If 5 men can clear a bush in 3hours , how many men can clear the same bush in 3/4 hours

We are given that 5 men can clear a bush in 3 hours. We need to find out how many men are required to clear the same bush in $\frac{3}{4}$ hours. This is an inverse proportion problem, meaning that as the time decreases, the number of men must increase.

Let $M_1$ be the initial number of men and $T_1$ be the initial time.
Let $M_2$ be the new number of men and $T_2$ be the new time.

Given:
$M_1 = 5$ men
$T_1 = 3$ hours
$T_2 = \frac{3}{4}$ hours

For inverse proportionality, the product of men and time remains constant:

$$M_1 T_1 = M_2 T_2$$

$$15 = M_2 \times \frac{3}{4}$$

To find $M_2$, divide both sides by $\frac{3}{4}$:

$$M_2 = \frac{15}{\frac{3}{4}}$$ $$M_2 = 15 \times \frac{4}{3}$$ $$M_2 = \frac{15 \times 4}{3}$$ $$M_2 = \frac{60}{3}$$ $$M_2 = 20$$

Therefore, 20 men can clear the bush in $\frac{3}{4}$ hours.

The number of men required is $\text{20 men}$.