160 \times 9\\ \\\Rightarrow 348 + 6x = 360\\ \\\Rightarrow 6x = 12\[\\ \\\Rightarrow x = 2\]\end{array}\]$

Hence, 2 more men should be added to the existing number of workers to finish the remaining work in 2 days.