## Solution.

The total amount of chickpeas in container $A$ is $45 \% \cdot 4=0.45 \cdot 4=1.8$ kilograms.

The total amount of chickpeas in container $B$ is $48 \% \cdot 5=0.48 \cdot 5=2.4$ kilograms.

The total amount of chickpeas in container $C$ is $k \% \cdot 1=\frac{k}{100}$ kilograms.

Let $x$ be the amount of chickpeas transferred from container $C$ to container $A$. Then, after the transfer, container $A$ contains $50 \% \cdot(4+x)=2+0.5 x$ kilograms of chickpeas.

The remainder, which is $1-x$ kilograms of chickpeas from container $C$, is transferred to container $B$, so after the transfer, container $B$ contains $50 \% \cdot(5+1-x)=3-0.5 x$ kilograms of chickpeas. Since the total amount of chickpeas is the same before and after the transfer, we have

$$
1.8+2.4+\frac{k}{100}=2+0.5 x+3-0.5 x
$$

which simplifies to

$$
4.2+\frac{k}{100}=5
$$

From this, we get $\frac{k}{100}=0.8$, so $k=80$.

Note: The solution can be written without introducing the unknown $x$. Specifically, the total amount of chickpeas is 10 kilograms, and at the end, $50 \%$ of this amount is chickpeas. Therefore, the total amount of chickpeas is 5 kilograms. This conclusion is worth 3 points. Calculating the amount of chickpeas in each container before the transfer, equating the amounts of chickpeas before and after the transfer, and determining the value of $k$ each earn 1 point.
## Task A-1.4.

In container $A$, there are four kilograms of chickpeas, of which $45 \%$ are chickpeas. In container $B$, there are five kilograms of chickpeas, of which $48 \%$ are chickpeas. In container $C$, there is one kilogram of chickpeas. A certain portion is transferred from this container to container $A$, and the remainder to container $B$, in such a way that the proportion of chickpeas in both parts is equal and amounts to $k \%$. After this, the proportion of chickpeas in both container $A$ and container $B$ is exactly $50 \%$. Determine $k$.