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.