Number of chocolate chip and butter cookies

Question

(3/5) of the cookies were chocolate chip cookies and the rest were butter cookies. After (2/5) of the chocolate chip cookies were sold, there were 24 more butter cookies than chocolate chip cookies. How many cookies were there at first?

I'm not sure how to create an algebraic expression for this.

Answer 1

$3/5$ of the cookies, which we'll call C, are chocolate chip so $Chocolate Chip=(3/5)C$. That means that $2/5$ of the cookies were butter cookies so $Butter Cookies=(2/5)C$.

Lastly, there were $24$ more butter cookies than chocolate chip cookies after $2/5$ of the chocolate chip cookies were sold so $(1-2/5)(Chocolate Chip)+24=Butter Cookies$.

Now solve the rest.

Answer 2

I'll make everything explicit; in practice you'll of course always do some shortcuts, but I think you'll understand better if you see it be done to pedantic detail.

Let's first determine what quantities occur in the problem, and give names to those.

To begin with, there's the total number of cookies. Let's call that $n$.
Next, there's the initial number of chocolate chip cookies. Let's call that $c$.
Next, there's the number of chocolate chip cookies sold. Let's call that $s$.
Next, there's the number of chocolate chip cookies remaining. Let's call that $r$.
Finally, there's the number of butter cookies. Let's call that $b$.

Note that since no butter cookies have been sold, we only need one quantity for them.

Now, having found the quantities involved, let's see what relations are given for them:

$3/5$ of the cookies were chocolate chip cookies

So the number of chocolate chip cookies is $3/5$ of the total number of cookies. So $$c = \frac{3}{5} n \tag{I}$$

and the rest were butter cookies.

So all the cookies that are not chocolate chip cookies are butter cookies. Thus if you remove the chocolate chip cookies from all cookies, the butter cookies are what is left. Therefore $$n-c=b \tag{II}$$

After $2/5$ of the chocolate chip cookies were sold,

So the number of chocolate chip cookies sold is $2/5$ of the initial number of chocolate chip cookies: $$s = \frac{2}{5} c \tag{III}$$

there were 24 more butter cookies than chocolate chip cookies.

So the number of butter cookies is 24 more than the number of remaining chocolate chip cookies, or $$b = r + 24 \tag{IV}$$

So now we have four equations for five quantities; that's usually (and indeed, also in this case) not enough to solve. So, what's the missing equation? Well, it's the fact that the chocolate chip cookies that remain are those of the initially available that were not sold, or $$r = c - s \tag{V}$$

Now let's look at what is asked for:

How many cookies were there at first?

So we are asked for the value of $n$.

One way to solve this is to insert $(III)$ into $(V)$ (elimiating $s$), the resulting equation into $(IV)$ (eliminating $r$), then the resulting equation into $(II)$ (elimiating $b$), then solving the resulting equation for $c$ and insert into $(I)$ (eliminating $c$). This then gives a single equation for $n$ that's easily solved.

Note that usually you'll do some of those eliminations already implicitly when writing down the equations (for example, you'll usually not write down equation $(V)$ explicitly, but use $c-s$ directly in $(IV)$; likely you'll also avoid introducing $s$ by writing $\frac{3}{5} c$ directly). But I intentionally did it in maximal detail here, so the correspondence between text and formulas is more immediately apparent.