, 2023

In 1 hour, Liang paints $\frac{1}{3}$ of the room.

Thus, in 2 hours, Liang paints $\frac{2}{3}$ of the room.

Edmundo needs to paint $1 - $\frac{2}{3} = \frac{1}{3}\$$ of the room.

In 1 hour, Edmundo paints $\frac{1}{4}$ of the room.

Since $\frac{1}{4} = \frac{3}{12}$
and $\frac{1}{3} = \frac{4}{12}$,
this means that Edmundo paints for $\$\frac{1}{3} \div \frac{1}{4} = \frac{4}{12} \div \frac{3}{12} = \frac{4}{3}\$$ of an hour.

Therefore, Edmundo paints for 80 minutes.
When converted to a fraction, $A\%$ is equal to $\dfrac{A}{100}$.

When an amount is increased by $A\%$, we can find its new value by
multiplying by $1 +
$\dfrac{A}{100}\$.$

When an amount is decreased by $A\%$, we can find its new value by
multiplying by $1 -
$\dfrac{A}{100}\$.$

When $400 is increased by$A\%$,
the amount becomes $\$\$400\left(1 + \dfrac{A}{100}\right)\$.$

When this value is decreased by $A\%$, the amount becomes $\$\$400\left(1 + \dfrac{A}{100}\right)\left(1 - \dfrac{A}{100}\right)\$.$

Therefore, $$\begin{align*} \$400\left(1 + \dfrac{A}{100}\right)\left(1 - \dfrac{A}{100}\right) & = \$391 \\ \left(1 + \dfrac{A}{100}\right)\left(1 - \dfrac{A}{100}\right) & = \dfrac{391}{400} \\ 1 - \dfrac{A^2}{100^2} & = 1 - \dfrac{9}{400} \\ \dfrac{A^2}{100^2} & = \dfrac{9}{400} \\ \dfrac{A^2}{100^2} & = \dfrac{3^2}{20^2} \\ \dfrac{A}{100} & = \dfrac{3}{20} ~~ \text{(since $A > 0$)} \\ A & = 100 \cdot \dfrac{3}{20} = 15\end{align*}$$ Therefore,
$A = 15$.