Difference between revisions of "2020 CIME II Problems/Problem 3"
(Created page with "In a jar there are blue jelly beans and green jelly beans. Then, <math>15\%</math> of the blue jelly beans are removed and <math>40\%</math> of the green jelly beans are remov...") |
|||
Line 2: | Line 2: | ||
==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: 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 |
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.