1. A bag contains counters, of which ten are coloured blue and $Y$ are coloured yellow. Two yellow counters and some more blue counters are then added to the bag. The proportion of yellow counters in the bag remains unchanged before and after the additional counters are placed into the bag.
Find all possible values of $Y$.
Let $X$ be the number of blue counters added to the bag. The proportion of yellow counters in the bag at the beginning is
$$
\frac{Y}{Y+10}
$$
and the proportion of yellow counters in the bag at the end is
$$
\frac{Y+2}{X+Y+12}
$$

Hence we have
$$
\frac{Y}{10+Y}=\frac{Y+2}{X+Y+12}
$$

Cross-multiplying, we obtain $Y(X+Y+12)=(Y+2)(10+Y)$, which gives
$$
X Y+Y^{2}+12 Y=10 Y+Y^{2}+20+2 Y
$$

This simplifies to $X Y=20$. Hence the number of yellow counters originally in the bag is either $1,2,4,5,10$ or 20 .
(Two other valid equations, $\frac{Y}{10}=\frac{Y+2}{X+10}$ and $\frac{Y}{10}=\frac{2}{X}$, also lead to $X Y=20$ )