The problem has two parts: (1) If $\frac{9}{8} m^2$ of a wall is painted with $\frac{3}{7} dL$ of paint, how much paint is used per $1 m^2$? (2) If $\frac{1}{4} m$ of wire weighs $\frac{3}{10} kg$, what is the weight of $1 m$ of the wire?

1. Problem Description

The problem has two parts:

(1) If $\frac{9}{8} m^2$ of a wall is painted with $\frac{3}{7} dL$ of paint, how much paint is used per $1 m^2$?

(2) If $\frac{1}{4} m$ of wire weighs $\frac{3}{10} kg$, what is the weight of $1 m$ of the wire?

2. Solution Steps

(1) To find the amount of paint used per $1 m^2$, we need to divide the amount of paint used by the area covered.

Amount of paint per $m^2$ = (Amount of paint) / (Area)

Amount of paint per $m^2 = \frac{3}{7} \div \frac{9}{8}$

$\frac{3}{7} \div \frac{9}{8} = \frac{3}{7} \times \frac{8}{9}$

$\frac{3}{7} \times \frac{8}{9} = \frac{3 \times 8}{7 \times 9} = \frac{24}{63} = \frac{8}{21}$

(2) To find the weight of $1 m$ of wire, we need to divide the weight of the wire by its length.

Weight per $m$ = (Weight) / (Length)

Weight per $m$ = $\frac{3}{10} \div \frac{1}{4}$

$\frac{3}{10} \div \frac{1}{4} = \frac{3}{10} \times \frac{4}{1}$

$\frac{3}{10} \times \frac{4}{1} = \frac{3 \times 4}{10 \times 1} = \frac{12}{10} = \frac{6}{5}$

3. Final Answer

(1) $\frac{8}{21} dL$

(2) $\frac{6}{5} kg$