Question

4 men and 10 women were put on a work.they completed 1/3rd of the work in 4 days. after this 2 men and 2 women were increased. they completed 2/9 more of the work in 2 days. if the remaining work is to be completed in 3 days, then how many more women must be employed?

Answer 1

4 women and 6 men should be increased

Answer 2

Solution:

Let total amount of work = x

→(4 Men + 10 Women )'s 4 day's work = $\frac{x}{3}$

→ (4 Men + 10 Women )'s 1 day's work = $\frac{x}{12}$ ------(1)

Remaining work = x - $\frac{x}{3}$ = $\frac{2x}{3}$

Now, 2 Men and  2 Women were increased.

→(6 Men + 12 Women )'s 2 day's work =$\frac{2x}{9}$

→(6 Men + 12 Women )'s 1 day's work = $\frac{x}{9}$ ---------(2)

Now remaining work = $\frac{2x}{3}- \frac{x}{9} = \frac{6 x - x}{9}=\frac{5 x}{9}$

Solving (1) and (2)

3 ×Equation.(1) - 2 × Equation.(2) → 6 Women = $\frac{3x}{12} - \frac{2x}{9} = \frac{x}{36}$

→ 1 Women = $\frac{x}{216}$

1 women can complete x amount of work in 216 days.

putting value of (1 Women) in equation (2), we get

→6 Men = $\frac{x}{9} - \frac{12x}{216} =\frac{x}{9} - \frac{x}{18}=\frac{x}{18}$

→ 1 Men = $\frac{x}{108}$

So, one man can complete x amount of work alone in 108 days.

Total work done = $\frac{x}{3} +\frac{2x}{9} = \frac{5x}{9}$

Remaining work = x - $\frac{5x}{9}$= $\frac{4x}{9}$

6 Men + 12 Women = $\frac{6x}{108}+ \frac{12x}{216}=  \frac{24x}{216} =\frac{x}{9}$ work

Let P number of Women's be increased to complete the work in 3 days.

(6 Men + 12 Women )  1 day's work = $\frac{x}{9}$

(6 Men + 12 Women )  3 day's work= $\frac{3x}{9}$

Remaining work = $\frac{4x}{9} -\frac{3x}{9}=\frac{x}{9}$

P women one day's work = $\frac{P}{216}$

P women three day's work= $\frac{3 P}{216}$

$\frac{3P}{216}=  \frac{x}{9}\\P = 8 x$

$\frac{P}{8}= x$

So, 8 women's should be increased to complete remaining work.