Grades 7-8 Video Solutions 2021-video 2021 7-8/24

Video Transcription

Question 24. A box contains only green, red, blue, and yellow counters. There is always at least one green counter among any 27 counters chosen from the box,  always at least one red among any 25 counters chosen, always at least one blue among any 22 counters chosen, and always at least one yellow among any 17  counters chosen. What is the largest number of counters that could be in the box? And start off by labeling these by letters. Green will be G, red R, blue B,  and yellow Y. Now we know when we do R plus B plus Y has to be less than or equal to 27 minus 1. We know this because there is always at least one green  counter among 27 counters. We can do the same with G plus B plus Y being less than or equal to 25 minus 1. Then we get R plus G plus Y is less than or equal to  22 minus 1. And finally, R plus G plus B is less than or equal to 17 minus 1. We can total these together up and get 91 minus 4, or 87. Now we have to take this  number and divide it by 3, since if we added all the colors we had, we would end up with 3 of each one. So we do 87 divided by 3, and this gives us our  answer of 29. So the answer is B, 29.

Video Summary

The video transcript outlines a problem involving counters of different colors in a box. It establishes constraints for the minimum number of each color in selected groups of counters: at least one green among 27, one red among 25, one blue among 22, and one yellow among 17. By defining equations and summing them, the total limit without one color is determined to be 87. Dividing this by 3, as each color is included three times across equations, results in a maximum of 29 counters fitting the conditions. Thus, the largest possible number of counters in the box is 29.

Keywords

counters

constraints

equations

maximum

colors