1. In box A, there are twice as many clickers as in box B. If one quarter of the clickers from A and one third of the clickers from B were moved to box C, then it would contain 180 clickers, which is one fifth more than it had before the transfer. How many clickers are in box A?
1. Let $x$ be the number of balls in box C before the transfer.

Then we have $x+\frac{1}{5} x=180$

$$
\begin{aligned}
& \frac{6}{5} x=180 \\
& x=150
\end{aligned}
$$

There were 150 balls in box C.

Since $180-150=30$, 30 balls would have been transferred.

Let $y$ be the number of balls in box B.

Then $2 y$ is the number of balls in box A, and we have $\frac{1}{4} \cdot 2 y+\frac{1}{3} \cdot y=30$

$$
\begin{aligned}
& \frac{1}{2} y+\frac{1}{3} y=30 \\
& \frac{5}{6} y=30 \\
& y=36
\end{aligned}
$$

that is, $2 y=72$.

There are 72 balls in box A.