# 2020 CIME II Problems/Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

In a jar there are blue jelly beans and green jelly beans. Then, $15\%$ of the blue jelly beans are removed and $40\%$ of the green jelly beans are removed. If afterwards the total number of jelly beans is $80\%$ of the original number of jelly beans, then determine the percent of the remaining jelly beans that are blue.

## Solution 1

Suppose there are $x$ jelly beans total at the beginning. Suppose further that there are $b$ blue jelly beans and $x-b$ green jelly beans. Then, after the removal, there will be $0.85b$ blue jelly beans and $0.6x-0.6b$ green jelly beans. Because the total number of jelly beans at the end is $80\%$ of the starting number, we can create an equation: $$0.6x+0.25b=0.8x$$ $$0.2x=0.25b$$ $$0.8x=b$$ This tells us there were originally $0.8x$ blue jelly beans and $0.2x$ green jelly beans at the beginning, so now there must be $0.68x$ blue and $0.12x$ green. The percent of the remaining jelly beans that are blue is $$\frac{0.68x}{0.68x+0.12x}=\frac{68}{80}=\frac{85}{100},$$ so the answer is $\boxed{085}$.