16 men can finish a piece of work in 8 days while 20 women and 4 men can do it in 16 days. In how many days can 10 women complete the entire work by themselves?

Correct option is C

Given:

16 men can finish the work in 8 days.

20 women and 4 men can finish the work in 16 days.

We need to find how many days 10 women would take to complete the work.

Formula Used:

Total work = Number of workers × Rate of work × Time

Solution:

Total work done = 1

​$$16 \times 8 \times m = 1 \\ \ \\m = \frac{1}{128}$$​​

So, each man does $$\frac{1}{128}$$​ of the work per day.

20 women and 4 men complete the work in 16 days. Hence, the total work is:

​$$(20 \times w + 4 \times m) \times 16 = 1$$​​

Substitute m $$= \frac{1}{128}:$$​​

​$$(20 \times w + 4 \times \frac{1}{128}) \times 16 = 1$$​​

​$$(20w + \frac{4}{128}) \times 16 = 1$$​​

​$$(20w + \frac{1}{32}) \times 16 = 1$$​​

​$$320w + \frac{1}{2} = 1$$​​

​$$320w = 1 - \frac{1}{2} = \frac{1}{2}$$​​

​$$w = \frac{1}{640}$$​​

So, each woman does $$\frac{1}{640}$$​ of the work per day.

Now, 10 women will complete the work, and the total work done by them per day is:

​$$10 \times w = 10 \times \frac{1}{640} = \frac{10}{640} = \frac{1}{64}$$​​

Therefore, the number of days required for 10 women to complete the work is:

Time $$= \frac{1}{\frac{1}{64}} = 64 \text{ days}$$​

10 women will complete the work in 64 days.

Alternate Method:

$$\frac{M_1D_1H_1}{W_1}=\frac{M_2D_2H_2}{W_2}$$

M = Man , D = Days , H = Hours , W= Work / Wages

16 M$$\times 8$$ = ( 20W + 4M )$$\times 16$$

128 M = 320 W + 64 M

64 M = 320 W

​$$\frac MW= \frac 51$$

Efficiency of Man = 5, Efficiency of Women =1

Total Work = 16 M$$\times 8$$ = 16$$\times 5 \times 8$$ = 640

Then, 640 = 10 W + D

640 = 10$$\times 1$$ + D

D = $$\frac {640}{10}$$ = 64 days​