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 21:02, 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: ![\[0.6x+0.25b=0.8x\]](http://latex.artofproblemsolving.com/e/e/2/ee29bccf754db7ff5bb9dc6449506e7974fc9fbb.png)
![\[0.2x=0.25b\]](http://latex.artofproblemsolving.com/e/e/0/ee0da4a2d551961808881fe432e40e27a8757221.png)
![\[0.8x=b\]](http://latex.artofproblemsolving.com/7/e/8/7e805abb0746b741e7ee69234ce2d6d0663640f7.png)
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 ![\[\frac{0.68x}{0.68x+0.12x}=\frac{68}{80}=\frac{85}{100},\]](http://latex.artofproblemsolving.com/d/9/6/d9644409a80bc9f3c379a959f59f1f1e87392c6c.png)
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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