**Problem of the Week**

Problem E and Solution

Candies Anyone?

**Problem**

A jar contains only small red and small yellow candies. Another 30 red candies are added to the candies already in the jar so that one-third of the total number of candies in the jar are red candies. At this point, 30 yellow candies are added to the jar, and now three-tenths of the total number of candies in the jar are red candies.

What fraction of the number of the candies originally in the jar were red candies?

**Solution**

Let \(r\) represent the number of red candies originally in the jar.

Let \(y\) represent the number of yellow candies originally in the jar.

Then \(r+y\) represents the total number of candies originally in the jar.

After adding 30 red candies to the candies already in the jar, there are \((r+30)\) red candies in the jar and a total of \((r+y+30)\) candies in the jar. Now one-third of the candies in the jar are red candies, so

\[\begin{aligned}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r+30}{r+y+30}&=&\frac{1}{3}\\
3(r+30)&=&1(r+y+30)\\
3r+90&=&r+y+30\\
2r+60&=&y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\[-7mm]\end{aligned}\]

After adding 30 yellow candies to the candies in the jar, there are \((r+30)\) red candies in the jar and a total of \((r+y+60)\) candies in the jar. Now three-tenths of the candies in the jar are red candies, so

\[\begin{aligned}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r+30}{r+y+60}&=&\frac{3}{10}\\
10(r+30)&=&3(r+y+60)\\
10r+300&=&3r+3y+180\\
7r-3y&=&-120\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\[-10mm]\end{aligned}\] Substituting (1) into (2),

\[\begin{aligned}
7r-3(2r+60)&=&-120\ \ \ \ \ \ \ \ \ \ \ \ \ \\
7r-6r-180&=&-120\\
r&=&60\\[-7mm]\end{aligned}\] Substituting \(r=60\) into (1), we get \(y=180\).

Therefore, there were originally 60 red and 180 yellow candies in the jar, and \(\dfrac{60}{60+180} = \dfrac{60}{240}=\dfrac{1}{4}\) of the candies originally in the jar were red.