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.