If 5 men and 3 women can reap 18 acre of crop in 4 days, 3 men and 2 women can reap 22 acre of crop in 8 days, then how many men are required to join 21 women to reap 54 acre of crop in 6 days ?

Answer: Option A

Acreage reaped by 5 men and 3 women in 1 day
$$\begin{array}{rl}& =\frac{18}{4}\\ & =\frac{9}{2}\end{array}$$
Acreage reaped by 3 men and 2 women in 1 day
$$\begin{array}{rl}& =\frac{22}{8}\\ & =\frac{11}{4}\end{array}$$
Suppose 1 man can reap x acres in 1 day and 1 women can reap y acres in 1 day
$$\begin{array}{rl}& \therefore 5x+3y=\frac{9}{2}\\ & \Rightarrow 10x+6y=9.....\text{(i)}\\ & 3x+2y=\frac{11}{4}\\ & \Rightarrow 9x+6y=\frac{33}{4}.....\text{(ii)}\\ & \text{Subtracting (ii) from (i),}\\ & \text{We get}:x=9-\frac{33}{4}=\frac{3}{4}\\ & \text{Putting x}=\frac{3}{4}\text{ in (i), we get}\\ & \Rightarrow 6y=9-\frac{15}{2}\\ & \Rightarrow 6y=\frac{3}{2}\\ & \Rightarrow y=\frac{1}{4}\end{array}$$
Acreage reaped by 21 women in 6 days
$$\begin{array}{rl}& =\left(\frac{1}{4}\times 21\times 6\right)\\ & =\frac{63}{2}\end{array}$$
Remaining acreage to be reaped
$$\begin{array}{rl}& =\left(54-\frac{63}{2}\right)\\ & =\frac{45}{2}\end{array}$$
Acreage reaped by 1 men in 6 days
$$\begin{array}{rl}& =\left(\frac{3}{4}\times 6\right)\\ & =\frac{9}{2}\end{array}$$
In 6 days, $$\frac{9}{2}$$ acre is reaped by 1 man
∴ In 6 days, $$\frac{45}{2}$$ acre is reaped by
$$\begin{array}{rl}& =\left(\frac{2}{9}\times \frac{45}{2}\right)\text{men}\\ & =5\text{ men}\end{array}$$