4. With one mower, it is possible to mow $2 \frac{1}{2}$ ha of meadow in 2 hours, and with another, smaller mower, $1 \frac{1}{4}$ ha can be mowed in 1 hour and 30 minutes. How much meadow area will both mowers mow together in 3 hours and 36 minutes?
First method: The first mower mows in 1 hour $2 \frac{1}{2}: 2=\frac{5}{2} \cdot \frac{1}{2}=\frac{5}{4}=1 \frac{1}{4}$ ha.

The second mower mows in 1 hour $1 \frac{1}{4}: 1 \frac{1}{2}=\frac{5}{4}: \frac{3}{2}=\frac{5}{4} \cdot \frac{2}{3}=\frac{5}{6}$ ha

Both mowers will mow in 1 hour $1 \frac{1}{4}+\frac{5}{6}=\frac{5}{4}+\frac{5}{6}=\frac{15}{12}+\frac{10}{12}=\frac{25}{12}=2 \frac{1}{12}$ ha.

For 3 hours and 36 minutes $=3 \frac{36}{60}=3 \frac{3}{5}$ hours

the mowers will mow $3 \frac{3}{5} \cdot 2 \frac{1}{12}=\frac{18}{5} \cdot \frac{25}{12}=\frac{15}{2}=7 \frac{1}{2}$ ha.

The second mower mows in 1 hour $1 \frac{1}{4}: 1 \frac{1}{2}=\frac{5}{4}: \frac{3}{2}=\frac{5}{4} \cdot \frac{2}{3}=\frac{5}{6}$ ha.

$1 \text{POINT}$

For 3 hours and 36 minutes $=3 \frac{36}{60}=3 \frac{3}{5}$ hours

$1 \text{POINT}$

the first mower will mow $3 \frac{3}{5} \cdot \frac{5}{4}=\frac{18}{5} \cdot \frac{5}{4}=\frac{9}{2}$ ha,

1 POINT

and the second $\frac{18}{5} \cdot \frac{5}{6}=3$ ha.

$1 \text{POINT}$

Both mowers together will mow $\frac{9}{2}+3=\frac{15}{2}=7 \frac{1}{2}$ ha.