2001 CEMC Gauss (Grade 8) Problems/Problem 19
- The following problem is from both the 2001 CEMC Gauss (Grade 8) #19 and 2001 CEMC Gauss (Grade 7) #20, so both problems redirect to this page.
Problem
Anne, Beth and Chris have 
candies to divide amongst themselves. Anne gets at least 
candies, while Beth and Chris each get at least 
. If Chris gets at most , the number of candies that Beth could get is
Solution 1
Because Chris must have at least candies but gets at most , he must have either candies or candies.
If Chris gets candies, then that means that Anne and Beth get 
candies in total. Anne gets at least , so the most that Beth can get is 
. However, he must have at least , so Beth can have 
candies.
~anabel.disher
Solution 2 (answer choices)
We can start by realizing that Chris must have or candies like in solution 1.
If Chris gets candies, then that means that Anne and Beth get candies in total. Anne gets at least , so the most that Beth can get is . We can see that eliminates all answer choices except D and E because D and E are the only answers that allow Beth to have candies.
We also see that Beth could have 
candies while Anne could have . D doesn't have as a possible option, so is the only option left.
| 2001 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| CEMC Gauss (Grade 8) | ||
| 2001 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| CEMC Gauss (Grade 7) | ||
