Problem of the Week
Problem D and Solution
Sticker Situation

Problem

Kendi has a large collection of vinyl stickers, and each sticker has an animal on it or an emoji on it (but not both). In this collection, $300$ of the stickers have animals on them, and $1$ out of $5$ of all of the stickers have animals on them. She would like to add more stickers to her collection so that there are $3$ stickers with animals on them out of every $10$ stickers.

If she can buy the stickers in packages of $60$ stickers where $21$ are animal stickers and the remaining are emoji stickers, how many whole packages does she need to buy?

Solution

There were initially $300$ animal stickers, and there was $1$ animal sticker for every $5$ stickers. This means that $4$ out of $5$ stickers were emoji stickers. Therefore, there were four times as many emoji stickers as animal stickers. That is, there were $4\times 300=1200$ emoji stickers and a total of $300+1200=1500$ stickers.

Each package contains $21$ animal stickers and $39$ emoji stickers, for a total of $21+39=60$ stickers.

Let $n$ represent the number of additional whole packages required to add to this collection so that there are $3$ animal stickers out of every $10$ of the stickers. By purchasing $n$ packages, she is adding $60n$ stickers to her collection, of which $21n$ are animal stickers. Thus, she will have a total of $1500+60n$ stickers, of which $300+21n$ are animal stickers.

If $3$ out of $10$ of the stickers in her collection are animal stickers, then we have $$\begin{array}{rl}\frac{\text{the number of animal stickers }}{\text{the total number of stickers}}& =\frac{3}{10}\\ \frac{300+21n}{1500+60n}& =\frac{3}{10}\\ 10(300+21n)& =3(1500+60n)\\ 3000+210n& =4500+180n\\ 30n& =1500\\ n& =50\end{array}$$ Therefore, $50$ additional packages of stickers must be purchased so that $3$ out of $10$ of the stickers in her collection are animal stickers.

We can check this. After purchasing $50$ additional packages of stickers, there would be $300+21(50)=1350$ animal stickers and a total of $1500+60(50)=4500$ stickers. Then, the ratio of animal stickers to the total number of stickers is $\frac{1350}{4500}=\frac{3}{10}$, as required.