15 men can finish a piece of work in 20 days, however it takes 24 women to finish it in 20 days. If 10 men and 8 women undertake to complete the work, then they will take ?

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{15 men}} = {\text{20 days}} \cr & {\text{300 men}} = 1{\text{ days}}.....{\text{(i)}} \cr & {\text{24 women}} = {\text{20 days}} \cr & {\text{480 men}} = 1{\text{ days}}......{\text{(ii)}} \cr & {\text{Compare equation (i) and (ii)}} \cr & {\text{300 men}} = 480{\text{ women}} \cr & {\text{5 men}} = 8{\text{ women}}.....{\text{(iii)}} \cr & {\text{10 men}} + 8{\text{ women}} = ? \cr & {\text{10 men}} + {\text{5 men}} = ? \cr & 15\,{\text{men}} = ? \cr} $$
$${\text{15 men}} \times {\text{20 days}}$$     = $${\text{15 men}}$$  $$ \times $$ $$x{\text{ days}}$$
$$x$$ = 20 days
Alternate
$$\eqalign{ & {\text{15m}} \times {\text{20 days}} = 24{\text{w}} \times 20{\text{ days}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{8}{5} \cr} $$
So, 1 man work 8 units work in one day
and 1 woman work 5 units work in one day
Total work = 15 × 8 × 20
Hence, (10 men + 8 women) work whole in D days
$$\eqalign{ & \left( {{\text{10m}} + {\text{8w}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{10}} \times {\text{8}} + {\text{8}} \times {\text{5}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{80}} + 40} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & {\text{D}} = 20{\text{ days}} \cr} $$