In the Shop of Merriments, a package of streamers costs $p$ percent more than a package of confetti. We could also say that a package of confetti is $q$ percent cheaper than a package of streamers. The difference between $p$ and $q$ is 90. How many packages of streamers can you get for the price of 10 packages of confetti?
Let $y$ be the price of one package of streamers, and $x$ be the price of one package of confetti. Based on the problem statement, we can write the following equations:

$$
\begin{aligned}
& x\left(1+\frac{p}{100}\right)=y \\
& y\left(1-\frac{q}{100}\right)=x
\end{aligned}
$$

and

$$
|p-q|=90
$$

Let $\frac{p}{100}=a$ and $\frac{q}{100}=b$. From the system of equations, we get that $y(1-b)(1+a)=y$, hence

$$
a-b-a b=0
$$

1) If $p>q$, then $p-q=100 a-100 b=90$. From this, $a-b=0.9$, so $a=0.9+b$. Substituting this into equation (4):

$$
\begin{aligned}
& 0.9-b(0.9+b)=0 \\
& b^{2}+0.9 b-0.9=0 \\
& b=\frac{-0.9 \pm \sqrt{0.81+3.6}}{2}=\frac{-0.9 \pm 2.1}{2}, \quad b_{1}=0.6, \quad b_{2}<0
$$

2) If $p<q$, then $q-p=90$. Thus, $100 b-100 a=90$, which means $b=a+0.9$. Substituting this into equation (4), we get

$$
a^{2}+0.9 a+0.9=0, \quad a=\frac{-0.9 \pm \sqrt{0.81-3.6}}{2}
$$

Since the discriminant is negative, the equation has no real solutions, so only the previous case can hold.

From equation (2),

$$
x=y\left(1-\frac{60}{100}\right)=\frac{4}{10} y
$$

From the price of 10 packages of confetti, $\frac{10 x}{y}=\frac{4 y}{y}=4$ packages of streamers can be bought.