If $50\%$ are removed, what fraction of the red pens are left? If $33\dfrac{1}{3}\%$ (one-third) of the blue pens are removed, what fraction of the blue pens are left?

After removal, the box has 68 pens and the same number of red and blue pens. If two equal groups add up to 68, how many are in each group?

Once you know how many blue pens are left and that this is $\dfrac{2}{3}$ of the original number, think about how to find the whole when you know $\dfrac{2}{3}$ of it (what do you multiply 34 by?).

In Desmos, type 68/2 to verify that if two equal groups add to 68, then each group (the remaining red pens and the remaining blue pens) has 34 pens.

Let $x$ represent the original number of blue pens. Since $\dfrac{2}{3}$ of them remain and that equals 34, type the two expressions as functions: y1 = (2/3)x and y2 = 34.

Look at the graph of y1 and y2 and tap on their intersection point. The x-coordinate of this intersection is the original number of blue pens before any were removed.

Let $R$ be the number of red pens and $B$ be the number of blue pens before any pens are removed.

• Removing $50\%$ of the red pens means half are taken away and half remain, so the number of red pens left is $\dfrac{R}{2}$.
• Removing $33\dfrac{1}{3}\%$ (one-third) of the blue pens means $\dfrac{1}{3}$ of them are taken away and $\dfrac{2}{3}$ remain, so the number of blue pens left is $\dfrac{2}{3}B$.

After the removals, there are a total of 68 pens and an equal number of each color.

If the remaining red pens and blue pens are equal and together make 68, then each color must have

So:

• Remaining red pens: $\dfrac{R}{2} = 34$
• Remaining blue pens: $\dfrac{2}{3}B = 34$

Focus on the blue-pen equation:

To undo multiplying by $\dfrac{2}{3}$, multiply both sides by its reciprocal, $\dfrac{3}{2}$:

Compute this:

• $34 \cdot 3 = 102$
• $102 \div 2 = 51$

So there were 51 blue pens in the box before any pens were removed.