Difference between revisions of "2020 CIME II Problems/Problem 3"
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==Solution 1== | ==Solution 1== | ||
− | Suppose there are <math>x</math> jelly beans total at the beginning. Suppose further that there are <math>b</math> blue jelly beans and <math>x-b</math> green jelly beans. Then, after the removal, there will be <math>0.85b</math> blue jelly beans and <math>0.6x-0.6b</math> green jelly beans. Because the total number of jelly beans at the end is <math>80\%</math> of the starting number, we can create an equation: <cmath>0.6x+0.25b=0.8x</cmath> <cmath>0.2x=0.25b</cmath> <cmath>0.8x=b</cmath> This tells us there were originally < | + | Suppose there are <math>x</math> jelly beans total at the beginning. Suppose further that there are <math>b</math> blue jelly beans and <math>x-b</math> green jelly beans. Then, after the removal, there will be <math>0.85b</math> blue jelly beans and <math>0.6x-0.6b</math> green jelly beans. Because the total number of jelly beans at the end is <math>80\%</math> of the starting number, we can create an equation: <cmath>0.6x+0.25b=0.8x</cmath> <cmath>0.2x=0.25b</cmath> <cmath>0.8x=b</cmath> This tells us there were originally <math>0.8x</math> blue jelly beans and <math>0.2x</math> green jelly beans at the beginning, so now there must be <math>0.68x</math> blue and <math>0.12x</math> green. The percent of the remaining jelly beans that are blue is <cmath>\frac{0.68x}{0.68x+0.12x}=\frac{68}{80}=\frac{85}{100},</cmath> so the answer is <math>\boxed{085}</math>. |
==See also== | ==See also== |
Latest revision as of 22:02, 5 September 2020
In a jar there are blue jelly beans and green jelly beans. Then, of the blue jelly beans are removed and
of the green jelly beans are removed. If afterwards the total number of jelly beans is
of the original number of jelly beans, then determine the percent of the remaining jelly beans that are blue.
Solution 1
Suppose there are jelly beans total at the beginning. Suppose further that there are
blue jelly beans and
green jelly beans. Then, after the removal, there will be
blue jelly beans and
green jelly beans. Because the total number of jelly beans at the end is
of the starting number, we can create an equation:
This tells us there were originally
blue jelly beans and
green jelly beans at the beginning, so now there must be
blue and
green. The percent of the remaining jelly beans that are blue is
so the answer is
.
See also
2020 CIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All CIME Problems and Solutions |
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