# 6. Option 1

Initially, there were 20 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{25}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{16}$ greater than the probability of drawing a blue ball at the beginning. How many white balls were in the box?
Answer: 4.

Solution: Let there be $a$ white balls, $b$ blue balls, and $c$ red balls in the box. We can set up the following equations:

$$
\begin{gathered}
a+b+c=20 \\
\frac{a}{20}=\frac{a}{20+b}+\frac{1}{25} \\
\frac{b}{20}+\frac{1}{16}=\frac{b}{20-a}
\end{gathered}
$$

Transform the second equation: $\frac{a b}{20(20+b)}=\frac{1}{25}$ and the third equation: $\frac{1}{16}=\frac{a b}{20(20-a)}$. Finally, $\frac{20+b}{25}=\frac{20-a}{16}$ and, substituting into the second equation, after simplification, we get a quadratic equation for $a$ (for $a \neq 20$): $\frac{a}{20}=\frac{16 a}{25(20-a)}+\frac{1}{25} \Leftrightarrow$ $100 a-5 a^{2}=64 a+80-4 a \Leftrightarrow a^{2}-8 a+16=0 \Leftrightarrow a=4$. Then, from the first and second equations, we have: $b=5, c=11$.

## Variant 2

Initially, there were 30 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{36}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{25}$ greater than the probability of drawing a blue ball at the beginning. How many white balls were in the box?

Answer: 5.

## Variant 3

Initially, there were 42 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{49}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{36}$ greater than the probability of drawing a blue ball at the beginning. How many white balls were in the box?

Answer: 6.

## Variant 4

Initially, there were 56 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{64}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{49}$ greater than the probability of drawing a blue ball at the beginning. How many white balls were in the box?

Answer: 7.

## Variant 5

Initially, there were 72 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{81}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{64}$ greater than the probability of drawing a blue ball at the beginning. How many white balls were in the box?

Answer: 8.