## Task B-1.6.

Nora has three necklaces with different numbers of beads. From them, she made three new necklaces, each with 50 beads. She achieved this by removing $\frac{2}{7}$ of the beads from the first necklace and moving them to the second necklace, then moving $\frac{2}{7}$ of the beads from the resulting second necklace to the third, and finally moving $\frac{2}{7}$ of the beads from the resulting third necklace to the first. How many beads were on each necklace before the transfer?
## Solution.

Let there be $x$ beads on the first necklace, $y$ beads on the second, and $z$ beads on the third.

When $\frac{2}{7} x$ beads are moved from the first necklace to the second, $\frac{5}{7} x$ beads remain on the first. 1 point

On the second necklace, there are now $y+\frac{2}{7} x$ beads.

When $\frac{2}{7}\left(y+\frac{2}{7} x\right)$ beads are moved from the second necklace to the third, 50 beads remain on the second, so $\frac{5}{7}\left(y+\frac{2}{7} x\right)=50$, or

$$
y+\frac{2}{7} x=70
$$

On the third necklace, there are now $z+\frac{2}{7}\left(y+\frac{2}{7} x\right)=z+\frac{2}{7} \cdot 70=z+20$ beads,
and after moving $\frac{2}{7}(z+20)$ beads to the first necklace, 50 beads remain, so we have

$$
\begin{gathered}
\frac{5}{7}(z+20)=50 \\
z=50
\end{gathered}
$$

On the first necklace, there are now $\frac{5}{7} x+\frac{2}{7}(z+20)=50$, which means $x=42$.

From $y+\frac{2}{7} x=70$ we get that there were $y=58$ beads on the second necklace.