Problem of the Week
Problem C and Solution
Marbles, Marbles

Problem

A box contains $6$ red marbles, $5$ blue marbles, $2$ yellow marbles, and $3$ green marbles. Several orange marbles are added to the box. All the marbles in the box are identical except for colour.

A marble is then randomly selected from the box, and the probability that a blue or green marble is selected is $\frac{2}{7}$.

How many orange marbles were added to the box?

Solution

The number of blue and green marbles in the box is $5+3=8$.

Let $n$ be the total number of marbles in the box after adding some orange marbles. Since the probability of picking a blue or green marble is $\frac{2}{7}$, we must have $\frac{8}{n}=\frac{2}{7}.$

If we multiply the numerator and denominator of the fraction $\frac{2}{7}$ by $4$, we obtain $\frac{2}{7}=\frac{2\times 4}{7\times 4}=\frac{8}{28}$. Therefore, $\frac{8}{n}=\frac{8}{28}.$ Since the fractions are equal and the numerators are equal, the denominators must also be equal. It follows that $n=28.$

In the beginning, there were $5+6+3+2=16$ marbles in the box. Since there were 16 marbles in the box and there are now 28 marbles in the box, then $28-16=12$ orange marbles were added to the box.

Therefore, $12$ orange marbles were added to the box.