Question

A particular work can be completed by 6 men and 6 women in 24 days whereas the same work can be completed by 8 men and 12 women in 15 days
FIND
1) according to the amount of work done one man is equivalent to how many women
2) the time taken by 4 men and 6 women to complete the same work

Answer 1

given,
6men + 6 women = 24 days
⇒ 6M + 6W = 24
and, 8M + 12W =15
1)Applying, M×D = M'×D'
(6M + 6W )×24= (8M + 12W)×15
⇒( M + W) ×6×24 = (2M +3W) ×4×15
⇒( M+ W)12 = (2M +3W)5
⇒12M + 12W = 10M +15W
⇒2M = 3 W
hence, 2 men is equal to 3 women
2)now, 4M +6 W = ??
Applying M×D =M'×D'
(6M + 6W )×24 = (4M +6W ) ×D'
⇒(9W + 6W) ×24 = (6W +6W)×D'   ∵2M = 3 W
⇒15W ×24 = 12W ×D'
⇒D' = 30 days
It will takes 30 days to complete the same work by 4 men and 6 women

Answer 2

1) The work done by one man is equal to $\frac{3}{2}$ of work done by woman.

2) 4 Men and 6 Women can complete the work in 30 days.

Given:

6 men + 6 women = 24 days

$\Rightarrow 6M + 6W = 24$

$\Rightarrow 8M + 12W =15$

Solution:

1) Applying, $M\times D = M' \times D'$

$(6M + 6W) \times 24 = (8M + 12W) \times 15$

$\Rightarrow (M + W) \times 6\times 24 = (2M +3W) \times 4\times 15$

$\Rightarrow (M + W)12 = (2M + 3W)5$

$\Rightarrow 12M + 12W = 10M +15W$

$\Rightarrow 2M = 3 W$

Hence, $1M = \frac{3}{2} \mathrm{W}$

2) Now, 4M + 6W = ?

Applying, $M\times D =M'\times D'$

$(6M + 6W) \times 24 = (4M + 6W) \times D'$

$(144 M+144 W) \times 24=(4 M+6 W) \times D^{\prime}$

Since, $1M = \frac{3}{2} \mathrm{W}$

$\left(\left(144 \times \frac{3}{2}\right) W+144 W\right)=\left(\left(4 \times \frac{3}{2}\right) W+6 W\right) \times D^{\prime}$

$216 W+144 W=(6 W+6 W) D^{\prime}$

$\Rightarrow D^{\prime}=\frac{216 W+144 W}{6 W+6 W}$

$D^{\prime}=\frac{360}{12}=30$

$\Rightarrow D' = 30 days$